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Thread: Manifolds

  1. #101  
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    DrRocket: Yeah, I've kind of picked up on that.

    I have what I feel to be a stupid question... What's the difference between and ?

    I understand now that is just the fiber at and consists only of vectors. has been called the local bundle over , so in this case does the preimage consist of elements of the form or just ? I could make use of a "feeling stupid/awkward" emoticon right about now.

    Guitarist, I guess at least you know I work at this stuff and I don't pretend to understand it when I don't...
    "There is a kind of lazy pleasure in useless and out-of-the-way erudition." -Jorge Luis Borges
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  2. #102  
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    I have what I feel to be a stupid question... What's the difference between and ? I understand now that is just the fiber at and consists only of vectors. has been called the local bundle over , so in this case does the preimage consist of elements of the form or just
    ?
    OK, DrRocket makes a very good point, which is in fact a bit of a hobby-horse of mine, so I will come back to it.

    The answer to your question is that the points in are the ordered pairs , say. The points in are vectors, which, for we may call . I think the point you are missing is in this choice of notation; if we say that is the local tangent bundle over , we mean to imply it is the union of all tangent vector spaces defined at all points in U. IOW, it's elements are vectors, pure and simple - there is nothing else in this set (you said this in your quote, so essentially you're home on that).

    Now there is a homeomorphism , with inverse. Let's review what this means. The definition of a homeomorphism is that it maps open sets onto open sets and back again. In other words, it is not especially interested in points. So there is no guarantee that, say, . Or to put it another way, our homeomorphism does not necessarily map , and and back again, though it may.

    However, the definition of a homeomorphism as a bijective continuous map does guarantee that for every there is at least one and at most one pre-image , which we take to define the continuous inverse mapping of an open set onto an open set. Note from the above, I not not require i = j. The upshot being, as I said the other day, the homeomorphism mixes up fibres.

    I decided not to mount my hobby horse after all, as it might confuse you.

    PS. I am glad to see you are thinking this through, and I am delighted that DrRocket has pitched in - we all have a slightly different take on the same subject, and his may well match yours better than mine does
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  3. #103  
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    OK, change of plan. We are going to put structure groups on the back burner for now - we can come back to this with a vengeance shortly, if anyone wants. We will then find the Lie groups.

    Heeding DrRocket's advise about building intuition first, I invite you think of the tangent bundle over the 1-sphere as follows; we have a circle, and a complete set of 1-dimensional tangent vector spaces, one at each point on the circle. In your imagination, rotate each vector space, a line-space, through 90 degrees, and imagine now the 2-cylinder formed thereby.

    Take an extremely sharp knife and cut out the thinnest possible sliver of this cylinder. If your hand was steady enough, you will have found that, for every point of the circle, you have selected exactly one element from it's associated vector space.

    You have created a "section of the tangent bundle", which is what is called a vector field. One vector at each point in the base manifold, IOW. This is, of course, how we normally think of vector fields.

    Now Chemboy once asked me why it is unusual and undesirable for the bundle manifold to be globally trivial. Here's why; there is a thm. that states that, if a bundle admits of a global section, it is a globally trivial bundle, and conversely. The implication here is that such a bundle allows me to define a global field on the base manifold in just one way

    There is also a thm. called (I think) the "Hairy Ball Thm." that states that the only manifold of dimension > 1 that admits of a globally defined vector field is the trivial manifold R^n. We may take "trivial" to mean boringly unilluminating
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  4. #104  
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    Quote Originally Posted by Guitarist
    ...Here's why; there is a thm. that states that, if a bundle admits of a global section, it is a globally trivial bundle, and conversely. ...
    There is also a thm. called (I think) the "Hairy Ball Thm." that states that the only manifold of dimension > 1 that admits of a globally defined vector field is the trivial manifold R^n. ...
    Do you have specific references for these theorems ? I think there must be more to them than just what you have stated.
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  5. #105  
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    Well of course there is! But..... I am sorry, I don't quite understand your post. Surely you're not suggesting I do other people's Google search for them? But by asking me to "provide references", I can only assume you mean something of the sort. Am I wrong?

    I had rather taken your previous post on this topic as a (very) mild rebuke to the effect that I was being somewhat too technical in an introductory exposition. I tried to take that on board. Did I misinterpret that post?

    Whatever. Are you now suggesting I give chapter and verse for my earlier assertions? I think I can.

    PS. Tell you what, one and all, a propos my first paragraph: The day this sub-forum becomes merely a forum for the exchange of Google search results is the day I quit.

    PPS. I am just closing on a truly gruesome work day, so please forgive my grumpiness.
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  6. #106  
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    Sorry I haven't gotten back to you since the last post until now. I'm just getting into the last three weeks of the semester and I have tons to do. I will certainly devote time to this when I am able, but I'm not sure when that will be as I really should be concentrating on school. I will do my best. Things should definitely settle down by the end of this week, but I should be back with you before then, hopefully.
    "There is a kind of lazy pleasure in useless and out-of-the-way erudition." -Jorge Luis Borges
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  7. #107  
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    Quote Originally Posted by Guitarist
    Well of course there is! But..... I am sorry, I don't quite understand your post. Surely you're not suggesting I do other people's Google search for them? But by asking me to "provide references", I can only assume you mean something of the sort. Am I wrong?

    I had rather taken your previous post on this topic as a (very) mild rebuke to the effect that I was being somewhat too technical in an introductory exposition. I tried to take that on board. Did I misinterpret that post?

    Whatever. Are you now suggesting I give chapter and verse for my earlier assertions? I think I can.

    PS. Tell you what, one and all, a propos my first paragraph: The day this sub-forum becomes merely a forum for the exchange of Google search results is the day I quit.

    PPS. I am just closing on a truly gruesome work day, so please forgive my grumpiness.
    No I wasn't asking you do Google work for me. I was trying to politely suggest that I think the theorems that you stated regarding bundle sections do not exist, and giving you an opportunity to clarify.

    And I really was not trying to intimate a rebuke, mild or otherwise, for your presentation. It strikes me as a good one given the limitations of this format. I thought that perhaps a bit of outside discussion might help to illuminate the topic by providing a slightly different (not necessarily better) point of view. There is a fine line to walk with this topic between being too loose and intuitive and being imprecise. I don't know precisely where that line is myself, so my approach is to make it a wide line and somewhat straddle it.

    In the introduction to Mike Spivak's 5-volume set on differential geometry he makes a statment to the effect that modern treatmenst, because they put a lot of work into the definitions, are clean, efficient and elegant, but nobody can understand where the definitions come from. It is a difficult subject to treat because of that feature. You are doing a good job, but sometimes a slightly different perspective is useful. The more viewpoints that you see the more likely one of them wil fit your perspective of the moment.
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  8. #108  
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    Know what? It was extremely lame of me to say "there is a theorem.. blah,blah" So let's see if I can explain myself.

    WLOG we may take , the tangent bundle over the real n-sphere. If this bundle is globally trivial, we will have a continuous map . The crucial point here is that Cartesian products are "exhaustive", in the sense that for every point and every vector there is an ordered pair of the form .

    It follows that the set for some fixed . Now this map is a homeomorphism, so the inverse sends this subset to , from which it follows that there is always a section - a vector field - that associates to each point in the same vector.

    Assume that . Then we will have a nowhere zero vector field on .

    But the Hairy Ball theorem, which is, after all, an application of the Fixed Point Theorem of Brouwer gives us that, provided only that the mappings above are continuous (they are), a nowhere zero field cannot exist on a non-trivial manifold.

    The converse is a little harder to find - lemme dwell on it a while. Unless anyone can help me out a bit?

    I did mis-speak earlier - any tangent bundle admits of at least one global section, the null section, but this doesn't mean it is thereby a trivial bundle.

    DrRocket: I agree completely with your remarks; as I said in an earlier post, I regard your contributions as very valuable in this thread, since I have my way of looking at this, you have yours, and hopefully our "tutees" can find the one that suits them best
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  9. #109  
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    Quote Originally Posted by Guitarist
    Know what? It was extremely lame of me to say "there is a theorem.. blah,blah" So let's see if I can explain myself.

