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Thread: Geometry Primer

  1. #101  
    Forum Professor river_rat's Avatar
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    Ok, sorry once again for the long wait times for these posts - ive promised myself to post more regularly from the 1st of august

    So lets try describe an abstact space where we can do geometry. For starters, as i mentioned earlier, we need it to look like normal euclidean space (at least locally) so that we can talk about calculus and thus lengths of curves and other fun things. So lets recall some topology

    A topological space is called hausdorff if for any two points x and y (with x not equal to y) of the space we can find two disjoint open neighbourhoods of those points.

    So we want our abstact spaces to be hausdorff (so the topology can distinguish points in it), for two important reasons. The first is easy to see, we want to model our abstract idea on normal euclidean geometry but the other more important reason will have to wait for a bit.

    Now do you know what it means for a set to be countable wallaby?
    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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  2. #102  
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    Quote Originally Posted by river_rat
    Ok, sorry once again for the long wait times for these posts - ive promised myself to post more regularly from the 1st of august

    So lets try describe an abstact space where we can do geometry. For starters, as i mentioned earlier, we need it to look like normal euclidean space (at least locally) so that we can talk about calculus and thus lengths of curves and other fun things. So lets recall some topology

    A topological space is called hausdorff if for any two points x and y (with x not equal to y) of the space we can find two disjoint open neighbourhoods of those points.

    So we want our abstact spaces to be hausdorff (so the topology can distinguish points in it), for two important reasons. The first is easy to see, we want to model our abstract idea on normal euclidean geometry but the other more important reason will have to wait for a bit.

    Now do you know what it means for a set to be countable wallaby?
    i havn't been taught much about sets, just the highschool definition of one and about unions and intersections.
    however a quick search elsewhere tells me that "a countable set is some set whose elements can be placed in one to one correspondance with a subset of the natural numbers." which makes sense.
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  3. #103  
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    I'm going to dive right into the maths here wallaby - please shout if something doesn't make sense!

    Now recall that a base B for a topological space X is a collection of subsets of X such that any open subset of X can be written as some union of those subsets in B. For example the collection of all open intervals in the real line form a base for the usual topology on the real line. Now we call a space second countable if its topology admits at least one countable base. Once again the real line is second countable.

    What is nice about second countable spaces is that there topology can be totally described by sequences! So instead of the old definition of a continuous function we can say a function is continuous iff for any convergent sequence a<sub>n</sub> -> a we have that f(a<sub>n</sub>) -> f(a)

    This is where hausdorffness really comes in handy, as a hausdorff space has unique limits - so this whole exercise is meaningful and well defined. So we want our spaces at least to be hausdorff and second countable!

    Any questions so far?
    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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  4. #104  
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    well i don't quite know how a topology can be represented as a sequence, well some form of function obviously, but i guess it's made easy by a definition.
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  5. #105  
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    Hi wallaby

    I've just got back from a fun week watching game in the wild - one of the joys of living in Africa i guess

    So on with the show - All you have to worry about here is that a continuous function preserves convergent limits. I doubt we will end up using that fact here, but it does explain some of the stranger requirements of a manifold.

    Our next tool is what we call a transition function. It is what captures the idea of our space looking like normal euclidean space up close. The idea is this, we have a family of homeomorhpisms (so 1-1, onto continuous functions with continuous inverses) f<sub>i</sub>: U<sub>i</sub> -> R<sup>n</sup> which fit together nicely. Think of each function giving you a different set of local coordinates, with the point x in X giving the co-ordinates (x<sub>1</sub>, x<sub>2</sub>, ... , x<sub>n</sub>). Our requirements are as follows:

    • The U<sub>i</sub>'s form an open cover of the space
    • Lets suppose that two of these open sets which are elements of our open cover overlap, lets call them U and V. Then the function f:U -> R<sup>n</sup> and g:V -> R<sub>n</sub> fit together nicely. By that i mean that the functions f o g<sup>-1</sup> : R<sup>n</sup> -> R<sup>n</sup> and g o f<sup>-1</sup> : R<sup>n</sup> -> R<sup>n</sup> are smooth. The functions f o g<sup>-1</sup> are called transition functions as they tell you how to change from one coordinate system to another.


    The collection of functions f<sub>i</sub> and open sets U<sub>i</sub> is called a smooth atlas. The functions f<sub>i</sub> are called local coordinate charts (or just local charts). As you will see, these give us a way of getting abstract problems back into normal concrete euclidean space where we know how to do calculus.

    Next stop is the idea of compatiable atlases and then the actual definition of a manifold (with some examples!).

    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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  6. #106  
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    Quote Originally Posted by river_rat
    Hi wallaby

    I've just got back from a fun week watching game in the wild - one of the joys of living in Africa i guess
    sounds like a pleasent way to spend the week, as the closest i came to the wild was in the act of booting a cane toad of my lawn.

