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On learning about how to define smooth vector fields on a manifold $M$, I learned that one should first define a tangent bundle , $T(M)$, as $\cup T_p(M)$ together with a topology(smooth structure). And then, a smooth vector field $X$ would be a smooth map from $M \rightarrow T(M)$ s.t. $\pi \circ X=id_M$

However, it is obvious that the use of the auxilary manifold $T(M)$ should not be confined to merely offering a good def. of "smooth" vector fields. Well, at least you would not be using the global topology of $T(M)$. (Since the checking of "smoothness" is done locally)

I am inclined to believe that the global topology of $T(M)$ may play a more important role in determining the properties of vectorfields on $M$. So here are my two questions:

  1. Can you give an example where the global topology of $T(M)$ is used to control certain properties(possibly about vec. fields) on $M$?

  2. If the global topology on $T(M)$ is indeed important, how do we set about determining it? I have only seen trivial examples where for $M=S^1 or \mathbb{R}^n$, $T(M)=M\times \mathbb{R}^n$. But surely, there are examples where $T(M)\neq M\times \mathbb{R}^n$. And how do we determine the topology in that case?

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Let's consider the tangent bundle $T(S^2)$ of the $2$-dimensional sphere $S^2$. Note that $T(S^2)$ is not trivial bundle, i.e. $T(S^2)\not\simeq S^2\times\mathbb{R}^2$.

To see this, let $X$ be a vector field of $T(S^2)$. Then, as you said, $X$ is a section of the tangent bundle $T(S^2)$, i.e. $X:S^2\rightarrow T(S^2)$ such that $\pi\circ X=id_{S^2}$. By Poincare-Hopf's theorem, $X$ must have a zero, i.e. there exists $p\in S^2$ such that $X(p)=0$. Therefore, $T(S^2)\neq S^2\times\mathbb{R}^2$; otherwise, if $T(S^2)\simeq S^2\times\mathbb{R}^2$, then there would exist a non-vanishing vector field $X$. More precisely, if $\pi: S^2\rightarrow T(S^2)\simeq S^2\times\mathbb{R}^2$, $X(p)=(p,(1,0))$ where $p\in S^2$ and $(1,0)\in\mathbb{R}^2$ is a non-vanishing vector field.

From the above example, we can see that the topology of $T(S^2)$ (the property that $T(S^2)$ is non-trivial) gives property on the vector field.

Note added: See also Zhen Lin's answer. He beats me by one minute.

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About the next to last sentence, it seems to me that the non-existence of not vanishing vector fields on $S^2$ implies the non-triviality of $TS^2$, rather than the viceversa. – Giuseppe Jan 21 '12 at 13:35
Ah, the Hairy Ball Theorem then confirms the non-triviality of $T(S^2)$. So I guess one could use the technique to find more manifolds with the "Hairy Ball" property given that if you could systematically calculate their tangent bundles! – Michael Luo Jan 21 '12 at 14:37

Of course the global topology of $T(M)$ is important. It controls, for example, the existence of globally-defined vector fields with certain properties. Conversely, the non-existence of certain vector fields tells us that the global topology of $T(M)$ is non-trivial.

Consider the sphere $S^2$. By the hairy ball theorem, there are no continuous vector fields $S^2 \to T(S^2)$ which are nowhere vanishing. But if $T(S^2) \cong S^2 \times \mathbb{R}^2$ then by choosing a continuous basis of the tangent space at each point, we could obtain a continuous nowhere-vanishing vector field $S^2 \to T(S^2)$ — a contradiction. So $T(S^2) \ncong S^2 \times \mathbb{R}^2$.

In the case of an embedded manifold $T(M)$ defined as the vanishing of some smooth function $F : \mathbb{R}^N \to \mathbb{R}^k$, it is easy to describe the tangent bundle as another embedded manifold $T(M)$: it is (diffeomorphic to) the submanifold

$$\{ (x, \vec{v}) \in \mathbb{R}^N \times \mathbb{R}^N : F(x) = 0, D_x F (\vec{v}) = 0 \}$$

where $D_x F : \mathbb{R}^N \to \mathbb{R}^k$ is the Jacobian matrix of $F$ evaluated at $x$. So, for example, $$T(S^2) = \{ (x, y, z, u, v, w) \in \mathbb{R}^6 : x^2 + y^2 + z^2 = 1, 2 x u + 2 y v + 2 z w = 0 \}$$ It is an amusing exercise to show that $$T(S^n) \cong \{ (z_0, \ldots, z_n) \in \mathbb{C}^{n+1} : {z_0}^2 + \cdots + {z_n}^2 = 1 \}$$ In other words, $T(S^n)$ has the structure of a complex manifold!

