Let V is a m-dimensional vector space and $V^{*}$ is dual vector space. How can define isomorphism between exterior algebra $Λ(V)$ and exterior algebra $Λ(V^{*})$ with use a volume element $f\in Λ^{m}(V) $?

  • $\begingroup$ What are your thoughts on this? $\endgroup$ – giobrach May 31 at 16:03
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    $\begingroup$ I'm not sure one can. $\endgroup$ – Lord Shark the Unknown May 31 at 16:03
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    $\begingroup$ I don't think so. Having such an isomorphism would entails an isomorphism $V\cong V^*$, and it is unclear how you can achieve that with just $f\in\Lambda^mV$. $\endgroup$ – user10354138 May 31 at 16:07
  • $\begingroup$ @user10354138 I don’t have any data about f but f nonzero. $\endgroup$ – Fidel gh. May 31 at 16:22
  • $\begingroup$ @Fidelgh. Since $\Lambda^mV$ is 1-dimensional, if such a construction is possible that means you have somehow a distinguished 1-parameter family of isomorphisms $V\to V^*$ labelled by $\Lambda^mV-\{0\}$. This is highly suspicious unless $m=1$. $\endgroup$ – user10354138 May 31 at 16:28

There is no way to get such an algebra isomorphism that is natural (more precisely, functorial with respect to isomorphisms). If $T$ is any automorphism of $V$, then $T$ multiplies $f$ by $\det T$. This means that the isomorphism $\Lambda(V)\cong \Lambda(V^*)$ must be invariant under any element of $SL(V)$. Note that any isomorphism $\Lambda(V)\to \Lambda(V^*)$ functorially induces an isomorphism $V\to V^*$ (if $N$ is the ideal of nilpotent elements of $\Lambda(V)$ there is a canonical isomorphism $N/N^2\cong V$), so we would need an isomorphism $S:V\to V^*$ such that $S=T^*ST$ for any $T\in SL(V)$. Or, interpreting $S$ as a nondegenerate bilinear form $\langle\cdot,\cdot\rangle$ on $V$, we must have $\langle v,w\rangle=\langle Tv,Tw\rangle$ for all $v,w\in V$ and $T\in SL(V)$. But this is impossible if $\dim V>2$ (and the scalar field has more than $2$ elements) since then $SL(V)$ acts transitively on pairs of linearly independent elements of $V$ and so this would mean $\langle\cdot,\cdot\rangle$ is constant on linearly independent pairs which yields a contradiction if you multiply one of the vectors by a scalar different from $0$ or $1$.

(Probably with a bit more work you can also show it is impossible for $\dim V=2$ as long as the scalar field is not too trivial, and that it is impossible for sufficiently large $\dim V$ even when the scalar field is $\mathbb{F}_2$.)

If you just want an isomorphism of vector spaces, then note that the exterior product is a perfect pairing $\Lambda^i(V)\times \Lambda^{m-i}(V)\to \Lambda^m(V)$ so picking a nonzero element of $\Lambda^m(V)$ gives an isomorphism $\Lambda^i(V)\cong (\Lambda^{m-i}(V))^*\cong \Lambda^{m-i}(V^*)$ and taking the direct sum of these isomorphisms gives degree-reversing vector space isomorphism $\Lambda(V)\cong \Lambda(V^*)$.

  • $\begingroup$ Your answer is complete, thank you Eric $\endgroup$ – Fidel gh. May 31 at 17:20

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