# Is it meaningful to take “exterior products” of vector fields?

Let $M$ denote a smooth manifold.

I've read that a differential $k$-form is a smooth section of the $k$th exterior power of the cotangent bundle of $M$. However I barely understand what this means, and I'm trying to understand it better by tinkering with the definition. It seems that there is a notion of "$k$-vectorfield" obtained by putting the tangent bundle in place of the cotangent bundle. As in:

Potentially Silly Definition. A $k$-vectorfield is a smooth section of the $k$th exterior power of the tangent bundle of $M$.

Following this line of thought, it seems that we can take wedge products of vector fields. As in:

$$f \frac{\partial}{\partial x} \wedge \frac{\partial}{\partial y} + g\frac{\partial}{\partial y} \wedge \frac{\partial}{\partial z}$$

Question. Is this, like, a thing? If not, why is it only the cotangent bundle whose exterior powers make sense and/or matter?

An example is the concept of Poisson tensor in classical mechanics: it's an antisymmetric tensor of type $(2,0)$, so it can be viewed as a bivector field.
• Would it be fair to say that "differential $k$-forms eat $k$-multivector fields"? – goblin Aug 18 '15 at 14:42
• Yes, instead of $\omega(X_1,\dots,X_k)$ you can view it as $\omega(X_1 \wedge \dots \wedge X_k)$. – Hans Lundmark Aug 18 '15 at 14:48
• Cool! $\;\!\;\!$ – goblin Aug 18 '15 at 15:26
• I'm tempted to use $\mho^k M$ for the space of $k$-multivector fields on $M$, by analogy with $\Omega^k M$ for the space of $k$-forms. Is there a more standard notation available? – goblin Aug 18 '15 at 15:33
• Nothing completely standard that I can recall. I searched the web and found $\mathcal{T}_k(M)$ in a few places, but I think you would have to explain whatever notation you choose to use. – Hans Lundmark Aug 19 '15 at 6:32