But of course some work still has to be done. It is not at all obvious from this calculation that $T(S^3) \cong S^3 \times \mathbb{R}^3$, which is a consequence of the existence of a Lie group structure on $S^3$. (All Lie groups have trivial tangent bundle.)

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Dear Zhen Lin In your second sentence, I think that the two occurrences of $S^2\to\mathbb{R}^2$ should be replaced by $S^2\to T(S^2)$. – Giuseppe Jan 21 '12 at 13:45
@Giuseppe: Quite right, thank you! – Zhen Lin Jan 21 '12 at 13:54
-Thanks for giving a way to actually compute $T(M)$, I can see how to make it rigorous.- Though I don't think I'm in a position to show $T(S^3)=S^3\times R^3$, I'll keep it in mind for future encounters. – Michael Luo Jan 21 '12 at 14:44

Every manifold with trivial tangent bundle is orientable: this is immediate from the definition.
Hence every non-orientable manifold is an example of a manifold with non-trivial tangent bundle.
Easy examples are the Möbius band, the real projective plane $\mathbb P^2(\mathbb R)$ and the Klein bottle.
The point of view of orientability has the advantage that it is completely elementary and self-contained: no input from topology id s needed.

(This is definitely not meant as a criticism of Paul's and Zhen's great answers which I have just upvoted: the theorems they allude to are quite interesting and should eventually be learned too)

Edit Let me show you how easy it is to prove non-orientability!

1) A useful remark
If $M$ is orientable and if $(U,x)$ and $(V,y)$ are two charts with connected domains $U,V$ , then the change of coordinates $\phi =y\circ x^{-1}: x(U\cap V) \to y(U\cap V)$ has a jacobian $Jac(\phi) =det (\frac {\partial y_i}{\partial x_j})$ which does not change sign on $x(U\cap V)\subset \mathbb R^n$.
[This is remarkable because $U\cap V$ and $x(U\cap V)$ need not be connected even though $U$ and $V$ are]

2) Application: $\mathbb P^2(\mathbb R)$ is not orientable
Consider the two charts $x :U\to \mathbb R^2:[1:v:w]\mapsto (v,w)$ and $y :V\to \mathbb R^2:[u:1:w]\mapsto (u,w)$ , where the domain $U$ (resp. $V$) is the set of $[u:v:w]\in \mathbb P^2(\mathbb R)$ with $u\neq 0$ (resp. $v\neq 0$).
The change of coordinates is the diffeomorphism
$$\phi=y\circ x^{-1}: \mathbb R^*\times \mathbb R \to \mathbb R^*\times \mathbb R:(u,v)\mapsto (\frac {1}{u},\frac {v}{u})$$
whose jacobian $(Jac (\phi))(u,v)= -\frac {1}{u^3}$ does change sign on its (disconnected!) domain $\mathbb R^*\times \mathbb R$.
Hence $\mathbb P^2(\mathbb R)$ is not orientable according to the useful remark above.

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I am a little hazy on the definition of orientability, is it: Given any loop starting and ending at P, a smooth change of tangent basis along the loop, the basis ending at P would have a positive determinant expressed as the basis starting at P? – Michael Luo Jan 21 '12 at 14:55
Dear @Michael: there are many definitions of orientable. The most elementary and operational is that you can find an atlas whose charts give rise to changes of coordinates with positive jacobian. I'll write an edit to show you how easily we can work with that definition. – Georges Elencwajg Jan 21 '12 at 16:45
Thanks for the elaboration! Turns out(from the Orientable and Hairy Ball thm example), the story is that properties of vec.fields on $M$ are somehow characterized in the topologies of $T(M)$ – Michael Luo Jan 21 '12 at 18:09

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