0
$\begingroup$

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) $?

$\endgroup$
  • $\begingroup$ What are your thoughts on this? $\endgroup$ – giobrach May 31 at 16:03
  • 1
    $\begingroup$ I'm not sure one can. $\endgroup$ – Lord Shark the Unknown May 31 at 16:03
  • 1
    $\begingroup$ 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
1
$\begingroup$

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 $\Lambda(V)\cong \Lambda(V^*)$.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.