Continuum limit for a discrete Hodge–Dirac operator on square lattices

Abstract

We study the continuum limit for Dirac–Hodge operators defined on the n dimensional square lattice $h\mathbb {Z}^n$ as h goes to 0. This result extends to a first order discrete differential operator the known convergence of discrete Schrödinger operators to their continuous counterpart. To be able to define such a discrete analog, we start by defining an alternative framework for a higher–dimensional discrete differential calculus. We believe that this framework, that generalize the standard one defined on simplicial complexes, could be of independent interest. We then express our operator as a differential operator acting on discrete forms to finally be able to show the limit to the continuous Dirac–Hodge operator.

Appendix A. Proof of Proposition 2.7

Because the definition of $\partial$ involves $({}_{}{i}{\hat{s}})^*$ we need a better understanding of how the involution and restriction are related before attempting to prove Proposition 2.7.

Lemma A.1

Let $\hat{s}$ be given by (9). Let $1\le i_0 \le j$ and $1\le i_1 \le j-1$. Then the following statements hold:

• (1) ${}_{}{i_0}{(\hat{s}^*})=({}_{}{j-i_0+1}{\hat{s}})^*$

• (2) $({}_{}{i_0}{(\hat{s}^*))^*}=({}_{}{j-i_0+1}{\hat{s}})$

• (3)
  $$\begin{aligned} {}_{}{i_1}{({}_{}{i_0}{\hat{s}})}={\left\{ \begin{array}{ll} {}_{}{(i_0-1)}{({}_{}{i_1}{\hat{s}})} &{}\text { if }i_1<i_0\ ;\ \\ {}_{}{i_0}{({}_{}{(i_1+1)}{\hat{s}})} &{}\text { if }i_0\le i_1\ . \end{array}\right. } \end{aligned}$$

Proof

Let $1\le i\le j-1$, then on one hand we have

$$\begin{aligned} {}_{}{i_0}{(\hat{s}^*})(i)=&{\left\{ \begin{array}{ll} \hat{s}^*(i)\quad &{}i<i_0\\ \hat{s}^*(i+1)\quad &{} i_0\le i \end{array}\right. }\\ =&{\left\{ \begin{array}{ll} \hat{s}(j-i+1)\quad &{}i<i_0\\ \hat{s}(j-i)\quad &{} i_0\le i \end{array}\right. } \end{aligned}$$

while in the other we can compute

$$\begin{aligned} ({}_{}{j-i_0+1}{\hat{s}})^*(i)=&{}_{}{j-i_0+1}{\hat{s}}(j-i)\\ =&{\left\{ \begin{array}{ll} \hat{s}(j-i)\quad &{}j-i<j-i_0+1\\ \hat{s}(j-i+1)\quad &{} j-i_0+1\le j- i \end{array}\right. }\\ =&{\left\{ \begin{array}{ll} \hat{s}(j-i)\quad &{} i_0\le i\\ \hat{s}(j-i+1)\quad &{}i<i_0 \end{array}\right. } \end{aligned}$$

which proves Item 1. Further, Item 2 is a direct consequence of Item 1. To prove Item 3 we fix $1\le i_0\le j$ and $1\le i_1\le j-1$ and first consider the case $i_1<i_0$. We first compute

$$\begin{aligned} {}_{}{i_1}{({}_{}{i_0}{\hat{s}})}(i)=&{\left\{ \begin{array}{ll} {}_{}{i_0}{\hat{s}}(i)\quad &{}i<i_1\\ {}_{}{i_0}{\hat{s}}(i+1)\quad &{} i_1\le i \end{array}\right. }\\ =&{\left\{ \begin{array}{ll} \hat{s}(i)\quad &{}i<i_1\\ \hat{s}(i+1)\quad &{}i_1\le i\le i_0-2\\ \hat{s}(i+2)\quad &{}i_0-1\le i \end{array}\right. } \end{aligned}$$

and then compare it with

$$\begin{aligned} {}_{}{i_0-1}{({}_{}{i_1}{\hat{s}})}(i) =&{\left\{ \begin{array}{ll} {}_{}{i_1}{\hat{s}}(i)\quad &{}i<i_0-1\\ {}_{}{i_1}{\hat{s}}(i+1)\quad &{} i_0-1\le i \end{array}\right. } \end{aligned}$$

which will coincide with the previous expression and hence show that Item 3 holds for $i_1<i_0$. The computations for $i_0\le i_1$ follow the same structure and are omitted. $\square$

