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I'm having trouble visualizing what the topology/atlas of a tangent bundle $TM$ looks like, for a smooth manifold $M$. I know that $$\dim(TM)=2\dim(M).$$

Do the tangent bundles of the following spaces have any "known form", i.e. can be constructed (up to diffeomorphism) from known spaces $\mathbb{R}^n$, $\mathbb{S}^n$, $\mathbb{P}^n$, $\mathbb{T}^n$ via operations $\times$, $\#$, $\coprod$?

  • $T(\mathbb{S}^2)=?$
  • $T(\mathbb{T}^2)=?$
  • $T(\mathbb{T}^2\#T^2)=?$
  • $T(k\mathbb{T}^2)=?$, $\;\;\;k\in\mathbb{N}$ ($k$-fold connected sum $\#$)
  • $T(\mathbb{P}^2)=?$
  • $T(\mathbb{P}^2\#\mathbb{P}^2)=?$
  • $T(k\mathbb{P}^2)=?$, $\;\;\;k\in\mathbb{N}$ ($k$-fold connected sum $\#$)
  • $T(\mathbb{S}^n)=?$
  • $T(\mathbb{T}^n)=?$
  • $T(\mathbb{P}^n)=?$

($\mathbb{S}^n$ ... n-sphere, $\mathbb{T}^n$ ... $n$-torus $\mathbb{S}^1\times\ldots\times\mathbb{S}^1$, $\mathbb{P}^n$ ... real projective $n$-space, $\#$ ... connected sum)

I'm making these examples up, so if there are more illustrative ones, please explain those.

BTW, I know that $T(\mathbb{S}^1)=\mathbb{S}^1\times\mathbb{R}$ by visually thinking about it.

P.S. I'm just learning about these notions...

ADDITION: I just realized that all Lie groups have trivial tangent bundle, so $T(\mathbb{T}^n)\approx\mathbb{T}^n\!\times\!\mathbb{R}^n$.

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Why did you add the Riemannian and symplectic tags? You didn't specify Riemannian structures and the tangent bundle is not naturally a symplectic manifold (as opposed to the cotangent bundle). –  t.b. Jun 19 '11 at 10:42
    
Isn't riemannian geometry an upgrade (additional structure) of differential topology? And isn't Symplectic Geometry an upgrade of Riemannian? If those tags are inappropriate, I can remove them... –  Leon Lampret Jun 19 '11 at 10:46
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I would say Riemannian geometry is a subdiscipline of differential geometry (not topology). Every (paracompact) manifold can be equipped with a Riemannian metric and indeed, this is additional structure. On the other hand, symplectic geometry is a different business: only even-dimensional manifolds admit a symplectic structure and among the spheres only $S^2$ does. As I said, the cotangent bundle of a manifold has a canonical symplectic structure. I think removing those tags would be better. –  t.b. Jun 19 '11 at 10:51
    
Also: what is $g$ in $T(g\mathbb{T}^2)$ and $T(g\mathbb{P}^2)$? I suppose you mean the $g$-fold connected sum, right? By the way: in differential geometry $g$ usually stands for a Riemannian metric and in differential topology for the genus of a $2$-manifold, so I would choose a different letter. –  t.b. Jun 19 '11 at 10:52
    
Yes, $g\mathbb{P}^2=\mathbb{P}^2\#\ldots\#\mathbb{P}^2$ $g$-times. I included the tags because smooth manifolds are the central object of all those fields. OK, I'll remove them. –  Leon Lampret Jun 19 '11 at 10:55
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3 Answers 3

If you want to describe tangent bundles, the appropriate language is classifying spaces.

A vector bundle $\mathbb R^k \to E \to B$ over a space $B$ is described by a homotopy-class of map

$$B \to Gr_{\infty,k}$$

where $Gr_{\infty,k}$ is the space of all $k$-dimensional vector subspaces of $\oplus_\infty \mathbb R$.

So for example, the tangent bundle of $S^2$ is a $2$-dimensional vector bundle over $S^2$, so described by a map

$$S^2 \to Gr_{\infty,2}$$

$Gr_{\infty,2}$ as a space would be called $B(O_2)$, the classifying space of the Lie group $O_2$, meaning that it is the quotient of a contractible space by a free action of $O_2$ (think of the associated Stiefel space). So an element of $\pi_2 Gr_{\infty,2}$ is equivalent (via the homotopy long exact sequence) to an element of $\pi_1 O_2$, which is isomorphic to $\mathbb Z$.

i.e. 2-dimensional vector bundles over $S^2$ are described by an integer.

There's another way to see the above construction. Decompose $S^2$ into the union of two discs, the upper and lower hemisphere. Via pull-backs this decomposes $TS^2$ into (up to an isomorphism) $D_u \times \mathbb R^2$ and $D_l \times \mathbb R^2$ where $D_u$ and $D_l$ are the upper and lower hemi-spheres respectively. $\partial D_u = \partial D_l = S^1$. So there's a gluing map construction

$$ TS^2 = (D_u \times \mathbb R^2) \cup (D_l \times \mathbb R^2) $$

There's is a map describing how point on $\partial D_l \times \mathbb R^2$ have to be glued to points on $\partial D_u \times \mathbb R^2$ and it has the form

$$(z,v) \longmapsto (z,f_z(v))$$

where

$$f : S^1 \to O_2$$

The homotopy-class of this map is again described by an integer. These are the same two integers. A fun calculation shows you it's two, the Euler characteristic.

