Compute the simplicial homology groups of the $\Delta$-complex obtained from $n+1$ $2$-simplices $\Delta_0^2,...,\Delta_n^2$ by identifying all three edges of $\Delta_0^2$ to a single edge, and for $i>0$ identifying the edges $[v_0,v_1]$ and $[v_1,v_2]$ of $\Delta_i^2$ to a single edge and the edge $[v_0, v_2]$ to the edge $[v_0, v_1]$ of $\Delta_{i-1}^2$.

My attempt:

Our space $X$ has $1$ vertex, $n+1$ edges and $n+1$ faces.

We have $\partial_0=0$ and $\partial_1=0$. So we conclude that $H_0(X) = \ker \partial_0 / \text{Im} \partial_1 \approx \mathbb{Z}$. Also $H_n(X)$ is trivial for $n>2$.

Now let's compute $H_1(X) = \ker \partial_1 / \text{Im} \partial_2$. Let's call the vertex $v$, the edges $a_i$ and the faces $U_i$ $(i=0,1,...,n)$. We then have $$\partial_2(U_0)= a_0, \qquad \partial_2(U_i) = 2a_i - a_{i-1} \quad (i\neq0).$$

So $\text{Im} \partial_2 = < a_0, 2a_1-a_o,...,2 a_n - a_{n-1}>$.

Also since $\partial_1=0$, we have $\ker \partial_1 = <a_0,...,a_n> \approx \mathbb{Z}^{n+1}$

Is this correct? How do we compute the factor group $H_1(X)$ from this?


1 Answer 1


One way to compute the factor group would be to view $\textrm{Im}(\partial_2)$ as the largest element in the following ascending chain:

  • $\langle a_0\rangle \subset \langle a_0, 2a_1 - a_0\rangle \subset \langle a_0, 2a_1 - a_0, 2a_2-a_1\rangle \subset \dots \subset \langle a_0, 2a_1 - a_0, \ldots, 2a_n - a_{n-1}\rangle $

Now you can inductively determine the isomorphism class (in the standard classification of finitely-generated abelian groups) of the quotient $\mathbb{Z}^{n+1} / \langle a_0, 2a_1 - a_0, \ldots, 2a_{i+1} - a_{i}\rangle$ for $i = 0, 1, \ldots, n-1$. It is clear, for instance, that $\mathbb{Z}^{n+1}/\langle a_0\rangle \cong \mathbb{Z}^n$. Hopefully that is enough get started.


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