Existence of differential form on a manifold

Question

I have a fundamental question about the existence of differential forms on manifolds.

A $k$-form on a manifold in local coordinates looks like $f(x_1,...,x_n)dx_{i_1}...dx_{i_k}$, where $f$ is a smooth function.

Given any smooth function $f$, is $f(x_1,...,x_n)dx_i$ a $1$-form for every $i$?

In general does the existence of a smooth function ($0$-form) imply the existence of a $k$-form for all $k<n+1$? I can just multiply the function by the desired number if $dx$'s.

Thanks

Answer 1

Unfortunately we can't multiply a given form or a function globally by $dx_j$. For one, this is because such a form is nowhere zero: in local coordinates it would be equal to $dx_j$, which is most certainly nonzero. However, there are manifolds on which any differential $1$-form must be zero at some point, like the $2$-dimensional sphere (see the hairy ball theorem).

In general, the existence of a nowhere zero differential form $u$ on a manifold $M$ means that its cotangent bundle contains a trivial subbundle: If $T^*_M$ is the cotangent bundle then it must contain the trivial line bundle $\mathbb R u$. Similarly, if we have $k$ differential forms $u_1, \ldots, u_k$ that are linearly independent everywhere on $M$, then $T^*_M$ has a trivial subbundle of rank $k$. Similar things apply to degree-$p$ nonzero forms, with $\bigwedge^p T^*_M$ instead of $T^*_M$.

Now, it is true that any manifold admits plenty of differential forms of any degree. For a cheap example, take a local coordinate chart $U$ of your manifold. Construct a smooth function $\theta$ supported on $U$, ie a smooth function that is not identically zero but whose support is contained in $U$. We can then extend $\theta$ by zero as a smooth function to all of $M$. Similarly, we can extend the smooth $p$-form $\theta(x) dx_1 \wedge \ldots \wedge dx_p$ defined on $U$ by zero to a smooth form on all of $M$.

This may look like cheating, but we do get in this way forms that are nonzero on most of $M$, by covering $M$ with charts on which we have these kinds of forms and taking linear combinations thereof.

For less artificial examples, you can look at the volume forms induced by (pseudo)-Riemannian metrics, symplectic forms on symplectic manifolds, Kahler and curvature forms on complex manifolds and so on.

Answer 2

Given a smooth function $f$ on $\mathbf R^n$, the expression $f(x_1, \dots, x_n) dx_i$ is a $1$-form on $\mathbf R^n$. However, if $f$ is a function on an arbitrary manifold, the expression $f(x_1, \dots, x_n) dx_i$ makes no sense because it depends on a choice of coordinates.

A manifold $M$ of dimension $n$ has $k$-forms of every degree $0 \leq k\leq n$. For instance, there is a zero form in every degree. For a less trivial example, choose a coordinate patch on $M$, take any $k$-form on $\mathbf R^n$ which has compact support, and push it forward to $M$ by extending it by zero outside the patch.