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Thread: Geometry Primer

  1. #1 Geometry Primer 
    Forum Professor river_rat's Avatar
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    I have been asked to start a thread on the basics of differential geometry/topology i.e. What the hell is a manifold?

    I will limit my talks to surfaces (i.e. 2 dimensional manifolds) as most of them are easy to handle, the theory is very rich and most "mathematical gems" people recognise come from this group - like the mobius band or klein bottle.

    Also, i can skip a lot of the machinery at the beginning and introduce a surface as a level set and not as an abstract maximal collection of atlases on some set

    Well first we need a definition -

    A level set of a smooth function f:U -> R where U is some open subset of the R<sup>3</sup> is the set f<sup>-1</sup>(c) = {(x<sub>1</sub>, x<sub>2</sub>, x<sub>3</sub>) in R<sup>3</sup> : f(x<sub>1</sub>, x<sub>2</sub>, x<sub>3</sub>) = c}

    An example is worth a thousand words (or one picture apparently) so consider the function f(x, y, z) = x<sup>2</sup> + y<sup>2</sup> + z<sup>2</sup> which is "nice" everywhere except at the origin which happens to be an open subset of R<sup>3</sup>. Now what is the level surface f<sup>-1</sup>(4)?

    Well it is the set of all points (x, y, z) such that x<sup>2</sup> + y<sup>2</sup> + z<sup>2</sup> = 4 i.e. the sphere of radius 2 centred at the origin.

    Any questions about level sets?


    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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  3. #2 Re: Geometry Primer 
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    Quote Originally Posted by river_rat
    Also, i can skip a lot of the machinery at the beginning and introduce a surface as a level set and not as an abstract maximal collection of atlases on some set
    phew
    Quote Originally Posted by river_rat
    Any questions about level sets?
    no questions from me


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  4. #3  
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    Sorry for the edit - i needed to fix some terminology else things would get hairy later.

    Now you should of been asking yourself why i had to stay away from the origin for my example - well lets take a look.

    The level set f<sup>-1</sup>(0) = {(x, y, z) : x<sup>2</sup> + y<sup>2</sup> + z<sup>2</sup> = 0} = {(0,0,0)} which you would agree is not a nice surface, in fact it is zero dimensional (intuitively, dont worry about the many definitions of dimension)! So something major broke down here and it would be nice if we could exclude those cases. Now i hope your multivariable calculus hat is on because that is where we are heading.

    Definitions:
    The gradient of a function f:U -> R is the vector function grad(f):U -> R<sup>3</sup> given by grad(f)(u, v, w) = (f<sub>x</sub>(u), f<sub>y</sub>(v) f<sub>z</sub>(w)) where f<sub>x</sub> denotes the partial derivative of f with respect to the variable x.

    A point (u,v,w) of a function f:U -> R is called a regular point if gradf(f)(u, v, w) is not the zero vector. A point that is not regular is called a singular point. A function is called regular if its domain contains no singular points.

    So lets see what happened to our function f(x, y, z) = x<sup>2</sup> + y<sup>2</sup> + z<sup>2</sup>. Its gradient is grad(f)(u, v, w) = (2u, 2v, 2w) which is non-zero provided we never look at the point (0, 0, 0) i.e. the origin. So there seems to be a link between having nice 2 dimensional level sets and our functions being regular - and that is the next stop on this bus.

    Any questions?
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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  5. #4  
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    well my multivariable calculus consists purely of partial derivatives so i don't know if theres other bits of calculus that i need to read into or whether my bases are covered.

    i'm still ok so far for understanding.
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  6. #5  
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    Ah well if you recall most of what guitarist covered in the Vector spaces thread and remember what a partial derivative is you should be fine - if we need anything that i don't think you have seen before i will cover it when we need it

    For starters, i don't think you know what i mean by a smooth function do you? A smooth function is one for which all its derivatives exist at every point of its domain. So for example the function f(t)=Sin(t) is smooth while the function f(t) = |t| is not smooth as f'(0) is not defined.

    Now on with the show... The gradient of a function is quite a handy beast, but to fully explain why i need to show you what went wrong with the level set i mentioned earlier and to do that i need to show you what a tangent vector is.

    Okay, lets start at the beginning:

    Where you are on a surface is important, so we need a way of keeping track of that information as we are going to construct a whole host of vector spaces but each one is different and you cant add the vectors of one vector space to the vectors of another vector space. Now a vector at a point p is an ordered pair (p, v) where v is an element of R<sup>n</sup>. Now we can add two different vectors at the same point in the obvious way ( (p, v<sub>1</sub>) + (p, v<sub>2</sub>) = (p, v<sub>1</sub> + v<sub>1</sub>) ) and take scalar multiples ( k(p, v) = (p, kv) ) so for each point we have a nice well defined vector space. Notice that we have not defined how to add the vectors of two different spaces! This is because the space we are looking at need not have any meaningful way of adding its points, this operation is only defined for the vector spaces.

    What is the derivative of a function f:R->R? Well geometrically it is the gradient of the tangent line to the function at that point. Now the vector (1, f'(t)) is a vector parallel to the tangent line of the function f(t). So we can see a derivative as being a way of defining a tangent vector for each point on the function f(t) given by v(t) = (t, f(t), 1, f'(t)) where the v is meant to remind you of velocity. So for every point (t, f(t)) of the graph of the function we can find a tangent vector at that point (p, v) where p = (t, f(t)) and v = (1, f'(t)).

    Okay, i have introduced a few new concepts here - any questions?
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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  7. #6  
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    Quote Originally Posted by river_rat
    Ah well if you recall most of what guitarist covered in the Vector spaces thread and remember what a partial derivative is you should be fine - if we need anything that i don't think you have seen before i will cover it when we need it
    excellent.

    Quote Originally Posted by river_rat
    For starters, i don't think you know what i mean by a smooth function do you? A smooth function is one for which all its derivatives exist at every point of its domain. So for example the function f(t)=Sin(t) is smooth while the function f(t) = |t| is not smooth as f'(0) is not defined.
    so to what extent does nice differ from smooth?


    Quote Originally Posted by river_rat
    Okay, i have introduced a few new concepts here - any questions?
    well i don't believe i have any questions on the introduced concepts but i was wondering if we are to reguard the set of all tangent vectors to be a space of its own, seperate from that of our function.
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  8. #7  
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    Quote Originally Posted by wallaby
    so to what extent does nice differ from smooth?
    Well that's where i am heading For starters a smooth function is quite nice but they are not nice enough to do geometry with. They don't preserve dimension in a real sense - and that preservation of dimension is what we need to start doing geometry.

    Quote Originally Posted by river_rat
    well i don't believe i have any questions on the introduced concepts but i was wondering if we are to reguard the set of all tangent vectors to be a space of its own, seperate from that of our function.
    Its better to keep the link between single variable functions and tangent vectors (it can be made even stronger but i think the link is a little abstract for you right now). Remember i have not told you what a tangent vector to surface is yet, just what a tangent vector to a curve is.

    let f:U->R be a nice function and let f<sup>-1</sup>(c) be a level set. We want the idea of a curve on this level set, so let I be some interval in the reals and a:I-> U given by a(t) = (a<sub>x</sub>(t), a<sub>y</sub>(t), a<sub>z</sub>(t)) be a smooth function. We call a(t) a smooth curve parametrised by t and if we have that for all t in U f(a(t)) = f(a<sub>x</sub>(t), a<sub>y</sub>(t), a<sub>z</sub>(t)) = c then we say that a(t) is a smooth parametrised curve in f<sup>-1</sup>(c). This makes sense as the image of the function a(t) definitely lies in f<sup>-1</sup>(c) (it only takes a line so prove it).

    Now we define a tangent vector at a point p in f<sup>-1</sup>(c) to be a tangent vector to any curve that goes through the point p when t = 0. So tangent vectors look like (p, ∂<sub>t</sub>a<sub>x</sub>, ∂<sub>t</sub>a<sub>y</sub>, ∂<sub>t</sub>a<sub>z</sub>) evaluated at t = 0.
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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  9. #8  
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    Quote Originally Posted by river_rat
    Its better to keep the link between single variable functions and tangent vectors (it can be made even stronger but i think the link is a little abstract for you right now). Remember i have not told you what a tangent vector to surface is yet, just what a tangent vector to a curve is.

    let f:U->R be a nice function and let f<sup>-1</sup>(c) be a level set. We want the idea of a curve on this level set, so let I be some interval in the reals and a:I-> U given by a(t) = (a<sub>x</sub>(t), a<sub>y</sub>(t), a<sub>z</sub>(t)) be a smooth function. We call a(t) a smooth curve parametrised by t and if we have that for all t in U f(a(t)) = f(a<sub>x</sub>(t), a<sub>y</sub>(t), a<sub>z</sub>(t)) = c then we say that a(t) is a smooth parametrised curve in f<sup>-1</sup>(c). This makes sense as the image of the function a(t) definitely lies in f<sup>-1</sup>(c) (it only takes a line so prove it).
    i'm having trouble here, but i'm having trouble pinning down what it is i don't understand. i think it starts at a:I -> U, looking purely at the notation i'd interpret that as a function of an interval in the reals to produce an open set, hene every element in the open set is a function of an interval which i'm guessing we labeled 't'?

    then what i can gather is that this function a(t)= (a<sub>x</sub>(t), a<sub>y</sub>(t), a<sub>z</sub>(t)), open set, is the function of an interval on each axis, (x,y,z), the function of an interval on each axis comprising an element of the open set.

    confusing
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  10. #9  
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    Maybe i should of been clearer here. a(t) is a function from a real interval to an open set U which is contained R<sup>3</sup>. So our function has three co-ordinate functions: a<sub>x</sub>(t), a<sub>y</sub>(t), a<sub>z</sub>(t).

    Lets look at a simpler example, suppose a : (0, 2) -> R<sup>2</sup> given by a(t) = ( cos(2*pi*t), sin(2*pi*t) ). The a(t) is a path that goes around the unit circle twice, starting on the x-axis and the co-ordinate functions are a<sub>x</sub>(t) = cos(2*pi*t) and a<sub>y</sub>(t) = sin(2*pi*t).

    Does that clear up what the functions a(t) looks like?
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  11. #10  
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    yes thats much clearer.

    so then the tangent vectors are (p, -2*pi*sin(2*pi*t), 2*pi*cos(2*pi*t) evaluated at t=0.
    (p, 0, 2*pi)
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  12. #11  
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    Quote Originally Posted by wallaby
    yes thats much clearer.

    so then the tangent vectors are (p, -2*pi*sin(2*pi*t), 2*pi*cos(2*pi*t) evaluated at t=0.
    (p, 0, 2*pi)
    Almost, remember the tangent vector at the point a(0) is given by (a(0), d/dt a(t)|<sub>t=0</sub>) so a tangent vector at the point (1, 0) is the vector (1,0, 0, 2*pi).

    Let me sketch this in plain language - we have a point p in some level set f<up>-1</sup>(c) and we want to find a vector (p, v) which is tangent to the level set at that point. Notice we are talking about a specific point on the level set, all these constructions are very local for reasons i will make clear later (in short, a nice level set locally looks flat). Now our level set lives in R<sup>n</sup> and we know what a tangent vector to a curve looks like in R<sup>n</sup> (namely something like (a(0), d/dt a(t)|<sub>t=0</sub>) where a(t) is a smooth curve that goes through the point p when t = 0) and we want to carry this notion over to the level set. What we do is find a smooth curve a(t) that is on the level set (and thus also in R<sup>n</sup>) and that goes through the point p in question when t = 0. We then call the vector (a(0), d/dt a(t)|<sub>t=0</sub>) a tangent vector to the level set at p.

