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Suppose $\varepsilon$ is a multilinear skew-symmetric function of its arguments with vales in $\mathbb R$. Is it true that:

$$\varepsilon(a_1+\theta_1, \ldots, a_p+\theta_p)=\varepsilon(a_1, \ldots, a_p)+\\ +(-1)^p\sum_{i=1}^{p-1}\sum_{1\leq j_1<\ldots< j_i\leq p} (-1)^{j_1+\ldots +j_i} \varepsilon(a_1, \ldots, \widehat{a_{j_1}}, \ldots, \widehat{a_{j_i}}, \ldots, a_p, \theta_{j_1}, \ldots, \theta_{j_i})\\ +\varepsilon(\theta_1, \ldots, \theta_p)$$ where the hats means omission?

Thanks.

PtF
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  • Could you elaborate on where you run into difficulty when you try to check this formula? – jgon Feb 22 '20 at 21:16
  • The sign on the term $\epsilon(a_1,\ldots,\widehat{a_{j_1}},\ldots,\widehat{a_{j_i}},\ldots,a_p,\theta_{j_1},\ldots,\theta_{j_i}$ should be $(-1)^\sigma$, with $\sigma = \sum_{k=1}^i p-k+1-j_{i+1-k} = ip + i -\frac{i(i+1)}{2} -\sum_{i=1}^k j_i $. – jgon Feb 22 '20 at 21:26
  • Taking the sign mod $2$, the claim becomes $ip+i+\frac{i(i+1)}{2} \equiv p+i+1$ when $i> 0$, and $ip+i+\frac{i(i+1)}{2} \equiv 0$ when $i=0$. The last claim is certainly true. The first is equivalent to $(i-1)p +\frac{i(i+1)}{2} \equiv 1$ for all $0<i\le p$, and all $p>0$. This seems false. E.g., take $p$ even, $i=4$. Then we get $0\equiv 1$. – jgon Feb 22 '20 at 21:31
  • But perhaps my calculations are wrong. These are just comments, not an answer. – jgon Feb 22 '20 at 21:31
  • As a matter of fact, the formula is a conjecture. I'll try to fix the signs.. – PtF Feb 22 '20 at 22:01
  • @jgon I fixed the signs. Do you agree with the formula now? – PtF Feb 22 '20 at 22:57
  • Is it true for $\epsilon(x,y)=x -y$? – ancient mathematician Feb 23 '20 at 08:00
  • @ancientmathematician your map is not bilinear.. – PtF Feb 23 '20 at 12:46
  • The question has changed. – ancient mathematician Feb 23 '20 at 13:29
  • I had missed the word multilinear. – PtF Feb 23 '20 at 13:34
  • @jgon the sign shouldn't be $ip-(j_1+\ldots+j_i)- \frac{i(i-1)}{2}$? – PtF Feb 24 '20 at 16:38
  • @PtF, I'm pretty sure that my sign is correct. You have to commute $\theta_{j_{i+1-k}}$ past $p-j_{i+1-k} - (k-1)$ things. – jgon Feb 24 '20 at 17:48
  • The first time $\varepsilon(a_1, \ldots, b_{j_1}, \ldots, b_{j_i}, \ldots, a_p)=(-1)^{p-j_1}\varepsilon(a_1, \ldots, b_{j_2}, \ldots, b_{j_i}, \ldots, a_p, b_{j_1})$. The next time you will have to commute $b_{j_2}$ past $p-j_2+1$. The last time you will have to commute $b_{j_i}$ past $p-j_i+(i-1)$, hence the sign should be $ip-(j_1+\ldots+j_i)+(1+\ldots+(i-1))=ip-(j_1+\ldots+j_i)+ \frac{i(i-1)}{2}$, nope? – PtF Feb 24 '20 at 17:56
  • As a matter of fact, $i-\frac{i(i+1)}{2}= \frac{2i-i^2-i}{2}=\frac{i-i^2}{2}=-\frac{i(i-1)}{2}$ so our signs are identical for $(-1)^{-\frac{i(i-1)}{2}}=(-1)^{\frac{i(i-1)}{2}}$. – PtF Feb 24 '20 at 18:09

1 Answers1

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Discussing with @jgon we got to the same conclusion, therefore I assume we have the right answer which I'm posting just in case someone else needs in the future.