    WLOG we may take , the tangent bundle over the real n-sphere. If this bundle is globally trivial, we will have a continuous map . The crucial point here is that Cartesian products are "exhaustive", in the sense that for every point and every vector there is an ordered pair of the form .

    It follows that the set for some fixed . Now this map is a homeomorphism, so the inverse sends this subset to , from which it follows that there is always a section - a vector field - that associates to each point in the same vector.

    Assume that . Then we will have a nowhere zero vector field on .

    But the Hairy Ball theorem, which is, after all, an application of the Fixed Point Theorem of Brouwer gives us that, provided only that the mappings above are continuous (they are), a nowhere zero field cannot exist on a non-trivial manifold.

    The converse is a little harder to find - lemme dwell on it a while. Unless anyone can help me out a bit?

    I did mis-speak earlier - any tangent bundle admits of at least one global section, the null section, but this doesn't mean it is thereby a trivial bundle.

    DrRocket: I agree completely with your remarks; as I said in an earlier post, I regard your contributions as very valuable in this thread, since I have my way of looking at this, you have yours, and hopefully our "tutees" can find the one that suits them best
    I think the problem is perhaps that you have extended the Hairy Ball theorem too far. As a reference for what I am going to say I submit the book Differential Topology by Morris Hirsch.

    Let us restrict attention to compact oriented manifolds without boundary. Then the tangent bundle admits a nonvanishing section if and only if the Euler charistic is 0. For spheres the Euler characteristic is 0 if the dimension is odd and it is 2 if it is even. So there exist non-vanishing sections of the tangent bundle if and only if the dimension of the sphere is odd. The Hairy Ball theorem is reflection of the fact that to comb your hair, you must somewhere have a "cowlick", and that is because your head is a topological 2-sphere.

    Also note that there is big difference between the existence of a nonvanishing (never 0) section, a section that is constant and 0, and simply any global smooth (or continuous) section without further qualification.
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  10. #110  
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    I see now that DrRockect, or more accurately, fact, has the better of me. So I urge all readers to disregard this flamboyant claim
    Quote Originally Posted by Guitarist
    there is a thm. that states that, if a bundle admits of a global section, it is a globally trivial bundle, and conversely. The implication here is that such a bundle allows me to define a global field on the base manifold in just one way

    There is also a thm. called (I think) the "Hairy Ball Thm." that states that the only manifold of dimension > 1 that admits of a globally defined vector field is the trivial manifold R^n.
    I quote from a text here:

    The tangent bundle always admits global sections, such as the null section,....... but this does not imply the {global} triviality of . If admits a global coordinate system, then is clearly {globally} trivial, but the converse is not true.....is {globally} trivial, but, as we know, the sphere has no global chart.

    I shan't try to hide my embarrassment by pretending it is an unimportant point; it's not, it is at the heart of what we are discussing.

    I apologize to all for misleading you.
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  11. #111  
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    Quote Originally Posted by Guitarist
    I see now that DrRockect, or more accurately, fact, has the better of me. So I urge all readers to disregard this flamboyant claim
    Quote Originally Posted by Guitarist
    there is a thm. that states that, if a bundle admits of a global section, it is a globally trivial bundle, and conversely. The implication here is that such a bundle allows me to define a global field on the base manifold in just one way

    There is also a thm. called (I think) the "Hairy Ball Thm." that states that the only manifold of dimension > 1 that admits of a globally defined vector field is the trivial manifold R^n.
    I quote from a text here:

    The tangent bundle always admits global sections, such as the null section,....... but this does not imply the {global} triviality of . If admits a global coordinate system, then is clearly {globally} trivial, but the converse is not true.....is {globally} trivial, but, as we know, the sphere has no global chart.

    I shan't try to hide my embarrassment by pretending it is an unimportant point; it's not, it is at the heart of what we are discussing.

    I apologize to all for misleading you.
    No apology necessary. Mistakes are one of the best avenues to learning. Everybody makes them. Mistakes offer learning opportunities.

    I once did a seminar using Kirillov's book representation theory. It was an interesting seminar, since it quickly became apparent that the book was simply full of mistakes and false theorems.
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  12. #112  
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    [quote="Guitarist"] I quote from a text here:

    The tangent bundle always admits global sections, such as the null section,....... but this does not imply the {global} triviality of . If admits a global coordinate system, then is clearly {globally} trivial, but the converse is not true.....is {globally} trivial, but, as we know, the sphere has no global chart.

    I/quote]

    It would be of interest to me for you to identify the text from which you take quotes such as this. That would me a better idea of the particular slant that you are taking with regard to the study of manifolds. There are different approaches to the subject, all good and all valid, but differing in emphasis and perspective. It is a big subject (a REALLY big subject), and very active from a research perspective.
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  13. #113  
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    DrRocket: You are too forgiving!

    The quote in question is from my current favourite Azcarraga & Izquirdo Lie groups, Lie algebras, cohomology and..... I also have on my shelf at home:

    Bishop & Goldberg Tensor analysis on manifolds

    Lovelock & Rund Tensors, variational principles and diff. forms

    Schutz Geometric methods in math. phys

    As is all too evident, I don't cross-check with them as often as I should.

    Let me continue this section on tangent bundles. The local section with the requirement that, for all that is called a vector field over . Obviously this implies . (Recall that )

    The quantity represents the field evaluated at the point , and is equally obviously an element - a vector - in . As such it is subject to the transformation law under a change of coordinates we saw earlier.

    Finally, it should be easy to see that the cotangent bundle and the resulting cotangent field are similarly defined, but let me insert a note of caution. Some authors refer to the cotangent field as a 1-form, whereas I started by using this term as a synonym for a covector.

    I used to think this different usage merely reflected a different choice of definition, but I later saw this is not quite right. There is a good reason to consider a 1-form as a field. The reason is only mildly technical, and of no real relevance here, but just check for context in your reading.

    And now I need some advice. I promised I would come back to structure groups, and I would especially like to scratch the surface of principal bundles. For this we will need to understand the Lie groups. Should I do this here as a slight digression, or would a companion thread be more appropriate? I can see both arguments

    Errr, who did you say moderates this sub-forum......?
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  14. #114  
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    Quote Originally Posted by Guitarist
    DrRocket: You are too forgiving!

    The quote in question is from my current favourite Azcarraga & Izquirdo Lie groups, Lie algebras, cohomology and..... I also have on my shelf at home:

    Bishop & Goldberg Tensor analysis on manifolds

    Lovelock & Rund Tensors, variational principles and diff. forms

    Schutz Geometric methods in math. phys

    As is all too evident, I don't cross-check with them as often as I should.

    Let me continue this section on tangent bundles. The local section with the requirement that, for all that is called a vector field over . Obviously this implies . (Recall that )

    The quantity represents the field evaluated at the point , and is equally obviously an element - a vector - in . As such it is subject to the transformation law under a change of coordinates we saw earlier.

    Finally, it should be easy to see that the cotangent bundle and the resulting cotangent field are similarly defined, but let me insert a note of caution. Some authors refer to the cotangent field as a 1-form, whereas I started by using this term as a synonym for a covector.

    I used to think this different usage merely reflected a different choice of definition, but I later saw this is not quite right. There is a good reason to consider a 1-form as a field. The reason is only mildly technical, and of no real relevance here, but just check for context in your reading.

    And now I need some advice. I promised I would come back to structure groups, and I would especially like to scratch the surface of principal bundles. For this we will need to understand the Lie groups. Should I do this here as a slight digression, or would a companion thread be more appropriate? I can see both arguments

    Errr, who did you say moderates this sub-forum......?
    I am not familiar with Azcarraga & Izquirdo. I do have Bishop & Goldberg and Lovelock & Rund in the form of Dover paperbacks.

