# Non-vanishing differential forms

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.

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.

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