    So lets look at my previous example, let f(x, y) = x<sup>2</sup> + y<sup>2</sup> and lets consider the level set f<sup>-1</sup>(1) which is the unit circle {(x, y) : x<sup>2</sup> + y<sup>2</sup> = 1}. Now i want to find a tangent vector to the point (Sqrt(2)/2, Sqrt(2)/2) which is on the circle. So i need to find a smooth curve on the circle that goes through that point when t=0, the curve a : (0,1) -> R<sup>2</sup> given by a(t) = (Cos(2*pi*t + pi/4), Sin(2*pi*t + pi/4)) works and we find that a tangent vector at the point (Sqrt(2)/2, Sqrt(2)/2) is given by (Sqrt(2)/2, Sqrt(2)/2, -Sqrt(2)/2, Sqrt(2)/2).

    Its actually an easy exercise to show that on the unit circle {(x, y) : x<sup>2</sup> + y<sup>2</sup> = 1} a tangent vector at the point (x, y) is given by (x, y, -y, x). See if you can show it. HINT - this takes nothing fancier then normal implicit differentiation.

    Any questions?
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  13. #12  
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    well i'm not satisfied with my method but i've delayed long enough.

    for the unit circle (x,y): x<sup>2</sup>+y<sup>2</sup>=1 we have a curve a(t)= (cos(2*pi*t+x), sin(2*pi*t+y)) where x and y are in radiuns.
    hence the curve at t=0 is a(0)=(cos(x), sin(y)). taking the derrivative of the function at t=0 we get, da(0)/dt = (a<sub>x</sub>(0), a<sub>y</sub>(0)) = (-sin(x), cos(y)) which means that the tangent to the point (x,y) at t=0 is (x,y,-y,x).

    shakey i'll bet, the curve a(t) isn't really parametrised anymore!

    i'm sure i'll have questions soon enough.
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  14. #13  
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    Quote Originally Posted by wallaby
    well i'm not satisfied with my method but i've delayed long enough.

    for the unit circle (x,y): x<sup>2</sup>+y<sup>2</sup>=1 we have a curve a(t)= (cos(2*pi*t+x), sin(2*pi*t+y)) where x and y are in radiuns.
    hence the curve at t=0 is a(0)=(cos(x), sin(y)). taking the derrivative of the function at t=0 we get, da(0)/dt = (a<sub>x</sub>(0), a<sub>y</sub>(0)) = (-sin(x), cos(y)) which means that the tangent to the point (x,y) at t=0 is (x,y,-y,x).

    shakey i'll bet, the curve a(t) isn't really parametrised anymore!

    i'm sure i'll have questions soon enough.
    Almost, why does your curve have two variables (x and y)? Let Phi be the angle measured from the x-axis to the point (x, y) on the unit circle in question and define a(t)= (cos(t+phi), sin(t+phi)) where t goes from (-pi, pi). Then a tangent vector at the point (x, y) is (x, y, -sin(phi), cos(phi)) but by the definition of phi we have that cos(phi) = x and sin(phi) = y which implies that our tangent vector is (x, y, -y, x).

    Next exercise - prove that the vector grad(f) is always orthogonal to any tangent vector to the level set f<sup>-1</sup>(c). Dont worry if you get stuck here, ill go through this idea carefully tomorrow.
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    Yummy! I missed a lot as I had a hard drive failure, so it will take me a while to catch up. This thread is looking good, though I might want to say something about differential operators in general (I had been tempted in the Vector Spaces thread, but decided against).

    Anyway, ignore me for now.....
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  16. #15  
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    Quote Originally Posted by river_rat
    Almost, why does your curve have two variables (x and y)? Let Phi be the angle measured from the x-axis to the point (x, y) on the unit circle in question and define a(t)= (cos(t+phi), sin(t+phi)) where t goes from (-pi, pi). Then a tangent vector at the point (x, y) is (x, y, -sin(phi), cos(phi)) but by the definition of phi we have that cos(phi) = x and sin(phi) = y which implies that our tangent vector is (x, y, -y, x).

    Next exercise - prove that the vector grad(f) is always orthogonal to any tangent vector to the level set f<sup>-1</sup>(c). Dont worry if you get stuck here, ill go through this idea carefully tomorrow.
    my curve had two variables because i hadn't conidered expressing it in terms of a rotation through an angle phi. so i probably just decided that since the fucntion x<sup>2</sup>+y<sup>2</sup>=1 has two variables that i'd include two in the curve, silly really.

    as for the next exercise, well i could easily show that the vector gradient is orthogonal to the tangent vector for the specific case we sighted above. but i don't know where to begin in the general case, int the specific case a simple inner product will reveal the orthogonal property but generally i don't know if theres a notation i can use which would make using the inner product to obtain the nessisary result or whether its a different method altogether.

    Quote Originally Posted by Guitarist
    Yummy! I missed a lot as I had a hard drive failure, so it will take me a while to catch up. This thread is looking good, though I might want to say something about differential operators in general (I had been tempted in the Vector Spaces thread, but decided against).

    Anyway, ignore me for now.....
    welcome back
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  17. #16  
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    Hey guitarist welcome back

    Well seeing that i am working in full generality here let me redo a few definitions :

    A level set of a smooth function f:U -> R where U is some open subset of the R<sup>n+1</sup> is the set f<sup>-1</sup>(c) = {(x<sub>1</sub>, x<sub>2</sub>,... , x<sub>n+1</sub>) in R<sup>n+1</sup> : f(x<sub>1</sub>, x<sub>2</sub>,... , x<sub>n+1</sub>) = c}

    The gradient of a function f:U -> R is the vector function grad(f):U -> R<sup>n+1</sup> given at the point p = (x<sub>1</sub>, x<sub>2</sub>,... , x<sub>n+1</sub>) by grad(f)(p) = (f<sub>x<sub>1</sub></sub>(p), f<sub>x<sub>2</sub></sub>(p), ..., f<sub>x<sub>n+1</sub></sub>(p)) where f<sub>x<sub>i</sub></sub> denotes the partial derivative of f with respect to the variable x<sub>i</sub>.

    A point p = (x<sub>1</sub>, x<sub>2</sub>,... , x<sub>n+1</sub>) of a function f:U -> R is called a regular point if grad(f)(p) is not the zero vector. A point that is not regular is called a singular point. A function is called regular if its domain contains no singular points.

    Ok, no major changes but i thought for fullness i should not limit this discussion to 2D surfaces so i have given the n-dimensional versions of our previous definitions.

    Now let me show that the gradient of a function f:U->R at a point p = (x<sub>1</sub>, x<sub>2</sub>,... , x<sub>n+1</sub>) is orthogonal to the tangent vector to the level set f<sup>-1</sup>(c) at the point p. To describe a tangent vector at a point p i need a curve, so let a(t) be a smooth curve on f<sup>-1</sup>(c) such that a(0) = p. We know that a tangent vector at the point p is then given by (p, v) = (p, d/dt a(t) |<sub>t=0</sub>) with d/dt a(t) = (d/dt a<sub>x<sub>1</sub></sub>(t), d/dt a<sub>x<sub>2</sub></sub>(t), ..., d/dt a<sub>x<sub>n+1</sub></sub>(t)) = a'(t). But the curve a(t) is on the level set so we know that f(a(t)) = c and thus by the chain rule we find that grad(f)(a(0)).a'(0) = 0. But this implies that grad(f)(p).v = 0

    i.e. the vectors (p, grad(f)(p)) and (p, v) at the point p are orthogonal. QED.

    I hope you could follow all that, if not please shout! As an exercise repeat this proof for the specific case of the circle we did earlier.

    Now with that little result under our belt lets define an n-dimensional surface.

    A n-dimensional, orientable, smooth surface is the level set of a smooth regular map f:U -> R where U is an open subset in R<sup>n+1</sup>.

    Dont worry about the orientable adjective, i will explain it in due course - lets just say it is to exclude the klein bottle or mobius strip from our discussion as you cant describe them as the level set of some smooth function and thus they fall outside of the area of geometry i want to explore here.

    As a final exercise, show that the unit sphere is a smooth, 2-dimensional orientable surface. HINT - the unit sphere is the level set of the function f(x, y, z) = x<sup>2</sup> + y<sup>2</sup> + z<sup>2</sup>. So what do we need to show about f?
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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    Hey! Well it's always good to get a different perspective on things, I guess, but river_rat has started from a completely different point than I would have, but he knows more than I. Moreover, I'm not sure I fully understand all of this.

    As I am rather rushed just now, I hope river_rat won't mind a coupla comments, and possibly a question or two.

    First, I hope you all noticed that, in the definition of smooth surfaces he gave, he said all derivatives exist at all point. That means infinitely differentiable everywhere. This condition is usually referred to as C<sup>∞</sup>.

    Second, a level set is sometimes referred to as a contour map. This is no mere analogy: to any surface S in R<sup>2</sup> given by f(x,y) = k, one can associate a contour map (= level set), whose element are those curves for which k takes on the same value. One calls these (in context) curves of equipotential for S.

    And just so we're all absolutely sure about this: mathematicians (bless them!) consider the sphere, say, to be a 2-dimensional object. That is, one can have as much (or as little) "curvature" as you like, but as long as there are only n degrees of freedom, the space in question is n-dimensional. (MagiMaster tried this as a definition of dimension, but, as was pointed out to him, this fails in all generality).

    Finally, there is slightly contentious issue regarding the grad operator. You will occasionally see this referred to as a vector. It's not, of course, it's an operator. So, let φ denote a scalar field. grad φ ≡ ∇φ defines a vector field by the requirement that ∇ · dφ = ds, where dφ is the incremental change in φ corresponding to some arbitrary spatial displacement ds.

    But now suppose f is a function on the scalar field φ, and let x ∈ φ. It follows that ∇f(x) is a vector in the the vector field ∇φ.

    There is a list of 5(?) such differential operators, most of which, I suspect, we shan't encounter here.

    Jeez, that's a bit garbled, sorry. And I see another post from river_rat came in meantime.
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  19. #18  
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    Quote Originally Posted by Guitarist
    Hey! Well it's always good to get a different perspective on things, I guess, but river_rat has started from a completely different point than I would have, but he knows more than I. Moreover, I'm not sure I fully understand all of this.
    Hey guitarist, no probs - i decided not to try and take the highbrow intro into abstract geometry as you end up spending 100 posts setting up the basic machinery before you even start looking at interesting things about manifolds. Don't worry, i will introduce atlases etc when we really need them but this way is easier to explain.

    What i like about this method is that it leads us immediately into a proof of the lagrange multiplier method - which is a handy thing to know. Also as we have an ambient space to put our vectors in i don't have to explain a vector as a derivative on a real valued function etc.

