Isomorphism between $\bigwedge^k V$ and $\bigwedge^{n-k} V^*$?

Question

This question states the existence of an isomorphism between $\bigwedge^k V$ and $\bigwedge^{n-k} V^*$. I am new to this subject and was wondering how such an isomorphism can be proved?

Answer 1

Let $V$ be an $n$-dimensional vector space, then we get $\dim (\Lambda^k V^*) = \binom{n}{k} = \binom{n}{n-k} = \dim (\Lambda^{n-k} V)$, since the binomial coefficients are symmetric. Since both have the same dimension, that already gives us that they are isomorphic as vector spaces. If you want to construct a rather complicated (algebra-)isomorphism, I would do it like this: Now, let $$\iota_v : \Lambda^k V \to \Lambda^{k-1} V$$ be the interior product for a $v \in V^*$. For a $w \in \Lambda^k V$ it is defined by $$\iota_v w(u_2, ..., u_k) := w(v, u_2, ..., u_k)$$ for all $u_2, ..., u_k \in V^*$. We define recursively $\iota_{v_1 \wedge v_2} = \iota_{v_2} \circ \iota_{v_1}$.

For a given volume form $\Omega \in \Lambda^n V$ you get an isomorphism $$\Lambda^k V^* \to \Lambda^{n-k} V, \; v \mapsto \iota_v \Omega,$$ which of course you have yet to prove. Does that help?

(Edit: I noticed that I switched the $V$ and $V^*$ in your question, but you can do pretty much the same on the dual spaces of both.)

Answer 2

Hint: Suppose that $V$ is an $n$-dimensional vector space and consider a volume form $\omega$ defined on $V$. Define $f:\Lambda^kV\rightarrow \Lambda^{n-k}V^*$ as follows: On the generator $e_{i_1}\wedge...\wedge e_{i_k}$, $f(e_{i_1},...,e_{i_k})$ is the $n-k$-alternated form defined on $V$ by $f(e_{i_1},...,e_{i_k})(u_1,...,u_{n-k})=\omega(e_1,..,e_k,u_1,..,u_{n-k})$.