# Isomorphism between exterior algebras

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 $$f\in Λ^{m}(V)$$?

• What are your thoughts on this? – giobrach May 31 at 16:03
• I'm not sure one can. – Lord Shark the Unknown May 31 at 16:03
• 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 $f\in\Lambda^mV$. – user10354138 May 31 at 16:07
• @user10354138 I don’t have any data about f but f nonzero. – Fidel gh. May 31 at 16:22
• @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 $m=1$. – 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 $$\Lambda(V)\cong \Lambda(V^*)$$.