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Consider a positive integer $n$ and integers $(c_i)_{1\le i \le 4}$, with $1 \le c_i \le n$. Conside the map:

$$f_n: (c_1,c_2,c_3,c_4) \mapsto \delta_{c_1,c_2}\delta_{c_3,c_4} - \# \{ |2n+1-2|x||, \ x \in \{c_1+c_2, c_3+c_4, c_1-c_2, c_3-c_4\} \},$$

where the notation $\#$ means cardinal. The problem is to show that this map is invariant by permutation, i.e. $\forall \sigma \in S_4$ $$ f_n(c_1,c_2,c_3,c_4) = f_n(c_{\sigma(1)},c_{\sigma(2)},c_{\sigma(3)},c_{\sigma(4)}). $$

I see a straightforward way to prove that by cases, but it should be quite long.

Question: Is there a short (conceptual) way to prove that?

Context: this problem poped up to prove that some rings (parametrized by $n$) are associative.

For the conveniance of the reader, here is the checking for $n<100$:

First of all, the symmetric group $S_r$ is generated by $\sigma_i = (i,i+1)$ for $i=1,\dots, r-1$. Here $r=4$ and the shape of the map provides immediately the invariance by $\sigma_1$ and $\sigma_3$. So it remains to show the invariance by $\sigma_2$:

Code

cpdef delta(int c1, int c2): if c1==c2: return 1 else: return 0 cpdef map(int c1, int c2, int c3, int c4,int n): cdef int x return delta(c1,c2)*delta(c3,c4)-len(set([abs(2*n+1-2*abs(x)) for x in [c1+c2,c3+c4,c1-c2,c3-c4]])) cpdef test(int N): cdef int n,c1,c2,c3,c4 for n in range(1,N): for c1 in range(1,n+1): for c2 in range(1,n+1): for c3 in range(1,n+1): for c4 in range(1,n+1): if map(c1,c2,c3,c4,n)!=map(c1,c3,c2,c4,n): print([c1,c2,c3,c4,n])

Computation

sage: test(100) sage:
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  • $\begingroup$ Is this particular to $4$ or might there be a generalization to $2k$? $\endgroup$ Commented Oct 8, 2020 at 7:54
  • $\begingroup$ @MartinRubey: Thanks to your comment, I realized that the cyclic permutation invariance of the map $(c_1, c_2, c_3) \mapsto ...$, extends immediately to a permutation invariance of the map $(c_1, c_2, c_3, c_4) \mapsto ...$ Such a map can immediately by generalized to $2k$ as you suggest, but for $k=3$ and $n=2$, $(1, 1, 1, 1, 2, 2) \mapsto -2$ whereas $(1, 1, 1, 2, 1, 2, 2) \mapsto -3$. So the direct generalization does not work. I do not exclude the existence of an alternative generalization, in fact, a conceptual proof of the case $k=2$ should help to find the good way to generalize. $\endgroup$ Commented Oct 8, 2020 at 10:08
  • $\begingroup$ @MartinRubey: to understand where $4$ comes from here, let me explain a bit more the context. It is about a finite dimensional algebra, with an orthonormal basis $x_1, \dots, x_m$. The permutation invariance of the map is equivalent to the associativity, i.e. $ \langle (x_{c_1} x_{c_2})x_{c_3} , x_{c_4} \rangle = \langle x_{c_1} (x_{c_2}x_{c_3}) , x_{c_4} \rangle$. $\endgroup$ Commented Oct 8, 2020 at 11:11
  • $\begingroup$ Clearly the permutations leaving your map invariant include the transpositions $(1,2)$ and $(3,4)$. Thus it suffices to show that $(2,3)$ also has this property, since these three transpositions generate $S_4$. $\endgroup$ Commented Oct 8, 2020 at 17:38
  • $\begingroup$ @RichardStanley Yes this is precisely what I checked by computer for n<100; but in general, I only see a long proof by cases, whereas I would like a short (conceptual) proof. $\endgroup$ Commented Oct 8, 2020 at 18:45

1 Answer 1

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One conceptual proof is to observe that it comes from the associativity of the character ring of $\mathrm{PSL}(2,q)$, interpolated (the ring) to every integer $q$. See Corollary 4.6 in this paper. Note that it corresponds to the case $q$ even (where $n=q/2$), and requires a tiny part of the associativity of the interpolated ring, then many other such combinatorial properties can be obtained similarly.

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