# Any example of manifold without global trivialization of tangent bundle

It is said for most manifolds, there does not exist a global trivialization of the tangent bundle. I am not quite clear about it.

The tangent bundle is defined as $$TM=\bigsqcup_{p\in M}T_PM$$

So is the above statement saying that generally $$\bigsqcup_{p\in M}T_PM\neq M\times\mathbb{R}^n?$$

But I think the tangent space is just attaching a $\mathbb{R}^n$ to every point on $M$, so I wonder what's the reason for it is not a product space?

Plus, when defining trivialization, we have a lot of constraint on the function $F:TM\rightarrow M\times V$, can anyone explain the necessity of those constraint?

At last, does $S^2$ has a global trivialization?

Update:

Following is my attempt to trivialize $S^2$, but meet some problem. I think it may reflect some aspect in the impossibility to trivialization of $S^2$, isn't it?

We want to define trivialization $F:TS^2\rightarrow S^2\times \mathbb{R}^2$. First of all, $F$ should be well-defined.

There are 3 different approach to define a tangent space, here I take the definition via chart.

So, an element in $TS^2$ is $\left[(p, v, (U,\varphi))\right]$, and of course I try to define its image to be $(p, v)$.

Then the problem comes. Because we need at least 2 chart to cover $S^2$, so when taking another representative $(p, w, (V,\phi))$ of the equivalent class $\left[(p, v, (U,\varphi))\right]$, we map it to $(p, w)$, which conflicts the previous image.

Of course it is only one attempt, but I think it may reflect some difficulty to define $F$ because it need to preserve coordinate transformation.

Right?

Eh.. I realized my attempt is too trivial. If I apply this method to any manifold, $F$ is never well-define...

Can anyone provide an manifold which can be trivialize? I think I may use it to get better understanding.

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The tangent bundle of $S^1$ is trivial. –  levap Oct 24 '12 at 14:35
As levap says, you can try $S^1$. If you master that, you can also try $S^3$. The only other sphere with a trivial tangent bundle is $S^7$. This is a very hard result due to Adams. (Well, the fact that $S^1$, $S^3$ and $S^7$ have trivial tangent bundles is not too hard, it's the fact that they are the only ones which is tough). You may also try products of things. For example, the product of an odd dimensional sphere and any other sphere does have trivial tangent bundle, while the product of any two even dimensional spheres has a nontrivial tangent bundle. –  Jason DeVito Oct 24 '12 at 15:11
Try trivializing the tangent bundle to a compact Lie group like $O(n)$. –  Neal Oct 24 '12 at 15:42
Or a non-compact Lie group :-) –  Mariano Suárez-Alvarez Oct 25 '12 at 2:12
OK. thank you all guys! –  hxhxhx88 Oct 25 '12 at 3:20

We define $$TM=\bigsqcup_{p\in M}T_PM$$ with a smooth structure pulled back from the projection map. This is a key point. The tangent bundle's topology and smooth structure capture some of the manifold's topology. You can find an easy bijection $TM\leftrightarrow M\times\mathbb{R}^n$, but you cannot in general find a fiber-preserving diffeomorphism between the two spaces.

When we trivialize, we require that $F:TM\to M\times V$ be not just a diffeomorphism but a diffeomorphism that is a fiberwise isomorphism. A tangent bundle isn't just some disjoint collection of vector spaces all floating off in abstract mathland - there's more structure than that. We need two additional things:

• a projection map $p:TM\to M$ taking a tangent vector to its basepoint, and that
• $M$ is covered by neighborhoods $U$ which obey two conditions:
• $p^{-1}U$ is diffeomorphic to $U\times\mathbb{R}^n$ (say via $\phi_U$) in a way that respects projection onto $U$ (i.e. $\pi_U\circ\phi_U = p$), and
• for two such neighborhoods $U$ and $V$, there is a family of vector space isomorphisms which govern the transformation of the fibers: $$U\cap V\ni x\mapsto\theta_{UV}(x):\{x\}\times \mathbb{R}^n\to\{x\}\times\mathbb{R}^n$$ (where the first $\{x\}\times\mathbb{R}^n\subset U\times\mathbb{R}^n$ and the second $\{x\}\times\mathbb{R}^n\subset V\times\mathbb{R}^n$). This condition is in place so that when we change coordinates, the new fiber still has the structure of a vector space.

These neighborhoods are called "local trivializations;" they're analogous to coordinate neighborhoods in a manifold. (In fact, one method of constructing $TM$ is by suitably patching together local trivializations from a cover of coordinate neighborhoods.)

