wedge product of dual vector spaces is 0

Question

I have a $m$ dimensional vector space $V$. And we define $\wedge^rV^*$ as the collection $r$ antisymmetric tensors. Why is $\wedge^rV^* = 0$ if $r>m$?

I have no idea.

Answer 1

Let $(e_1,\ldots,e_m)$ be a basis of $V$, then for any integer $r$, $\Lambda^rV^*$ is spanned by elements of the form: $$\mathrm{d}e_{i_1}\wedge\ldots\wedge\mathrm{d}e_{i_r},$$ where $i_1,\ldots,i_r$ are elements of $\{1,\ldots,m\}$. Now, if $r>m$, for each choice of $i_1,\ldots,i_r$, at least twice the same index appears, so that $\mathrm{d}e_{i_1}\wedge\cdots\wedge\mathrm{d}e_{i_r}$ is identically vanishing. Whence the result.

Answer 2

Because:

• $V^*$ is also an $m$-dimensional vector space.
• $\Lambda^r U \cong 0$ for any vector space $U$ of dimension less than $r$

An easy way to see the latter fact is to use (multi-)linearity decompose any wedge product into a linear combination of wedges of basis vectors. If $U$ has dimension less than $r$, then any collection of $r$ basis vectors must have a repeated vector, and thus the wedge is zero.