# simplex summation formula

Let $D$ be a simplex and $F$ be any face of $D$ and $G$ be any other face that contains $F$

Prove $\sum_{F\subseteq G}(-1)^{\dim(G)} =0$

I know that $\sum_{F\subseteq D}(-1)^{\dim(F)} =1$ but how can I use this to prove the statement above ?

Would appreciate any help

• This is basically $(1-1)^d=0$ where $d=\dim G-\dim F$. – Lord Shark the Unknown Jun 1 '18 at 20:47

So $D$ is a simplex, and $F$ is a face in $D$. Our goal is to compute $$\sum_{F \subseteq G \subseteq D} (-1)^{{\rm dim}(G)},$$ where the summation is taken over all simplices $G$ that contain $F$ and are contained within $D$.
To make progress, let $d = {\rm dim}(D)$ and $f = {\rm dim}(F)$. We will count how many simplices $G$ there are of dimension $g$ that contain $F$ and are contained in $D$; here, $g$ is some integer between $f$ and $d$.
Any such $G$ must contain $g + 1$ of the $d + 1$ vertices of $D$. $f + 1$ of these vertices coincide with the $f + 1$ vertices of $F$, while the remaining $g - f$ vertices are freely chosen from the $d - f$ vertices of $D$ that are not in $F$. So altogether, there are $\binom{d - f}{g - f}$ possible $G$ of dimension $g$.