    I suggest that you maintain a single thread. I think that one thread on Lie groups and another on manifold might be a bit much to handle. Both are big topics. And they are related, so you would not only have to manage two separate threads you would have to coordinate them. That would be asking a lot. Besides, William has stayed away from this one. Why tempt him with a different target ?
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  15. #115  
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    At long last I am caught up. I feel like I should mention that there are some other things I'd love to see if you're willing to cover them that I know tie in with manifolds, so if it would make sense to make a foray into them and then back into manifolds that would be fine. Specifically I'm thinking group theory and tensors. So if more knowledge of group theory for example would help me understand some of the more high-level manifold stuff then I wouldn't mind going in that direction for a while...but as its own thread, just on group theory (but perhaps a slight emphasis on its applications in manifolds). Just an idea.
    "There is a kind of lazy pleasure in useless and out-of-the-way erudition." -Jorge Luis Borges
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  16. #116  
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    Quote Originally Posted by Chemboy
    At long last I am caught up. I feel like I should mention that there are some other things I'd love to see if you're willing to cover them that I know tie in with manifolds, so if it would make sense to make a foray into them and then back into manifolds that would be fine. Specifically I'm thinking group theory and tensors. So if more knowledge of group theory for example would help me understand some of the more high-level manifold stuff then I wouldn't mind going in that direction for a while...but as its own thread, just on group theory (but perhaps a slight emphasis on its applications in manifolds). Just an idea.
    Books on group theory tend to emphasize very general stuff and finite groups. For manifolds what you need to know is that a group is a set with an associative binary operation, usually written as multiplication, an identity and inverses for each element. Think of a set of non-singular matrices closed under multiplication and inversion. For work with manifolds the groups involved are generally Lie groups. You can give an abstract definition for a Lie groups (it is basiclly a group that is also a manifold with the operations being smooth) but when you look for examples the most common Lie groups are just groups of matrices.

    Tensors and differential forms will come up naturally in discussing differentiable manifolds. I think that if you try to treat them as distinct subjects and cover them simultaneously you will find yourself overwhelmed.

    The most likely way for your study of manifolds to become derailed is to try to get into too many topics all at once.
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  17. #117  
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    What you mentioned on group theory I already know, so I guess I'm ok there. I was just wondering since an was thrown into that definition back there so I wondered if it wouldn't hurt to get a better grasp of s and s and s and whatnot. And I'm just kind of anxious to see other things... Not that I'm tired of manifolds mind you, but I've taken an interest in group theory lately and I'm looking forward to it, and I just don't know where we'll stop with manifolds (I imagine we could go on pretty much forever). But at the same time I don't want to stop with manifolds. Maybe you see my dilemma.
    "There is a kind of lazy pleasure in useless and out-of-the-way erudition." -Jorge Luis Borges
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  18. #118  
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    Quote Originally Posted by Chemboy
    What you mentioned on group theory I already know, so I guess I'm ok there. I was just wondering since an was thrown into that definition back there so I wondered if it wouldn't hurt to get a better grasp of s and s and s and whatnot. And I'm just kind of anxious to see other things... Not that I'm tired of manifolds mind you, but I've taken an interest in group theory lately and I'm looking forward to it, and I just don't know where we'll stop with manifolds (I imagine we could go on pretty much forever). But at the same time I don't want to stop with manifolds. Maybe you see my dilemma.
    Yes I see your dilemma. And it is not going to get better. There is a lot of mathematics out there to be learned. You can't do it all at once. You can't even do it all in a normal lifetime.

    I saw in the chemistry forum that you are a sophomore. That is actually pretty early to be doing deep manifold theory.

    Tell me a few things, and maybe I can give you some advice on what to look at and in what order.

    1. What is your intended college major ?

    2. What mathematics courses have you taken ?

    3. What other mathematics have you studied on your own and feel that you understand well ? In particular, do you think you understand any of the following at a fairly good level:

    Basic calculus of one variable
    Basic calculus of several variables
    Linear algebra
    Ordinary differential equations
    Basic real analysis/advanced calculus
    Basic abstract algebra (groups and rings)
    Point set topology
    Complex analysis

    4. Do you plan to go to graduate school ? If so, in what discipline ?

    5. What is your objective in studying advanced mathematics ?

    6. What advanced mathematics is of primary interest to you at the moment, in order of importance ?

    7. Strictly optional. What school are you attending and to which school do you intend to transfer ? (I might know some faculty members in the math departments and knowing their specialties may color my suggestions).
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  19. #119  
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    Mathematics is really just a hobby to me. Yes, I'm learning about manifolds strictly for fun. I cannot express how happy this type of thing makes me. That's just the way I am, knowledge and academics is my 'thing.' I would love to obtain a degree in mathematics someday but it's not my current path. Part of my motivation for learning advanced mathematics is my interest in QM/theoretical physics which I hope to have a very good understanding of someday. I know this is all quite far out there, but I trust that I can make it work given enough time and given my passion for it.

    1. My college major right now is conservation biology and I intend to obtain a degree in chemistry as well. Unfortunately I can't double major but I may do chemistry as grad work, or however it works out.

    2. I have taken Calc I, Calc II, and am in Calc III right now.

    3. I'm good with basic calculus of one and several variables. I worked through a good set of linear algebra notes online and feel like I have at least a good foundation in it, and I am taking Intro to Linear Algebra next semester. I'm taking a course in ODEs next semester as well and intend to learn the material over winter break before then. I don't know anything really of real analysis or advanced calculus beyond what random items I've read on Wikipedia. Sadly I know little to nothing of basic abstract algebra. I've never gotten far into rings or fields or anything of the sort on my own. I know the very basics of groups, as stated previously. Guitarist ran a thread on point set topology here (at my request) and I feel I have a fairly good foundation in it. I've been reading about complex analysis in Wikipedia lately and find it very interesting but haven't formally learned any of it.

    4. I plan to go to graduate school for conservation biology and ideally chemistry.

    5. As far as my objective in studying advanced mathematics goes, I simply have a passion for it (which started while taking Calc I) and just want to learn as much as possible because I love it. Also, as stated above, I hope to learn QM/theoretical physics and know I need the advanced mathematics for that.

    6. It's very difficult to say what's of primary interest to me. It all is, really. I'm very happy to be learning about manifolds right now, but really I'm always willing to learn anything at any time. More advanced group theory has been more at the forefront of my mind lately because I know it has significant applications in QM.

    7. The answer to this one is scary...I'm at a community college right now but transferring in the fall. The college I'm transferring to does not have a math department. It's SUNY-College of Environmental Science and Forestry in Syracuse, NY. They're very environmentally focused and so stick largely to biology and chemistry. However it is located right next to Syracuse University and I will have their facilities available to me (though of course with schedule-oriented and financial constraints).

    I realize my desire to learn a lot of advanced mathematics is very far out there, but I feel it's wrong to say it's unrealistic to learn it when I have the passion for it that I do. Right now my main issue is just lack of resources. I cannot afford expensive books (or any books at all really) and that is why I truly appreciate the knowledge that you, Guitarist, JaneBennett, and others are able to bring to the forum and to me.
    "There is a kind of lazy pleasure in useless and out-of-the-way erudition." -Jorge Luis Borges
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  20. #120  
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    Mod note: Off-topic post deleted

    William - I warn you that further deliberate vandalism will have rather serious consequences
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  21. #121  
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    DrRocket...I think you jinxed us...
    "There is a kind of lazy pleasure in useless and out-of-the-way erudition." -Jorge Luis Borges
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  22. #122  
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    Quote Originally Posted by Chemboy
    DrRocket...I think you jinxed us...
    Sorry about that. I just sent a formal complaint via PM to guitarist.
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    I hope he will forgive my making it public, but this entire thread was started at Chemboy's request. So if he wants to talk about groups, then so be it - he's the boss. I suggest the Lie groups as a compromise. But first me make something absolutely plain:

    I make no claims that my attempts to explain manifolds is in any way thorough. Rather my intention was to give my friend, and other interested readers, just enough of a feel for the subject that, should they wish to dig deeper in textbooks (good Heavens - what are those?) or on sites like Wolfram or PlanetMath, they will at least have a fighting chance.

