Tangent Bundles §2.9

[20:04] <@breeden> this was just Tangent Bundles, and hochs request Sard's theorem
[20:04] <@breeden> There were a few things that I wanted to get to that I really didn't get the preparation for.
[20:05] <@breeden> Ok so let's start with tangent bundles.
[20:06] <@breeden> What's the question here?  We've constructed manifolds as a generalization of curves and surfaces in euclidean space, by insisting that they locally look like a euclidean space.
[20:07] <@breeden> The down side to this is that we've now lost a meaning for differentiation.
[20:07] <@breeden> So tangent bundles were designed to fix this gap.
[20:07] <@breeden> Essentially, differentiation in a euclidean space amounts to finding a linear approximation for a smooth function.
[20:08] <@breeden> But we no longer live in a vector space, so linear no longer makes sense.
[20:08] <DaMancha> take the best linear fit  :)
[20:09] <@breeden> yup, the best linear fit... which is important, this gives us uniqueness of the derivative.
[20:09] <@breeden> and our best estimates.
[20:09] <@breeden> now, at first you might say.  Hey, wait a minute breeden.  Manifolds are locally euclidean, and there's your vector space right there.
[20:10] <@breeden> However, this perspective would fail to give us a well defined concept once we start thinking about changing coordinates.
[20:11] <@breeden> So instead, we will need to attach a vector space to each point of the manifold in a way that is invariant with the coordinates we decide to look at it.
[20:12] * @breeden (~breeden@66-188-126-198.dhcp.mdsn.wi.charter.com) Quit (Read error: Connection reset by peer�)
[20:12] <Capso> and what better way to do that than to pass curves...
[20:12] <Capso> oh
[20:13] <hochs> someone call 911 
[20:13] <i_c-Y> haha
[20:13] <Capso> <intermission> ?
[20:13] * breeden_ (~breeden@66-188-126-198.dhcp.mdsn.wi.charter.com) has joined #mathematics
[20:13] <hochs> he's breathing again 
[20:13] <breeden_> heh
[20:13] <breeden_> odd
[20:13] * breeden_ is now known as breeden
[20:13] * Evariste sets mode: +o cyby
[20:13] <Capso> don't do that to me again.
[20:13] * cyby sets mode: +o breeden
[20:13] * cyby sets mode: -o cyby
[20:14] <@breeden> sorry, internet died
[20:14] <hochs> got cutoff at <@breeden> So instead, we will need to attach a vector space to each point of the manifold in a way that is invariant with the coordinates we decide to look at it. 
[20:15] <@breeden> Ok good
[20:15] <@breeden> so we need to keep track of three things with constructing our tangent bundle.
[20:15] <@breeden> The chart, the point in the manifold, and the vectors.
[20:16] * Andrew- (~Andrew@c-76-29-130-219.hsd1.fl.comcast.net) has joined #mathematics
[20:16] <@breeden> We need to keep track of that charts, because we'll need our construction to ultimately be invariant of a change in coordinates.
[20:17] <DaMancha> can i interrupt for a second?
[20:17] <@breeden> so for any point p in a manifold M and a chart x containing p, we can consider the set { (x,v)_p | v in R^n }
[20:17] <@breeden> sure
[20:17] <DaMancha> if you weren't going to already, is it possible for you to pick a simple manifold and construct some tangent bundle and show why it is immune or invariant to coordinate changes?
[20:18] <Capso> :-)
[20:18] <Capso> I like DaMancha.
[20:18] <@breeden> we'll, this construction will make it blantly obvious as to why it's invariant to coordinate change :)
[20:18] <@breeden> but, hmmm....
[20:19] <Capso> an example would be nice.
[20:19] <Capso> the construction in bredon is a tad abstract.
[20:19] <@breeden> I didn't read bredon
[20:19] <@breeden> heh
[20:19] <@breeden> I just picked up my notes from when i was studying for my quals
[20:19] * Ino-2 (~Ino@41.34.17.35) has joined #mathematics
[20:20] <Capso> but it isn't much more than passing curves and taking differentials at the point.
[20:20] <@breeden> Ok.  So if we have a nice manifold, in fact any manifold that is contractible, we can do something really nice.  We can just take the product, M x R^n.
[20:20] <@breeden> there for each point p, we have { p } x R^n, as our vector space to work with, but this doesn't always work
[20:21] <@breeden> It's hard to visualize as well, since the only two 1-dimensional manifolds are the circle and a line, and their tangent bundles are the cyclinder and plane respectively
[20:21] <DaMancha> pick one, say our planet and your point is nyc (home of the best pizzas)
[20:21] <@breeden> and we've already hit dimension 2
[20:21] <@breeden> now if you go to dimension 2 manifolds, the tangent bundle will be dimension 4
[20:22] <@breeden> which is hard to visualize.
