# On the entries of a matrix representation for a boundary operator of a persistence module

In equation 6 of Computing Persistent Homology (page 8), the authors put forward the following identity: $$\deg \hat{e_i}+\deg M_k (i,j)=\deg e_j$$ Where $$\hat{e_i}$$ and $$e_j$$ are elements of homogeneous bases for the space of $$(k-1)$$-chains and $$k$$-chains, respectively, and $$M_k$$ is a matrix representation of $$\partial_k$$ with respect to these bases.

This identity is intuitive and makes it easy to come up with matrix representations for the boundary operators. However, it's the only part of the paper that I've never felt fully comfortable with. The authors treat this identity as obvious, saying that the reader may verify it "using this example as intuition" (referring to an example introduced at the beginning of the paper). I'm not sure if this means verify that the identity holds in the case of the example or verify that it holds in general. I'm not sure how to approach the latter.

## 1 Answer

One can certainly verify Equation 6 on several instances, but this holds true in general as a consequence of Theorem 3.1 (Correspondence Theorem) on page 7.

Indeed, $\mathcal{M}_k \Doteq \{C_k^{l},\, f^l\}$ and $\mathcal{M}_{k - 1} \Doteq \{C_{k - 1}^{l},\, f^l\}$ are persistent modules in the sense of Definition 3.2 on page 7 and $\partial_k: \mathcal{M}_k \rightarrow \mathcal{M}_{k - 1}$ is a homomorphism in the category of persistence modules. By the Correspondence Theorem, $\alpha(\partial_k): \alpha(\mathcal{M}_k) \rightarrow \alpha(\mathcal{M}_{k - 1})$ is homomorphism of $\mathbb{N}$-graded $R[t]$-modules, where $R$ is the ground ring of homology coefficients and $\alpha$ is the correspondence of Section 3.1 (this is an equivalence of categories). The matrice $M_k$ is the matrix of $\alpha(\partial_k)$ with respect to some homogeneous bases $\{e_j\}$ and $\{\hat{e}_i\}$ (see definition in Lemma 1) of $\alpha(\mathcal{M}_k)$ and $\alpha(\mathcal{M}_{k - 1})$ respectively.

The fact that the coefficients of $M_k$ are unique and homogeneous is explained in Lemma 1 below, as well as the fact that $\alpha(\mathcal{M}_{k - 1})$ is free over $R[t]$ with basis $\{\hat{e}_i\}$ in the category of (non-graded) $R[t]$-modules.

By definition, $\alpha(\partial_k)(e_j) = \sum_i M_k(i, j)\hat{e}_i$. Since $\alpha(\partial_k)$ is a graded $R[t]$-module homomorphism, it maps a homogeneous element to a homogeneous element with the same degree. Therefore, every non-zero term in the previous sum is homogeneous of degree $\deg e_j$, that is $$\deg e_j = \deg M_k(i, j)\hat{e}_i = \deg M_k(i, j) + \deg\hat{e}_i$$ whenever $M_k(i,j) \neq 0$.

Let us explain now why $M_k$ is well-defined by unique homogeneous coefficients. Let $\Lambda = \Lambda_0 \oplus \Lambda_1 \oplus \cdots$ be an $\mathbb{N}$-graded ring and let $f:M \rightarrow N$ an $\mathbb{N}$-graded homomorphism of $\mathbb{N}$-graded $\Lambda$-modules with $M$ and $N$ finitely generated. The coefficients $\lambda_{ij}$ of a matrix that represents $f$ with respect to sets of generators $\{e_j\}$ and $\{\hat{e}_i\}$ of minimal cardinalities are not necessarily unique but we can always assume that they are homogeneous (consider the homogeneous term of degree $\deg e_j$ in $\lambda_{ij}\hat{e}_i$ when it exists). In the context of persistent homology, uniqueness of coefficients is guaranteed by

Lemma 1 Let $\Lambda = R[t]$ where $R$ is commutative ring with identity. Let $N = \alpha(\mathcal{N})$ where $\mathcal{N} = \{N^j, \varphi^j\}$ is a persistent module over $R$. Assume that $N$ is finitely generated over $\Lambda$, that $\varphi^i$ is injective, $\varphi^j(N^j)$ and $N^{j + 1}/\varphi^j(N^j)$ are free over $R$ for every $j$. Then there is a generating set $\{\hat{e}_i\}$ of minimal cardinality whose elements are homogeneous. In addition, $N$ is free over $\Lambda$ with basis $\{\hat{e}_i\}$ in the category of (non-graded) $\Lambda$-modules for any such generating set. We call such a set a homogeneous basis of $N$.

Beware: the module $N$ of Lemma 1 is not necessarily a free $\mathbb{N}$-graded module over $\Lambda$.

Proof of Lemma 1. Since $N$ is finitely generated over $\Lambda$, the existence of $\{\hat{e}_i\}$ is immediate. Because of our assumptions on $N$, the set of elements of the form $t^l\hat{e}_k$ of degree $j$ is a generating set of $N^j$ over $R$ with minimal cardinality. Since $N^j$ is free over $R$, these elements form an $R$-basis of $N^j$ [1, Theorem 2.4]. Now take coefficient $\lambda_i \in \Lambda$ such that $\sum_i \lambda_i \hat{e}_i = 0$. By splitting $\lambda_i \hat{e}_i$ in a sum of homogeneous terms and by using the previous fact and the injectivity of $\varphi^j$, it readily follows that $\lambda_i = 0$ for every $i$.

[1] H. Matsumura, "Commutative Ring Theory", 1989.