    As DrRocket raid, it is a truly massive subject. But I hope we have gained some insight here. So will it be with the Lie groups. Warning: this will be devilishly long for a single post: I excuse it on the grounds it is mostly words, and a weekend.

    Now this will be something of a digression, but you will see that maybe it's not too far afield if I offer a quick definition: A Lie group is a manifold endowed with the structure of a group. The converse will serve equally as a definition, obviously.

    My initial approach may raise a few eyebrows, but it has, I think, the advantage of being moderately intuitive. So we will start with a pair of garden-variety real vector spaces, and consider the totality of linear maps (transformations, operators) .

    It is relatively easy to show that this collection of maps is itself a vector space over the reals (use etc), but since this fact is at present of no interest to us, we may pass over it. Let's write this vector space as .

    By my notation, the space has as elements (vectors) all the maps . Of course, we introduce no ambiguity by writing this as .

    Now we know that any linear transformation a real space can be written as a matrix, whose entries are real; specifically, for a n-dimensional real space, these transformation matrices are real n x n matrices. We now want to restrict our attention to those maps which are invertible i.e. they are "isomorphisms", though here, the correct term is automorphism.

    In terms of the matrix representation of our automorphisms, this must mean that our matrices have non-zero determinant (this is because the formula for calculating the inverse of a matrix includes the factor ). By stipulating the existence of all inverses, and recognizing the identity matrix we will find that the vector space of all automorphisms, written as matrices, is in fact a group. This group we will call as , the general linear group.

    So the group has matrix multiplication as the group operation, the identity , say, and all inverse matrices such that . Let us observe that, in general .

    Now we will ultimately be rewarded by concentrating, not on what groups do but rather by what they are. The obvious abstraction is to rewrite this group as to signify this is the group of all real n x n matrices with non-zero determinant (we call such a matrix "non-singular", btw).

    This is an example of a Lie group, though to show this, we need to convince ourselves it is indeed a manifold.

    So, let be a group element, a matrix as defined above. is an n x n matrix, i.e. there are entries in , which we will write as , where i labels the rows, and j labels the columns, say. For ease of notation let's set Then evidently there is a mapping .

    Now provided we are consistent in our mappings, say the i-th element in our m-tuple maps onto the jk-th entry in our matrix, we see that the the m-tuple uniquely determines a matrix in . Thus we have an isomorphism .

    If we can show continuity of this map, we will have a homeomorphism. Without attempting any degree of rigour, I offer this. Let be the matrix with entries . Then whenever the point is arbitrarily close to so the matrix whose entries are is arbitrarily close the matrix whose entries are .

    This would imply a continuous isomorphism - a homeomorphism - so this group would indeed be a manifold.
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    Quote Originally Posted by Guitarist
    I hope he will forgive my making it public, but this entire thread was started at Chemboy's request. So if he wants to talk about groups, then so be it - he's the boss. I suggest the Lie groups as a compromise. But first me make something absolutely plain:

    I make no claims that my attempts to explain manifolds is in any way thorough. Rather my intention was to give my friend, and other interested readers, just enough of a feel for the subject that, should they wish to dig deeper in textbooks (good Heavens - what are those?) or on sites like Wolfram or PlanetMath, they will at least have a fighting chance.

    As DrRocket raid, it is a truly massive subject. But I hope we have gained some insight here. So will it be with the Lie groups. Warning: this will be devilishly long for a single post: I excuse it on the grounds it is mostly words, and a weekend.

    Now this will be something of a digression, but you will see that maybe it's not too far afield if I offer a quick definition: A Lie group is a manifold endowed with the structure of a group. The converse will serve equally as a definition, obviously.

    My initial approach may raise a few eyebrows, but it has, I think, the advantage of being moderately intuitive. So we will start with a pair of garden-variety real vector spaces, and consider the totality of linear maps (transformations, operators) .

    It is relatively easy to show that this collection of maps is itself a vector space over the reals (use etc), but since this fact is at present of no interest to us, we may pass over it. Let's write this vector space as .

    By my notation, the space has as elements (vectors) all the maps . Of course, we introduce no ambiguity by writing this as .

    Now we know that any linear transformation a real space can be written as a matrix, whose entries are real; specifically, for a n-dimensional real space, these transformation matrices are real n x n matrices. We now want to restrict our attention to those maps which are invertible i.e. they are "isomorphisms", though here, the correct term is automorphism.

    In terms of the matrix representation of our automorphisms, this must mean that our matrices have non-zero determinant (this is because the formula for calculating the inverse of a matrix includes the factor ). By stipulating the existence of all inverses, and recognizing the identity matrix we will find that the vector space of all automorphisms, written as matrices, is in fact a group. This group we will call as , the general linear group.

    So the group has matrix multiplication as the group operation, the identity , say, and all inverse matrices such that . Let us observe that, in general .

    Now we will ultimately be rewarded by concentrating, not on what groups do but rather by what they are. The obvious abstraction is to rewrite this group as to signify this is the group of all real n x n matrices with non-zero determinant (we call such a matrix "non-singular", btw).

    This is an example of a Lie group, though to show this, we need to convince ourselves it is indeed a manifold.

    So, let be a group element, a matrix as defined above. is an n x n matrix, i.e. there are entries in , which we will write as , where i labels the rows, and j labels the columns, say. For ease of notation let's set Then evidently there is a mapping .

    Now provided we are consistent in our mappings, say the i-th element in our m-tuple maps onto the jk-th entry in our matrix, we see that the the m-tuple uniquely determines a matrix in . Thus we have an isomorphism .

    If we can show continuity of this map, we will have a homeomorphism. Without attempting any degree of rigour, I offer this. Let be the matrix with entries . Then whenever the point is arbitrarily close to so the matrix whose entries are is arbitrarily close the matrix whose entries are .

    This would imply a continuous isomorphism - a homeomorphism - so this group would indeed be a manifold.
    An alternate way to get to your conclusion that is a Lie group goes something like this. The full set of nxn matrices is a manifold of dimension using the coordinatizatoin that you suggested. One can look at the determinant as a map from that manifold to the real line. It is a polynomial in the coordinates and so is clearly clearly continuous and in fact smooth. The set of points on which it is zero is closed. So being an open subset of that manifold is clearly a manifold itself. Group multiplication and inversion are again determined by rational functions in the coordinates so the group operations and are therefore smooth. So is a Lie group.
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    I thought if I showed what I know about manifolds. That someone would say either, "Yes" that is what we were talking about. Or "No" that is not what we are talking about.

    No, that is not what we are talking about. As Chemboy said, the briefest glance at this thread should have shown you this.

    William - it is not in my power to ban you, but I shall certainly be making that recommendation to those with this power, since you seem incapable heeding my warnings

    -G-
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  26. #126  
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    William, I agree partly with what you say, but if you had looked at the previous 7 pages of this thread at all it should have been obvious that we are dealing with strictly mathematical manifolds which does not resemble the pictures you posted, so it was hardly necessary to ask the question...
    "There is a kind of lazy pleasure in useless and out-of-the-way erudition." -Jorge Luis Borges
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    Quote Originally Posted by Chemboy
    Mathematics is really just a hobby to me. Yes, I'm learning about manifolds strictly for fun. I cannot express how happy this type of thing makes me. That's just the way I am, knowledge and academics is my 'thing.' I would love to obtain a degree in mathematics someday but it's not my current path. Part of my motivation for learning advanced mathematics is my interest in QM/theoretical physics which I hope to have a very good understanding of someday. I know this is all quite far out there, but I trust that I can make it work given enough time and given my passion for it.

    1. My college major right now is conservation biology and I intend to obtain a degree in chemistry as well. Unfortunately I can't double major but I may do chemistry as grad work, or however it works out.

