Take the 2-minute tour ×

Does anybody know a proof (or a reference to proof) of the following combinatorial question: Let $a,b,c,\ldots=1,\ldots,n$ enumerate elements of a set of size $n$, e.g. vertices of an $n-1$ symplex. Let $C_{ab}$ be a cocycle, i.e. an object satisfying $C_{ab}=-C_{ba}$ and $$C_{ab}+C_{bc}+C_{ca}=0.$$ This can be interpreted as the sum over edges of a triangle $abc$, where the weight that each edge is assigned is the cocycle. Then the following identity holds: $$\sum_a (-1)^{a+1} \prod_{b<c,\, b,c\not=a} C_{bc}=0.$$ This can be interpreted as a sum over $n-1$ simplices of an $n$ simplex, with what is summed over being the product of the cocycles for all edges of the corresponding $n-1$ symplex (with signs). Note that for $n=3$ the above identity is just the cocycle condition itself.

This identity was checked (with Mathematica) for some low $n$'s, so I am reasonably confident it holds for a general $n$. It is also so simple that it must be known (if true). This is why I am hoping that somebody will point out a reference. Yet another question is: Is there any cohomological interpretation of this identity?

share|improve this question
add comment

1 Answer

up vote 12 down vote accepted

Choose an arbitrary number $D_n$ and define $D_i = C_{in}+D_n$ for $i \neq n$. So $C_{ij} = D_i - D_j$.

The identity we want to prove is $$\sum_{a=1}^n (-1)^{a+1} \prod_{\begin{matrix} 1 \leq b < c \leq n \\ b,c \neq a\end{matrix}} (D_b-D_c) = 0.$$

The product is the Vandermonde determinant $\det (D_j^k)_{ 1 \leq j \leq n,\ j \neq a}^{0 \leq k \leq n-2}$. So the sum is the row expansion of $$ \det \begin{pmatrix} 1 & 1 & 1 & \cdots & 1 \\ 1 & 1 & 1 & \cdots & 1 \\ D_1 & D_2 & D_3 & \cdots & D_n \\ D_1^2 & D_2^2 & D_3^2 & \cdots & D_n^2 \\ \vdots \\ D_1^{n-2} & D_2^{n-2} & D_3^{n-2} & \cdots & D_n^{n-2} \\ \end{pmatrix}.$$

Since this matrix has a duplicate row, its determinant is $0$.

share|improve this answer
Thank you! It helps a lot. Just want to add that the power in the last row must be $n-2$. –  Kirill Krasnov Feb 21 '13 at 17:59
Oops! Fixed now, thanks. –  David Speyer Feb 21 '13 at 18:32
add comment

Your Answer