[20:23] <@breeden> Normally you would think of attaching a tangent plane to new york, and saying that is the tangent space to new york
[20:23] <koro> i don't think it's clear a priori what is meant by "invariant by coordinate changes" when talking about tangent spaces
[20:23] <@breeden> but this isn't entirely accurate, because attaching a tangent plane to new york doesn't yield a manifold.
[20:23] <@breeden> koro, i was talking about wanting to define the derivative
[20:23] <koro> maybe this is why it is not easy to provide an example before giving a definition
[20:24] <DaMancha> and it is exterior to our manifold so it'll give problems, right?
[20:24] * onceler (~fhdfjdfgg@adsl-99-16-57-239.dsl.lsan03.sbcglobal.net) has joined #mathematics
[20:24] <DaMancha> okay fair enough, i don't want to derail this. maybe my question was anachronistic.
[20:24] <@breeden> DaMancha, well actually, it's because at the point in new york, you no longer have a "smooth" surface, it's no longer a manifold because it's not locally homeomorphic to R^k
[20:24] * Ino-2 (~Ino@41.34.17.35) has left #mathematics
[20:25] <@breeden> like if you drew a line tangent to the circle, you'd get something that looks like an X locally.
[20:26] <@breeden> But, rest assure.  Your question will be made clear in the construction.
[20:26] <DaMancha> thanks :)
[20:26] <@breeden> So we need to track 3 things, charts, points, and vectors.
[20:27] <@breeden> So given a point p in a manifold M^n (^n means "of dimension n"), we will consider the collection of vectors { (x,v)_p | v in R^n }
[20:28] <@breeden> Now there are many charts that can cover p, we need a way to say when a vector (x,v)_p is identified with (y,v)_p for two charts x and y.
[20:28] * oncelerr (~fhdfjdfgg@adsl-99-16-57-239.dsl.lsan03.sbcglobal.net) has joined #mathematics
[20:28] <@breeden> we say (x,v)_p ~ (y,w)_p if w = D(y o x^{-1})(x(p)) v 
[20:28] <@breeden> Ok, so let's say that in words
[20:29] <@breeden> recall, y o x^{-1} is the transition map from going to what image of x to the image of y.
[20:29] <@breeden> We are taking the derivative of this map, a the point x(p), and sending v through it
[20:30] <@breeden> (just to try to clear up notation, this stuff get notoriously bad for irc)
[20:30] <@breeden> Anyways, this relation will define an equivalence class which we can denote by [x,v]_p.
[20:32] <@breeden> Ok, the set of equivalence classes { [x,v]_p | v in R^n } will be our tangent space to p in M denoted by T_p M
[20:32] <@breeden> so before I go on with more abstract none sense, we'll see how we use this to meet our original goal -- to make sense of differentiation.
[20:33] <@breeden> So, suppose that we have a smooth map f: M -> N.
[20:33] <@breeden> Now we define Df_p : T_p M -> T_f(p) N in the follow way:
[20:34] <@breeden> Df_p ([x,v]_p) = [y, D(y o f o x^{-1}) v]_f(p)
[20:35] <@breeden> Now this is the part where we needed to make sure that our map is well-defined
[20:35] <@breeden> ie: does it matter which charts x and y i use?
[20:35] <@breeden> and this is why we've constructed the tangent bundle the way we did:
[20:35] * onceler (~fhdfjdfgg@adsl-99-16-57-239.dsl.lsan03.sbcglobal.net) Quit (Ping timeout: 504 seconds�)
[20:36] <@breeden> Given charts x and z of p, we have: Df_p([x,v]_p) = [y, D(y o f o x^{-1}) v]_f(p)
[20:36] <@breeden> now D(y o f o x^{-1}) v = D(y o f o z^{-1} o z o x^{-1}) v
[20:37] <@breeden> but the chain rule gives,  D(y o f o z^{-1} o z o x^{-1}) v = D(y o f o z^{-1}) o D(z o x^{-1}) v
[20:37] <@breeden> and this is really Df_p([z, D(z o x^{-1}) v]_p)
[20:39] <@breeden> right, and [x,v]_p = [z, D(z o x^{-1}) v]_p by our equivalence class relation
[20:39] <@breeden> And the same calculation shows that it doesn't depend on our choice of charts for y either.
