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Let $V$ be a finite-dimensional vector space.

Is there an isomorphism between $\Lambda^k(V^\ast)$ and $\left(\Lambda^k(V)\right)^\ast$?

I was able to prove this with the additional requirement of an inner product on $V$ (and thus subsequently on $\Lambda^k(V)$) via $$ \require{AMScd} \begin{CD} \left(\Lambda^k(V)\right)^\ast @>\mathcal{J}^{-1}>> \Lambda^k(V) @>\Lambda^kJ>> \Lambda^k(V^\ast) \end{CD} $$ where $J: V \to V^\ast$ and $\mathcal{J}: \Lambda^k(V) \to \left(\Lambda^k(V)\right)^\ast$ are the isomorphisms given by the Riesz representation theorem and $\Lambda^kJ$ is the map given by $v_1\wedge \cdots \wedge v_k \mapsto J(v_1) \wedge \cdots \wedge J(v_k)$.

Is there another way to identify these two spaces without the requirement of an inner product on $V$? I read Qiaochu Yuan's comment to his answer on a similar question but did not really understand it I fear. Thank you very much.

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2 Answers 2

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An isomorphism is given by the non-degenerate pairing $\Phi \colon \Lambda^k(V^{*}) \times \Lambda^k(V) \rightarrow \mathbb{F}$ defined by

$$ \Phi(\varphi^1 \wedge \ldots \wedge \varphi^k, v_1 \wedge \ldots \wedge v_k) = \det (\varphi^i(v_j))_{i,j=1}^n $$

and extended bilinearly. Sometimes, when working over $\mathbb{R}$ or $\mathbb{C}$, people use a slightly different pairing $\Phi' = \frac{1}{k!} \Phi$ which differs from $\Phi$ by a constant factor.

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    $\begingroup$ Thanks! That is in fact a very natural isomorphism and not as complicated as mine. For other people who search for this question: To show that $\Psi(\sigma)(\tau) := \Phi(\sigma,\tau)$ an isomorphism consider a basis $e_1,\ldots,e_n$ and its dual basis $e_1^\ast,\ldots,e_n^\ast$. One only needs to show that all basis $k$-forms map to basis vectors of $\left(\Lambda^k(V)\right)^\ast$, i.e. show that $e_{i_1}^\ast\wedge \ldots \wedge e_{i_k}^\ast$ maps to $\left(e_{i_1}\wedge \ldots \wedge e_{i_k}\right)^\ast$. $\endgroup$ Dec 28, 2015 at 19:18
  • $\begingroup$ Dear @Arrow, I'm not sure I can explain the pairing geometrically. If you put an inner product $\left< \cdot, \cdot \right>$ on $V$ and take $\varphi^i$ to be the dual elements to $v_i$ (that is, $\varphi^i(u) = \left< v_i, u \right>$) then the determinant above is just the Gram determinant which gives you the volume of the parallelotope spanned by $v_1, \dots, v_k$. I'm not aware of other interpretations $\endgroup$
    – levap
    May 22, 2019 at 13:18
  • $\begingroup$ @levap I understood what I wanted. Thanks anyway! I'll delete my comments in a couple of hours. $\endgroup$
    – Arrow
    May 22, 2019 at 14:45
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The natural isomorphism $I$ between these spaces is defined by $$ (I(\alpha_1\wedge\ldots\wedge\alpha_k))(v_1\wedge\ldots\wedge v_k):=\sum_\pi \mathrm{sgn}(\pi)\alpha_1(v_{\pi(1)})\ldots \alpha_k(v_{\pi(k)}). $$

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