    Quote Originally Posted by river_rat
    The functions f o g<sup>-1</sup> are called transition functions as they tell you how to change from one coordinate system to another.
    does that make the Lorentz Transformations an example of transition functions?
    well i guess if i try to answer my own question like a big kid,
    yes, because two observers co-ordinate systems can be considered to be two (overlapping?) open sets in minkowski space and the lorentz transformations provide a mapping between these two open sets, which if shown to be homeomorphic results in a transition function.


    Quote Originally Posted by river_rat
    The collection of functions f<sub>i</sub> and open sets U<sub>i</sub> is called a smooth atlas.
    at the risk of sounding silly i'll ask if there is any sort of structure to this collection, i'm thinking of it as more of a set right now but i can't quite come to a conclusion on whether or not it would be meaningfull to talk about the elements of such a set if the set is a collection of open sets and functions. most likely i'd go with probably not, which leaves me mildly stuck on this point.
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  7. #107  
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    Quote Originally Posted by wallaby
    does that make the Lorentz Transformations an example of transition functions?
    Not exactly, the lorentz stuff is a level higher then where we are now.

    Quote Originally Posted by wallaby
    at the risk of sounding silly i'll ask if there is any sort of structure to this collection, i'm thinking of it as more of a set right now but i can't quite come to a conclusion on whether or not it would be meaningfull to talk about the elements of such a set if the set is a collection of open sets and functions. most likely i'd go with probably not, which leaves me mildly stuck on this point.
    I'm not sure what you are on about here wallaby, could you explain it differently.

    Ok, something more for you to think about. Two atlases are called comparable if their union is still an atlas. Think of it like this, if you have a map book of stereographic projections of the earth and another map book of Mercator projections of the earth then putting the two map books together does not change the shape of the earth (and you can move from one projection to the other).
    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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  8. #108  
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    Quote Originally Posted by river_rat
    Not exactly, the lorentz stuff is a level higher then where we are now.
    damn my imagination strikes again.

    Quote Originally Posted by river_rat
    I'm not sure what you are on about here wallaby, could you explain it differently.
    well re-reading it a week later i can tell you it was just my mind not grasping a simple concept, it's happening all too frequently lately.
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  9. #109  
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    Ok here comes the faith based lesson (i.e. i don't want to branch off into the bowls of set theory to justify the next claim unless you are certain you want to head off in that direction).

    Definition : A smooth manifold is a second-countable, hausdorff topological space X with a smooth atlas.

    The faith based part is this, every manifold has a maximal atlas (which comes from the axiom of choice) i.e. an atlas such that any other compatible atlas on the manifold is contained in our maximal atlas. A maximal atlas is called a differential structure.

    Here is an easy example of a manifold:

    Consider the set C = { (x, y) in R<sup>2</sup> : x<sup>2</sup> + y<sup>2</sup> = 1} (i.e. the unit circle) and consider the following two transition functions

    f<sub>1</sub> : C \ (0, 1) -> R given by f<sub>1</sub>(x, y) = x/(1-y) and f<sub>2</sub> : C \ (0, -1) -> R given by f<sub>2</sub>(x, y) = x/(1+y).

    Then C with the atlas {(f<sub>1</sub>, C \ (0, 1)), (f<sub>21</sub>, C \ (0, -1))} is a smooth manifold (you can check the details). Also, it shows that the circle is one-dimensional. In fact in a topological sense the circle is merely the real line with one extra point added (the so called one-point compactification). The transition functions i gave are the stereographic projections on the circle onto the real line. You can do a similar thing for any sphere and thus all spheres are smooth manifolds.
    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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  10. #110  
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    Quote Originally Posted by river_rat
    i don't want to branch off into the bowls of set theory to justify the next claim unless you are certain you want to head off in that direction).
    yea that's probably another time and another thread.

    Quote Originally Posted by river_rat
    f<sub>1</sub> : C \ (0, 1) -> R given by f<sub>1</sub>(x, y) = x/(1-y) and f<sub>2</sub> : C \ (0, -1) -> R given by f<sub>2</sub>(x, y) = x/(1+y).
    i'm curious to know how those transition functions were arrived at, or a better question might be, "how do we find transition functions for a particular set?"
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  11. #111  
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    that is not a trivial question Wallaby and the transition functions are intrinsically tied up with the manifold in question. The transition functions for the normal circle come from the idea of stereographic projection - can you think of another one (or if you feel daring two)?

    Next definition: A function f from a manifold M to the reals R is smooth iff for every point x of M we can find a transition function phi : M -> R<sup>n</sup> such that the function f o phi<sup>-1</sup> is smooth at phi(x) in the normal 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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  12. #112  
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    hmm maybe it's about time i learnt some projective geometry then.
    Quote Originally Posted by river_rat
    Next definition: A function f from a manifold M to the reals R is smooth iff for every point x of M we can find a transition function phi : M -> R<sup>n</sup> such that the function f o phi<sup>-1</sup> is smooth at phi(x) in the normal sense.
    so functions like covariant derrivatives and vector functions?
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  13. #113  
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    Hey wallaby, so much for posting regularly! My damn thesis is not playing ball at the moment - but i guess it is my fault leaving all the write up till the end and trying to cram it!