    2. I have taken Calc I, Calc II, and am in Calc III right now.

    3. I'm good with basic calculus of one and several variables. I worked through a good set of linear algebra notes online and feel like I have at least a good foundation in it, and I am taking Intro to Linear Algebra next semester. I'm taking a course in ODEs next semester as well and intend to learn the material over winter break before then. I don't know anything really of real analysis or advanced calculus beyond what random items I've read on Wikipedia. Sadly I know little to nothing of basic abstract algebra. I've never gotten far into rings or fields or anything of the sort on my own. I know the very basics of groups, as stated previously. Guitarist ran a thread on point set topology here (at my request) and I feel I have a fairly good foundation in it. I've been reading about complex analysis in Wikipedia lately and find it very interesting but haven't formally learned any of it.

    4. I plan to go to graduate school for conservation biology and ideally chemistry.

    5. As far as my objective in studying advanced mathematics goes, I simply have a passion for it (which started while taking Calc I) and just want to learn as much as possible because I love it. Also, as stated above, I hope to learn QM/theoretical physics and know I need the advanced mathematics for that.

    6. It's very difficult to say what's of primary interest to me. It all is, really. I'm very happy to be learning about manifolds right now, but really I'm always willing to learn anything at any time. More advanced group theory has been more at the forefront of my mind lately because I know it has significant applications in QM.

    7. The answer to this one is scary...I'm at a community college right now but transferring in the fall. The college I'm transferring to does not have a math department. It's SUNY-College of Environmental Science and Forestry in Syracuse, NY. They're very environmentally focused and so stick largely to biology and chemistry. However it is located right next to Syracuse University and I will have their facilities available to me (though of course with schedule-oriented and financial constraints).

    I realize my desire to learn a lot of advanced mathematics is very far out there, but I feel it's wrong to say it's unrealistic to learn it when I have the passion for it that I do. Right now my main issue is just lack of resources. I cannot afford expensive books (or any books at all really) and that is why I truly appreciate the knowledge that you, Guitarist, JaneBennett, and others are able to bring to the forum and to me.
    I have to admit that this is not what I expected. You clearly have a strong interest in advanced mathematics. You are also clearly pursuing a formal academic program in which advanced mathematics is not a priority.

    You have your work cut out for you. Obviously your main priority will be your primary interest, conservation biology and chemistry. Nothing wrong with that.
    SUNY-ESF looks like an ideal place for such studies.

    Syracuse University is where you would take your mathematics. They seem to have a good, though perhaps small department. I looked at the catalog of courses that they offer. It looks pretty good, and for instance they have solid courses in topology and differential geometry -- which is where you would see manifolds studied in some detail. But in the normal course of things in your chosen academic fields you will never see such classes, and they would be sufficiently time-consuming that you cannot take them on a lark while pursuing your main interests.

    So, assuming that you stay in conservation biology, and there is no reason why should not unless your interests change. it strikes me as almost impossible for you to pursue advanced mathematics intensely while pursuing your majors. It would be equally impossible for a budding differential geometer to simultaneously pursue deep studies in conservation biology.

    The implication is that extra-curricular study of advanced mathematics will have to play second fiddle to your main studies and therefore you will have to be patient in learning that material. It is quite simply fairly deep material that requires a lot of concentration and effort to learn. But since it is not central to your career goals, you have all the time in the world to learn it.

    So, here is my advice. Take it slowly and concentrate on your immediate career goals. If you decide to change careers and pursue advanced mathematics, then SUNY Stony Brook is as good as any place in the world. If it is good enough to have John Milnor on the faculty, it is good enough for anyone. If you stay with conservation biology, then study mathematics at your leisure, but pick one area at a time to study and work through it thoroughly. Don't try to do too much at once, despite the temptation to chase a lot of interesting ideas.

    Trying to learn an area in detail from the internet whether from Wikipedia or through a forum is probably not going to work well. Text books are polished presentations prepared by experts. They have selected key topics to be covered and have arranged them in an order that permits logical progression and development of complete proofs. That is critical. Forums and Wikipedia might help, and can provide a "flavor" of the subject, but they are no substitute for a unified treatment put together by a subject matter expert and polished through classroom use and comments from other experts.

    You need good texts and you need to go through them is some reasonable order. One problem is that the topics in which you have expressed interest basically cover all of the material one would expect of someone with an advanced degree in mathematics. In fact about the only ingredient, admittedly the key ingredient, missing for a Ph.D. is original significant research.

    Perhaps it would be good to back off a bit and concentrate on a few basics. If we do that, and don't get too fancy, but still cover material that ought to be of real interest to you, then you might want to look at the following books, in roughly the following order:

    1. Introduction to Real Analysis -- Bartle

    2. Algebra -- Michael Artin

    3. Calculus on Manifolds -- Michael Spivak

    4. General Topology -- John Kelley or (and this might interest you) Topology; An Outline for a First Course, by Lewis E. Ward which is designed for the student to provide all of the proofs. This a book for what is called a "Moore Method" topology course and is a lot of fun (and hard work).

    Ward's book is available on the used book market, try Alibris, for something under $10. It would be a wonderful way for you to work on your own and see if you really like theoretical, proof-oriented mathematics. It is a set of definitions, statements of theorems and statements of examples for which you are to provide the proofs and work out the examples. It also has a few mistakes for you to correct. Only a few of the proofs are really hard, and might require you to get help or go to a reference.

    If you understand the basic material in these books (not all of it but just the basics) you will head and shoulders beyond nearly everyone in the general populace and you could reasonably continue mathematics studies formally in a graduate school if you chose to do that.
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  28. #128  
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    Quote Originally Posted by DrRocket
    An alternate way to get to your conclusion that is a Lie group goes something like this........ .
    Thank you, that's much better.

    Actually you are making me feel a bit guilty about dragging Chemboy away from his proper studies. But, that's his affair I suppose, and besides it appears he is not the only reader of this thread.

    So, recall we have our first Lie group, the general linear group of real non-singular n x n matrices, which we called . First thing to note is that often, not always, when we see an "R" given for any Lie group we may replace it with a "C" to denote the fact we are working over the complex numbers.

    Second that the groups are the mummy and daddy of all the matrix Lie groups, since all matrix Lie groups come to us as subgroups of one of them. Let's have a look at one.

    As a manifold is not connected, which for now we can take to mean that there is no matrix which "bridges the gap" between those matrices whose determinant is strictly positive and those with strictly negative determinant.

    But if we concentrate on the subset of matrices with positive determinant, we will find we have included the identity matrix, so this is in fact a subgroup. This is the group called the special real linear group. We may also have it's big sister .

    The groups of particular interest to physicists are the groups whose elements are the matrices that satisfy where the "dagger" means the transpose of the matrix whose entries have been complex-conjugated. Since complex conjugation has no meaning for real numbers, we recognize 2 general constructions, the unitary groups and the orthogonal groups which simply satisfy . Notice this;

    In the first case I have no need to specify the field, as the term "unitary" includes it. In the second case I simply chose not to; although orthogonal groups can be imposed on the complex field, in the words of my tutor "only a madman or a masochist would study them".

    So the group consists of the unitary matrices with entries from the complex field and determinant . As before, we find the subgroup with . The group is the group of real orthogonal matrices with with subgroup .

    Not only are these groups important in physics, they are relatively easy to work with, but not so easy as to render them trivial. For this reason, we shall restrict our attention to them.

    Another monster post I fear. Sorry
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  29. #129  
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    If you're really against taking a little journey into Lie group-land we don't have to. I was just simply saying that if a little work with groups would assist our study of manifolds then I'm willing to see some group stuff, to benefit our manifolds. If you feel we'd get along better with our manifolds without having the support of knowledge around Lie groups then I'm fine with sticking to manifolds.

    I'm good with everything so far...it looks very nice.

    So the issue of there being a gap between positive-determinant matrices and negative-determinant matrices...do we find a way to bridge that gap or is that why we look at the special groups, and solve the problem just by sticking to the positive-determinant matrices?
    "There is a kind of lazy pleasure in useless and out-of-the-way erudition." -Jorge Luis Borges
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  30. #130  
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    Quote Originally Posted by Chemboy
    If you're really against taking a little journey into Lie group-land we don't have to. I was just simply saying that if a little work with groups would assist our study of manifolds then I'm willing to see some group stuff, to benefit our manifolds. If you feel we'd get along better with our manifolds without having the support of knowledge around Lie groups then I'm fine with sticking to manifolds.

