Is the wedge product $\wedge : \Lambda^{p}(V) \otimes \Lambda^{q}(V) \to \Lambda^{p+q}(V)$ surjective, for $V$ a real vector space of finite dimension? What dimension does its kernel have?


Let $(v_{1} , \ldots , v_{n})$ be a basis for $V$. Then a basis for $\bigwedge^{p}V$ is given by $(v_{I}|I \; p\text{-multi-index})$, where a multi-index is ordered (i.e. $i_{1} < \dots <i_{p}$) and for $I=(i_{1}, \ldots , i_{p})$ we set $v_{I} = v_{i_{1}} \wedge \dots \wedge v_{i_{p}}$.

Then you can see the map is surjective. Suppose you want to see you can reach $v_{I}$ where $I$ is a $(p+q)$-multi-index. You can split it in $I_{1}=(i_{1} , \ldots , i_{p})$ and $I_{2}=(i_{p+1} , \ldots , i_{p+q})$. Then $v_{I}$ is the image of $v_{I_{1}} \otimes v_{I_{2}}$.

For dimensions, you know (it follows from the basis we exhibited) that $\mathrm{dim}\bigwedge^{p}V= \binom{n}{p}$, where $n = \dim V$. Hence the dimension of the kernel is $\binom{n}{p}\binom{n}{q}-\binom{n}{p+q}$.

  • $\begingroup$ Thanks, this is what i guessed. This should then imply that the category with objects $V^{\wedge i}$ together with the bifunctor $\wedge$ is monoidal, right? $\endgroup$ – user83496 Apr 26 '14 at 13:45
  • $\begingroup$ Unfortunately I am not that good in category theory. You'd better edit the question including this point. @TedShifrin Once you can show that a basis is in the image of a linear map, then by linear combinations you get that the map is surjective. $\endgroup$ – Stefano Apr 26 '14 at 14:04

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