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Recall that a manifold $M$ of dimension $n$ is parallelizable if there are $n$ vector fields that form a basis of the tangent space $T_x M$ at every point $x \in M$. This is equivalent to the tangent bundle $TM$ being trivial or the frame bundle $FM$ having a global section.

I know of some conditions (both necessary and sufficient), as well as counter-examples, both of which I provided in an answer to this recent question on tangent bundles.

But the results I am familiar with are rather disparate and I was wondering whether some more coherent theory is known. In the ideal case giving a set of necessary and/or sufficient conditions (e.g. in terms of cohomology groups) or at least completely characterizing some nice class of manifolds (like for Lie groups, which are parallelizable; or for compact manifolds with non-zero Euler characteristic, which are not).

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2 Answers 2

up vote 21 down vote accepted

(Disclaimer. The following describes some cohomological conditions that formally answer the question. But they are 1) usually too hard to compute; 2) don't really explain what is the class of parallelizable manifolds geometrically.)


The problem of existence of section is answered (well, in a sense) by obstruction theory. Namely, there is the first obstruction $o_1\in H^1(X,\pi_0(F))$ and there is a section on $sk_1(X)$ iff $o_1=0$; if $o_1=0$ each section on $sk_1(X)$ defines an obstruction $o_2\in H^2(X,\pi_1(F))$ and so on (and if all obstructions are trivial, the bundle has a section).

(Well, actually one should be careful with $H^1(X;\pi_0(F))$: in general, $\pi_0(F)$ is not a group, so this $H^1$ just doesn't make sense, and the story starts a step (or two, if $\pi_1(F)$ is not abelian) later. But in the cases we're interested in, $o_1$ is well-defined.)

In the case of frame bundle of $n$-dimensional vector bundle, the fiber is $O(n)$, so obstructions lie in groups $H^i(M,\pi_{i-1}O(n))$. In stable range homotopy groups of orthogonal groups are given by Bott periodicity; in particular, if we're talking about tangent bundle, we care only about $\pi_{i-1}(O(n))$ for $i\leq n$ and these groups definitely lie in stable range.

A (toy) example: for $S^3$ only nontrivial (reduced) cohomology group is $H^3(S^3)$; but $\pi_2 O(n)=0$ — so any vector bundle on $S^3$ is trivial (well, not the simplest proof of the fact, but still).


In case of vector bundles these obstructions can be also described more geometrically (in the spirit of characteristic classes theory).

  1. First obstruction $o_1\in H^1(M;\pi_0 O(n))=H^1(M;\mathbb Z/2\mathbb Z)$ is nothing else but $w_1$, the first Stiefel-Whitney class. It gives the obstruction to orientability — i.e. to reducing the structure group of the bundle from $O(n)$ to $SO(n)$.
  2. If the bundle is oriented, second obstruction $o_2\in H^2(M;\pi_1 O(n))=H^1(M;\mathbb Z/2\mathbb Z)$ is defined. It coincides with $w_2$ and gives the obstruction to the existence of a spin struction — i.e. to lifting structure group of the bundle from $SO(n)$ to its universal cover, $Spin(n)$.
  3. Next obstruction is defined for a spin bundle; $\pi_2O(n)=0$, so first non-trivial obstruction here is $o_4\in H^4(M;\pi_3 O(n))=H^4(M;\mathbb Z)$. In fact, it coincides with $\frac12p_1$ (where $p_1$ is the first Pontryagin class of oriented bundle). And it is the obstruction to lifting the structure group from $Spin(n)$ to (infinite-dimensional) topological group $String(n)$.

...And so on: the sequence of obstructions corresponds to the Postnikov tower $$ O(n)\gets SO(n)\gets Spin(n)\gets String(n)\gets FiveBrane(n)\gets... $$ (this is a kind of duality: one can think either about sequence of extensions of the section through the filtration of $M$ by skeleta, or about sequence of lifts through the Postnikov tower of $O(n)$).


Some references. Obstruction theory in general is discussed in the section 4.3 of Hatcher — but Hatcher uses the Postnikov-towers-approach (like in the second part of the answer), AFAIR. And more classical approach + obstruction-theoretic POV on characteristic classes is explained e.g. in section 12 of Milnor-Stasheff, I believe.


One more remark. As it is explained in the other answer, ordinary ("primary") characteristic classes can't answer the question, since they coincide for stably equivalent vector bundles. What obstruction theory gives is, in a sense, a theory of higher characteristic classes: secondary class (a priori) defined only if primary one is zero and so on.