Proof of Proposition 2.7

Let us first show that $\partial (\overline{s})=\overline{\partial (s)}$. By (10) and using Items 1 and 2 from Lemma A.1 we can check that:

$$\begin{aligned} \partial (\overline{s}) =&\cup _{i=1}^j \{(-1)^{j-i} (\lfloor \overline{s} \rfloor ;{}_{}{i}{(\hat{s}^*)}) \}\bigcup \cup _{i=1}^j\{(-1)^{i} (\lceil \overline{s} \rceil ;({}_{}{i}{(\hat{s}^*)})^*) \}\\ =&\cup _{i=1}^j \{(-1)^{j-i} (\lceil s \rceil \lfloor ;({}_{}{j-i+1}{\hat{s}})^*) \}\bigcup \cup _{i=1}^j\{(-1)^{i} (\lfloor s \rfloor ;{}_{}{j-i+1}{\hat{s}}) \}\\ =&\cup _{m=1}^j \{(-1)^{m-1} (\lceil s \rceil \lfloor ;({}_{}{m}{\hat{s}})^*) \}\bigcup \cup _{m=1}^j\{(-1)^{j-m+1} (\lfloor s \rfloor ;{}_{}{m}{\hat{s}}) \}\\ =&\cup _{m=1}^j \overline{\{(-1)^{m} (\lceil s \rceil \lfloor ;({}_{}{m}{\hat{s}})^*) \}}\bigcup \cup _{m=1}^j\overline{\{(-1)^{j-m} (\lfloor s \rfloor ;{}_{}{m}{\hat{s}}) \}}=\overline{\partial (s)}\ . \end{aligned}$$

It remains to show that $\partial (\partial (s))=\overline{\partial (\partial (s))}$. For this we introduce the following notation for $1\le i \le j$ and $s\in X^j$

$$\begin{aligned} A_i(s):=(-1)^{j-i}(\lfloor s \rfloor ; {}_{}{i}{\hat{s}})\quad B_i(s):=(-1)^{i}(\lceil s \rceil :({}_{}{i}{\hat{s}})^*)\ . \end{aligned}$$

It follows that

$$\begin{aligned} \partial (\partial (s))=\cup _{l=1}^{j-1}\cup _{i=1}^{j}\{A_l(A_i(s))\cup B_l(A_i(s))\cup A_l(B_i(s))\cup B_l(B_i(s))\}\ . \end{aligned}$$

Let us now consider $s\in X^j$. Without lose of generality we will assume that $s=(\lfloor s \rfloor ; \delta _1,\dots ,\delta _j)$. From the proof of Lemma A.1 on can see that if $i_1<i_0$, then ${}_{}{i_1}{({}_{}{i_0}{\hat{s}})}$ skips the $i_0$ and $i_1$ terms. We start by identifying the faces $r \in \partial (\partial (s))$ that satisfies $\lfloor r \rfloor = \lfloor s \rfloor$. There are given for $1\le i\le j$ and $1\le l\le j-1$ such that $j-i$ is even and $j-l$ is odd by

$$\begin{aligned} A_l(A_i)(s)=A_l((-1)^{j-i}(\lfloor s \rfloor ; {}_{}{i}{\hat{s}})=(-1)^{j-1-l}(\lfloor s \rfloor ; {}_{}{l}{({}_{}{i}{\hat{s}})})= {\left\{ \begin{array}{ll}(\lfloor s \rfloor ; {}_{}{l}{({}_{}{i}{\hat{s}})}) &{}\text { if }l<i\\ (\lfloor s \rfloor ; {}_{}{i}{({}_{}{l+1}{\hat{s}})}) &{}\text { if }i\le l \end{array}\right. }\ ; \end{aligned}$$

and for $1\le i\le j$ and $1\le l\le j-1$ such that $j-i$ is odd and l is even by

$$\begin{aligned} B_l(A_i)(s)=B_l((-1)^{j-i}(\lfloor s \rfloor ; {}_{}{i}{\hat{s}})=&B_l((\lceil s \rceil -\delta _i ; ({}_{}{i}{\hat{s}})^*)\\ =&(-1)^{l}(\lfloor s \rfloor ; ({}_{}{l}{(({}_{}{i}{\hat{s}})^*}))^*)\\ =&(\lfloor s \rfloor ; {}_{}{j-l}{({}_{}{i}{\hat{s}})})= {\left\{ \begin{array}{ll}(\lfloor s \rfloor ; {}_{}{j-l}{({}_{}{i}{\hat{s}})}) &{}\text { if }j-l<i\\ (\lfloor s \rfloor ; {}_{}{i}{({}_{}{j-l+1}{\hat{s}})}) &{}\text { if }i\le j-l \end{array}\right. }\ . \end{aligned}$$