    If you see anything i have not explained properly or at all please shout - its easy to miss something when you are typing out your thoughts.
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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    Quote Originally Posted by river_rat
    I hope you could follow all that, if not please shout! As an exercise repeat this proof for the specific case of the circle we did earlier.
    had to read it a few times but i do believe i understand.

    in the set f<sup>-1</sup>(1) a point {(x,y): x<sup>2</sup>+y<sup>2</sup> = 1} has a tangent vector to the point p=(x,y) which is represented by the smooth curve a(t), where a(t)=p when t=0.
    the curve a(t)=(cos(t+phi), sin(t+phi)) is one such curve. evaluating at t=0 we find that a(0)=(cos(phi), sin(phi)). taking the derrivative of a(t) and evaluating at t=0 we find that the tangent to the point p, (p,v), is given by (cos(phi), sin(phi), -sin(phi), cos(phi)).
    now finding the gradient function for our level set f<sup>-1</sup>(1) at a point p, we have that grad(f)(p)= (f<sub>x</sub>(p), f<sub>y</sub>(p)) which gives us grad(f)(p)=(2x, 2y)= (2cos(phi), 2sin(phi)) when we substitute our values for p into a(t).

    well that part is just to see if i have a grip on my terms and use of language, "mathematical manners" i believe it may have been called once.
    the real proof comes from the inner product of grad(f)(p and a'(t) evaluated at t=0.
    (2cos(phi), sin(phi) <sup>.</sup> (-sin(phi), cos(phi)) = -2sin(phi)cos(phi) +2sin(phi)cos(phi)= 0
    therefore grad(f)(p) is orthogonal to a'(0). QED

    Quote Originally Posted by river_rat
    As a final exercise, show that the unit sphere is a smooth, 2-dimensional orientable surface. HINT - the unit sphere is the level set of the function f(x, y, z) = x<sup>2</sup> + y<sup>2</sup> + z<sup>2</sup>. So what do we need to show about f?
    we need to show that it can be represented by a smooth curve that contains no singular points, i've seldom done trigonometry in 3 dimensions but i'm guessing we don't need to do that, something to do with treating the surface as 2 dimensional i suppose...i'll give it some thought, don't let that stop you from telling me though.
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    Well the first proof looks good - but i think you missed the definition in your second proof. What you need to show is that the unit sphere is a level set of some nice smooth regular function from some open subset of R<sup>3</sup> to R. Well we know that the unit sphere is the set of all points (x, y, z) such that x<sup>2</sup> + y<sup>2</sup> + z<sup>2</sup> = 1 which suggests we look at the function f(x, y, z) = x<sup>2</sup> + y<sup>2</sup> + z<sup>2</sup>. Now the unit sphere is definitely a level set of this function (namely the level set f<sup>-1</sup>(1)) and being a "polynomial" we know that the function is smooth so all we need to do is show that it is regular on some open set containing the unit sphere.

    Well grad(f)(x, y, z) = (2x, 2y, 2z) is non-zero provided x, y, z is not equal to zero. But the subset R<sup>3</sup> \ {(0,0,0)} is definitely open (as the singleton is closed) and the point (0,0,0) does not lie on the unit sphere so f<sup>-1</sup>(1) is contained in R<sup>3</sup> \ {(0,0,0)}.

    So we have found a smooth regular function from R<sup>3</sup> \ {(0,0,0)} to R which has the unit sphere as a level set, so the unit sphere is a smooth, 2-dimensional orientable surface
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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    that easy huh.
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    Hmm, except we haven't been told yet what it means for a surface to be orientable. There are in fact several definitions of differing degrees of technicality, but so far, we have only been given the tools to understand the most intuitive. I hope I don't disrupt river_rat's plan too much by offering this extremely loose guide.

    Let S<sup>2</sup> be the 2-sphere embedded in R<sup>3</sup>. At any point p in S<sup>2</sup> apply a vector normal v<sub>p</sub>, that is, a vector perpendicular to the tangent plane T<sub>p</sub> at p. Clearly we have a choice of two such vectors - outward normal and inward normal. Let's call these v<sub>p<sub>o</sub></sub> and v<sub>p<sub>i</sub></sub>

    Now I can transport T<sub>p</sub> through any path of my choosing, and always find that v<sub>p<sub>o</sub></sub> is outward normal and v<sub>p<sub>i</sub></sub> is inward normal.

    A little geometric agility (possible even a twisted strip of paper) should convince you that this is not true of, say, the Moebious strip.

    So: a surface S is said to be orientable if, for any pair of neighbourhoods U and W of S I can always choose them such that the outward normals of U are the outward normals of W, likewise the inward normals. For this condition, one says that a surface is orientable iff it has a continuous non-zero normal vector field.

    The somewhat more technical variants of this definition are of much more utility in respect of transition functions - but we haven't yet been told what these are!
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    after reading that i can't stop thinking of my toe nail, and thats not because its ingrown and sore.

    i'm thinking that the tangent vector to a point on the curve of my toe nail can be translated down the length of it and rotated to the annoying part which needs to be removed with one definition of inbound and outbound vectors applicable to both cases. so my toe nail is a 2 dimensional orientable surface in R<sup>3</sup> is it not?
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    Wallaby, hi! I don't fully understand your toenail analogy, but yes it is an orientable 2-surface in R<sup>3</sup>, but not (as far as I can judge) from what you said.

    First please, please don't go transporting tangent vectors across a curved surface, you will see why in due course (I assume).

    Let me try again. A surface is orientable iff to find a reflection symmetry (i.e. mirror image) one has to pass to the "other side" over some "edge". Which extremely roughly means we have two sides to our surface, and know the difference between them.
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    Ah i see guitarist went running a head here so let me pull everyone back and firstly give a "rough" idea of want the tangent space at every point of the surface is.

    Lets recap some definitions:

    A n-dimensional, orientable, smooth surface is the level set of a smooth regular map f:U -> R where U is an open subset in R<sup>n+1</sup>.

    Now don't worry about the orientable adjective yet, i will get to that in due course. I just wanted to clear up that i am not dealing with this subject in full generality (mainly because the audience would get lost very quickly in all the notation!) and to make sure you don't start asking about that favorite surface the klein bottle!

    Now we have shown that any tangent vector is perpendicular to the gradient of the function at that point on the level set. Now the definition of a tangent vector i gave earlier is quite painful to work with in this setting and it would be great if there was an easier way of talking about tangent vectors! Well luckily there is, but i wont prove this next statement as I'm not sure you know enough about integral curves for it to make sense.

    Fact: Any vector that is orthogonal to the gradient of a regular function at a point is a tangent vector to the level sets of that function at that point.

    If you feel adventurous what you need to do here is to show that for any vector that is orthogonal to the gradient there exists a smooth curve on the surface that has that vector as a tangent vector to it self at that point. This boils down to solving a differential equation as you could guess.

    So if we restrict our selves to regular functions we get an easy way of finding tangent vectors at a point, which is handy to say the least. Lets do an example, going back to our friend the unit sphere. Now it is the level set of the function f(x, y, z) = x<sup>2</sup> + y<sup>2</sup> + z<sup>2</sup> (namely f<sup>-1</sup>(1)) and as i stated earlier our function f is regular on R<sup>3</sup> / {(0,0,0)}. Now as you mentioned earlier wallaby, find curves on this surface is not an entirely trivial problem and finding all curves on the sphere through a point seems almost impossible to do so how can we find all the tangents at a point on the sphere?

    Well lets look back at what i stated earlier, Any vector that is orthogonal to the gradient of a regular function at a point is a tangent vector to the level sets of that function at that point. So lets look at the gradient of the function f which is grad(f)(x, y, z) = (2x, 2y, 2z) and consider the north pole of the sphere i.e. the point (0,0,1) so we are trying to find all vectors which are orthogonal to the vector (0,0,1). Now its an easy exercise to see that any vector of the form a(1, 0, 0) + b(0, 1, 0) where a and b are real numbers is a vector which is orthogonal to the gradient vector given. Also notice that the dimension of this plane happens to be the same as the surface in question. Now look what we have just done, we have found all the tangent vectors to the sphere at the point (0,0,1) with out even mentioning a curve on the sphere!

    This family of vectors defines a plane of tangent vectors at that point and it suggests the following definition:

    The tangent plane at a point p of a smooth, orientable surface f<sup>-1</sup>(c) is the set of all vectors (p, v) where v . grad(f)(p) = 0.

    Now as an exercise prove that the tangent plane is actually a vector space, and for more credit prove that it is n-dimensional at each point
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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    well the properties of vector addition and scalar multiplication on tangent vectors v to a point p, (p,v), are such that.

    (p, v) + (p, w) = (p, v+w)
    =(p, w+v)
    hence the axioms of vector addition apply. hence addition of tangent vectors to a point p forms an abelian group.

    a(p,v)= (p,av)
    hence the axioms of scalar multiplication apply.
    hence scalar multiplication on tangent vectors forms a (monoid i've heard it refered to) group of sorts.

    hence the collection of all vectors v satisfying the conditions, v<sup>.</sup> grad(f)=0, is a vector space. since the tangent plane is a collection of such vectors, the tangent plane is a vector space.

    further more we can define a vector v in this space such that v = a<sup>i</sup>e<sub>j</sub>, where a certain summation convention is used.

    looking back over this i think its time for bed.
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    Quote Originally Posted by river_rat
    Ah i see guitarist went running a head here so let me pull everyone back and firstly give a "rough" idea of want the tangent space at every point of the surface is.
    Well "ahead" is slightly to flatter me, but I apologize for interrupting your flow.

    I have to say, though, I find your coordinate-free approach intriguing, unfamiliar and slightly intimidating. I am pleased to see that Wallaby is keeping up rather better than I, perhaps because I have more baggage here than he has.
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    Almost Wallaby, you used the right facts but did not show what you needed to show! Remember, mathematics is all about showing things properly so its important not to skip important steps.

    So consider two vectors u and v from the tangent plane and let a and b be real numbers. We want to show that a u + b v is also an element of the tangent plane. Now this is the part you missed, you need to show that the inner product of a u + b v and grad(f)(p) is zero. To save us some notation lets call grad(f)(p) = w.

    Then w.(a u + b v) = a(w.u) + b(w.v) [as the inner product is linear as you mentioned] = a.0 + b.0 [as u and v are elements of the tangent plane] = 0 which shows that a u + b v is an element of the tangent plane and thus the tangent plane is a vector subspace of R<sup>n+1</sup> and is thus a vector space in its own right. This is a standard trick to show something is a vector space btw, you find a larger vector space which contains the set you are interested in and then show that this set is actually a subspace of the larger vector space.

    Now to show that the tangent plane is n-dimensional needs the projection theorem (and that is why it is an extra credit question) but you can motivate the result as follows. We can find an orthonormal set of basis vectors for R<sup>n+1</sup> rather easily (just take the normal basis vectors for example) and we can rigidly rotate this system that the basis vector e<sub>1</sub> points in the direction of the gradient which we have used to define the tangent plane. Now the remaining rotated basis vectors are all still orthogonal to the vector parallel to the gradient vector, span the tangent plane and there are n of them which shows that the tangent plane is n-dimensional.

    So lets continue and give an important theorem which is very handy to know. Now lets consider a function g : R<sup>n+1</sup> -> R and suppose we want to find the maximums and minimums of this function on some smooth surface f<sup>-1</sup>(c). For example let f(x, y, z) = xyz and g(x, y, z) = 2xy + 2yz + 2xz then the surface f<sup>-1</sup>(1) is the surface of all rectangular box dimensions which give a total area of 1 and g(x, y, z) is the total surface area of that box. Now you can guess that you want to find the box with the minimum surface area as that would use up the least material to build so the question is how do we minimize the function g on the surface f<sup>-1</sup>(1)? What we need is the following theorem:

    Theorem : Let f<sup>-1</sup>(c) be a smooth n-dimensional surface and let g:U -> R be a smooth function. Now if the point p is an extremal point of the function g on the surface f<sup>-1</sup>(c) then there exists a real number k such that grad(f)(p) = k grad(g)(p).

    The number k is called a Lagrange multiplier and this method can be generalized to more and more constraining surfaces.

    Well lets use this theorem to solve the problem i gave earlier. Now the grad(f)(x, y, z) = (yz, xz, xy) and the grad(g)(x, y, z) = (2y + 2z, 2x + 2z, 2y + 2x) so we have four equations we need to solve
    • yz = k(y+z)
    • xz = k(x+z)
    • xy = k(x+y)
    • xyz = 1

    Where the last equation defines the surface we are looking at. You can solve this system and find that x=y=z i.e. the minimum is given by a cube. (As an exercise, solve this system and make sure i haven't made a mistake)

    Now using the relationship between tangent vectors, curves on a surface and our characterization of tangent vectors can you see why this works?
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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    Quote Originally Posted by Guitarist
    Well "ahead" is slightly to flatter me, but I apologize for interrupting your flow.