    Projective geometry is a different beast entirely, so don't worry about that and you don't need it for what we are doing here. Could you solve that problem i asked last time? Finally, the covariant derivative needs a metric and we need a lot more stuff before we can get there (and it is not a real valued function) but we should be getting to vectors soon.

    Here is an easy real valued function example - consider the earth as your manifold and a function f which assigns to every point on the earth its height above sea level.
    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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  14. #114  
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    Quote Originally Posted by river_rat
    Hey wallaby, so much for posting regularly! My damn thesis is not playing ball at the moment - but i guess it is my fault leaving all the write up till the end and trying to cram it!
    sounds like a familiar predicament.


    Quote Originally Posted by river_rat
    Could you solve that problem i asked last time
    well i've got something, but i'm trying to think about whether my imagination has taken me away or if i've actually got something. i'll post it anyway but right now i have to run...
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  15. #115  
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    well i got majorly distracted since thursday, my appologies if you were waiting.

    ok so i need to produce some transition functions,
    f<sub>1</sub> : C\(0,1) -> R given by f<sub>1</sub>(x,y) = 2x/1-y
    f<sub>2</sub> : C\(0,-1) -> R given by f<sub>2</sub>(x,y) = 2x/1+y

    not much different to what you had mind you, so then i decided to go for more,

    f<sub>1</sub> : C\(0,1) -> R given by f<sub>1</sub>(x,y) = x/y
    f<sub>2</sub> : C\(0,-1) -> R given by f<sub>2</sub>(x,y) = -x/y
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  16. #116  
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    Hey Wallaby

    First question - can you see why your second answer is not an atlas for the circle. I think your first answer was correct - will have to check though. I'm surprised you missed the obvious answer though - parametrise the circle in polar coordinates! I'll leave the details to you.

    The second definition we need is a curve on a manifold - as this is how we introduce tangent vectors to the space. Let f : R -> M be a function and let m be in f(R) a subset of M. So m = f(s) for some real s. Let g be a local co-ordinate system at m. The function f is smooth at m iff the function g o f : R -> R<sup>n</sup> is smooth at s. A function f : R -> M is smooth if it is smooth at every point in its image. Such a smooth function is called a smooth curve.

    The polar co-ordinate trick i mentioned earlier also gives a whole host of smooth curves on the circle - see if you can think up a few.
    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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  17. #117  
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    Quote Originally Posted by river_rat
    Hey Wallaby

    First question - can you see why your second answer is not an atlas for the circle.
    the function is undefined at y=0, thus the two open sets are disjoint and we don't have an atlas.
    out of curiosity, is it still a stereographic projection?

    Quote Originally Posted by river_rat
    I'm surprised you missed the obvious answer though - parametrise the circle in polar coordinates! I'll leave the details to you.
    i did think about that a little, only a little though.
    ok so i'll take a crack at the polar co-ordinates buisness;
    f<sub>1</sub> : C\(0, 180) given by f<sub>1</sub>(r, Θ) = (pi/180)*Θ
    f<sub>2</sub> : C\(180, 360) given by f<sub>2</sub>(r, Θ) = (pi/180)*Θ
    probably wrong but it's midnight here and almost time to go to sleep.
    EDIT: ok now that i've had some sleep i thought i might add this to increase my chances of atleast getting part marks;
    f<sub>1</sub> : C\(0, pi) given by f<sub>1</sub>(r, Θ) = cos Θ
    f<sub>2</sub> : C\(0, -pi) given by f<sub>2</sub>(r, Θ) = -cos Θ
    i imagine something similar could be done with sin Θ.

    Quote Originally Posted by river_rat
    The second definition we need is a curve on a manifold - as this is how we introduce tangent vectors to the space. Let f : R -> M be a function and let m be in f(R) a subset of M. So m = f(s) for some real s. Let g be a local co-ordinate system at m. The function f is smooth at m iff the function g o f : R -> R<sup>n</sup> is smooth at s. A function f : R -> M is smooth if it is smooth at every point in its image. Such a smooth function is called a smooth curve.
    so if i've interpreted this correctly, we're mapping a function from a Euclidean space, R for example, onto a subset of the manifold, which i'm guessing is an atlas.
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  18. #118  
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    Quote Originally Posted by river_rat
    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.
    Hi. I spotted this absorbing thread just now and I've been spending half my day reading it up to this post of yours. (It's on page 2 in case you're looking for it.)

    My question is, how do you know that

    the remaining rotated basis vectors are all still orthogonal to the vector parallel to the gradient vector, span the tangent plane
    I'm totally at a loss here. Can you please explain?
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  19. #119  
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    Quote Originally Posted by river_rat
    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.
    WTF! I don't get this either! If

    as a basis is linearly independent
    then it should be the a<sub>i</sub> = 0, not w.v<sup>i</sup> = 0, shouldn't it! What is it you're talking about? I'm sorry you've lost me completely.
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