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Maybe my question has an answer here, but I don't understand that. I will pose mine differently. The following definitions are from [1, Sec 4.1].

Consider a simplicial complex. The $k$-th chain group $C_k$ is the set of all $k$-chains $$\sum_i n_i \, \text{[oriented $k$-simplex $i$]}$$ with integer coefficients $n_i$. The boundary homomorphism $\partial_k : C_k \to C_{k-1}$ between chain groups is defined by $$ \partial_k [v_0, \ldots, v_k] = \sum_i (-1)^i [v_0, \ldots, \hat{v}_i, \ldots, v_n]. $$

The $k$-th homology group is $$H_k := \ker \partial_k / \operatorname{im} \partial_{k+1}.$$

As a finitely generated group, it is isomorphic to a direct sum of primary cyclic groups and infinite cyclic groups, or it decomposes into into the torsion subgroup and the free subgroup. The Betti number $\beta_k$ is the rank of the free subgroup.

Let $A_k \in \mathbb{R}^{m \times n}$ be the matrix representation of $\delta_k$ with respect to some basis. Consider it as an operator between the vector spaces $\mathbb{R}^n$ and $\mathbb{R}^m$ (over the reals). We can compute $$\tilde{\beta}_k := \dim \ker A_k - \dim \operatorname{im} A_{k+1}.$$ I think this is the dimension of the homology group obtained when in the chain group we admit with real coefficients(?).

How are $\beta_k$ and $\tilde{\beta}_k$ related?

[1] Zomorodian, Compuational topology, Notes, 2009.

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I'll try showing $\beta_k=\tilde\beta_k$ by expanding out the relevant portions of the proof of the universal coefficient theorem.

The first thing is let's define $A_k$ instead to be the $\mathbb{Z}$-valued matrix for $\partial_k:C_k\to C_{k-1}$, with respect to the basis of oriented simplices. Assume the simplicial complex has only finitely many simplices for this matrix to be finite-dimensional. Let $\tilde{A}_k$ be the same matrix, but thought of as being $\mathbb{R}$-valued. So, using $\operatorname{rank}$ to mean free rank, \begin{align} \beta_k&=\operatorname{rank}(\ker A_k/\operatorname{im} A_{k+1})\\ \tilde{\beta}_k&=\dim(\ker \tilde{A}_k/\operatorname{im} \tilde{A}_{k+1})=\dim\ker\tilde{A}_k-\dim\operatorname{im}\tilde{A}_{k+1} \end{align} (where the latter equality is just the rank-nullity theorem).

A main way to change coefficients in algebra is tensor products. $\mathbb{Z}^n\otimes\mathbb{R}\cong\mathbb{R}^n$. If you are not familiar with the concept, just take this axiomatically for now. One can say that $\tilde{A}_k=A_k\otimes \mathbb{R}$ to represent the change of the coefficient ring. Tensor product properties:

  • $\mathbb{Z}\otimes\mathbb{R}\cong\mathbb{R}$.
  • $(\mathbb{Z}/n\mathbb{Z})\otimes\mathbb{R}\cong 0$.
  • If $A,B$ are abelian groups, $(A\oplus B)\otimes\mathbb{R}\cong (A\otimes\mathbb{R})\oplus(B\otimes\mathbb{R})$.
  • So: If $A$ is a finitely generated abelian group, $A\otimes\mathbb{R}\cong \mathbb{R}^{\operatorname{rank} A}$.

