Computing the Simplicial Homology of a Circle

March 29, 2010

Here we will compute the simplicial homology of a circle with coefficients in $\mathbb{Z}$. It's the simplest example I could come up with except for a point, which isn't at all interesting.

Since we are using simplicial homology, we will need a simplicial complex for the circle. Although there are infinitely many of these, let us choose a simple one: a triangle. That is, our triangle will have vertices (0-simplicies) A,B, and C. The 1-simplicies will be AB,BC, and CA. It looks like this:

Since we have 0-simplicies and 1-simplicies, we have two abelian groups ($\mathbb{Z}$-modules) in our associated chain complex. That is, $C_1 = \mathbb{Z}(AB,BC,CA)$, which I denote to mean the free Abelian group generated by our 1-simplicies, and $C_0 = \mathbb{Z}(A,B,C)$, which denotes the free Abelian group generated by $A,B,C$.

Remember the formula for the boundary map $\partial$? In general it is expressed by

where the $\hat{x}$ notation means the $x$ is excluded from the summation. It's not difficult to see that this map is indeed a group homomorphism, and that $\partial_{n-1}\partial_n = 0$. Thus we get a complex of Abelian groups, which in our case looks like this:

where the map $C_1\rightarrow C_0$ is just $\partial_1$, etc. If you work out this formula for our example, we get:

• $\partial_1(AB) = B - A$,
• $\partial_2(BC) = C - B$, and
• $\partial_3(CA) = A - C$.

The first homology is $H_1(S_1) = \ker \partial_1/\text{im} \partial_2$. However, the image of $\partial_2$ is just {0} since $C_2$ has no generators! So we just have to compute the kernel of $\partial_1$. For this example it's not hard to just look and see that $-AB -BC + CA\in C_1$ generates the kernel, but for more complicated examples I suggest using a matrix. We can use familiar linear algebra techniques since any rational solution we get for the nullspace can be converted into a (smallest) integer solution. Thus we just write the matrix of $\partial_1$ in the basis we are given:

We can use Gaussian elimination to find the nullspace of this matrix, which will confirm the intuition that $-AB-BC+CA$ is a basis of it, so we have $H_1(S^1) \cong \mathbb{Z}(-AB-BC+CA) \cong \mathbb{Z}$. This is the first homology of $S^1$!

Now that we've seen the first homology and that the kernel is spanned by a single element, we can look at what we have and can guess that the zeroth homology is probably 1-dimensional as well. But let's compute it to make sure. What is the kernel of $\partial_0$? Since the codomain of this map is just 0 the kernel is the whole space $C_0$.

Earlier we computed the matrix of our homomorphism $\partial_1$, if you took the time to row-reduce it, you alread have the image, which is spanned by the nonzero rows in the row-reduced matrix! Thus computing the kernel of $\partial_1$ also gives us the image for free. Here are the spanning vectors: $\{ [-1,0,1],[1,-1,0]\}$. If we now define the group homomorphism $\alpha:C_0\rightarrow\mathbb{Z}$ by $\alpha(z_1A+z_2B + z_3C) = z_1 + z_2 + z_3$, then you should verify that

• The image of $\alpha$ is $\mathbb{Z}$, and
• The kernel of $\alpha$ is exactly the image of $\partial_1$.

From this we can conclude from the first isomorphism theorem that $H_0(S^1)\cong \mathbb{Z}$. This was expected since if $X$ is a path-connected space then $H_0(X)$ where coefficients are taken in $\mathbb{Z}$ is always isomorphic to $\mathbb{Z}$.

For more information there is a nice chapter in Crossley's "Essential Topology" (introductory), or in Kozlov's book, "Combinatorial Algebraic Topology" (more advanced).