Let $K=[v_0,v_1,\dots,v_n]$ be a $n$-simplex, and let $\partial_i$ denote the $i$th boundary map.
I am trying to find out the rank of $\ker\partial_i$, for $0\leq i\leq n$.
What I have tried so far is:
For $i=0$, $\ker\partial_0$ is generated by $v_0,\dots, v_n$. Hence, $\text{rank}\ker\partial_0=n+1$.
For $i=n$, $\ker\partial_n=0$ and hence $\text{rank}\ker\partial_n=0$.
The more difficult situation seems to be $1\leq i\leq n-1$.
I have tried some calculations, and it seems that $\text{rank}\ker\partial_i$ seems to be 1 less than the number of faces of dimension $i+1$. (May be wrong.)
That is, my conjecture is $$\text{rank}\ker\partial_i={{n+1}\choose{i+2}}-1$$
For instance, for $n=3$, the rank of $\ker\partial_1$ seems to be $3={4\choose 3}-1$.
How do I prove it rigorously?
Thanks a lot.
Since $\ker\partial_i=\text{Im}\partial_{i+1}$ for $1\leq i\leq n-1$ for the case of $n$-simplex, equivalently it is possible to calculate the rank of $\text{Im}\partial_{i+1}$ instead. However, I am not sure how to proceed in this direction either.