    I have to say, though, I find your coordinate-free approach intriguing, unfamiliar and slightly intimidating. I am pleased to see that Wallaby is keeping up rather better than I, perhaps because I have more baggage here than he has.
    Its more an embedded version of differential geometry then the even scarier coordinate free approach! I like it as a way of starting as we can get to some interesting things quickly and relatively rigorously - for example the next major stop is orientations and then geodesics (and i haven't had to even mention covariant differentiations or local coordinate charts yet). Also the tangent vectors are very concrete - which is in my view a major bonus to the more modern but abstract introductions
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    is there a proof of the theorem, or is it too technical, because i'm fine with the maths of it but i don't know why that it is the case that grad(f)=k grad(g). i can't seem to visualise why its the case either.

    as far as the system of equations goes, x=y=z is indeed correct. i did type out the working out but somewhere between here and preview post it was lost. i'll type it out another time if you want it but i'm out of time for now.
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    I can give a proof but its easier to sketch the details and leave the technical stuff out.

    So lets consider a function g:R<sup>n+1</sup>->R which we want to maximize on a smooth surface given by the level set of a smooth function f:R<sup>n+1</sup>->R (say f<sup>-1</sup>(c)). Now if p is a maximum on the surface of the function g then it is a maximum for any curve that goes through that point. What we are doing here is to convert the n-dimensional problem to a one dimensional problem.

    So let a(t) be a curve on the surface with a(0) = p. Then the function h(t) = g(a(t)) achieves a local maximum at t = 0 (as then the function g has its maximum at p = a(0)) and so we know that h'(0) = 0. This implies that grad(g)(a(0)).a'(0) = 0 (write this out and see it is just the multi variable chain rule in action if you are uncomfortable with this) but we know that any tangent vector at the point p is of the form a'(0). So we know that for any vector v in the tangent plane at p that grad(g)(a(0)).a'(0) = grad(g)(p).v = 0. i.e. The vector grad(g)(p) is orthogonal to any vector in the tangent plane. Now this implies that the vector grad(g)(p) is parallel to the normal to the tangent plane grad(f)(p) and so there is a real number k with grad(f)(p) = k grad(g)(p).

    Does that make sense?
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    it's making sense. once i know more about the chain rule, after a bit of reading of the pile of books i dig out for occasions like these, i'm sure it will become even clearer.

    EDIT: yep its a lot clearer now that i've read further into partial differentiation.
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    Ah ok - its a handy result to know in its own right and you can easily extend it to multiple constraints (i.e. the intersection of multiple surfaces).

    The next stop on our journey would be vector fields and orientations but before we get there do you have any questions, comments or problems you would like to discuss?
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    no problems or queries at this point. i think.
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    Okay, first stop here would be smooth tangent vector fields but to talk about tangent fields we need to introduce a new concept - that of the tangent bundle. Now the tangent bundle (denoted by TS) in this setting is nothing more then the union of all the tangent planes of the surface in question.

    i.e. TS = U<sub>p in S</sub> T<sub>p</sub> where T<sub>p</sub> is the tangent plane at the point p.

    The interesting thing here is that the tangent bundle is in fact a more general kind of surface (a 2n - manifold) but I will get to manifolds, atlases and transition functions in due course.

    A smooth tangent vector field on a smooth orientable surface S or a smooth section of the tangent bundle is a function phi : S -> TS given by phi(p) = (p, phi(p)) where phi is a smooth function which picks a tangent vector from each tangent plane at the point p.

    I hope that definition made sense, for example lets consider the unit circle again and define the function phi(p) = (p, phi(p)) = (x, y, -y, x). You can check that this is a smooth tangent vector field on the unit circle.

    Now here is something for you to think about - how would you define a smooth normal vector field on a surface S? And remember, if you have any questions please shout!
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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    so how would i define length or magnitude in a smooth vector field, i'm guessing its pretty close to the way i'd define it in a vector space, since we wish to define a norm of the tangent vectors and hence on the tangent bundle.

    the norm in this case would be some unitary function on a vector to produce a scalar. hence for some tangent plane, V, and the associated scalar field, F. we have that f:V -> F, where the function satisfies the following conditions:
    f(v) + f(w) = f(v+w)
    f(av) = a*f(v)
    for vectors v,w in V and some scalar a in F.

    if the tangent plane to any point p is a normed vector space then this implies that the vector field is normed.

    now the question i'm pondering right now is how to define length between two points on a surface, is there a way of defining this for all surfaces?
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    Hi, Wallaby!
    Well, it's not really my place to check river_rat's assignments, but I think you mis-read what you were being asked. I think the relevant words in the question are "normal" and "field", rather than "normed" and "space" in your answer. These concepts are not really transferable. (Hint: think products).

    Sorry for butting in, I'm way behind on this thread, through lack of time.
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    Thanks guitarist. Wallaby, do you want to take another bash at the problem?
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    well that was rather silly of me.

    i don't think i'll try to assault this any further as i'm not sure i know whats ment by normal. unless its got something to do with the direction the field runs in i just don't know.
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    Hey wallaby.

    A normal vector field is a vector field on a surface such that at each point p of the surface we assign a vector of (p, v) such that v is normal to the tangent plane at the point p. A vector v is normal to a plane if it is orthogonal to any vector that lies in that plane.

    For example lets consider the unit circle again. The vector field f(x, y) = (x, y, x, y) is an example of a normal vector field on the unit circle.

    Does that make sense? I went for a more casual definition here in case you were wondering, if you want you can try turn it into a more rigorous definition be my guest.
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    Hey river_rat: My first instinct for the normal field to a differentiable surface was to use the vector product, hence my hint to Wallers. I strongly suspect, though, that I can use the gradient, in fact you said as much:
    Quote Originally Posted by river_rat
    So we know that for any vector v in the tangent plane at p that grad(g)(a(0)).a'(0) = grad(g)(p).v = 0. i.e. The vector grad(g)(p) is orthogonal to any vector in the tangent plane.
    it merely requires to iterate across all points p<sub>i</sub> of S, surely?
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    The gradient gives you one example guitarist - its what I used to create my example In fact a simple dimensional argument gives you that any normal vector field is of the form (p, g(p)*grad(f)(p)) where g : S -> R is a smooth function. Using that fact you can prove that any surface only admits two unit normal vector fields (which is our next stop).
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    spent the weekend reading up on the cross product, still don't understand how the definition of a normal vector field is made rigourous though.
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    Hey Wallers! I'll leave the rigour to river_rat, but, as we're doing geometry here, a mind-picture might be helpful.

    You were given a surface S. To each point p in S you can associate a vector space, T<sub>p</sub>, this being the space of all vectors tangent to S at p. Clearly this generates a tangent plane at the point p in S.

    A vector normal at p can only be defined with reference to some plane, in this case the tangent plane at p, the vector normal being a vector that is orthogonal to each v<sub>p</sub> in T<sub>p</sub>.

    Then for each T<sub>p<sub>i</sub></sub> in S, we may have a vector normal at each p<sub>i</sub>. Assume for now there is one and only one. Then for the tangent bundle ∪<sub>i ∈I</sub> T<sub>p<sub>i</sub></sub>, the class of vectors normal to each T<sub>p<sub>i</sub></sub> at p<sub>i</sub> constitutes a normal vector field on S.

    I know that this is easily shown using the vector (cross) product for a differentiable surface S, and I think, as r_r and I were discussing, I'm pretty comfortable with the idea that the gradient will do nicely too.

    I have a bit of arithmetic to do to show this, but see how this sits with you.
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    Are you still with us Wallaby or should i back track a bit? Please if you are feeling lost shout.
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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    yea i'm still with you, just had to take care of an assignment over the weekend.

    i have to say i don't fully know if i'm on track or not, its pretty easy to just say that the cross product of two vectors on the tangent plane is a normal vector which implies that there is a normal vector at every point p where a tangent exists and then from there you can define a normal vector field. it makes sense but i don't know if i'm missing something or if i'm ready to roll along.
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    You can't use cross products though Wallaby - the cross product is a tool which is only available when you are in R<sup>3</sup> which we are not.
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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    so how do we find the normal vector at a point p in R<sup>2</sup> or R<sup>n</sup>? is there a way or does it vary with the kind of object or the dimensions?
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    Well the trick is to remember the definition

    Now a general plane is a n dimensional subspace of R<sup>n+1</sup> so we can find a basis for the plane consisting of n vectors (say v<sup>1</sup>,..,v<sup>n</sup>). Now a vector w is a normal to a plane if for any vector v in the plane we have that w.v = 0.

    Now any vector v on the plane is the linear combination of the basis vectors for that plane. So v = a<sub>i</sub>v<sup>i</sup> (where i have assumed the Einstein summation convention) so substituting this into our original definition shows that w.v<sup>i</sup> = 0 (as a basis is linearly independent etc). So if w is a normal to the plane then w.v<sup>i</sup> = 0 for i = 1..n. The reverse implication is also true and i leave it as an exercise.

    So w is a normal <=> w.v<sup>i</sup> = 0 for i = 1..n.

    Now w can be represented as (w<sub>1</sub>, w<sub>2</sub>, ... , w<sub>n+1</sub>) in the standard basis for R<sup>n+1</sup>.
    Similarly each basis vector v<sup>i</sup> can be represented as (v<sup>i</sup><sub>1</sub>, v<sup>i</sup><sub>2</sub>, ... , v<sup>i</sup><sub>n+1</sub>). So we have n linear equations for n+1 variables. So the can solve the equation and find a vector that spans the solution space. If we normalize that vector then we have found a unit normal to the plane.

    Perhaps an example would help? Consider the xy plane in R<sup>3</sup> which is spanned by the vectors v<sup>1</sup> = (1, 0, 0) and v<sup>2</sup> = (0, 1, 0). We want to find a unit normal to the plane, so let w = (w<sub>1</sub>, w<sub>2</sub>, w<sub>3</sub>).

    w.v<sup>1</sup> = (1, 0, 0).(w<sub>1</sub>, w<sub>2</sub>, w<sub>3</sub>) = w<sub>1</sub> = 0.

    w.v<sup>2</sup> = (0, 1, 0).(w<sub>1</sub>, w<sub>2</sub>, w<sub>3</sub>) = w<sub>2</sub> = 0.

    So the solution space is the set of all vectors {(0, 0, t) : t in R}. Its easy to find a unit spanning vector (0, 0, 1) which is the obvious unit normal to the xy plane.

    Any questions?
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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    Quote Originally Posted by river_rat
    Well the trick is to remember the definition

    Now a general plane is a n dimensional subspace of R<sup>n+1</sup> so we can find a basis for the plane consisting of n vectors (say v<sup>1</sup>,..,v<sup>n</sup>). Now a vector w is a normal to a plane if for any vector v in the plane we have that w.v = 0.

    Now any vector v on the plane is the linear combination of the basis vectors for that plane. So v = a<sub>i</sub>v<sup>i</sup> (where i have assumed the Einstein summation convention) so substituting this into our original definition shows that w.v<sup>i</sup> = 0 (as a basis is linearly independent etc). So if w is a normal to the plane then w.v<sup>i</sup> = 0 for i = 1..n. The reverse implication is also true and i leave it as an exercise.
    ok so we're given a pair of vectors w and v such that w<sup>.</sup>v = 0, since a vector is a combination of its basis vectors we have that w<sup>j</sup><sup>.</sup>v<sup>i</sup> = δ<sup>ij</sup> = 0 for i=1,2,3....n and some j not equal to i. for a vector v in the tangent plane R<sup>n</sup>, subspace of R<sup>n+1</sup>, then the vector w who's basis is orthogonal to v<sup>i</sup> must lie in the remaining subspace, if i had to call it something i'd say: R<sup>n+1-n</sup>, hence a vector w is orthogonal to the tangent plane and is a normal vector.

    i know theres bound to be a simpler way of doing that...but seeing as i've gone and done it, or done something anyway, i'll stick with that.