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thanks. In that case I am even more lost than I initially thought :) Actually, I am not even sure what $\pi_0(X)$ is. Wikipedia article on homotopy groups (e.g. of spheres) omits them altogether and starts at $\pi_1(X)$. –  Marek Jun 19 '11 at 20:35
    
$\pi_0 X$ is homotopy-classes of based maps $S^0 = \{-1,1\} \to X$. So in effect, it's the path-components of $X$. –  Ryan Budney Jun 20 '11 at 6:26
    
@Ryan: aah, I am so used to dealing with path-connected manifolds that it didn't even occur to me this might contain some information. Thank you. –  Marek Jun 20 '11 at 9:02
    
many thanks for the update with explicit objects that I am familiar with; now I can see what's going on more or less. –  Marek Jun 20 '11 at 19:23
    
What does $sk_1(X)$ mean? –  Mikola Aug 30 '12 at 19:21

Although characteristic classes give very nice necessary conditions for parallelizability and also some partial sufficient conditions, they are doomed to fail in general. Here is why. If the tangent bundle $TM$ of a manifold $M$ of dimension $n$ is stably trivial i.e. has the property that $TM\oplus \theta^r\simeq \theta^{n+r}$ for some integer $r$ (where $\theta $ is the rank one trivial bundle) , then all characteristic classes of $TM$ will be trivial (if they satisfy Whitney's axiom) . But there is no reason that $TM$ itself should be trivial.
For example the tangent bundle to any sphere $S^n$ is trivially (!) stably trivial [add the trivial normal bundle to get the trivial rank $n+1$ vector bundle] but only $S^1,S^3$ and $S^7$ are parallelizable.
An impressive theorem of Adams is that the maximum number of independent vector fields on $S^n$ or $\mathbb P^n$ is $8a+2^b-1$ where $n+1=k.2^{4a+b}$ with $k$ odd, $a\geq0$ and $0\leq b\leq 3$ .

Some positive results
a) Steenrod has proved that every orientable 3-manifold is parallelizable.
b) Forster has proved some incredibly strong results on the analytic parallelizability (which is of course much stronger than differentiable parallelizability) of Stein manifolds.
For example every analytic submanifold of codimension one (=hypersurface) of $\mathbb C^n$ is analytically parallelizable. And a codimension two analytic submanifold of $X\subset \mathbb C^n$ is analytically parallelizable if and only if the first Chern class of its tangent bundle is zero: $c_1(TX)=0$

The real analogues of Forster's results are of course completely false: every sphere $S^n$ is a hypersurface of $\mathbb R^{n+1}$ and a submanifold of codimension 2 of $\mathbb R^{n+2}$ with vanishing Stiefel-Whitney classes (since its tangent bundle is stably trivial). However it is not parallelizable if $n\neq 1,3,7$ , as already mentioned.

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On the other hand, the Euler class is unstable so it at least has a chance at catching these things (e.g. even dimensional spheres). –  Dylan Wilson Jun 19 '11 at 19:48
    
Very true, Dylan! (That's why I added a caveat about Whitney's axiom.) –  Georges Elencwajg Jun 19 '11 at 21:58
    
I'm way late to the party, but I think I have a misconception I'd like to get cleared up. I have, perhaps incorrectly, always referred to "Whitney's axiom" the fact that $c(V\oplus W) = c(V)\cup c(W)$ where $c$ is whatever characteristic class we're looking at. If this is what you meant, then the Euler class does satisfy Whitney's axiom. The bigger issue is that the Euler class is not of the form $1 + v$ with $1\in H^0$ and $v\in H^{>0}$, so it's easy for $e$ to be a $0$ divisor. –  Jason DeVito Jun 20 '13 at 17:00
    
Dear @Jason: don't worry, the party is still in full swing! What Dylan probably meant is that $e(V\oplus 1)\neq e(V)$, so that subreptitiously adding a trivial line bundle will go unnoticed by Stiefel-Whitney, but will not fool Euler. –  Georges Elencwajg Jun 20 '13 at 18:26
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I think we are all on the same page. I was just remarking that $e$ does satisfy the Whitney axiom, but $e(1)\neq 1$ (unlike the other standard characteristic classes). This is what allows it to (potentially) detect nontrivial stably trivial bundles. –  Jason DeVito Jun 21 '13 at 13:32

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