I would like to get a more concrete understanding of a general isomorphism I have read about. I apologize if this is too basic, but I was not satisfied with the references at my disposal.

Let $K$ be a field of characteristic zero and $V$ a finite dimensional vector space over $K$.

Let $Alt_k(V,K)$ denote the set of k-multilinear alternating forms on $V$ and let $\Lambda ^k (V)$ be the k-th exterior power, defined as usual as a quotient of the tensor algebra $V^{\otimes k}$.

We have an isomorphism $Alt_k(V,K) \cong \Lambda ^k (V^*)$ induced by the analogous one at the level of the tensor algebra.

Therefore it makes sense to take some functionals $\varphi_1,\dots,\varphi_k \in V^*$ and consider $\varphi_1 \wedge\dots \wedge\varphi_k$ as a multilinear alternating form on $V$. So, take $v_1,\dots,v_k \in V$; how should the following object (=number in $K$) be defined?

$$(1) \quad \quad \quad \quad \quad \quad (\varphi_1 \wedge\dots \wedge\varphi_k)(v_1,\dots,v_k) $$

In particular, if I fix some basis $e_1, \dots, e_n$ for $V$, how should I compute the following?

$$(2) \quad \quad \quad \quad \quad \quad (e^*_{i_1} \wedge\dots \wedge e^*_{i_k})(e_{j_1},\dots,e_{j_k})$$

I can carry out the case $k=2$, so I suspect some kind of determinant is involved in (2), maybe the det of the submatrix obtained from the rows of indices $i_1,\dots,i_k$ in the matrix having the $e_{j_1},\dots,e_{j_k}$ in columns. But I cannot check this, since I don't know the right (=coordinates-free) definition for (1).

To explain what I'm looking for, I'll sketch the same things for the case of $V^{\otimes k} \cong \{ $ k-multilinear forms on $ V \}$. In this case if I take $\varphi_1,\dots,\varphi_k \in V^*$ and $v_1,\dots,v_k \in V$ then it's ok to set

$$ (\varphi_1 \otimes \dots \otimes \varphi_k)(v_1,\dots,v_k)= \varphi_1 (v_1)\cdot \textrm{ $\dots$ } \cdot \varphi_k (v_k) $$

and this gives a k-multilinear form. If I fix some basis $e_1, \dots, e_n$ for $V$, then I get

$$ (e^*_{i_1} \otimes \dots \otimes e^*_{i_k})(e_{j_1},\dots,e_{j_k})=\delta_{i_1,j_1} \cdot \textrm{ $\dots$ } \cdot \delta_{i_k,j_k}$$

So, I think one should properly reduce these expression in a way which is compatible with the ideal defining the exterior power, but I don't know how to do this.

  • $\begingroup$ Well $\omega \wedge \eta = \frac{(k+l)!}{k!l!}$ Alt($\omega \otimes \eta$). Is this what you were looking for? $\endgroup$ – Sandeep Thilakan Feb 12 '14 at 21:00
  • $\begingroup$ I suppose Alt is defined so that this gives the same formula as the one in Peter Crook's answer below, so yes, thank you. Anyway, are all these coefficients necessary? $\endgroup$ – Lor Feb 12 '14 at 23:28
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    $\begingroup$ The reason why the coefficients are defined this way is that you get $\epsilon ^I \wedge \epsilon ^J = \epsilon ^{IJ}$ $\endgroup$ – Sandeep Thilakan Feb 13 '14 at 6:17
  • $\begingroup$ If I'm not wrong $(\varphi_1\wedge\ldots\wedge \varphi_k)(v_1, \ldots, v_k)=\det(\varphi_i(v_j))$, which is just a more familiar way of writing @Peter Crooks answer. $\endgroup$ – PtF Mar 5 at 1:13

I think you want $$(\varphi_1 \wedge\dots \wedge\varphi_k)(v_1,\dots,v_k)=\frac{1}{k!}\sum_{\sigma\in S_k}sgn(\sigma)\varphi_1(v_{\sigma(1)})\varphi_2(v_{\sigma(2)})\cdots\varphi_k(v_{\sigma(k)}).$$

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  • $\begingroup$ yes, thank you! maybe with some sign? I think one should get for example $(e^*_1 \wedge e^*_2)(u,v)=u_1v_2-u_2v_1$, so maybe something like $sign(\sigma)$. $\endgroup$ – Lor Feb 12 '14 at 23:22

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