7
$\begingroup$

Let $M$ be a differentiable manifold of dimension $n$. If the tangent bundle is trivial, then the cotangent bundle is trivial, and so are its exterior powers. In other words, on a parallelizable manifold we can also find parallel (and so, non-vanishing) differential forms of any degree up to $n$.

I have the "converse" question. Suppose we have a globally non-vanishing form of degree $k\le n$. Does this imply any condition on the bundles (either cotangent, or its exterior powers)?

What if we have $n\choose k$ independent forms?

For example, it is clear that if $k=n$ the only condition is that the manifold is orientable (what does it mean in terms of bundles?). What happens for the generic order $k$?

Thanks.

$\endgroup$
3
$\begingroup$

In general, let $V$ be a real or complex vector bundle of dimension $n$. What can we say about $V$ if it admits $k$ independent nonvanishing global sections? This is equivalent to admitting a splitting $V \cong W \oplus \mathbb{R}^k$ resp. $V \cong W \oplus \mathbb{C}^k$, and so it implies some conditions on characteristic classes. If $V$ is real, then

$$w(V) = w(W)$$

where $W$ has dimension $n-k$, and hence the $k$ top Stiefel-Whitney classes $w_n(V), \dots w_{n-k+1}(V)$ vanish. Similarly, if $V$ is complex, then

$$c(V) = c(W)$$

and hence the $k$ top Chern classes $c_n(V), \dots c_{n-k+1}(V)$ vanish. In fact it is possible to define the Stiefel-Whitney and Chern classes in this way in terms of obstructions to admitting nonvanishing global sections. If $V$ is an oriented real vector bundle, then even when $k = 1$ it immediately follows that the Euler class vanishes; for tangent bundles this is the Poincaré–Hopf theorem.

Now we can ask what happens if some bundle constructed from $V$, which in your case is $\Lambda^k(V^{\ast})$, admits some number of nonvanishing global sections. The answer is the same as above: some number of top characteristic classes of $\Lambda^k(V^{\ast})$ vanishes. This indirectly implies something about the characteristic classes of $V$, which you can express the characteristic classes of $\Lambda^k(V^{\ast})$ in terms of. I explain how to do this in general using the splitting principle here. More specifically, we can say the following.

First, note that if $V$ is real, then $V^{\ast} \cong V$, and in particular the two have the same characteristic classes. If $V$ is complex, then $V^{\ast} \cong \overline{V}$, the conjugate bundle, which satisfies $c_k(V^{\ast}) = (-1)^k c_k(V)$. From now on I am going to talk about the exterior powers of $V$ and not of $V^{\ast}$.

Now let's look at top exterior powers. If $V$ is real, then

$$w_1(\Lambda^n(V)) = w_1(V)$$

so $\Lambda^n(V)$ admits a nonvanishing global section iff $w_1(V)$ vanishes iff $V$ is orientable iff its structure group reduces from $O(n)$ to $SO(n)$. Similarly, if $V$ is complex, then

$$c_1(\Lambda^n(V)) = c_1(V)$$

so $\Lambda^n(V)$ admits a nonvanishing global section iff $c_1(V)$ vanishes iff $V$ is "complex orientable" iff its structure group reduces from $U(n)$ to $SU(n)$.

More interestingly, suppose $n = 3$ and let's look at $\Lambda^2(V)$. Via the splitting principle let's write $V \cong L_1 \oplus L_2 \oplus L_3$, so that the Stiefel-Whitney resp. Chern classes are the elementary symmetric polynomials in the first Stiefel-Whitney resp. Chern classes of the $L_i$; for both the real and complex cases call these $\alpha_1, \alpha_2, \alpha_3$. Then

$$\Lambda^2(V) \cong L_1 L_2 \oplus L_2 L_3 \oplus L_3 L_1$$

where to save space I have omitted the tensor product symbol. This tells us that the Stiefel-Whitney resp. Chern classes of $\Lambda^2(V)$ are the elementary symmetric polynomials in $\alpha_1 + \alpha_2, \alpha_2 + \alpha_3, \alpha_3 + \alpha_1$. Expressing the latter in terms of the former is an exercise in symmetric function theory. When $V$ is real this gives, if I haven't made a computational error,

$$w_1(\Lambda^2(V)) = 2 w_1(V) = 0$$ $$w_2(\Lambda^2(V)) = w_1(V)^2 + w_2(V)$$ $$w_3(\Lambda^2(V)) = w_1(V) w_2(V) - w_3(V)$$

and similarly when $V$ is complex this gives

$$c_1(\Lambda^2(V)) = 2 c_1(V)$$ $$c_2(\Lambda^2(V)) = c_1(V)^2 + c_2(V)$$ $$c_3(\Lambda^2(V)) = c_1(V) c_2(V) - c_3(V).$$

Hence, for example, if $V$ is real and $\Lambda^2(V)$ admits a nonvanishing global section then $w_3(V) = w_1(V) w_2(V)$.

Of course, these characteristic class arguments aren't the complete answer: for example, they tell you very little on spheres.

$\endgroup$
  • $\begingroup$ This is very interesting. Thanks a lot! $\endgroup$ – geodude Nov 10 '14 at 16:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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