The above story is worked-out in more detail in Steenrod's book on fibre bundles. Also Milnor and Stasheff.

By-the-way, many people have trouble initially thinking about tangent bundles. They're fairly delicate objects.

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is $X$ always a strong deformation retract of $TX$? –  Leon Lampret Jun 26 '11 at 6:21
    
more generally, does every vector bundle $E$ over $X$ strongly deformation retract to $X\!\times\!\{0\}\!\approx\!X$? I'm thinking (for $x\!\in\!X$ and $v\!\in\!E_x\!=\!π^{−1}(x)$) that the deformation retraction is the homotopy $H(x,v,t):=(x,(1−t)v)$. Is this correct? –  Leon Lampret Jun 26 '11 at 9:53
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@Leon: you're right. –  Miha Habič Jun 26 '11 at 12:19
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I think operations other than $\times$ are irrelevant here because bundles are locally product structures. So let's restrict just to this case. Then you are in fact asking whether the tangent bundle can be trivial and this has to do with topology. The obstruction to being trivial is coming from the fact that the bundle can "wind" around the manifold in a non-trivial way, if the manifold is compact, containts holes, etc. More precisely, if the Euler characteristic is non-zero and the manifold is compact then the bundle can't be trivial by Poincaré-Hopf index theorem which shows that the vector field on such a manifold must have at least one zero. This is the case e.g. for $S^{2n}$ (the case $n=1$ being the famous Hairy ball theorem).

So, if we want to get non-trivial bundles on compact manifolds, we better look at manifolds with Euler characteristic zero, e.g. tori ${\mathbb T}^n$ which is indeed trivial. For ${\mathbb R}^n$ and it's easy to show that the tangent manifold is ${\mathbb R}^{2n}$ (note that the Euler characteristic here is 1 but the above theorem doesn't apply because this is not compact). Similarly for any one-dimensional manifold we get a trivial tangent bundle (essentially because only one-dimensional manifolds are ${\mathbb R}$ and $S^1$).

Both of the above constructions are special cases of more general families. Specifically, every Lie group has a trivial tangent bundle (this can be seen from the isomorphism between Lie algebra and left-invariant vector fields). Also, every contractible open subset $U$ of some manifold will have $TU = U \times {\mathbb R}^n$.

I am not sure about the general case though (i.e. whether vanishing Euler characteristic is also a sufficient condition for compact manifolds, etc.). But I suspect the situation can be quite non-trivial and one will need tools of algebraic topology to resolve it.

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Thank you, this was very helpful. Hmm, I guess the tangent bundle functor $T$ is much more complicated than I thought. I didn't expect $T(M)$ to produce complicated manifolds for such simple $M$s. But I still don't understand, why is $\#$ an irrelevant operation? –  Leon Lampret Jun 19 '11 at 12:49
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@Leon: I am not actually sure it is irrelevant, that's just my gut feeling. I'll think about it. –  Marek Jun 19 '11 at 13:00
    
@Marek: One more thing, is $T(M\times N)\approx T(M)\times T(N)$? –  Leon Lampret Jun 19 '11 at 14:07
    
@Leon: well, in this case we have again a Lie group, so the bundle is indeed trivial. Also, $T(M \times N) \cong TM \times TN$ because the coordinates don't mix. –  Marek Jun 19 '11 at 14:09
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It is not true that vanishing euler class implies trivial tangent bundle. Counter examples are all odd dimensional spheres except those in dimension 1,3, and 7. This is related to the fact that the only real division algebras occurring in dimensions 2, 4, and 8 (reals, quaternions, octonions). –  Eric O. Korman Jun 19 '11 at 14:41
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Here is a concrete perspective on tangent bundles that might help. Take an embedding of your manifold $M$ into $\mathbb{R}^N$ (by Whitney we can always do this). This gives an embedding of $TM$ into $T\mathbb{R}^N = \mathbb{R}^N\times \mathbb{R}^N$ (it's injective and you can write down the differential to check it is an immersion). This already gives you your tangent space inside a manifold with one chart.

Suppose that $M = f^{-1}(0)$ where $f:\mathbb{R}^N\rightarrow \mathbb{R}^k$ is a smooth function and 0 is a regular value. Then $M$ is a $N-k$ dimensional submanifold of $\mathbb{R}^N$ and $TM \subset T\mathbb{R}^N$ is $Df^{-1}(0,0)$.

For example, consider the function $f(x_1,\ldots ,x_n) = x_1^2+\cdots +x_n^2$. 1 is a regular value and $f^{-1}(1)$ is the $n-1$-sphere with its standard embedding. The derivative of $f$ at $(x_1,\ldots ,x_n)$ is the $1\times n$ matrix $Df_x = 2( x_1 \, x_2\, \cdots \, x_n)$ and the kernel of this map is the set $y\in \mathbb{R}^n$ such that $y\cdot x = 0$ (standard inner product). Therefore, we have an identification of $TS^{n-1}$ with $$\{(x,y) \in \mathbb{R}^{2n} | x\cdot y = 0, |x| = 1 \}.$$

The advantage to this perspective is that things are computable now (such as geometric structures). The disadvantage is that this is computationally practical only if you have a simple embedding into euclidean space, the description is not intrinsic (although the tangent bundle is), and it does not say nice abstract things about the tangent bundle.

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I realize this doesn't answer your question directly, but I hope it helps you understand what tangent bundles 'look like'. –  AnonymousCoward Feb 16 at 4:20
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