    I'm good with everything so far...it looks very nice.

    So the issue of there being a gap between positive-determinant matrices and negative-determinant matrices...do we find a way to bridge that gap or is that why we look at the special groups, and solve the problem just by sticking to the positive-determinant matrices?
    This is in part why I worry a bit about your going too quickly into deep waters. Lie groups are a large and very deep subject. But we are here, so here is one way to look at the issue.

    Look at the determinant as a function defined on some open subset of GL(n,R). It is given by a polynomial in the entries of the matrix and is therefore clearly continous. On the entire manifold GL(n,R) there are two classes of matrices, ones for which the determinant is positive and ones for which it is negative. Those classes comprise two open and disjoint subsets of GL(n,R) and itis not too hard to see that theyare precisely the connected components of GL(n,R) viewed as just a topological space. So one usually looks at only one or the other component at a particular time.

    This is one reason why one does not normally study manifolds in detail until one has a solid background in the basics of point set topology. The issue is not group theory, but topology. Lie groups are first topological groups and secondly manifolds.
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  31. #131  
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    Quote Originally Posted by DrRocket
    This is in part why I worry a bit about your going too quickly into deep waters. Lie groups are a large and very deep subject..
    Then you must blame me for this. I picked up on Chemboy's interest in learning some group theory and also his interest in QFT. Putting these together I thought that principal bundles would be the way to go, which as you know, in QFT are Lie group bundles.

    But here, Chemboy - I'm afraid you have it the wrong way around. The study of Lie groups will not aid your understanding of manifolds, if anything they will impede it; they are a rather specialized application of the theory and involve concepts that most certainly do not generalize.

    I suggest that, having been told what a Lie group is, maybe we should leave like that for now? I confess I am at a loss to know where to go from here, if anywhere.
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    Quote Originally Posted by Guitarist
    Quote Originally Posted by DrRocket
    This is in part why I worry a bit about your going too quickly into deep waters. Lie groups are a large and very deep subject..
    Then you must blame me for this. I picked up on Chemboy's interest in learning some group theory and also his interest in QFT. Putting these together I thought that principal bundles would be the way to go, which as you know, in QFT are Lie group bundles.
    If you can explain quantum field theory in terms of rigorous mathematics then I am all ears.

    There is a 2-volume set published by the American Mathematics Society called Quantum Fields and Strings: A Course for Mathematicians, It is a set of notes and lectures from a special year at the Institute for Advanced Studies and the creators are some of the best mathematicians on the planet -- Deligne, Witten, Kashdan, Eitingof, MOrgan, Freed, Morrison, and Jeffrey. These books contain some of the hairiest mathematics that I have ever seen. I can't begin to follow all of it or in fact any significant fraction of it.

    The subject is still not on rigorous mathematical footing. If you can explain Feynman path integrals or renormalization in rigorous mathematical terms then I can assure you of lasting fame.
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  33. #133  
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    Now where did that come from? I made neither if the claims you attribute to me.

    Look, this site is, at the end of the day, an internet chat-room. Some people take this to mean it is a place just to mess around. In the context of this sub-forum, I take it to mean it is a place where interested folk can share their limited (my case) or extensive (your case) knowledge.

    I have at my disposal some, mind only some, of the mathematics used in QFT, which I had thought about sharing. It seems I have over-stepped the mark, and tried to appear more knowledgeable than I really am; for which I am sorry. Truly
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    It sounds as if sticking with manifolds is going to be best here so I'd like to do that.
    "There is a kind of lazy pleasure in useless and out-of-the-way erudition." -Jorge Luis Borges
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  35. #135  
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    Quote Originally Posted by Guitarist
    Now where did that come from? I made neither if the claims you attribute to me.

    Look, this site is, at the end of the day, an internet chat-room. Some people take this to mean it is a place just to mess around. In the context of this sub-forum, I take it to mean it is a place where interested folk can share their limited (my case) or extensive (your case) knowledge.

    I have at my disposal some, mind only some, of the mathematics used in QFT, which I had thought about sharing. It seems I have over-stepped the mark, and tried to appear more knowledgeable than I really am; for which I am sorry. Truly
    I was not trying to offend you. Quite the contrary. And I did not intend to make it sound as though you had made any specific claim. I was expanding on the theme of QFT in order to make a general point to an audience that I believe includes quite a few lurkers. It was not specifically directed at you, since I think you understood the points that I raised before I raised them.

    I was trying to make clear to any interested reader the extreme difficulty in making rigorous or indeed understanding at almost any level the mathematics of quantum field theory. In my opinion that understanding represents the most formidable challenge to both mathematics and physics, jointly and severally, on the planet. It is truly daunting stuff.

    You haven't overstepped anything. But folks reading the thread might come to expect some sort of crystal clear explanation, when none exists. That is not because of any personal shortcoming on your part. It is because NOBODY has such an understanding. And that nobody includes the smartest creatures on two legs.
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    Looks like I have a history of taking offense where none was intended. Am I prickly or is it cultural?
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    Quote Originally Posted by Guitarist
    Looks like I have a history of taking offense where none was intended. Am I prickly or is it cultural?
    Everybody is prickly sometimes. You ought to see me get prickly. Much more from William McCormick and I may start to resemble a saguaro cactus.

    I don't know if it is cultural. I don't think I can describe my culture very succintly. I think I understand that you are British, but beyond that I have no insight. Are your Anglo-Saxon, Norman, Scottish, Druid :-) ?

    I know a Scottsman locally who seems pretty reasonable. He shoots nice classic shotguns anyway. I was OK when he showed up on 4 July wearing a kilt, even if the dirk in his sock was dull. But I did have a problem when he offered to reveal what we was wearing under the kilt -- there are some things that are better left a mystery.
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  38. #138  
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    ...there is officially a large picture of a saguaro cactus in the thread on differentiable manifolds...

    ...that's all I have to say.
    "There is a kind of lazy pleasure in useless and out-of-the-way erudition." -Jorge Luis Borges
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    Quote Originally Posted by Chemboy
    ...there is officially a large picture of a saguaro cactus in the thread on differentiable manifolds...

    ...that's all I have to say.
    A manifold............. with singularities ?
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  40. #140  
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    haha. I guess so.
    "There is a kind of lazy pleasure in useless and out-of-the-way erudition." -Jorge Luis Borges
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    I sort of went off the boil a bit - see if I can rectify.

    Recall I was, quite rightly, pulled up for defining an abstract manifold as a topological space with a "shape". The reason was because without some notion of length, distance and angle, the word has no meaning. So we now want to find a manifold were this word does have meaning. For this we will need something called the "metric tensor". Since this arises as a rather general construction, I shall go through it in some detail.

    First: a function is said to be bilinear if it is linear in each argument considered separately. Now a definition:

    A bilinear mapping on a vector space is said to be a tensor product iff for some there is an induced mapping such that . Like this:



    This is the most general definition I know, and is called the "universal mapping property". (You will see why, I hope, I chose that one - I will give a more concrete definition later) When this is satisfied, I will use the notation that, for where, by definition, .

    Now we have maps which we called covectors i.e. elements in V*. If the condition in our definition is satisfied, I may have . Of course we recognize this as the inner product on V!

    The above defines as a tensor of type (0,2), and an element in the vector space . It's called a "metric tensor".

    And now I can explain what the notation means more concretely. Suppose is a basis for . Then let . Then the components of on this basis are simply the products taken individually, which is of course a matrix rather than a tuple.
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  42. #142  
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    Quote Originally Posted by Guitarist
    A bilinear mapping on a vector space is said to be a tensor product iff for some there is an induced mapping such that .
    This didn't click with me but everything below it makes sense. I'll keep working on it but for now I need to get some work done.
    "There is a kind of lazy pleasure in useless and out-of-the-way erudition." -Jorge Luis Borges
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    Ya, well, don't worry about it - that was just me pretending to be clever.

