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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

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

Hence the sum is \begin{align} \sum_{F \subseteq G \subseteq D} (-1)^{{\rm dim}(G)} & = \sum_{g = f}^{d} \binom{d - f}{g - f} (-1)^g \\ &= (-1)^f\sum_{i=0}^{d - f} \binom{d-f}{i}(-1)^i \\ &= (-1)^f(1-1)^{d-f} \\ &=0\end{align}

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