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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?

iwriteonbananas
  • 6,102

2 Answers2

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This is a full answer, regardless of how far the OP got on their own (it’s an old post and I’m surprised a full answer doesn’t exist yet). It is clear that the reduced homology of these spaces is trivial in all degree but degree $1$ and perhaps degree $2$ (though this is always trivial as shown later). I claim that: $$H_1(X_n)\cong\Bbb Z_{2^n}$$(the cyclic group, not the -adics).

Label the shared edge of $\Delta_0^2$ as $e_0$. Label the new edge introduced by $\Delta_1^2$ as $e_1$ (so its boundary is $(e_1,e_0,e_1)$ according to the identifications) et cetera. Our reduced chain complex is: $$\cdots\to0\to0\to\Bbb Z^{n+1}\overset{\partial}{\longrightarrow}\Bbb Z^{n+1}\overset{0}{\longrightarrow}\Bbb Z\overset{\epsilon}{\longrightarrow}\Bbb Z\to0$$So it remains to compute $H_1(X_n)\cong\Bbb Z^{n+1}/\mathrm{im}\,\partial$ and $H_2(X_n)\cong\ker\partial$.

$\partial(\Delta_j^2)\sim e_j-e_{j-1}+e_j=2e_j-e_{j-1}$ for $1\le j\le n$ and $\partial(\Delta_0^2)\sim e_0-e_0+e_0=e_0$. Here the $(n+1)$-coordinate axes of $\Bbb Z^{n+1}$ have been associated with the $e_j$ for convenience of notation.

Thus for $x=(x_0,\cdots,x_n)$ in the domain: $$\partial x\sim x_0e_0+x_1(2e_1-e_0)+\cdots+x_{n-1}(2e_{n-1}-e_{n-2})+x_n(2e_n-e_{n-1})\\=(x_0-x_1)e_0+(2x_1-x_2)e_1+\cdots+(2x_{n-1}-x_n)e_{n-1}+2x_n\cdot e_n$$The kernel of this thing is zero; if this evaluates to zero, then in particular $2x_n$ and $x_n$ are zero. Then (reading off the coordinates) $2x_{n-1}-0$ is also zero hence $x_{n-1}$ is zero; et cetera, we find $x=0$ by induction. That proves $H_2(X_n)\cong0$.

We can define a map: $$G:\Bbb Z^{n+1}\to\Bbb Z_{2^n},\,x\mapsto[x_n+2x_{n-1}+\cdots+2^{n-1}x_1]$$Which is clearly a surjective homomorphism. To get the claim about first homology, it suffices to show this map has kernel equal to the image of $\partial$. The definition of $G$ is motivated by toying around with low dimensional cases. The case $n=0$ is rather trivial, so assume $n\ge1$ here.

If $Gx=0$ then $x_n+\cdots=2^n a_1$ for some integer $a_1$. Reducing modulo $2$, it’s clear $x_n=2a_n$ for some integer $a_n$. Then $2^{n-1}a_1=a_n+x_{n-1}+2x_{n-2}+\cdots$ hence $a_n+x_{n-1}$ is even, equal to $2a_{n-1}$ for some integer $a_{n-1}$. That is, $x_{n-1}=2a_{n-1}-a_n$, $x_n=2a_n$ so far. Now, dividing by $2$ again, $x_{n-2}+a_{n-1}+2x_{n-3}+\cdots=2^{n-2}a_n$. The same logic finds $a_{n-2}$ with $x_{n-2}=2a_{n-2}-a_{n-1}$. Do you see the pattern? Arguing like this inductively, we can produce integers $a_1,\cdots,a_n$ with $x_j=2a_j-a_{j+1}$ for $1\le j<n$ and $x_n=2a_n$.

What about the first coordinate? $x_0$ could be absolutely anything, but by defining $a_0:=x_0+a_1$ we get $\partial(a_0,\cdots,a_n)=x$, as desired. To see the other inclusion just unwind this induction in the other direction; $G\circ\partial\equiv0$ is not so hard to check if you understand the above process.

Now we are done.

FShrike
  • 40,125
0

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.

Rolf Hoyer
  • 12,399