MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
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? – Hideyuki Kabayakawa 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
up vote 7 down vote accepted

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: $$e_i \in M\mapsto \Big(e_1 \wedge e_2 \wedge \ldots \wedge e_m \mapsto e_1 \wedge \ldots \wedge \hat{e_i} \wedge \ldots \wedge e_m \Big) \in \mathrm{Hom}_R(\Lambda^m M, \Lambda^{m-1} M)$$

share|cite|improve this answer
Why is it an isomorphism? You don´t specify inverse morphism – Hideyuki Kabayakawa 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.)

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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