# Homomorphism between exterior powers of a free module of finite rank

I´m looking for homomorphisms between exterior powers of a free module M of rank m

ΛmR M → Λm-1R M

Exactly, I´m looking for an explicit isomorphism

M → Hom RmR M , Λm-1R M)

I compare the ranks and the things go, but I can not imagine a concrete expression.

Suggestions are welcome

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It would feel more natural if you asked for the homomorphism M → Hom(Λ^{m-1}M, Λ^m M) which is induced by the graded algebra structure. –  user2035 Dec 8 '09 at 8:53
How do you ask this question using exterior algebra? –  Francisco Perdomo Dec 8 '09 at 16:17
The homomorphism is given by m ↦ (ω ↦ m ∧ ω). Using a basis you can see that it is an isomorphism if M is free of rank r. –  user2035 Dec 8 '09 at 18:04
@a-fortiori: the isomorphism you are giving takes (m-1)-forms to m-forms, not the other way around. I guess what Francisco wants would be m ↦ (ω ↦ m ∟ ω) (contraction). –  Alberto García-Raboso Dec 8 '09 at 20:03
@Alberto: I was explaining my previous comment. You need to change the question in some way to get a canonical map, contraction does not work for m ∈ M and ω ∈ Λ^m M. (Please apologize my using the letter m twice.) –  user2035 Dec 8 '09 at 21:15

If $R$ is commutative, let $e_1, \ldots, e_m$ be an $R$-basis of $M$. Then $\Lambda^m M$ is a free $R$-module of rank one generated by the vector $e_1 \wedge e_2 \wedge \ldots \wedge e_m$ and $\Lambda^{m-1}M$ is free of rank $m$ with basis $$\lbrace e_1 \wedge \ldots \wedge \hat{e_i} \wedge \ldots \wedge e_m \rbrace_{i=1}^{m}$$ where the hat denotes an omitted basis vector. Here is an explicit isomorphism: $$e_i \in M\mapsto \Big(e_1 \wedge e_2 \wedge \ldots \wedge e_m \mapsto e_1 \wedge \ldots \wedge \hat{e_i} \wedge \ldots \wedge e_m \Big) \in \mathrm{Hom}_R(\Lambda^m M, \Lambda^{m-1} M)$$

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Why is it an isomorphism? You don´t specify inverse morphism –  Francisco Perdomo Dec 8 '09 at 12:52
If R is a field, then M and the Hom are vector spaces of the same dimension. To specify an isomorphism, you should give the image of a basis of M, which must be a basis of the Hom. The argument is the same in the general case: you should think of free modules over any ring as if they were vector spaces, since the concept of a basis works. –  Alberto García-Raboso Dec 8 '09 at 13:33
@Alberto: You need the Invariant Basis Number property to hold for the ring $R$. Commutative rings, noetherian rings, have IBN. –  Sammy Black Dec 8 '09 at 19:49
@Sammy: thanks for pointing that out! –  Alberto García-Raboso Dec 8 '09 at 19:59
@Sammy: I edited my answer to include the IBN property. –  Alberto García-Raboso Dec 10 '09 at 14:43

Here is a basis free expression.

Let the rank of $M$ be $r$. Pick an isomorphism $\phi: M \to M^\*=\hom_R(M,R)$. Now, if $m\in M$, define a map $f_m:\Lambda^rM\to\Lambda^{r-1}M$ by contracting with $\phi(m)$, so that if $m_1$, $\dots$, $m_r\in M$, then $$f_m(m_1\wedge\cdots\wedge m_r)=\sum_{i=1}^r\;(-1)^i\;\langle\phi(m),m_i\rangle\; m_1\wedge\cdots\wedge\hat m_i\wedge\cdots\wedge m_r.$$ Here $\langle\mathord-,\mathord-\rangle:M^*\times M\to R$ is the evaluation map. It is not hard to show that $m\in M\mapsto f_m\in\hom(\Lambda^rM, \Lambda^{r-1}M)$ is an isomorphism

(This isomorphism is not natural, because it depends on the choice of $\phi$. Of course, there is a natural isomorphism $\Psi_M:M^\*\to\hom(\Lambda^rM, \Lambda^{r-1}M)$ given by essentially the formula, and there is no natural isomorphism $M\to\hom(\Lambda^rM, \Lambda^{r-1}M)$, because such a thing would, when composed with the inverse of the natural isomorphism $\Psi_M$, give a natural isomorphism $M\to M^\*$, which does not exist.)

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