The Euler Characteristic of a Simplicial Complex is defined to be $\sum (-1)^i\alpha_i$, where $\alpha_1$ is the number of $i$-simplices in the complex. Using this formula, we can see that the Euler Characteristic of any $n$ simplex is $1$.

Define the Barycentric subdivision of a complex $K$ to have as its vertex set the set of centers of simplices in $K$ (so each vertex in the subdivision corresponds to a simplex in the original complex $K$), and say that $k+1$ vertices form a $k$-simplex if, in some order, their corresponding simplices in $K$ form a flag (meaning that each simplex is a "face" (i.e., a subsimplex) of the next). I want to calculate the Euler Characteristic of the subdivision of an $n$-simplex.

Now, flags of length $k+1$ correspond to chains of length $k+1$ in the set of subsets of the $n+1$ vertices of the $n$-simplex. Therefore, the number of $i$-simplices in the subdivision is $(i+2)^{n+1}$, since we can choose which of the $i+2$ initial segments to place each of the $n+1$ vertices in. Thus, the desired Euler Characteristic is $$\displaystyle\sum_{i=0}^n (-1)^{i}(i+2)^{n+1}.$$

Now, is there a way of simplifying this? I want to show that it in fact equals $1$, since I'm trying to prove that the Euler Characteristic of a simplicial complex is invariant under subdivision.

Edit: The above formula is wrong, since it calculates the number of chains of length $k+1$, but here, the subsets in chains are required to all be nonempty and not equal to each other.

  • $\begingroup$ I am having difficulty verifying your formula for the number of $i$-simplices, e.g. in the cases $n=1$ and $n=2$. $\endgroup$ – Tyler Lawson Nov 11 '14 at 5:38
  • $\begingroup$ Oh, I think I see my mistake: we require each subset in the chain to be nonempty, whereas my formula allows them to be empty... $\endgroup$ – Nishant Nov 11 '14 at 5:43
  • $\begingroup$ Also, the chain has to be strict, meaning that none of the subsets in the chain can equal each other... $\endgroup$ – Nishant Nov 11 '14 at 5:44

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