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Let $\varepsilon$ and $\eta$ be multilinear, skewsymmetric forms taking their values on $\mathbb R$. How can I write the sum:

$$(\varepsilon\wedge \eta)(a_1, \ldots, a_{k}, a_{k+1}, \ldots, a_{k+l})=\sum_{\sigma\in Sh(k, l)} \textrm{sign}(\sigma) \varepsilon(a_{\sigma(1)}, \ldots, a_{\sigma(k)})\eta(a_{\sigma(k+1)}, \ldots, a_{\sigma(k+l)})$$ without using shuffles?

That is, I'm look to write something like $$\sum_{1\leq j_1<\ldots< j_k \leq k+l\atop 1\leq i_1<\ldots<i_l\leq k+2} (-1)^{j_1+\ldots+j_k}(-1)^{i_1+\ldots+i_l} \varepsilon(a_{j_1}, \ldots, a_{j_k})\eta(a_{i_1}, \ldots, a_{i_l}).$$ The above sum is not correct though, because there are overlapping indices.

The point is that once we know $1\leq j_1<\ldots<j_k\leq k+l$ the remaining indices $i_1<\ldots<i_l$ are uniquely determined, but I don't know to use this information properly.

Thanks.

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This is probably not a satisfying answer, but I give it a try.

Assume $\varepsilon$ and $\eta$ to be a $k$- and an $h$-form respectively. As you say, when you fix $a_{j_1},\dots,a_{j_k}$, with $j_1 < j_2 < \dots < j_k$, the remaining indices are uniquely determined and appear in the sequence of vectors in $\eta$ in increasing order due to the process of contraction. So there should be no sum over the $i$s, rather a remark that the $i$s are the indices which complete the sequence $j_1,\dots,j_k$ to $1,\dots,k+h$, and that they appear in increasing order.

Also, $$ \begin{align} (-1)^{j_1+\dots+j_k}(-1)^{i_1+\dots+i_h} & = (-1)^{1+\dots+(k+h)}\\ & = (-1)^{(k+h)(k+h+1)/2}. \end{align} $$ I did not check whether the signs are ok in general, but they look right in special cases where $h$ and $k$ are small.

Using contractions I end up with the expression you want to avoid, which is the only one I have always seen around in the general case.

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  • $\begingroup$ Thanks for the help, I'll try to develop it further.. $\endgroup$
    – PtF
    Mar 4 '20 at 22:01
  • $\begingroup$ Do you need an expression like that for a particular reason? $\endgroup$
    – Gibbs
    Mar 6 '20 at 11:41
  • $\begingroup$ As a matter of fact, it is for sake of clarity. Shuffles allow us to write the sum compactly, but for explicit computations it may be more obscure than using indices explicitly. $\endgroup$
    – PtF
    Mar 7 '20 at 13:13
  • $\begingroup$ how about $$\varepsilon\wedge \eta(a_1, \ldots, a_{p+q})= \sum_{1\leq i_1<\ldots< i_{k_1}<\ldots< i_{k_q}< \ldots< i_{p+q}\leq p+q\atop k_1, \ldots, k_q\in \{1, \ldots, p+q\}} \varepsilon(a_{i_1}, \ldots, \widehat{a_{i_{k_1}}}, \ldots, \widehat{a_{i_{k_q}}}, \ldots, a_{i_{p+q}}) \eta( a_{i_{k_1}}, \ldots, a_{i_{k_q}}),$$ up to signs? Is the above sum clear to you? $\endgroup$
    – PtF
    Mar 7 '20 at 13:13
  • $\begingroup$ I find it much harder to read than the one with permutations. $\endgroup$
    – Gibbs
    Mar 7 '20 at 13:15

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