    Quote Originally Posted by river_rat
    Any questions?
    don't believe so.
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    Almost wallaby, let me just clear up your proof a bit.

    Let v be any vector in the plane and suppose w is another vector such that w.v<sup>i</sup> = 0 for any basis vector v<sup>i</sup> for the plane.

    We have to show that w.v = 0. Now v is a vector in the plane so we can write it as a linear combination of the basis vectors v<sup>i</sup>, say v = a<sub>1</sub>v<sup>1</sup> + a<sub>2</sub>v<sup>2</sup> + ... + a<sub>n</sub>v<sup>n</sup>.

    But then w.v = w.(a<sub>1</sub>v<sup>1</sup> + a<sub>2</sub>v<sup>2</sup> + ... + a<sub>n</sub>v<sup>n</sup>) = a<sub>1</sub> w.v<sup>1</sup> + a<sub>2</sub> w.v<sup>2</sup> + ... + a<sub>n</sub> w.v<sup>n</sup> (as the inner product is linear) = 0 + 0 + ... + 0 = 0.

    i.e. w is a Normal to the plane spanned by the basis vectors {v<sup>1</sup>, v<sup>2</sup>, ... , v<sup>n</sup>}

    Okay, now our next stop is to show that any surface as we have defined one only admits two unit normal fields. To do this though needs a little bit of topology. For starters can you recall the definition of a continuous function?
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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    differentiable at any point x in the domain.

    EDIT: well when i think about it some more thats actually a smooth function.

    the only thing about continuity i've ever been taught is that a continuous function can be drawn without lifting your pencil from the page, ohh how useful a definition that is .
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    Quote Originally Posted by wallaby
    differentiable at any point x in the domain.

    EDIT: well when i think about it some more thats actually a smooth function. :oops:

    the only thing about continuity i've ever been taught is that a continuous function can be drawn without lifting your pencil from the page, ohh how useful a definition that is :roll:.
    As ratters seems to be bogged down with work, let me try and help you out.

    The standard definition in school texts is something like this: suppose f: R → R, then for x in R, f(x) in R, for every ε > 0 in R there is some δ > 0 in R such that for each x<sub>1</sub>, where x - x<sub>1</sub> < δ, it is the case that f(x<sub>1</sub>) - f(x) < ε.

    Which loosely says that, as x<sub>1</sub> gets "closer" to x, so does f(x<sub>1</sub>) get arbitrarily "close" to f(x). Or to use your pencil analogy, the graph of f(x) vs. x, there are no gaps or jumps, except possibly at x itself.

    I suspect, though, you being were asked about the topological definition (but I'm not sure) But, that's simply a generalization of the above, like:

    Let S, T be topological spaces, s in S, t in T. Then f: S → T is continuous at s iff, for every open neighbourhood ε of f(s) in T there is an open neighbourhood δ of S such that δ subset f<sup>-1</sup>, (or f(δ) subset ε).

    Can you see the connection? Do you know what f<sup>-1</sup> denotes? Say if not, it may not be what it first appears to be!
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    Quote Originally Posted by Guitarist
    As ratters seems to be bogged down with work, let me try and help you out.

    The standard definition in school texts is something like this: suppose f: R → R, then for x in R, f(x) in R, for every ε > 0 in R there is some δ > 0 in R such that for each x<sub>1</sub>, where x - x<sub>1</sub> < δ, it is the case that f(x<sub>1</sub>) - f(x) < ε.

    Which loosely says that, as x<sub>1</sub> gets "closer" to x, so does f(x<sub>1</sub>) get arbitrarily "close" to f(x). Or to use your pencil analogy, the graph of f(x) vs. x, there are no gaps or jumps, except possibly at x itself.
    i wish i had those school texts, not that my school texts are bad but they seem to prefer an intuitive explanation at the cost of formalism.

    Quote Originally Posted by Guitarist
    I suspect, though, you being were asked about the topological definition (but I'm not sure) But, that's simply a generalization of the above, like:

    Let S, T be topological spaces, s in S, t in T. Then f: S → T is continuous at s iff, for every open neighbourhood ε of f(s) in T there is an open neighbourhood δ of S such that δ subset f<sup>-1</sup>, (or f(δ) subset ε).

    Can you see the connection? Do you know what f<sup>-1</sup> denotes? Say if not, it may not be what it first appears to be!
    i remember a certain instrumentalist hinted that i should look in the Topology Primer thread for some more reading on the topological definition, well before the post vanished and i was left wondering if i had seen anything at all, so i've been reading all that i can understand for a while now.

    if i had an element r of R<sup>1</sup> where r is any element on the interval say [0,1], i guess whether it's a closed or open interval will affect this but ok, would the neighbourhood of r be all s: 0<s<1, where s is an element of R<sup>1</sup>. could be said clearer i guess but if its not clear then i guess i'm really just asking for clarification on neighbourhoods and probably open sets.

    as for the f<sup>-1</sup> denotation i'm reminded of its use during the introduction to level sets. in this way i think it symbolises either the set S or the topological space S,T. i'm betting on it being S though.
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    Quote Originally Posted by wallaby
    i remember a certain instrumentalist hinted that i should look in the Topology Primer thread for some more reading on the topological definition, well before the post vanished
    Ha! yes, I deleted it pretty quick as it looked uncomfortably like a conceited intrusion into ratty's excellent thread. In the same spirit, I won't say too much because it's his thread and anyway, he knows far more topology that I do.

    if i had an element r of R<sup>1</sup> where r is any element on the interval say [0,1], i guess whether it's a closed or open interval will affect this but ok, would the neighbourhood of r be all s: 0<s<1, where s is an element of R<sup>1</sup>.
    Yes, more or less. River_rat and I had some discussion about this a while back - some people, me included, define a neighbourhood as an open set containing the point r. River_rat prefers the alternative, perfectly legitimate definition which doesn't specify openness. so to satisfy him, you're better saying "open neighbourhood". Anyway, you don't really need to be specific about 0<s<1; to use (0,1) subset R with the standard topology is sufficient to define (0,1) as an open set containing r, i.e. an open neighbourhood of r.

    So f is continuous at r if for every open set containing f(r), there is an open set containing r.
    A more general form of words is often used: f is continuous if for every open set U in R that f<sup>-1</sup>(U) is open in R. Or, if you prefer, f is continuous if it maps open sets to open sets.

    It pays to be a bit careful here. R is certainly a topological space but, like any such space, what we think of as the real line (I called it the standard topology above) is just one of several possible topologies on the set of real numbers. Also, in most topologies, the definition of open set does not coincide with that usually given for R, i.e. (a,b) open, [a,b] closed.

    as for the f<sup>-1</sup>
    Well it's used generally in math. One thing it doesn't mean is f<sup>-1</sup>(x) = 1/f(x)! It's called the pre- image. So let f:X → Y. X is the domain, Y is the codomain, f(x) is the image of x under f. f<sup>-1</sup>(x) is the set of values x may take to deliver a certain f(x). Given x, we have a machine f that allows us to compute f(x): given f(x), we have a machine f<sup>-1</sup> that allows us to recover a set of x's. Only where this set has at most and at least one element (this is called bijectivity) may we think of f<sup>-1</sup> as an inverse function.

    Like this. Suppose f(x) = x<sup>2</sup> = 4. x = ±2, so f(x) has no inverse. Suppose g(x) = x<sup>3</sup> = 8. x = 2, so g(x) has an inverse g<sup>-1</sup> and g<sup>-1</sup>(g(x)) = x, reciprocally. In topology, this condition has a special name - did you come across it, it's in there somewhere, I believe?

    To sum up this typically over-long post (I see I broke my undertaking above, ah well). f is continuous if, for every open set U containing the image object f(x), there is an open set f<sup>-1</sup>(U) containing the pre-image object(s) f<sup>-1</sup>(x).
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    Quote Originally Posted by Guitarist
    Ha! yes, I deleted it pretty quick as it looked uncomfortably like a conceited intrusion into ratty's excellent thread. In the same spirit, I won't say too much because it's his thread and anyway, he knows far more topology that I do.
    well then until river_rat returns i guess its back to the learning about conics, differential equations and homogeneous linear equations.

    well that and reading through what i can understand of the topology thread. which will probably bring the odd question or two to mind.
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    Hey Wallerby! Keep this to yourself, OK? I was in our college book-store yesterday and found a 2nd hand copy of Stewart's Calculus. I didn't buy it, but he gives a very thorough treatment of continuity, limits and all that, but of more interest in the present context, he goes into some detail about level sets, surfaces, the grad operator and all that.

    If you can afford it, buy it (this 2nd hand copy was 35 GBP, er, what? 50 euro, about the same USD?) But it's a big bastard, so page for page it's a steal. Just a thought.

    Like I say, don't breathe a word about this
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    thanks Guitarist i'll have an inconspicuous look around for it.
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    Hey wallaby - sorry for the long delay here.

    So are you ready to continue? What we need from topology is the idea that a continues function cannot tear connected space into two pieces. Well the first thing you should be asking is what is a connected space!

    Definition: A connected space X is one that cannot be written as the disjoint union of two open subsets.

    Its an exercise in the basic properties of the real numbers to show that the intervals are the connected subsets of the real line for example. From now on I will assume our surfaces are connected (which is not a huge drawback actually - can you think why?)

    Now it is an exercise in topology to show that the two point set {a, b} with a<>b and the topology it inherits from the real line (i.e. the discrete topology) is not connected. Also it is easy to see (can you show it?) that the continuous image of a connected space is connected. These two facts together show us that we cannot have a continuous map from a connected space onto the two point set . This is the bit of topology we need for the next theorem.

    Theorem: Let S = f<sup>-1</sup>(c) be a connected smooth surface then S admits only two smooth unit normal fields.

    Proof: It is relatively easy to see (see above posts) that the normal fields v<sub>1</sub>(x) = (x, grad(f)(x)/||grad(f)(x)||) and v<sub>2</sub>(x) = (x, -grad(f)(x)/||grad(f)(x)||) are two smooth unit normal fields on S and that v<sub>1</sub>(x) <> v<sub>2</sub>(x). So we have to show that these are the only two such normal fields.

    So suppose v<sub>3</sub>(x) was another smooth (thus continuous) unit normal field. Then as the normal vectors at a point are spanned by v<sub>1</sub>(x) we know that v<sub>3</sub>(x) = g(x). v<sub>1</sub>(x) where g : R -> R is a smooth function. Also as this is a normal field we know that g(x) = ±1 for all x in S. Now g is smooth and so it is continuous and thus it cannot be onto the two point set {-1, 1} as S is connected. Thus g(x) = +1 for all x and so v<sub>3</sub>(x) = v<sub>1</sub>(x) or g(x)=-1 for all x and so v<sub>3</sub>(x) = v<sub>2</sub>(x) for all x.

    Thus S admits only two smooth unit normal vector fields. QED

    Now for each surface you can choose a unit normal field, and we call that choice of unit normal field an orientation. So any smooth connected surface S only admits two orientations. You should think of the two ways you can orientate the x-y-z axes in R<sup>3</sup> depending if you use your right or left hand rule.