One part to this is that we have a chain complex with $\mathbb{Z}$-coefficients $$\cdots\xrightarrow{A_{k+1}} \mathbb{Z}^{n_k}\xrightarrow{A_k} \mathbb{Z}^{n_{k-1}}\xrightarrow{A_{k-1}}\cdots$$ which can be converted to $\mathbb{R}$-coefficients by tensoring with $\mathbb{R}$ to get $$\cdots\xrightarrow{\tilde A_{k+1}} \mathbb{R}^{n_k}\xrightarrow{\tilde A_k} \mathbb{R}^{n_{k-1}}\xrightarrow{\tilde A_{k-1}}\cdots$$

Now, the difficulty is to find a way to relate the homology groups for each of these chain complexes. Let $H_k$ denote the $k$th homology group of the $C_k$ complex. By definition of $H_k$, there are short exact sequences $$0\to \operatorname{im} A_{k+1}\hookrightarrow \ker A_k\to H_k\to 0$$ Since $\operatorname{im} A_{k+1}$ and $\ker A_k$ are free abelian groups, $(\operatorname{im} A_{k+1})\otimes\mathbb{R}\cong \operatorname{im}\tilde A_{k+1}$ and $(\ker A_k)\otimes\mathbb{R}\cong \ker \tilde A_k$, and the inclusion map tensored with $\mathbb{R}$ remains injective. This ends up meaning there is a short exact sequence $$0 \to \operatorname{im}\tilde A_{k+1} \hookrightarrow \ker \tilde A_k \to H_k\otimes \mathbb{R}\to 0$$

By the first isomorphism theorem, $H_k\otimes \mathbb{R}\cong \ker \tilde A_k / \operatorname{im}\tilde A_{k+1}$. This latter quotient is the definition of the homology with real coefficients, so $$\tilde\beta_k=\dim(H_k\otimes\mathbb{R})=\dim(\mathbb{R}^{\operatorname{rank} H_k})=\operatorname{rank} H_k=\beta_k$$

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  • $\begingroup$ Thanks for the tensor product bit, that'll definitely help. I don't understand the last equality: $\operatorname{rank} H_k = \beta_k$. Isn't it $\operatorname{rank} H_k = \beta_k + [\text{# of torsion subgroups}]$? But I think, using $(\mathbb{Z} / n \mathbb{Z}) \otimes \mathbb{R} \cong 0$, I can see... $\endgroup$ – user66081 Oct 18 '17 at 5:04
  • $\begingroup$ Here I'm meaning free rank rather than number of generators. Sorry for the confusion. $\endgroup$ – Kyle Miller Oct 18 '17 at 5:09
  • $\begingroup$ Ok, thanks for the clarification. The haze is lifting. $\endgroup$ – user66081 Oct 18 '17 at 5:10
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Better notation would be $A_k$ for what you just call $A$. Then indeed $\dim\ker A_k-\dim\text{Im}\,A_{k+1}$ is $\beta_k$. This follows from the Universal Coefficient Theorem for homology, but can probably be proved from first principles.

ADDED IN EDIT

I outline a more naive approach.

I'm assuming that the simplicial complex is finite. The matrix $A_k$ represents a map $\partial_k:C_k\to C_{k-1}$ as well as a linear map $\overline\partial_k:C_k\otimes\Bbb R\to C_{k-1}\otimes\Bbb R$. The groups $C_k$ are free Abelian. Therefore the kernel and image of $\partial_k$ are free Abelian groups. Each free Abelian group has a rank and so we may consider $\text{rk}\ker(\partial_k)$ and $\text{rk}\,\text{im}(\partial_k)$. Now these are the same as the nullity and rank of the linear map $\overline\partial_k$. This follows since the rank of an integer matrix is the same as the rank of the $\Bbb Z$-span of its rows (to be pretentious one might say that $\Bbb R$ is a flat $\Bbb Z$ modules). From this observation, $\beta_k=\tilde\beta_k$ is immediate.

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  • $\begingroup$ That is a sensible comment which I took to heart. Now, /how/ does $\beta_k = \tilde{\beta}_k$ follow from the Universal Coefficient Theorem? I don't understand the statement of the theorem. And I wrote first thing that I don't understand this answer. $\endgroup$ – user66081 Oct 18 '17 at 2:29
  • $\begingroup$ @user66081 The UCT is easy to apply in this case, as the Tor term vanishes, and we get $H_k(C;\Bbb R)\cong H_k(C)\otimes\Bbb R$. $\endgroup$ – Lord Shark the Unknown Oct 18 '17 at 17:51

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