# Determining $1$-th simplicial homology group of the Möbius strip.

Let $$\underline {\mathcal M}$$ be the underlying s$$\Delta$$-set of the Möbius strip. I am trying to compute the simplicial homology groups of $$\underline {\mathcal M}.$$ I find that the $$0$$-th simplicial homology group $$H_0^{\Delta} \left (\underline {\mathcal M} \right ) \cong \Bbb Z$$ and the $$n$$-th simplicial homology group $$H_n^{\Delta} \left (\underline {\mathcal M} \right ) = 0,$$ for $$n \geq 2.$$ While computing the $$1$$-th simplicial homology group $$H_1^{\Delta} \left (\underline {\mathcal M} \right )$$ I find that $$H_1^{\Delta} \left (\underline {\mathcal M} \right ) = \langle a - b + c, d \rangle / \langle a + b -d, a - c + d \rangle.$$ But I can't able to simplify it further. Would anybody please help me in this regard?

EDIT $$:$$

• I think it's $\Bbb Z^2.$ Because we have $a + b = d$ and $a - c = -d$ and $2a = c - b.$ So $\langle a - b + c, d \rangle = \langle 3a, a + b \rangle \cong \mathbb Z^2,$ since $3a$ and $a + b$ are linearly independent. Commented May 27, 2021 at 7:14
• Could you elaborate on how you got the generators of $Z_1$? I can come up with a triangulation that yields the generators of $B_1$ you wrote down, but then $a-b+c$ is not a cycle. Commented May 27, 2021 at 7:15
• @Christoph is my triangulation correct? I have added a picture. I have found two $0$-simplices, four $1$-simplices and two $2$-simplices in the triangulation of $\underline {\mathcal M}.$ Commented May 27, 2021 at 7:22
• Yes, but $\partial(a-b+c)=3v-3w\neq 0$. You should get $Z_1=\langle d,a+b,a-c\rangle$. Commented May 27, 2021 at 7:22
• I have computed the face maps $d_0$ and $d_1.$ I have found that $$d_0 (a) = v, d_0(b) = w, d_0(c) = v, d_0(d) = w$$ and $$d_1 (a) = w, d_1(b) = v, d_1(c) = w, d_1(d) = w.$$ Are they correct @Christoph? Then the map $d^1 : \mathbb Z \{a,b,c,d\} \longrightarrow \mathbb Z \{v,w\}$ is defined as $$d^1 : = d_0 - d_1.$$ Commented May 27, 2021 at 7:29

The matrix of $$d^1\colon \mathbb Z\{a,b,c,d\}\to\mathbb Z\{v,w\}$$ is obtained from your triangulation as $$\begin{pmatrix} 1 & -1 & 1 & 0 \\ -1 & 1 & -1 & 0\end{pmatrix},$$ which has rank $$1$$ and hence the kernel is free of rank $$3$$. It is generated by $$d$$, $$a+b$$ and $$a-c$$. Hence, we get $$H_1 = \frac{\langle d,a+b,a-c\rangle}{\langle a+b-d,a-c+d\rangle} = \frac{\langle d,a+b-d,a-c+d\rangle}{\langle a+b-d,a-c+d\rangle} \cong\langle d\rangle\cong\mathbb Z.$$
• Oh! My bad. I have now computed. If $pa + qb + rc + sd \in \text {ker} d ^1.$ Then I find that $p-q + r = 0.$ Hence $p = q - r.$ That will give us $\text {ker} (d ^1) = \langle a + b, c - a, d \rangle.$ Sorry for my horrible algebraic manipulation. Commented May 27, 2021 at 7:51
• Can you give a triangulation of the genus-$2$-surface which makes the simplicial homology groups easier to compute. I have found one such by joining the opposite edges of the octagon whose edges are identified in a certain way. But this figure contains one $0$-simplex, twelve $1$-simplices and eight $2$-simplices which makes things complicated. Can it be simplified in a certain way? This problem is given in our lecture notes as one of the exercises but I am unable to figure it out. Commented May 27, 2021 at 17:15