For $F:TM\to M\times\mathbb{R}^n$ to be a global trivialization, we need not just that $F$ is a diffeomorphism, but that $F$ preserves all of this structure. In particular, when restricted to a single tangent space, $F$ must be an isomorphism. This is much stronger than simply requiring $F$ be a diffeomorphism between $TM$ and $M\times\mathbb{R}^n$.

The standard counterexample against the idea that all tangent bundles are trivializable is $T\mathbb{S}^2$. The "hairy ball" theorem states that there is no nonvanishing vector field on $\mathbb{S}^2$. You can see this from the Poincare-Hopf index theorem:

On any closed smooth manifold $M$, for any nondegenerate vector field $V$ on $M$, the Euler characteristic $$\chi(M) = \sum_{x\in\{\mbox{zeros of }V\}} \iota_v(x)$$ where $\iota_v(x)$ is the index of $v$ at $x$, the degree of the vector field when restricted to a small circle about $x$ and normalized.

Now it's clear that $T\mathbb{S}^2$ is not trivializable: $\chi(\mathbb{S}^2) = 2$, and if we had a trivialization, then we would have a nonvanishing vector field which would force the Euler characteristic to $0$.

In fact, we can see from this much more than that $T\mathbb{S}^2$ is nontrivializable: the Euler characteristic is an obstruction to the trivializability of the tangent bundle of a manifold. In order for the tangent bundle to be trivializable, we must be able to find $n$ global sections which are a pointwise basis for the tangent spaces. Each of these sections would be a nonvanishing vector field, which would imply that the Euler characteristic of the manifold is $0$.

(Note that, as Jason DeVito points out below, a zero Euler characteristic is necessary but not sufficient for a trivializable tangent bundle.)

This is an edited response to your attempt to trivialize $T\mathbb{S}^2$. Let's be a little more concrete: represent $\mathbb{S}^2$ as $\widehat{\mathbb{C}}$. Charts are the identity $\widehat{\mathbb{C}}-\{\infty\}\to\mathbb{C}$, and inversion $\widehat{\mathbb{C}}-\{0\}\to\mathbb{C}$ where $p\mapsto \frac{1}{p}$. (We define $\frac{1}{\infty}=0$). Note that transition maps are given by inversion, $w = z^{-1}$.

Each of these neighborhoods is a trivialization of $T\mathbb{S}^2$, so in each of them we can represent a tangent vector as $(v,z)$ where $v$ is the vector and $z$ is the basepoint. Let's start with $\mathbb{C}$. Define on this neighborhood $F(v,z) = (v,z)$. This takes care of the map $F$ for all of $\widehat{\mathbb{C}}-\{\infty\}$.

To extend to infinity, we need to define $F$ on $\widehat{\mathbb{C}}-\{0\}$ so that it agrees with the definition we have given on $\mathbb{C}$. Note that the differential of the transition function $\frac{1}{z}$ is $\frac{-1}{z^2}$. For every $w\in\widehat{\mathbb{C}}-\{\infty\}$, we need to define $F(v,w) = (\frac{-v}{w^2},w^{-1})$ so that it is well-defined under coordinate changes.

Now how should we define $F(v,\infty)$? We see the problem: We'll have to map $(v,\infty)\mapsto 0$ in order for $F$ to be continuous at $\infty$. This prevents $F$ from being an isomorphism on $T_\infty\widehat{\mathbb{C}}$, so it's not possible to use this method to trivialize $F$. (In fact, it's not possible for reasons discussed above.)

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It may be worth pointing out that having an Euler characteristic of $0$ is necessary for the tangent bundle to be parallelizable, but it is far from sufficent. For example, the manifolds $S^1\times \mathbb{C}P^n$ and $S^1\times\mathbb{H}P^n$ all have Euler characteristic $0$, but none are parallelizable unless $n=1$. –  Jason DeVito Oct 24 '12 at 12:31
Excellent answer! I have updated my post, describing my attempt to trivialize $S^2$. Is my thought right? –  hxhxhx88 Oct 24 '12 at 13:24
I got your point, thx! –  hxhxhx88 Oct 25 '12 at 3:27
You're welcome! I've been thinking a little more about this -- check out one of the early chapters of Thurston's book on 3-manifold topology and geometry for a nice visual proof sketch of the Poincare-Hopf theorem. It might help with understanding why a trivializable manifold must have zero Euler characteristc. –  Neal Oct 25 '12 at 3:33
An example and an answer to your last question: $S^2$ does not admit a global trivialization of $TS^2$.
By the Hairy Ball Theorem every vector field on $S^2$ has at least one zero. If $TS^2$ had a global trivialization $F \colon S^2 \times \mathbb{R}^2 \to TS^2$ then $X(p) = F(p,v)$ would be a globally non-vanishing vector field for any $0 \neq v \in \mathbb{R}^2$.