    Now it is customary to consider a non-orientable surface about now - the easiest example would be the mobius strip. Take a rectangular piece of paper and give it a half twist and glue the free ends together. The surface you have now is called a mobius strip. Can you see why it is not orientable? As a hint consider what happens to a unit normal vector as you move it around the mobius strip.
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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    Quote Originally Posted by river_rat
    So are you ready to continue? What we need from topology is the idea that a continues function cannot tear connected space into two pieces. Well the first thing you should be asking is what is a connected space!

    Definition: A connected space X is one that cannot be written as the disjoint union of two open subsets.

    Its an exercise in the basic properties of the real numbers to show that the intervals are the connected subsets of the real line for example. From now on I will assume our surfaces are connected (which is not a huge drawback actually - can you think why?)
    welcome back.
    Our surfaces, well the ones we've talked about so far anyway, are imbeded in R<sup>2</sup> or R<sup>3</sup>, which can't be defined by the disjoint union of two open subsets. or is it more to do with the surfaces being continuous.

    Quote Originally Posted by river_rat
    Now it is an exercise in topology to show that the two point set {a, b} with a<>b and the topology it inherits from the real line (i.e. the discrete topology) is not connected. Also it is easy to see (can you show it?) that the continuous image of a connected space is connected. These two facts together show us that we cannot have a continuous map from a connected space onto the two point set . This is the bit of topology we need for the next theorem.
    easy to see... not easy for me to show, don't know how i would show it.

    Quote Originally Posted by river_rat
    Now it is customary to consider a non-orientable surface about now - the easiest example would be the mobius strip. Take a rectangular piece of paper and give it a half twist and glue the free ends together. The surface you have now is called a mobius strip. Can you see why it is not orientable? As a hint consider what happens to a unit normal vector as you move it around the mobius strip.
    at any point on the strip there are two normal vectors both of which are defined by v(x)=(x, grad(f)(x)), which doesn't make much sense for a one sided object.
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    Quote Originally Posted by wallaby
    Our surfaces, well the ones we've talked about so far anyway, are imbeded in R<sup>2</sup> or R<sup>3</sup>, which can't be defined by the disjoint union of two open subsets. or is it more to do with the surfaces being continuous.
    Hey wallaby, you need to be careful here. Even though the larger spaces in question are connected does not imply that any subspace is connected - just think of any two disjoint open balls in R<sup>3</sup> as your space. Nothing in the definitions i gave really forces the surfaces to be connected. It does not take a lot of maths to show that every surface is just the union of its largest connected pieces though. So instead of working on the surface as a whole we just split it up into those connected pieces and work on them instead - which gets us around the connected space problem.

    easy to see... not easy for me to show, don't know how i would show it.
    First question - the real line is hausdorf so we can find two disjoint open sets U and V such that a is in U and b is in V. Well then {a} is open in {a, b} and {b} is also open and {a, b} = {a} union {b}

    Second question - Let X be a connected space and wlg let f:X->Y be a continuous onto map and let us suppose that Y is not connected. Well then we can write Y = A union B where A and B are two disjoint open sets. However f is continuous so f<sup>-1</sup>(A) and f<sup>-1</sup>(B) are also open and X = f<sup>-1</sup>(A) union f<sup>-1</sup>(B). But X is connected so f<sup>-1</sup>(A) intersect f<sup>-1</sup>(B) cannot be empty, so let t be in f<sup>-1</sup>(A) intersect f<sup>-1</sup>(B). But then f(t) is in A and f(t) is in B so A and B are not disjoint, a contradiction. So Y must be connected

    at any point on the strip there are two normal vectors both of which are defined by v(x)=(x, grad(f)(x)), which doesn't make much sense for a one sided object.
    Close, you can't use the grad operator here though as you don't have any function to take the gradient of! It is easier to see the problem if you actually make a mobius strip and look at what happens to a normal vector as you go around the circle which cuts the strip into two.
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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    the proofs almost look easy

    Quote Originally Posted by river_rat
    Close, you can't use the grad operator here though as you don't have any function to take the gradient of! It is easier to see the problem if you actually make a mobius strip and look at what happens to a normal vector as you go around the circle which cuts the strip into two.
    so we're left with rotations are we?
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    Ah as my computer sketching skills leave a lot to be desired i hoped onto google and found the followig websites which draw the picture i was trying to

    http://www.math.umn.edu/~nykamp/m237...s/surfmoebius/ (uses java)

    http://www1.kcn.ne.jp/~iittoo/us27f_mobi.htm (no java)

    The just of the matter is that going once around a mobius strip inverts your normal vector - which means you cannot have a smooth non-vanishing normal vector field on the mobius strip (though you can have a smooth non-vanishing tangent vector field, something the standard sphere does not allow and any hairdresser knows)

    Next stop should be covariant derivatives of vector fields and geodesics
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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    Quote Originally Posted by river_rat
    Ah as my computer sketching skills leave a lot to be desired i hoped onto google and found the followig websites which draw the picture i was trying to

    http://www.math.umn.edu/~nykamp/m237...s/surfmoebius/ (uses java)

    http://www1.kcn.ne.jp/~iittoo/us27f_mobi.htm (no java)

    The just of the matter is that going once around a mobius strip inverts your normal vector - which means you cannot have a smooth non-vanishing normal vector field on the mobius strip (though you can have a smooth non-vanishing tangent vector field, something the standard sphere does not allow and any hairdresser knows)

    Next stop should be covariant derivatives of vector fields and geodesics
    i'm amazed at how much time i just spent playing around with the slider in that java link.

    time to go look back on covariance i guess.
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    Quote Originally Posted by wallaby
    i'm amazed at how much time i just spent playing around with the slider in that java link.

    time to go look back on covariance i guess.
    Lol, you and me both

    Okay - so how did the reading go? We want to reach the point where we can do calculus on vector fields but this is a multi step process and we first start with curves.

    Let a:I -> S where I is a subset of R be some parametrised curve on our surface S. Now a vector field along a parametrised curve is a function which assigns a vector (p, v) at the point p = a(t) for every t in I. So instead of giving you a vector at each point of the surface we are only looking at a specific curve and the points that are on that curve which lies on the surface. We also need to define functions along curves, luckily these are easy to define. A function on a curve a:I->S is any function f:I->R.

    For example think of the vector field along a parametrised curve a:I->R^2 where a(t)=(t, 1/2 t^2) and I = [0, 1] which assigns the velocity vector v = (1, t) of that path to every point a(t) along the path. Then the function f:I->R defined by f(t) = ||(1, t)|| defines a function along that path and is called the speed of the particle

    Hopefully that all makes sense, if not shout.
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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  67. #66  
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    Quote Originally Posted by river_rat
    Lol, you and me both

    Okay - so how did the reading go? We want to reach the point where we can do calculus on vector fields but this is a multi step process and we first start with curves.
    the reading has had some mixed results.

    Quote Originally Posted by river_rat
    Hopefully that all makes sense, if not shout.
    makes sense.
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    If you have any questions from the reading just shout

    Now we want to be able to take the derivative of these vector functions along curves on the surface in question. Now this vector field assigns a vector v(t) = (v<sub>1</sub>(t), v<sub>2</sub>(t), ... , v<sub>n</sub>(t)) at the point p = a(t) along the curve a(t). We call a vetor field along a curve smooth if the component functions v<sub>i</sub> are smooth in the normal sense. So we can define the derivative of this vector field to be the function that assigns the vector v'(t) = (v'<sub>1</sub>(t), v'<sub>2</sub>(t), ... , v'<sub>n</sub>(t)). Notice what we have done here, we have extended the usual ideas of multi variable differentiation to surfaces in R<sup>n</sup>.

    It would be great if this new derivative behaved like the one we are used to, it is your exercise to show that it does. i.e. Show that if X(t) and Y(t) are two vector fields along the curve a(t) and f(t) is a function along the same curve then

    1. (X + Y)' = X' + Y'
    2. (f X)' = f' X + f X'
    3. (X.Y)' = X'.Y + X.Y'

    Also, can you define the acceleration vector field of a curve a(t) on a surface S?
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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    Quote Originally Posted by river_rat
    If you have any questions from the reading just shout
    i'll start making a list then

    Quote Originally Posted by river_rat
    Show that if X(t) and Y(t) are two vector fields along the curve a(t) and f(t) is a function along the same curve then

    1. (X + Y)' = X' + Y'
    for some vector field X and Y we have that (X + Y)' = (X + Y)'(t), when X and Y have been definied on some parametrised curve. by vector addition we have that (X + Y)'(t) = X'(t) + Y'(t) => (X + Y)' = X' + Y'

    Quote Originally Posted by river_rat
    2. (f X)' = f' X + f X'
    3. (X.Y)' = X'.Y + X.Y'
    reminds me of the product rule, so why i don't think i can prove it is something that escapes me right now.

    Quote Originally Posted by river_rat
    Also, can you define the acceleration vector field of a curve a(t) on a surface S?
    hmmm... now if we have that a velocity vector field on the curve a(t) is given as the derrivative of the function a: I -> S, a'(t), then there exists a vector field that is the second derrivative of the function a: I -> S, a''(t), which we call acceleration.
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    Hey Wallaby, sorry for the long radio silence again - i finally made some progress on my masters so i was stuck there.

    Well the proofs i asked for last time pretty much involve writing everything in component form and using the normal product rules etc so i will leave it to you to fill in the details.

    Okay now its time to define the first interesting thing on a surface, namely a geodesic. A geodesic on a surface S is a curve a : I -> S such that the acceleration of the curve (a''(t)) is parallel to the normal at the surface at the point a(t). There is a "nicer" characterisation of geodesics but it will have to wait for me to construct a new sort of derivative on a surface.

    For example, the curve a(t) = (cos(t), sin(t), t) is a geodesic on the cylinder in R<sup>3</sup> given by the equation x<sup>2</sup> + y<sup>2</sup> = 1.

    Now it is your job to show that a geodesic has a constant speed along its path
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    i'm patient, so however long it takes doesn't bother me.

    2. (f X)' = f' X + f X'
    since the function f: I -> R is a scalar valued function it has no components, thus:
    (fX)' = [ f( X<sub>1</sub>, X<sub>2</sub>, ... , X<sub>n</sub>)]'

    = f'( X<sub>1</sub>, X<sub>2</sub>, ... , X<sub>n</sub>) + f(X'<sub>1</sub>, X'<sub>2</sub>, ... , X'<sub>n</sub>

    = f'X + fX' QED

    3. (X.Y)' = X'.Y + X.Y'
    breaking into components,
    (X<sup>.</sup>Y)' = [(X<sub>1</sub>, X<sub>2</sub>, ... , X<sub>n</sub>) <sup>.</sup> (Y<sub>1</sub>, Y<sub>2</sub>, ... , Y<sub>n</sub>)]'

    = (X<sub>1</sub><sup>.</sup>Y<sub>1</sub>, X<sub>2</sub><sup>.</sup>Y<sub>2</sub>, ... , X<sub>n</sub><sup>.</sup>Y<sub>n</sub>)'

    = ((X'<sub>1</sub><sup>.</sup>Y<sub>1</sub> + X<sub>1</sub><sup>.</sup>Y'<sub>1</sub>), (X'<sub>2</sub><sup>.</sup>Y<sub>2</sub> + X<sub>2</sub><sup>.</sup>Y'<sub>2</sub>), ... , X'<sub>n</sub><sup>.</sup>Y<sub>n</sub> + X<sub>n</sub><sup>.</sup>Y'<sub>n</sub>))

    = X'<sup>.</sup>Y + X<sup>.</sup>Y' QED

    unfortunately in my flu infected state i havn't even been able to show that a geodesic maintains a constant speed along its path for the example given. so far i've been thinking that if i just took the derrivative of the curve to find the tangent or velocity vector then i should use the function f: I -> R to find the speed along the curve.
    however when doing this in the example the function f(t) is non-constant, well for me it is anyway.
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    Ah hope the flu is doing better. Well lets take a look at that proof i asked :

    Let c : R -> S be a geodesic curve (i.e. its acceleration vector is always parallel to the unit normal field on the surface S) and consider the velocity vector field along the curve c given by V : R -> S x R<sup>n</sup> (if the notation seems new this is just an easy way of writing the set of all vectors at all points on S, which introduces the idea of a vector bundle that I don't want to go into yet). Now the speed function is given by the function along the curve s : R -> R where s(t) = ||V(t)|| = (V(t).V(t))<sup>1/2</sup>. Now this function is constant iff the function ss(t) = (s(t))<sup>2</sup> is (proof left for you).

