Chain complex and cohomology

Question

I am trying to understant some ddetails from physical article.

In article there are two statements, which I wanna to understand.

1. We have complex $$ \mathcal{C}_* = \mathbb{Z}[1]\oplus \mathbb{Z}[2] $$ With differential operator: $$ \Delta = \begin{pmatrix} 0&1\\ 0&0 \end{pmatrix} $$ Cohomologies: $$ H_*(\mathcal{C}_*, \Delta) \cong 0 $$

2. Second example: $$ \Delta = \begin{pmatrix} 0& 0 & 0 &1\\ 0& 0 & 0 &1\\ 0& 0 & 0 &1\\ 0&0 & 0 & 0 \end{pmatrix} $$ $$ \mathcal{C}_* = \mathbb{Z}[1]\oplus \mathbb{Z}[1]\oplus \mathbb{Z}[1]\oplus \mathbb{Z}[2] $$ $$ H_*(\mathcal{C}_*, \Delta) \cong \mathbb{Z}[1] \oplus \mathbb{Z}[1] $$

3. There is also third example: $$ \Delta = \begin{pmatrix} 0& 1 & 1 &0\\ 0& 0 & 0 &1\\ 0& 0 & 0 &1\\ 0&0 & 0 & 0 \end{pmatrix} $$ $$ \mathcal{C}_* = \mathbb{Z}[0]\oplus \mathbb{Z}[1]\oplus \mathbb{Z}[1]\oplus \mathbb{Z}[2] $$ $$ H_*(\mathcal{C}_*, \Delta) \cong 0 $$

I am not familar with such calculation. I will be very glad if somebody clarify such calculation and give me answer with all details .

Answer 1

If I understand your notation right, the first complex is $$\cdots\to 0\to\Bbb Z\stackrel{\phi}\to\Bbb Z\to0\to\cdots$$ where $\phi(a)=a$ and your second complex is $$\cdots\to 0\to\Bbb Z^3\stackrel{\psi}\to\Bbb Z\to0\to\cdots$$ where $\psi(a,b,c)=a+b+c$.

I'll look at the second example. The cohomology at a position in the sequences is the kernel of the map emerging by the image of the map entering. Then $H^1(C^*)$ is the kernel of $\psi$, factored by the image of the zero map, so really just the kernel of $\psi$. This is a free subgroup of $\Bbb Z^3$ generated by $(1,-1,0)$ and $(0,1,-1)$ so is isomorphic to $\Bbb Z^2$.

Also $H^2(C^*)$ is the kernel of the zero map (so $\Bbb Z$), factored by the image $\psi$. But $\psi$ is surjective, so $\psi$ has image $\Bbb Z$ and so $H^2(C^*)$ vanishes.