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Let $V$ be a finite dimensional real vector space, let $\mathrm{Alt}^k(V)$ denote the space of alternating $k$-linear forms on $V$ and let $\wedge^k(V)$ denote the $k^{th}$ exterior power of $V$. I am trying to see why the algebraic dual $\wedge^k(V)^* := (\wedge^k(V))^*$ is isomorphic to $\mathrm{Alt}^k(V)$. Here are my thoughts:

By the universal property of the exterior power, for any alternating $k$-linear form $f$ with domain $V^k$ there exists a unique linear form $\phi$ with domain $\wedge^k(V)$ such that $$ \phi(v_1 \wedge \cdots \wedge v_k) = f(v_1, \dots, v_k). $$ The universal property thus provides a mechanism to produce elements in $\wedge^k(V)^*$ from elements in $\mathrm{Alt}^k(V)$ and this mechanism of production is unique.

Thus, we have an injection $$ \Phi: \mathrm{Alt}^k(V) \longrightarrow \wedge^k(V)^* $$

What I'm not sure about is how to argue surjectivity; what is the best way to approach this?

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Since everything is finite dimensional, you can just count dimensions to show surjectivity. –  Joe Johnson 126 Jun 12 '12 at 21:22
    
@JoeJohnson126 Yes, I see this, but counting the dimensions of these spaces is a fair amount of work. One has to pick bases, show that they are, in fact, bases and then count....Is there a more direct way from the universal property to do this that doesn't involve counting? –  ItsNotObvious Jun 12 '12 at 21:25
    
I think the two are switched up in the paragraph that begins "The universal property..." –  Dylan Moreland Jun 12 '12 at 21:34
    
@ItsNotObvious I can see what the dimension of the exterior power is immediately, but not the alternating forms. I imagine that it's standard though, and if the dimensions match you are already done. –  rschwieb Jun 12 '12 at 21:46
    
@rschwieb Yes, the dimension of both spaces, assuming $V$ is of dimension $n$, is given by $\frac{n!}{k!(n - k)!}$, but would like to prove this result without using dimension arguments, if possible... –  ItsNotObvious Jun 12 '12 at 21:50
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1 Answer

up vote 6 down vote accepted

The canonical map $\psi\colon V^k \to \bigwedge^k V$ given by $\psi(v_1, \ldots, v_k) = v_1 \wedge \cdots \wedge v_k$ is alternating. If you have an element $g \in (\bigwedge^k V)^*$ then $g \circ \psi\colon V^k \to \mathbf R$ is also alternating, and I believe that this assignment $g \mapsto g \circ \psi$ gives you an inverse to $\Phi$.

This is simpler than you might fear, and has little to do with the ground ring or the finiteness of $V$. And maybe we should expect that, since the universal property of $\bigwedge^k V$ more or less says that it represents the functor $W \mapsto L^k_a(V, W)$.

Extra trouble and hypotheses seem to enter once you try to write things such as an isomorphism $(\bigwedge^k V)^* \approx \bigwedge^k (V^*)$, an embedding $\bigwedge^k V \hookrightarrow V^{\otimes k}$, or the wedge product induced on alternating forms. Some examples of this are discussed in the back of Fulton and Harris.

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There's a bit more to say (and check!). I'll try to find time after walking the dog/doing other stuff. –  Dylan Moreland Jun 12 '12 at 21:54
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What you have said is already correct and interesting : +1. (And don't forget the ball for the dog !) –  Georges Elencwajg Jun 12 '12 at 22:06
    
Nice; thanks for your help. –  ItsNotObvious Jun 13 '12 at 13:41
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