In order to find the correct expression, it suffices to find $\sigma$ in:

$$\varepsilon(a_1, \ldots, a_{j_1-1}, b_{j_1}, a_{j_1+1}, \ldots, a_{j_i-1}, b_{j_i}, a_{j_i+1}, \ldots, a_p)=(-1)^\sigma \varepsilon(a_1, \ldots, \widehat{a_{j_1}}, \ldots, \widehat{a_{j_i}}, \ldots, a_p, b_{j_1}, \ldots, b_{j_i}).$$

Let us focus on $b_{j_1}$. For it to appear beside $a_p$ on the RHS of the above equality, it will have to go past $p-j_{1}$ positions. Therefore:

$$\varepsilon(a_1, \ldots, a_{j_1-1}, a_{j_1+1}, \ldots, a_{j_i-1}, b_{j_i}, a_{j_i+1}, \ldots, a_p)\\ = (-1)^{p-j_1} \varepsilon(a_1, \ldots, \widehat{a_{j_1}}, \ldots, a_{j_i-1}, b_{j_i}, a_{j_i+1}, \ldots, a_p, b_{j_1}).$$

Now we focus on $b_{j_2}$. It will have to go past $p-j_2+1$ places, therefore:

$$\varepsilon(a_1, \ldots, a_{j_1-1}, a_{j_1+1}, \ldots, a_{j_2-1}, b_{j_2}, a_{j_2+1}, \ldots, a_{j_i-1}, b_{j_i}, a_{j_i+1}, \ldots, a_p)\\= (-1)^{p-j_1} (-1)^{p-j_2+1} \varepsilon(a_1, \ldots, \widehat{a_{j_1}}, \ldots, \widehat{a_{j_2}}, \ldots, a_{j_i-1}, b_{j_i}, a_{j_i+1}, \ldots, a_p, b_{j_1}, b_{j_2}). $$

The last $b_{j_i}$ will have to go past $p-j_i+(1+\ldots+(i-1))$ places, therefore:

$$\varepsilon(a_1, \ldots, a_{j_1-1}, a_{j_1+1}, \ldots, a_{j_i-1},b_{j_i}, a_{j_i+1}, \ldots, a_p)=(-1)^{p-j_1}(-1)^{p-j_2+1}\cdots (-1)^{p-j_i+(i-1)}\varepsilon(a_1, \ldots, \widehat{a_{j_1}}, \ldots, \widehat{a_{j_i}}, \ldots, a_p, b_{j_1}, \ldots, b_{j_i}).$$ Now notice:

$$(-1)^{p-j_1}(-1)^{p-j_2+1}\cdots (-1)^{p-j_i+(i-1)}=(-1)^{pi-(j_1+\ldots+j_i)+ \frac{i(i-1)}{2}}$$ where I used the standard fact $$1+\ldots+n=\frac{n(n+1)}{2}$$ with $n=i-1$. Therefore:

$$\varepsilon(a_1+b_1, \ldots, a_p+b_p)= \varepsilon(a_1, \ldots, a_p)\\ + \sum_{i=1}^{p-1} (-1)^{ip} (-1)^{i(i-1)/2} \sum_{1\leq j_1<\ldots<j_i\leq p} (-1)^{j_1+\ldots+j_i} \varepsilon(a_1, \ldots, \widehat{a_{j_1}}, \ldots, \widehat{a_{j_i}}, \ldots, a_p, b_{j_1}, \ldots, b_{j_p})\\ +\varepsilon(b_1, \ldots, b_p).$$ I know this can be written in a straightforward manner using shuffles, but I wanted to avoid that.

PtF
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