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Mariano Suárez-Álvarez
Mariano Suárez-Álvarez
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Here is a basis free expression.

Let the rank of $M$ be $r$. Pick an isomorphism $\phi: M \to M^*=\hom_R(M,R)$$\phi: M \to M^\*=\hom_R(M,R)$. Now, if $m\in M$, define a map $f_m:\Lambda^rM\to\Lambda^{r-1}M$ by contracting with $\phi(m)$, so that if $m_1$, $\dots$, $m_r\in M$, then $$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.)

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.

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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Mariano Suárez-Álvarez
Mariano Suárez-Álvarez
  • 47.3k
  • 14
  • 147
  • 264

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 this$m\in M\mapsto f_m\in\hom(\Lambda^rM, \Lambda^{r-1}M)$ is an isomorphism.

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 this is an isomorphism.

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.

Source Link
Mariano Suárez-Álvarez
Mariano Suárez-Álvarez
  • 47.3k
  • 14
  • 147
  • 264

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 this is an isomorphism.