    Now our metric tensor, acting on vector spaces, is behaving simply as an inner product; hand the guy a pair of vectors, and he'll hand you back a scalar. The nice thing is this; if you hand him a single vector, he'll hand you back a covector. We can picture it like this.

    Suppose for some fixed we form . Then for any arbitrary we will have , so that is, in effect, acting as a covector.

    Now recall I said the components of is the matrix . In fact, most people write this matrix as Furthermore, it is almost universal to use this matrix to refer to the metric tensor itself, that is is in common usage.

    Purists may perhaps say this is wrong; let them, we all know what is meant. As my mentor was fond of saying "notation is arbitrary". So, adopting the same sort of notation for arbitrary vector and covectors, the above becomes

    Note there is a convention, due to none other that Einstein, that suppresses the summation symbol, and simply assumes summation over repeated indices. I imagine you have encountered this in your reading? Thus, . This called "index lowering"

    So what happens if you hand this guy a covector, say ? Well, he has no way of processing this (by definition, essentially), so we add the rider to the "summation convention" that the indices that are implicitly summed over must be in the up/down positions.

    However, even though our friend cannot digest covectors, he knows a man who can. But now I have to run.

    Later (L8er?? Ugh!)
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  44. #144  
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    Yes, I have seen the Einstein summation notation before somewhere.

    ...I was just thinking about it and I think I answered my own question, but here it is anyway... Will vectors always have an upper index and covectors a lower index? In any case I suppose I'll get a good feel for the notation as we go on.

    A question I have...and you should know that I think this way, I don't let details go...how does the metric tensor actually act on the two vectors? I understand its formation from the two covectors, I've seen how the outer product of two (co)vectors is taken, which is what's happening here, right? A matrix is formed from the (co)vectors. But how does this matrix now actually 'physically' act on the two vectors, how does one work it out?
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  45. #145  
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    Quote Originally Posted by Chemboy
    ...I was just thinking about it and I think I answered my own question, but here it is anyway... Will vectors always have an upper index and covectors a lower index?
    Yes, but it is merely a convention; you may change it if you want, but why bother? There is only one alternative.

    how does the metric tensor actually act on the two vectors?
    Umm, I'm not sure how to answer this question as posed. I tend to think of this tensor as more of a "rule" - it is certainly not an operator, since its "action" results in a scalar. Let's see if I can explain.

    Let's say that on Cartesian basis, i.e. a Euclidean space. Then the "dot product" is given by , which is clearly scalar. We can see a rule that says "multiply together only those components that have a matching superscript, then add the result".

    We can formalize this by using the rule "offer all pairwise products a coefficient of 1 when i = j in and a coefficient of 0 otherwise, and then add the result". Obviously this implies the multiplicative Kronecker delta on this sum of products.

    We will take this to define a Euclidean "flat" space". The general metric tensor need not be so picky, and we may assume from the above that, for some other "rule" , a different assignment of coefficients defines a non-Euclidean space, for which the metric tensor tells us how much it deviates from flat space.

    Thus: If our space is Euclidean and flat, otherwise it has some sort of curvature.

    I hope this clarifies - we had a houseful of weekender's so I haven't much of chance to think about this response!
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  46. #146  
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    ok, that works. I'm good to go.
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  47. #147  
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    OK. So we have a differentiable manifold and at each point a tangent vector space we called . Then it is not hard to see that the assertion that the metric tensor defines an inner product on every tangent space endows our manifold with a rather specific structure. (Recall that, in general, not all vector spaces have an inner product defined).

    Neither is it hard to see that this implies the existence of a field of metric tensors, i.e. one metric tensor at each point . This field, of course arises as a result of some arbitrary section of a tensor bundle, just as for vector fields (tensors are, after all, just sort of de luxe vectors).

    I remind you that an inner product space is, by definition, a normed space. That is, for some fixed vector, the norm ("length") of v.

    Now this. The vector subspace for which for any w is called the "kernel" of g. By requiring that the kernel is uniquely the zero vector I am requiring the metric on a manifold to be non-degenerate, i.e. for and any w, and I will have defined a pseudo-Riemann manifold. If the metric on a manifold is non-degenerate and invariably positive or zero, then this is said to be a Riemann manifold.

    So we will now want to think about distance on a Riemann manifold. I confess to having a slight problem here (I think I see a way to explain - lemme see if it checks out OK), so while I think about it, let me point out something rather obvious.

    Suppose are antipodal points on the manifold . Then there are two "shortest" distances between p and q. In the case of there is an infinity of them, so we need to proceed with some caution
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  48. #148  
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    Makes sense to me.
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    Is that it?? Not what I'd call a discussion, but I suppose I should grateful for any tangible evidence that my considerable efforts are being read.

    Anyway, we want to talk about the concept of distance on a manifold. Having done that, it will be no work at all to show that (a) this concept only has meaning on a Riemann manifold and (b) why we require the matrix representation of the metric tensor to be invertible.

    Now, we have an intuition about distance in Euclidean space, and we want this to carry over to our manifold, as far as is possible. Some of what follows may appear overly pedantic, but it will required to place an intuitive concept in a decidedly non-intuitive context.

    So, the distance between two points on a manifold is the length of the shortest curve connecting them. This is already slightly problematical, at least at first sight - what's meant by "shortest"?. I thought at first I might use a variational principle, but I couldn't make it work, so let's use the notion of a line integral from calculus. Let me remind you how this works in the Euclidean case:

    On our curve, place an arbitrary number of points and connect all neighbouring points by a chord. This called a "partition" btw. Then the sum of the lengths of all these chords (which are found using Pythagoras) approximates the length of our curve. As we allow the number of these points to approach infinity, so the length of the chords approaches zero, and their sum goes over to the integral. This is the line integral, very roughly indeed; it gives the true length of our curve.

    So, let be a curve . We will find it useful to think of this curve as the image of some line segment in , that is, at each point on the curve I may associate a different real number. This is called a "parametrization" of our curve. Call a "parameter point" on as

    Now we have at each point a tangent vector space, therefore, for any curve , a vector tangent to at each point. This is, of course a differential operator, acting now on 1 variable , which for brevity I will write as .

    So, from our preceding "discussions" I guess the rest is automatic, right? Fill the rest if you can, otherwise wait for me to get in a less prickly mood.
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  50. #150  
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    Sorry...I just didn't really have any questions on that...

    So when we look at our curve , we want to think of it as a function of the sort? And then it makes sense that our tangent vector at any point on gamma is .
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  51. #151  
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    Umm, if I understand your notation correctly, this doesn't look quite right. That is, I take it that the guys with hats are unit vectors? What are x and y? Points? Here's a tip: Whenever you introduce notation, explain exactly how you are using it. In fact, I'll go out on a limb; there is no such thing as standard notation!

    Anyway. here goes. If is a differentiable manifold, we lose no generality by defining a curve in as a continuous function under the condition that this function is strictly injective (this means that for any is unique).

    Nice thing about this definition is it gives us a 1-dimensional sub-manifold of - it's just a real line through our manifold. Let's insist. then that for a curve that

    As I thought we had agreed, at each and every point there is a tangent space which is tangent to some curve (there may be gazillions of such curves, of course). Then I define the vector tangent to my curve to be . We also know that the vector is totally oblivious to the existence of the vector , so that this provides us with a ready-made infinitesimal partition of our curve Note: I have edited out a silly mistake

    So our house of cards falls in on itself: We know that the norm of this vector is defined as

    Then by the definition of the line integral I gave earlier, we merely have that the length of any curve is given by

    So very easily we see that the distance between p and q is the greatest lower bound of all curves between these points. This is in complete accord with our intuition about distance in the Euclidean case
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  52. #152  
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    Now that I think about it the vector valued function for our curve doesn't really make sense, since we don't have s and s here.

    Quote Originally Posted by Guitarist
    Then I define the vector tangent to my curve to be .
    What exactly is , the rate of change of the vector with respect to ? That's what I take from that, but "rate of change of the vector" doesn't seem to make sense...