    Now ss'(t) = d/dt(V(t).V(t)) = 2 V(t).V'(t) by those things i asked you to prove. But the acceleration vector field is parallel to the normal vector field so V'(t) = k(t) N(s(t)) and N(s(t)).V(t) = 0 by the theorem a few posts back so ss'(t) = 0 i.e. ss(t) is constant which implies that s(t) is constant i.e. Geodesics are constant speed curves

    Any questions? Secondly, do you know how to work out the projection of one vector onto another vector? As a hint think of the 2d case and try generalise.
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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  73. #72  
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    sorry about the lateness of my reply but i've had a bit of a busy week, finally .

    anywho time to hand in my homework i guess,
    if we have some function s(t) = (V(t)<sup>.</sup>V(t))<sup>1/2</sup>
    then the function ss(t), which i've interpreted as (ss)(t), = s<sup>2</sup>(t)
    = [(V(t)<sup>.</sup>V(t))<sup>1/2</sup>] <sup>.</sup> [(V(t)<sup>.</sup>V(t))<sup>1/2</sup>]
    = (V(t)<sup>.</sup>V(t))<sup>
    = (s(t))<sup>2</sup>

    as for projections, in the 2d case a projection would be the mapping of a 2d vector to one dimension, a function say P:R<sup>2</sup> -> R<sup>1</sup>.
    so i guess a generalisation of this would be that the projection maps any vector in a vector space (V) onto a subspace of V.
    am i close?
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  74. #73  
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    Don't think of the general case here wallaby as the definition of a projection in that case is a bit subtle (and doesn't look like what you would expect at all!). You have a a vector in R<sup>n+1</sup> and we want to project any given vector of R<sup>n+1</sup> onto that vector.

    Think of the 2 dimensional case here, think angles and inner products if you are really stuck!
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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  75. #74  
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    ok so the projection of a vector v in R<sup>2</sup> is given by vcos(theta). so would it be fitting to say that the projection in this case is just the scalar multiplication of a scalar and a vector. using cos(theta) is pretty restrictive to 2 dimensions though. havning said that i now get the sneaking suspision that transformations or operators may be involved, but my imagination has been known to run away with me.
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    Hey wallaby, yep your imagination is running riot

    Let me rephrase the question, I am giving you two non-zero vectors a and b. Now the vector b spans a one dimensional linear subspace [b] = {k b : k is in R} and i want you to project the vector a into this subspace. You are on the right track, now think of how you define the angle between two vectors in general...
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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  77. #76  
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    Quote Originally Posted by river_rat
    Hey wallaby, yep your imagination is running riot

    Let me rephrase the question, I am giving you two non-zero vectors a and b. Now the vector b spans a one dimensional linear subspace [b] = {k b : k is in R} and i want you to project the vector a into this subspace. You are on the right track, now think of how you define the angle between two vectors in general...
    i'm assuming that by, "in general," you are refering to the 2 dimensional case. in which case the angle can be expressed as,
    Θ = arcos (adjacent component/vector) or in this case,
    =arcos (k<sup>1</sup>e<sub>1</sub>/a)
    =arcos (k<sup>1</sup>b/a) (where e = b)
    thus cos Θ = k<sup>1</sup>b/a
    hence acos Θ = k<sup>1</sup>b
    which is a vector in the one dimensional linear subspace we are projecting onto.
    therefore acos Θ is the projection of a onto b QED?
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    I'm not too sure what you are getting at there Wallaby - I must confess.

    Let me give you a hand up:

    Consider two vectors a and b in an inner product space. Then the angle between them (theta) is defined to make the following true:

    a.b = |a| |b| cos(theta)

    See what to do now with your original idea?
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    well i was aiming to show that acos Θ was an element of the subspace spanned by b and hence the projection of a onto b but i guess that was unsucessful, and that would be because i was only working with one vector and not two when considering the angle.

    to tell you the truth i can't even think of what the answer is ment to look like anymore, all i'm doing now is aimlessly rearranging equations.

    if a.b = |a| |b|cos Θ
    = |a| kb cos Θ
    thus bcos Θ = a.b / |ka|
    which doesn't mean much to me.
    so i really don't know
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  80. #79  
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    ah, i was trying to find an easy way of drawing the picture but i don't have any of that software on this pc.

    Now the projection of a vector a onto a vector b is the vector |a| cos(theta) b where theta is the angle between the vectors a and b. Now a.b = |a| |b| cos(theta) so we can substitute for cos(theta) and get

    proj<sub>b</sub>(a) = a.b / |b| b

    So here k = a.b / |b| which is definitely a scalar so proj<sub>b</sub>(a) is an element of [b]

    Make sense?

    Edit : I shouldn't post at 1 in the morning, please note the corrections in red
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    yes it makes perfect sense, much to my relief.
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    lol, sadly it should not have made sense (please note corrections)

    Now here is a question for you, consider the circle x<sup>2</sup> + y<sup>2</sup> = 1 and the curve s(t) = (cos(t<sup>3</sup>/3), sin(t<sup>3</sup>/3)). Firstly work out the velocity vector field along this curve and determine if it is a tangent vector field or not. Then work out the acceleration vector field and determine if it is also a tangent vector field.

    What does this tell you about the normal derivative of tangent vector fields in general?
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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    ok so we have a circle given by the equation x<sup>2</sup> + y<sup>2</sup> = 1 and a curve on this surface given by S(t) = [ cos (t<sup>3</sup>/3), sin (t<sup>3</sup>/3)].
    thus the velocity vector at any point, t, along the curve is given by the derrivative of the curve at that point:
    v(t) = (-t<sup>2</sup> sin (t<sup>3</sup>/3), t<sup>2</sup> cos (t<sup>3</sup>/3))
    = t<sup>2</sup>( -sin (t<sup>3</sup>/3), cos (t<sup>3</sup>/3))

    for this vector to be a tangent vector then it must be orthogonal to the gradient function on the surface:
    grad(f)= (2x, 2y)
    = 2( cos (t<sup>3</sup>/3), sin (t<sup>3</sup>/3) (on the curve S(t))

    so since two vectors are orthogonal if the inner product of the two is zero, the velocity vector and the gradient of the surface will be orthogonal if:
    v(t).grad(f) = 0
    so v(t).grad(f) = t<sup>2</sup>( -sin (t<sup>3</sup>/3), cos (t<sup>3</sup>/3).2( cos (t<sup>3</sup>/3), sin (t<sup>3</sup>/3))
    = 2t<sup>2</sup>(-sin(t</sup>3</sup>/3)cos(t<sup>3</sup>/3 + cos(t<sup>3</sup>/3)sin(t<sup>3</sup>/3))
    =2t<sup>2</sup>(0)
    =0
    thus v(t).grad(f)=0 <=> v(t) is orthogonal to grad(f)
    therefore the velocity vector is also a tangent vector.
    (just as a note: didn't know if it was appropriate to evaluate it at t=0 or not but as this hasn't changed the result really so i decided to use t=t)

    now as for the acceleration.
    the acceleration vector is the second derrivative of the curve at any point t.
    therefore a(t) = -t<sup>4</sup>( cos(t<sup>3</sup>/3), sin(t<sup>3</sup>/3))
    now using the inner product of the acceleration vector and the gradient, or gradient of the gradient, function i ran into a bit of a dead end. however the inner product of the acceleration vector and the velocity vector shows that.
    a(t).v(t) = -t<sup>4</sup>( cos(t<sup>3</sup>/3), sin(t<sup>3</sup>/3)).t<sup>2</sup>(-sin(t<sup>3</sup>/3), cos(t<sup>3</sup>/3))
    = -t<sup>6</sup>(0)
    =0
    therefore a(t).v(t) = 0 <=> a(t) is orthogonal to v(t)
    hence a(t) is not a tangent vector but a normal vector.

    this tells me that when the velocity vector is the tangent vector at some point p then the acceleration vector will be the normal vector. but incase you were searching for something deeper i'll say that this could be telling me that the above conditions are satisfied on any curve that is a geodesic along the surface, maybe the word any is a bit too general as there always seems to be a counter example or other conditions.

    so hows all that look?

    EDIT: wow i just saw how long this post actually was, which is usually a good indicator that it could be done an easier way.
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    Hey wallaby, you made a few mistakes with your derivatives for the acceleration part of the question but the velocity part looks great Just remember that a. geometry calculations are usually long and messy, which motivates all the notation to hide that fact and b. don't forget the product rule!

    Okay, they point of that exercise was two fold. First to see if you remember how to check if something is a tangent vector (which you did) and secondly to show that the normal derivative is not nice. By that i mean it does not take tangent vectors and spit back tangent vectors at you. Now it would be nice if this actually was the case so i am going to define a new type of derivative called the covariant derivative.

    Defn : The covariant derivative of a tangent vector field X along a curve a(t) is given by X<sub>|t</sub> = X'(t) - (X'(t).N(a(t))) N(a(t)) where N is a unit normal vector field on the surface S.

    So what the covariant derivative does is subtract that part of the derivative of the tangent vector which does not lie in the tangent plane at the point a(t).

    The next thing for me to do would be to show that this actually acts like a derivative and is independent of the choice of normal vector field (remember we had two choices for the unit normal field), but first i want you to show me that the covariant derivative of a tangent vector field along a curve a(t) gives another tangent vector field along a(t).
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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  85. #84  
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    Quote Originally Posted by river_rat
    b. don't forget the product rule!
    i forget that too often.

    Quote Originally Posted by river_rat
    Defn : The covariant derivative of a tangent vector field X along a curve a(t) is given by X<sub>|t</sub> = X'(t) - (X'(t).N(a(t))) N(a(t)) where N is a unit normal vector field on the surface S.

