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?


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.