    Quote Originally Posted by Guitarist
    We also know that the vector is totally oblivious to the existence of the vector , so that this provides us with a ready-made infinitesimal partition of our curve
    How is this? I'm not quite getting it...

    Quote Originally Posted by Guitarist
    So very easily we see that the distance between p and q is the greatest lower bound of all curves between these points.
    What do you mean by 'greatest lower bound'?
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  53. #153  
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    Quote Originally Posted by Chemboy

    What exactly is , the rate of change of the vector with respect to ? That's what I take from that, but "rate of change of the vector" doesn't seem to make sense...
    You're teasing me, right? Maybe not....

    What if v is velocity and t is time? What do you have now? Does that make sense? What sort of quantity do you have now?

    What do you mean by 'greatest lower bound'?
    Looks like there is work to be done here. Stay tuned
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  54. #154  
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    Well don't I feel unintelligent now. I considered velocity but doubted myself because we haven't had any mention of it before. Ah well.

    I did some reading on the greatest lower bound and I'm fine now...should've done that before.
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  55. #155  
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    So, I have a sneaky suspicion I am wasting my time here, but here goes.........

    Note that in what follows, I am using the notation is a type (1, 0) tensor, i.e. a vector, and is a type (0, 1) tensor, i.e. a covector; similarly for higher rank tensors. I am also using the summation convention discussed earlier. So.

    Recall we have a vector space , with the metric tensor .

    We saw that, for that , We also claimed that where is a covector.

    Recall also that the expression has no meaning, that is our type (0, 2) metric tensor cannot process covectors. I hinted he had a companion who can do so, however. Let's see.....

    The set is, of course, a perfectly respectable vector space, and, as such, is entitled to its own dual space. Is it ? Well, yes and no.

    Let's call it for now, Now, in the finite dimensional case, we know there is an isomorphism (how? Exercise!). Since this depends entirely upon an arbitrary choice of basis for , one says this isomorphism is "not natural", which conjures up an image or two.

    It turns out that, again in the finite dimensional case, that the isomorphism (by transitivity) is natural in the sense it does not depend on a choice of coordinates. The proofs (I know of two) are rather subtle, so I ask you to take my word for it; alternatively you can dazzle us with your own versions. In this circumstance, any time I see some ,say, I am free to replace it with some .

    (This relates to a comment by another poster that topology defines equality only up to isomorphism; this is a hobby-horse of mine, as I believe it applies generally. It's a radical view I know. I intend to say some more on this elsewhere)

    Thus I may have that for any (although I do not insist on equality with ). And so I form the vector space of type (2, 0) tensors .

    I define the element by as a metric tensor, and most particularly, .

    Now since we had that this implies that our two metric tensors, are mutual inverses.

    Now I think this is nice. But I am a little frustrated there is so little feed-back here.
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  56. #156  
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    ok, so the dual space to is and not , but since is isomorphic to it doesn't matter which one we use?

    In general, will all dual spaces be isomorphic to a lower one? So for example, do we have , etc.?

    I feel like I need a lesson on what ""up to" isomorphism" means... or "up to" some morphism in general... maybe that would be too off topic and would be more appropriate for another thread...

    I'm going to think about the proof for an isomorphism , but I'm not so sure I'll get anywhere... I'd like to see it though if I can't figure it out...

    Will we be seeing a sort of or something of the sort that creates a correspondence between vectors and covectors...? That's what I thought of when you said that and are inverses...
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  57. #157  
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    Quote Originally Posted by Chemboy
    ok, so the dual space to is and not , but since is isomorphic to it doesn't matter which one we use?
    "Naturally isomorphic" is the key here, but in this circumstance, yes

    I feel like I need a lesson on what ""up to" isomorphism" means... or "up to" some morphism in general... maybe that would be too off topic and would be more appropriate for another thread...
    Coming soon. mea culpa, I used jargon without explanation - bad boy!

    Will we be seeing a sort of or something of the sort that creates a correspondence between vectors and covectors...? That's what I thought of when you said that and are inverses...
    Hmm. Check your understanding of inverse mappings. Specifically, if are both bijective, what is ? What is ?
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  58. #158  
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    I see my problem... and produce scalars, but take in vectors and covectors, so one of them would have to be the inverse of itself, take in the scalar produced by the other and map it to a vector or covector. So if anything it would be more like (I hope). Is the order correct though, would it be ?
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  59. #159  
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    Nah, you are drifting way off the point. Let me remind you that our metric tensors "take in" a pair of (co)vectors and deliver a scalar. If offered a single (co)vector, they will offer a (co)vector. Like this: . Likewise the type (2, 0) metric tensor.

    Re-read my relevant post.

    Now I beg you to try this exercise. If the bijections , what would you call the map ? Hint: It is a map ; give it a name.

    Now give a name to
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  60. #160  
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    oh, yeah, sorry...I just didn't have that in mind, though I should have.

    is the identity mapping?

    is also the identity mapping, but .

    and , right?
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  61. #161  
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    Quote Originally Posted by Chemboy
    is the identity mapping?
    Yes, good work! A common notation is that Notice this is true only "up to isomorphism", about which I have promised to say more (but I hear Ms, G. barking in the background - so later)

    is also the identity mapping, but .
    Yup, again good - let's call this as

    and , right?
    Aaargh. Try to fix this up. You may compose maps but you may not compose images. Tidy this up for full credit.

    Hint: RHS of your first equality is correct, RHS of your second is not; both LHS's are meaningless, by the above
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  62. #162  
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    ...?
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  63. #163  
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    There you go again, composing images. Look, I am willing to assume full responsibility for your confusion, so let me try and make amends.

    If , then, for any its image is an element, a point, perhaps, or whatever sort of elements belong to .

    Similarly, if , then, for any its image is an element.

    Now can you see why the construction makes no sense; first because you cannot compose elements and second, even if you could (say by adding or whatever) they live in different places.

    What you seem to be missing is that the composite is to be thought of as a single map, like this



    You saw this already by equating this composite with the identity. In this circumstance, the correct interpretation is . Note, however, one doesn't require that these composites always equate to identity maps, even when maps "go in opposite directions"

    This will be true in general, so if the composite .
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  64. #164  
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    Quote Originally Posted by Guitarist
    There you go again, composing images. Look, I am willing to assume full responsibility for your confusion, so let me try and make amends.

    If , then, for any its image is an element, a point, perhaps, or whatever sort of elements belong to .

    Similarly, if , then, for any its image is an element.

    Now can you see why the construction makes no sense; first because you cannot compose elements and second, even if you could (say by adding or whatever) they live in different places.

    What you seem to be missing is that the composite is to be thought of as a single map, like this



    You saw this already by equating this composite with the identity. In this circumstance, the correct interpretation is . Note, however, one doesn't require that these composites always equate to identity maps, even when maps "go in opposite directions"

    This will be true in general, so if the composite .
    Everything that you said here is correct.

    But something is amiss with respect to this overall discussion. If there is confusion on the points involved in composing functions, particularly with respect to things like distinguishing between images and functions, then I suspect that there has been less than complete communication with respect to concepts such as manifolds, tangent bundles, sections of bundles, tensors, etc.

    The overall conversation is just a bit surreal. I don't think everyone is on the same page, but am at a loss as to how to correct this situation at this point.
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  65. #165  
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    ok, I'm fine now and I understand the notation. That was one of those very, very dull moments. And I've understand the concept perfectly well, I swear, I just wasn't thinking and screwed up the notation.

    Actually...I got a book on real and complex analysis and it stated in it the fact that is so often used to refer to the function but that's it's really a big abuse of notation because the function is just , and is its image. But functions are nearly always written as in schools and I see the difference now but it's kind of wired in my brain, but now I'll make the effort to stop thinking that way.
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  66. #166  
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    Quote Originally Posted by DrRocket
    The overall conversation is just a bit surreal. I don't think everyone is on the same page, but am at a loss as to how to correct this situation at this point.
    Please check your PM Inbox
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