How to see directly that $A^*(V) \cong A(V)^*$?

Question

Let $V$ be a vector space of finite dimension $n$, say over the field of real numbers. Now, I am aware that there is a canonical isomorphism $A^k(V^*) \cong A^k(V)^*$ between the space of alternating $k$-linear maps on $V^*$, and the vector dual of the space of alternating $k$-linear maps on $V$, but for some reason I am having trouble seeing how to define it directly at the moment.

If one goes to the trouble of defining the exterior powers of $V$ and then proves that $A^k(V) \cong \Lambda^k(V^*)$, then I see how to do this because I know how to prove the corresponding statement about exterior powers.... but surely the isomorphism isn't so hard to define directly? I would like to know: given $\alpha \in A^k(V)$ and $\beta \in A^k(V^*)$, how should I define the pairing $\langle \alpha, \beta \rangle$? I tried to to something by selecting an arbitrary basis $e_1, \ldots, e_n$ for $V$, but didn't really get anywhere with that.

Answer 1

I think I figured it out in the end. The point is that there is a canonical, at least up to choice of normalization constant, alternating $k$-linear map $(V^*)^k \to A^k(V)$ given by the wedge product $$ (\varphi_1,\ldots,\varphi_n) \mapsto \varphi_1 \wedge \cdots \wedge \varphi_k.$$ Personally, I take the wedge product to be defined by $$(\varphi_1 \wedge \cdots \wedge \varphi_n)(v_1,\dots,v_k) = \frac{1}{k!} \sum_{\sigma \in S_k} \varphi_1(v_{\sigma(1)}) \cdots \varphi_k(v_{\sigma(k)}).$$ Anyway, once you have chosen your wedge product, it is not so hard to see how to go from an element of $\psi \in A^k(V)^*$, i.e. a linear map $\psi : A^k(V) \to k$, to an element $\omega \in A^k(V^*)$, i.e. an alternating map $(V^*)^k \to k$. Just compose the wedge product $(V^*)^k \to A^k(V)$ with $\psi$. In other words, define $\psi \mapsto \omega_\psi : A^k(V)^* \to A^k(V^*)$ by $$ \omega_\psi (\varphi_1,\ldots,\varphi_k) := \psi( \varphi_1 \wedge \cdots \wedge \varphi_k).$$ This can be checked to be an isomorphism.

As an application, given a nondegenerate bilinear form on $V$, viewed as vector space isomorphism $v \mapsto v^* : V \to V^*$, we can give a formula for the induced bilinear form $\omega \mapsto \omega^* : A^k(V) \to A^k(V)^*$. First, we have the functorially induced isomorphism $A^k(V^*) \to A^k(V)$ given by precomposing alternating maps $(V^*)^k \to k$ with $(v_1,\ldots,v_k) \mapsto (v_1^*,\ldots,v_k^*) : V^k \to (V^*)^k$. Applying then the above isomomorphism, we get that $\omega \mapsto \omega^* : A^k(V) \to A^k(V)^*$ is determined by the formula $$\omega(v_1,\ldots,v_k) = \omega^*( v_1^* \wedge \cdots \wedge v_k^*).$$

Another question would be, how to show that the isomorphism $A^k(V)^* \cong A^k(V^*)$ is natural? It seems to me that you need to know this for the above application where you might want to prove that properties of $v \mapsto v^* : V \to V^*$ become properties of $\omega \to \omega^* : A^k(V) \to A^k(V)^*$. So, for instance, what happens if $v \mapsto v^*$ is symmetric? Anyway, the naturality of $A^k(V)^* \cong A^k(V^*)$ does hold, though it is a kind of confusing calculation. In the end, the key point seems to be that the wedge product commutes with pullbacks. That is, if $T : V \to W$ is a linear map and $\varphi_1,\ldots,\varphi_k \in W^*$, then $$T^*(\varphi_1) \wedge \cdots \wedge T^*(\varphi_k) = T^*(\varphi_1 \wedge \cdots \wedge \varphi_k).$$