MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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|cite|improve this question
up vote 13 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|cite|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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.