Let $p$ integer prime, $f$ a function of $A=\mathbb F\_p^n$ to $\mathbb F\_p$, with $n\geq p+1$.


Is it true that : for all $x\in A, \sum\limits\_{\sigma \in S\_n} s(\sigma) \times f(x\_\sigma) =0$?


$s$ the signature


$S\_n$ is the group of all bijection of $U\_n=\{1,...,n\}$ to $U\_n$.


If $x=(x\_1,...x\_n)$ then $x\_\sigma=(x\_{\sigma(1)},...,x\_{\sigma(n)})$.



> 
> The answer is yes, if $p=2$ but what about the other case?
> 
> 
> 



A new convolution, on function of $\mathbb F_p^n$ to $\mathbb F_p$ still zero?
As long as $n \geq p+1$, two of the entries of $x$ must be the same by the pigeonhole principle. Let $\tau$ be a transposition fixing those two entries. Then $s(\sigma \circ \tau) = -s(\sigma)$ but $x\_{\sigma \circ \tau} = x\_\sigma$ so $f(x\_{\sigma \circ \tau} ) =f(x\_\sigma)$. Thus the terms in your sum for $\sigma$ and $\sigma \circ \tau$ cancel.


Since ${}\circ \tau$ acts on $S\_n$ with $n!/2$ orbits of size 2, we can divide the sum into $n!/2$ pairs of terms which cancel, so the whole sum vanishes.


