Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

my question is very basic (i don't know too much of differential geometry): given a fiber bundle, is there a necessary and sufficient condition for its tangent bundle to be trivial?

i have some ideas, but submitted to some conditions on the cohomology ring of the bundle; (i apologize if it is trivial)

share|improve this question
9  
@fpqc: given the 'dg' tag, I think we can safely assume that we are in the category of smooth manifolds. –  José Figueroa-O'Farrill Apr 9 '10 at 13:43
1  
fpqc and Andrew: Thanks for the clarifications. I guess I do use "tangent space" informally for the bundle and "tangent space at x" for a particular fiber. And the question probably should include an explicit statement that the fiber bundle and its base are assumed to be smooth manifolds. –  Deane Yang Apr 9 '10 at 14:03
3  
krolik- This kind of question is very difficult to answer with no context. What sort of situation are you working in? What sort of things do you know that you are hoping imply this? –  Ben Webster Apr 9 '10 at 14:06
1  
I guess the question is: "When is the total space of a smooth fiber bundle a parallelizable manifold?" –  Qfwfq Apr 9 '10 at 14:15
3  
The question is still somewhat ambiguous. Do you want to bundle of tangent spaces to the fibres of your fibre bundle, or the tangent bundle to the total space of your fibre bundle? I suspect you want the latter. –  Ryan Budney Apr 9 '10 at 14:36
show 12 more comments

3 Answers 3

up vote 3 down vote accepted

This is far from being a complete answer, but there is a case when one construct a parallelizable bundle (meaning its total space has trivial tangent bundle) from a given (geometric) bundle.

The context is that of $G$-structure, which are a formalization of the concept of geometric structures. A $G$-structure is a set of data including a fiber bundle on a smooth manifold, which shall be thought of as the bundle of admissible frames. For example, in the Riemannian case ($G=O(n)$) the bundle is that of orthonormal frames. If the group $G$ has a finite-order rigidity property, then one can construct a sequence of bundles, the total space of each one being the base space of the next one, so that after a finite number of steps one gets a bundle whose total space is parallelizable. This is a tool to prove that the group of automorphisms of the $G$-structure is a Lie group. As an example, if I remember well the total space of the bundle of orthonormal frames on a Riemannian manifold is parallelizable.

All details are available in Kobayashi's transformation groups in differential geometry.

share|improve this answer
    
I really like this answer; it gives a nice generalization to the fact that the tangent bundle of a Lie group is trivial. Indeed, the tangent bundle of the principal $O(n)$-bundle of orthonormal frames of a Riemannian manifold is parallelizable. This is a standard fact in exterior differential systems, explained often by Robert Bryant. It seems to me that the same argument shows that any principal $G$-bundle constructed from tangent frames (not necessarily orthonormal) has a parallelizable tangent bundle. You just need to construct a connection on that bundle to get the trivialization. –  Deane Yang Apr 10 '10 at 14:24
add comment

As the comments indicate, the level of generality of the question is optimistic. Here's a special case: the product $M\times N$ of smooth manifolds.

Recall that a manifold $X$ is stably parallelizable if $TX\oplus \mathbb{R}$ is trivial. By a standard argument in obstruction theory, this is so as soon as $TX\oplus \mathbb{R}^n$ is trivial for some $n\geq 1$.

I assume $M$ and $N$ connected, positive-dimensional but not necessarily compact. The product $M\times N$ is parallelizable iff $M$ and $N$ are stably parallelizable and one of them has vanishing Euler characteristic.

Euler characteristics $\chi$ are relevant because $\chi(M\times N)=\chi(M)\chi(N)$ and because $\chi$ is precisely the obstruction to having one nowhere-zero vector field.

The "if" direction has a short, elementary proof that I'll leave to you, but you can also look it up in the extremely short paper of E. B. Staples, Proc. A.M.S. 18 no. 3 (1967). Conversely, if $M\times N$ is parallelizable then, choosing a trivialisation of $T(M\times N)$, and a point $y\in N$, we get by restriction to $M\times y$ a trivialization of $TM\oplus T_y N$. Hence $M$ (and similarly $N$) is stably parallelizable.

Part of this goes over to smooth fibre bundles $E \to B$ with connected base $B$ and fibre $F$: if $TE$ is trivial then $0=\chi(E)=\chi(B)\chi(F)$, and $F$ is stably parallelizable. But the pullback of the Hopf fibration $S^5\to \mathbb{CP}^2$ to $\mathbb{CP}^2\times S^1$ is an example where the total space $S^1\times S^5$ is parallelizable but the base is not stably parallelizable (it has non-vanishing $p_1$).

share|improve this answer
add comment

I just wanted to elaborate on Benoît Kloeckner's answer, so if you like what I say, please upvote his answer.

By a frame, I mean a basis of the tangent space at a point on a smooth manifold $M$. The space $F$ of all possible frames, called the frame bundle, is a principal $GL(n)$-bundle over the manifold, $n$ is the dimension of the manifold. A point in $F$ is given by $(x, e)$, where $x \in M$, $e = (e_1, \dots, e_n)$, and $e_i \in T_xM$. Associated with each point is the dual frame $\omega^1, \dots, \omega^n \in T_x^*M$. Let $\pi: F \rightarrow M$, $\pi(x,e) = x$, denote the natural projection.

There is a natural set of $n$ $1$-forms $\hat\omega^1, \dots, \hat\omega^n$ on $F$, which are called either "tautological" or "semi-basic" and act as follows: If $v \in T_{(x,e)}F$, then $ \langle \hat\omega^i,v\rangle = \langle\omega^i,\pi_* v \rangle, $ where $\omega^1, \dots, \omega^n \in T^*_xM$ form a dual basis to the basis $e_1, \dots, e_n \in T_xM$. These forms have the universal property that given any section $s: M \rightarrow F$, $s^*\bar\omega^i$ are $1$-forms on $M$ dual to the moving frame given by the $e_i$.

You can check that any connection $\nabla$ on $T_*M$ determines a set of global $1$-forms $\hat\omega^i_j$ on $F$, such given any section $s = (s_1, \dots, s_n): M \rightarrow F$, $\nabla s_j = s_is^*\hat\omega^i_j$. Therefore, a connection on $F$ gives a set of global $1$-forms $\hat\omega^1, \dots, \hat\omega^n, \hat\omega^1_1, \dots, \hat\omega^n_n$ that trivialize $T^*F$. The dual vector fields trivialize $T_*F$.

Since there always exists a connection on $T_*M $, this shows that $F$ has a parallelizable tangent bundle. The same argument can be extended to any principal $G$-bundle of tangent frames. As observed by Hoeckner, the case $G = O(n)$ corresponds to a Riemannian structure.

This, of course, does not answer the original question, but it is a important case where the answer is yes. These global $1$-forms are extremely useful in many contexts; the work of Robert Bryant illustrates this.

share|improve this answer
1  
People also call the collection $\omega^i$ the soldering form' You may as well take the $\omega^i _j$ to be the one-forms encoding the Levi-Civita connection. It is nice to view the $\omega^i$ pointwise as a linear identification of the tangent bundle with a standard' $R^n$. The $\omega^i_j$ are $so(n)$-valued. Taken together these forms take values in the isometry group of $R^n$. –  Richard Montgomery Apr 11 '10 at 21:09
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.