    So what the covariant derivative does is subtract that part of the derivative of the tangent vector which does not lie in the tangent plane at the point a(t).
    EDITED: (don't do maths when recovering from general anaesthetic)

    so for some tangent vector field X(t) on the surface defined by the equation x<sup>2</sup>+y<sup>2</sup> = 1 such that:
    X(t) = t<sup>2</sup>(-sin(u), cos(u)) where u = t<sup>3</sup>/3
    the derivative of this vector field along the curve a(t) = (cos(u), sin(u)) is defined by the equation,
    X'(t) = (-2tsin(u) - t<sup>4</sup>cos(u), 2tcos(u) - t<sup>4</sup>sin(u))
    now since the derivative of a tangent vector field is not nice we will now find the covarient derivative and show that it is a tangent vector field.
    X<sub>|t</sub> = X'(t) - (X'(t).N(a(t)))N(a(t))
    if N(a(t)) = (x,y) = (cos(u), sin(u)) along the curve a(t)
    then X'(t).N(a(t)) = (-2tsin(u) - t<sup>4</sup>cos(u), 2tcos(u) - t<sup>4</sup>sin(u)).(x,y)
    = (-2txsin(u) - t<sup>4</sup>xcos(u) + 2tycos(u) - t<sup>4</sup>ysin(u))
    = (-2t(xsin(u) - ycos(u)) -t<sup>4</sup>(xcos(u) + ysin(u))
    = -t<sup>4</sup> [along the curve]

    thus X<sub>|t</sub> = (-2tsin(u) - t<sup>4</sup>cos(u), 2tcos(u) - t<sup>4</sup>sin(u)) + t<sup>4</sup>(x,y)
    = (-2tsin(u) -t<sup>4</sup>cos(u) + t<sup>4</sup>x, 2tcos(u) - t<sup>4</sup>sin(u) + t<sup>4</sup>y)
    which along the curve is
    = (-2tsin(u), 2tcos(u))
    = 2t(-sin(u), cos(u))

    now if grad(f) is 2(cos(u), sin(u)) along the curve a(t) then we mearly need to show that X<sub>|t</sub> . grad(f) = 0
    so X<sub>|t</sub> . grad(f) = 2t(-sin(u), cos(u)) . 2(cos(u), sin(u))
    =4t(-sin(u)cos(u) + cos(u)sin(u))
    =4t(0)
    =0
    thus X<sub>|t</sub> . grad(f) = 0 <=> X<sub>|t</sub> is orthogonal to grad(f).
    thus X<sub>|t</sub> is a tangent vector field along the curve a(t).
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  86. #85  
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    Hey wallaby, i was thinking of the general case here (which is a lot easier to show)
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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    well in that case...

    since we've been talking about projections recently i guess first mention should be given to the fact that,
    (X'(t).N(a(t)))N(a(t)) = |N(a(t))| Proj<sub>N(a(t))</sub>(X'(t))
    as you mentioned this represents the part of the vector not in the tangent plane.

    this implies that the derivative of the vector field can be represented as the sum of the parts both in and out of the tangent plane,
    X'(t) = |T(a(t))| proj<sub>T(a(t))</sub>(X'(t)) + |N(a(t))| Proj<sub>N(a(t))</sub>(X'(t))
    where T(a(t)) is the tangent field, or is it unit tangent field?

    thus X'(t) - (X'(t).N(a(t)))N(a(t)) = |T(a(t))| proj<sub>T(a(t))</sub>(X'(t)) <=> X<sub>|t</sub> is an element of the tangent plane.
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    There is an even easier way Wallaby

    Remember that a vector at a point is a tangent vector at that point if it is orthogonal to the normal vector at that point. So that is what we have to check:

    X<sub>|t</sub>.N(a(t)) = X'(t).N(a(t)) - (X'(t).N(a(t))) N(a(t)).N(a(t)) = X'(t).N(a(t)) - X'(t).N(a(t)) (as N.N = 1) = 0

    Back to your question on unit tangent fields, sadly these do not exist in general. Its an interesting proof in topology that the sphere does not admit a non-vanishing tangent field i.e. given any tangent vector field X on the sphere there must exist a point a on the sphere such that X(a) is the zero vector. This zero vector stops you being able to define unit tangent fields on the sphere.
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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  89. #88  
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    hmm ok that is much easier.
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  90. #89  
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    I'm feeling lazy, so ill show that the covariant derivative is independent of the choice of unit normal vector field

    Well remember we proved that you could only have two unit normal vector fields, with them related by N<sub>1</sub> = - N<sub>2</sub> so suppose instead of choosing N to define the covariant derivative we decided to pick -N

    Well then our new derivative

    X<sub>||t</sub> = X'(t) - (X'(t). -N(a(t))) (-N(a(t))) = X'(t) + (X'(t). -N(a(t))) N(a(t)) = X'(t) - (X'(t).N(a(t))) N(a(t)) = X<sub>|t</sub>

    So the choice made no difference. Next we need to see if our derivative is linear:

    (aX + bY)<sub>|t</sub> = (aX + bY)'(t) - ((aX + bY).N(a(t))) N(a(t)) = aX'(t) + bY'(t) - a(X'(t).N(a(t))) N(a(t)) - b(Y'(t).N(a(t))) N(a(t)) = aX<sub>|t</sub> + bY<sub>|t</sub>

    You can show the product rule also still holds for this new derivative. So we still have an operator that behaves like a normal derivative, with the added benifit that the covariant derivative of a tangent vector field is another tangent vector field. Also i can now give you the geodesic equation :

    A curve a(t) is a geodesic on a surface S if its velocity vector field V(t) along a(t) satisfies the equation V<sub>|t</sub> = 0.

    Can you show that this equation is the same as the definition i gave earlier?
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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  91. #90  
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    so we wish to show that (aXbY)<sub>|t</sub> = ab(X<sub>|t</sub>Y + XY<sub>|t</sub>)

    so if X<sub>|t</sub> = X'(t) - (X'(t).N(a(t)))N(a(t))
    then (aXbY)<sub>|t</sub> = (aXbY)'(t) - ((aXbY)'(t).N(a(t))N(a(t))
    = abX'Y(t) + abXY')(t) - ((abX'Y(t) + abXY'(t)).N(a(t)))N(a(t))
    = abX'Y(t) + abXY'(t) - (abX'Y.N(a(t)) + abXY'.N(a(t)))N(a(t))
    = abX'Y(t) - (abX'Y(t).N(a(t)))N(a(t)) + abXY'(t) - (abXY'(t).N(a(t)))N(a(t))
    = abX<sub>|t</sub>Y + abXY<sub>|t</sub>
    = ab(X<sub>|t</sub>Y + XY<sub>|t</sub>) QED

    if we define a geodesic to be:
    Quote Originally Posted by river_rat
    A curve a(t) is a geodesic on a surface S if its velocity vector field V(t) along a(t) satisfies the equation V<sub>|t</sub> = 0.
    then since V<sub>|t</sub> = 0
    we have that 0 = V'(t) - (V'(t).N(a(t))N(a(t))
    thus V'(t) = (V'(t).N(a(t)))N(a(t))
    since proj<sub>N(a(t))</sub>(V'(t)) = V'(t).N(a(t))/ |N(a(t))| N(a(t))
    then V'(t) = proj<sub>N(a(t))</sub>(V'(t)) (as |N(a(t))| = 1)
    thus V'(t) is parrallel to the unit normal to the surface S, and the two definitions are equivalent.
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  92. #91  
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    Quote Originally Posted by wallaby
    so we wish to show that (aXbY)<sub>|t</sub> = ab(X<sub>|t</sub>Y + XY<sub>|t</sub>)

    so if X<sub>|t</sub> = X'(t) - (X'(t).N(a(t)))N(a(t))
    then (aXbY)<sub>|t</sub> = (aXbY)'(t) - ((aXbY)'(t).N(a(t))N(a(t))
    = abX'Y(t) + abXY')(t) - ((abX'Y(t) + abXY'(t)).N(a(t)))N(a(t))
    = abX'Y(t) + abXY'(t) - (abX'Y.N(a(t)) + abXY'.N(a(t)))N(a(t))
    = abX'Y(t) - (abX'Y(t).N(a(t)))N(a(t)) + abXY'(t) - (abXY'(t).N(a(t)))N(a(t))
    = abX<sub>|t</sub>Y + abXY<sub>|t</sub>
    = ab(X<sub>|t</sub>Y + XY<sub>|t</sub>) QED
    Careful Wallaby, I have not defined a multiplication of tangent vectors that gives you other tangent vectors and remember the covariant derivative is defined on tangent vectors. Instead look at the derivative of f(t) X(t) where f is a smooth function from R -> R.

    Did i define all the operations on the tangent bundle?
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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  93. #92  
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    using the function h(t) = f(t)X(t) i don't seem to be able to get an answer... maybe i'm just tired or maybe i'm missing something completely.

    Quote Originally Posted by river_rat
    Did i define all the operations on the tangent bundle?
    you gave a definition of the tangent bundle and the function phi: S -> TS.
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  94. #93  
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    Ah ok, let me fix that: You define the operations by just working "pointwise" in each tangent space.

    For example given two tangent vector fields X(t) and a function f : R -> R we can define (f X)(t) by (f X)(t) = (a(t), f(t) x<sub>1</sub>(t), f(t) x<sub>2</sub>(t), ... , f(t) x<sub>n</sub>(t)) where X(t) = (a(t), x<sub>1</sub>(t), x<sub>2</sub>(t), ... , x<sub>n</sub>(t))

    Make sense?
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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  95. #94  
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    we shall see,

    so for some function f:R -> R and a vector field X(t) on the curve a(t) then since (fX)(t) = (a(t), f(t)X<sub>1</sub>, f(t)X<sub>2</sub>,..., f(t)X<sub>n</sub>) we have that;
    (fX)<sub>|t</sub> = (fX)' - ((fX)'.N(a(t)))N(a(t)) by the definition of the covarient derivative.
    since (fX)' = (a'(t), f'(t)X<sub>1</sub>(t) + f(t)X<sub>1</sub>'(t), f'(t)X<sub>2</sub>(t) + f(t)X<sub>2</sub>'(t),..., f'(t)X<sub>n</sub>(t) + f(t)X<sub>n</sub>'(t))
    we have that (fX)'.N(a(t)) = (a'(t).N(a(t)), f'(t)X<sub>1</sub>(t).N(a(t) + f(t)X<sub>1</sub>'(t).N(a(t), f'(t)X<sub>2</sub>(t).N(a(t) + f(t)X<sub>2</sub>'(t).N(a(t),..., f'(t)X<sub>n</sub>(t).N(a(t) + f(t)X<sub>n</sub>'(t).N(a(t))
    substituting back into the equation for the covarient derivative and skipping a few steps due to a lack of time we now have that;
    (fX)<sub>|t</sub> = f'(t)X(t) - (f'(t)X(t).N(a(t))N(at(t)) + f(t)X'(t) - (f(t)X'(t).N(a(t))N(a(t)
    = f<sub>|t</sub>X(t) + f(t)X<sub>|t</sub>
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  96. #95  
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    So in short we have that (fX)<sub>|t</sub> = (fX)'(t) - ((fX)'(t).N(a(t)))N(a(t)) = (f'(t)X(t) + f(t)X'(t)) - (f'(t)X(t) + f(t)X'(t)).N(a(t)) N(a(t))
    =f'(t)( X(t) - (X(t).N(a(t))) N(a(t)) ) + f(t)( X'(t) - (X'(t).N(a(t))) N(a(t)) ) = f'(t)( X(t) + 0) + f(t) X<sub>|t</sub>(t) = f'(t)X(t) + f(t)X<sub>|t</sub>(t)

    Notice the use of the two different derivatives at the end (which is what you missed in your proof - the covariant derivative only works for tangent vectors!)
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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  97. #96  
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    ohh hell i should have seen that.
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  98. #97  
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    Ok, now what i want to do, wallaby, is start over and rebuild all our nice constructions abstractly. I want to do this for two reasons:

    a. I think you have enough of an idea of a surface for me to do this meaningfully and
    b. So i can talk about tensors (and perhaps GR) without all this excess baggage i created here.

    So first off, can you recall what it means for a function from R<sup>n</sup> to R<sup>n</sup> to be smooth?
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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  99. #98  
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    differentiable at any point in the domain of the function.
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  100. #99  
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    Close, a function is smooth if it has derivatives of all orders (so you can take second, third, etc derivatives) at any point in its domain. For example the function f(x) = sin(x) is smooth. Smooth functions are nice for a whole bunch of reasons (for starters you can make taylor series from them - which introduces the idea of analytic functions if you are interested) but they are nice as they give you a way of talking about a space looking roughly like normal euclidean space R<sup>n</sup> up close.
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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  101. #100  
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    reminds me of a book that illustrated how pythagoras' theorem applied to the sphere.
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