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Let $n$ be a positive integer and $S_n$ be the symmetric group on $\{1,2,\ldots,n\}$. Let $\mathcal{A}_n$ be the anti-symmetrization operator on $\mathbb{Z}[x_1,x_2,\ldots,x_n]$ such that for any $f(x_1,x_2,\ldots,x_n)\in \mathbb{Z}[x_1,x_2,\ldots,x_n]$, $$\mathcal{A}_n(f)=\sum_{w\in S_n}\varepsilon(w)f(x_{w(1)},x_{w(2)},\ldots,x_{w(n)}),$$ where $\varepsilon(w)=(-1)^{l(w)}$ is the sign of $w$.

My question:

For any integer $n>1$ with $n(n-1)\equiv 0 \pmod 4$, is it true that we can always choose $\displaystyle \frac{n(n-1)}{4}$ different polynomials $$x_{i_k}+x_{j_k},\ 1\leq k\leq \frac{n(n-1)}{4}$$ from the $\displaystyle \frac{n(n-1)}{2}$ polynomials $$x_i+x_j,\ 1\leq i<j\leq n,$$ such that $$\mathcal{A}_n\left(\prod_{1\leq k\leq \frac{n(n-1)}{4}}(x_{i_k}+x_{j_k})^2\right)\neq0\ ?$$

For example, \begin{align*} n=4,\ \ &\mathcal{A}_4\left((x_1+x_2)^2(x_2+x_3)^2(x_3+x_4)^2\right)\neq0;\\ n=5,\ \ &\mathcal{A}_5\left((x_1+x_2)^2(x_2+x_3)^2(x_3+x_4)^2(x_4+x_5)^2(x_5+x_1)^2\right)\neq0. \end{align*}

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    $\begingroup$ Is not it true that the span of all such products is the whole space of polynomials in $x_1,\dots,x_n$ of degree $n(n-1)/2$? $\endgroup$ – Fedor Petrov May 6 '18 at 16:35
  • $\begingroup$ Fedor Petrov: When $n=4$, the dimension of the space of homogenous polynomials in $x_1,x_2,x_3,x_4$ of degree $6$ is $\binom{9}{3}$ which is larger than the number of all such products $\binom{6}{3}$. I am not sure whether it is true or not when $n$ is large. $\endgroup$ – user173856 May 6 '18 at 18:08
  • $\begingroup$ For large $n$ the number of products is much greater than the dimension. $\endgroup$ – Fedor Petrov May 6 '18 at 20:28
  • $\begingroup$ Fedor Petrov: Yes, I mean I do not know whether the span of all such products is the space of homogenous polynomials in $x_1,\cdots,x_n$ of degree $n(n-1)/2$ for large $n$. $\endgroup$ – user173856 May 6 '18 at 21:05
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    $\begingroup$ If you take a polynomial in the span if these products and set $x_i$ to $1$ for all $i\neq 1$, then the result will be a polynomial in $(x_1+1)^2$. The same operation sends $(x_1+x_2)x_2^{n(n-1)/2-1}$ to $x_1+1$, so it doesn't lie in the span. $\endgroup$ – MTyson May 7 '18 at 4:03
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Hope that now it works, but please check.

For $n=4k$ or $n=4k+1$ we consider the following graph $H_n$ on the set $V_n$, $|V_n|=n$ with $n(n-1)/4$ edges: $V_n=\{a_i,b_i,c_i,d_i,1\leqslant i\leqslant k\}$ for $n=4k$ and $V_n=V_{n-1}\cup \{e\}$ for $n=4k+1$; the edges of $H_n$ are the edges of $H_{n-4}$ plus $a_kb_k,b_kc_k,c_kd_k$ and all edges from $b_k,c_k$ to $V_{n-4}$.

The graph $H_n$ has $2^k$ automorphisms generated by simultaneous changes $a_i\leftrightarrow d_i$, $b_i\leftrightarrow c_i$. It is important that these automorphisms are even permutations of $V_n$. Note that $H_n$ has unique directed Hamiltonian path up to automorphisms.

For each $u\in V_n$ take a variable $x_u$ and consider the polynomial $F=\prod_{(u,v)\notin H_n} (x_u+x_v-n)(x_u+x_v-(n-1))$. It has degree $n(n-1)/2$. I claim that its antisymmetrization is non-zero. It coincides with antisymmetrization of the polynomial $G=\prod_{(u,v)\notin H_n} (x_u+x_v)^2$, since the difference of two polynomials has degree less than $n(n-1)/2$, therefore its antisymmetrization equals to 0 (there are no other antisymmetric polynomials of such small degree).

For proving this we let the variables $\{x_v\}$ take the values $0,1,\dots,n-1$. Note that when $y_1,\dots,y_n$ is a permutation of $0,1,\dots,n-1$, and $F(y_1,\dots,y_n)\ne 0$, we get $y_i+y_j\ne n$ and $y_i+y_j\ne n-1$ for $(i,j)\notin H_n$, thus all pairs $(i,j)$ for which $y_i+y_j\in \{n,n-1\}$ belong to $H_n$. Such pairs form a Hamiltonian path in $H_n$. But all Hamiltonian paths of $H_n$ are obtained one from another by an even automorphism of $H_n$, thus the corresponding summands are equal and non-zero.

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  • $\begingroup$ I think that example of $H_8$ has 2 unoriented Hamiltonian paths: $a'b'abcdc'd'$ and $a'b'dcbac'd'$, where the unprimed vertices are also in $H_4$. $\endgroup$ – MTyson May 9 '18 at 19:01
  • $\begingroup$ oh indeed, my bad $\endgroup$ – Fedor Petrov May 9 '18 at 19:15
  • $\begingroup$ but maybe all paths have the equal contribution to the sum? $\endgroup$ – Fedor Petrov May 9 '18 at 19:17
  • $\begingroup$ Looks solid to me. $\endgroup$ – MTyson May 9 '18 at 20:34
  • $\begingroup$ Fedor Petrov: This is very good! The construction of $H_n$ is ingenious! Thank you very much for your answer! $\endgroup$ – user173856 May 10 '18 at 11:38

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