[20:40] <@breeden> DaMancha, so essentially, we defined our equivalence relation like that just so we can make our differentiation operator D to be well defined in using the most natural definition we can define.
[20:42] <@breeden> So we define our tangent bundle of a manifold M, denoted by TM, to be the disjoint union of T_p X for each p in M.
[20:42] <DaMancha> yes, i see this now.
[20:42] <@breeden> We need two things to make TM a manifold
[20:42] <@breeden> We need to impose a topology on TM and define some charts.
[20:43] <@breeden> Luckily, we'll be able to kill two birds with one stone on this one.
[20:43] <@breeden> but, i hope the notation doesn't get to bad
[20:43] <@breeden> Se TM is the disjoint union of T_p X for each p in M, and we can define a projection map, pi: TM -> M, that takes [x,v]_p |-> p.
[20:44] <@breeden> To put some charts on TM, we will use the charts that we already have on M.
[20:45] <@breeden> Given any chart (x,U) of M, we can define, t_x : pi^{-1}(U) -> UxR^n by [x,v]_p -> (q, v).
[20:45] <@breeden> We can insist that all of these maps are homeomorphisms (since we've yet to define a topology on TM).
[20:45] <@breeden> Further more, these will become our charts.
[20:46] <DaMancha> can i ask a question"?
[20:46] <@breeden> yes
[20:46] <Capso> so this is a sort of a "differential" surface to the original.
[20:46] <@breeden> (sorry, i kind of started speeding things up)
[20:46] <DaMancha> you define the TM as the disjoin union of T_pX for different p's on M
[20:46] <@breeden> yup
[20:47] <DaMancha> i see they're disjoint but aren't they by virtue of for p,q being differnte points on M, T_pX and T_qX are already disjoint
[20:47] <@breeden> well, yeah, i guess so :)
[20:47] <Capso> it's just a term
[20:48] <DaMancha> ok, that's good. was hoping it wasn't an additional restriction already not covered by uniqueness of points
[20:48] <DaMancha> thanks
[20:48] <@breeden> :)
[20:49] <Capso> no, the term "disjoint union" is used pretty liberally when we want to take the union of a collection of objects that should be seen as having nothing in common
[20:49] <Capso> whether they already do or do not, is not of concern.
[20:49] <@breeden> Right, so the only thing you'd have to check to see if these t_x's are in fact charts is showing that their transition maps are smooth.
[20:50] <@breeden> and i'll try to avoid using any notation for this one.
[20:51] <@breeden> but t_x o (t_y)^{-1} acts on U x R^n for some open set U, and sends it to V x R^n for some open set V.
[20:51] <DaMancha> if you don't give it a topology i will request a refund :)
[20:51] <@breeden> and does so by, (w,v) -> ( x o y^{-1}(w), D(x o y^{-1}) v )
[20:52] <@breeden> DaMancha, i did :)
[20:52] <@breeden> DaMancha, we gave the weak topology on TM to make all the t_x's continuous
[20:53] <DaMancha> yes, i missed it by babbling about dijointedness, i am going to be quiet now.
[20:53] <@breeden> Ie, open sets in TM are locally just t_x^{-1}(U) for open sets U.
[20:54] <@breeden> Right so the point is t_x o (t_y)^{-1} acts smoothly on the first component, and is just a linear map on the second component, and so these are indeed charts.
[20:55] <@breeden> and that's the tangent bundle
[20:55] <@breeden> it really is just a bunch of abstract nonesense at first, but there are some good uses for it
[20:56] <@breeden> For instance
[20:56] <@breeden> We can use the tangent bundle to find sub-manifolds.
[20:57] <@breeden> And 'similar' version of the implicit function theorem applies to manifolds saying, if f: M^n -> N^k, has constant rank k around a neighborhood of f^{-1}(p) in N, then f^{-1}(p) is a submanifold of M with rank n - k (or empty).
[20:58] <@breeden> oh right, so when I say if f has constant rank k, i mean the linear map Df has constant rank k at each point.
[20:58] <@breeden> and once you have that, you have more examples of manifolds than you'd know what to do with.
[20:59] <@breeden> For instance, f: R^n -> R, defined by (x_1, x_2, ..., x_n) -> x_1^2 + x_2^2 + ... + x_n^2
[21:00] <@breeden> Now the derivative of this is 2x_1 + 2x_2 + ... + 2x_n, which is not 0, and hence has rank 1, when (x_1, ..., x_n) are not 0.
[21:01] <@breeden> then this theorem says that f^{-1}(1) is a submanifold of R^n, and this is just S^{n-1}