In both cases we have faces that start at $\lfloor s \rfloor$ but have two directions less than s. To check that we have all possible $\tfrac{j*(j-1)}{2}$ combinations we first consider the case $j=2k$ for $k\in \mathbb {N}$. We have that the faces $r\in \partial (\partial (s))$ that satisfies $\lfloor r \rfloor =\lfloor s \rfloor$ are given by

$$\begin{aligned}&\left\{ \cup _{p=1}^k \cup _{m=1}^k A_{2p-1}(A_{2m}(s))\right\} \bigcup \left\{ \cup _{p=1}^{k-1}\cup _{m=1}^k B_{2p}(A_{2m-1}(s))\right\} \\&=\left\{ \cup _{p=1}^k \cup _{m=p}^k (\lfloor s \rfloor ; {}_{}{2p-1}{({}_{}{2m}{\hat{s}})}) \right\} \bigcup \left\{ \cup _{p=1}^k \cup _{m=1}^{p-1} (\lfloor s \rfloor ; {}_{}{2m}{({}_{}{2p}{\hat{s}})}) \right\} \\&{\quad }\bigcup \left\{ \cup _{p=1}^{k-1}\cup _{m=k-p+1}^{k}(\lfloor s \rfloor ; {}_{}{2k-2p}{({}_{}{2m-1}{\hat{s}})})\right\} \bigcup \left\{ \cup _{p=1}^{k-1}\cup _{m=1}^{k-p} (\lfloor s \rfloor ;{}_{}{2m-1}{({}_{}{2k-2p+1}{\hat{s}})})\right\} \\&=\left\{ \cup _{p=1}^k \cup _{m=p}^k (\lfloor s \rfloor ; {}_{}{2p-1}{({}_{}{2m}{\hat{s}})}) \right\} \bigcup \left\{ \cup _{p=1}^k \cup _{m=1}^{p-1} (\lfloor s \rfloor ; {}_{}{2m}{({}_{}{2p}{\hat{s}})}) \right\} \\&{\quad }\bigcup \left\{ \cup _{p=1}^{k-1}\cup _{m=p+1}^{k}(\lfloor s \rfloor ; {}_{}{2p}{({}_{}{2m-1}{\hat{s}})})\right\} \bigcup \left\{ \cup _{p=1}^{k-1}\cup _{m=1}^{p} (\lfloor s \rfloor ;{}_{}{2m-1}{({}_{}{2p+1}{\hat{s}})})\right\} \\&=\left\{ \cup _{l=1}^{j-1} \cup _{i=l+1}^j (\lfloor s \rfloor ; {}_{}{l}{({}_{}{i}{\hat{s}})}) \right\} . \end{aligned}$$

Similar computations work for $j=2k+1$ as well for the treatment of the other types of faces. We do not enter into the detail of each one but we will discuss where the terms come from with the aid of the following table.

	r	i such that	l such that	$\lfloor r \rfloor$
1	$A_l(A_i(s))$	$j-i$ is even	$j-l$ is odd	$\lfloor s \rfloor$
2	$B_l(A_i(s))$	$j-i$ is odd	l is even	$\lfloor s \rfloor$
3	$A_l(A_i(s))$	$j-i$ is even	$j-l$ is even	$\lceil s \rceil - \delta _{1} - \delta _{2}$
4	$B_l(A_i(s))$	$j-i$ is odd	l is odd	$\lceil s \rceil - \delta _{1} - \delta _{2}$
5	$A_l(A_i(s))$	$j-i$ is odd	$j-l$ is even	$\lfloor s \rfloor + \delta _{1}$
6	$B_l(A_i(s))$	$j-i$ is even	l is odd	$\lfloor s \rfloor + \delta _{1}$
7	$A_l(B_i(s))$	i is odd	$j-l$ is odd	$\lfloor s \rfloor + \delta _{1}$
8	$B_l(B_i(s))$	i is even	l is even	$\lfloor s \rfloor + \delta _{1}$
9	$A_l(A_i(s))$	$j-i$ is odd	$j-l$ is odd	$\lceil s \rceil - \delta _{1}$
10	$B_l(A_i(s))$	$j-i$ is even	l is even	$\lceil s \rceil - \delta _{1}$
11	$A_l(B_i(s))$	i is odd	$j-l$ is even	$\lceil s \rceil - \delta _{1}$
12	$B_l(B_i(s))$	i is even	l is odd	$\lceil s \rceil - \delta _{1}$
13	$A_l(B_i(s))$	i is even	$j-l$ is even	$\lfloor s \rfloor + \delta _{1} + \delta _{2}$
14	$B_l(B_i(s))$	i is odd	l is odd	$\lfloor s \rfloor + \delta _{1} + \delta _{2}$
15	$A_l(B_i(s))$	i is even	$j-l$ is odd	$\lceil s \rceil$
16	$B_l(B_i(s))$	i is odd	l is even	$\lceil s \rceil$

The first two rows corresponds to the calculations we have already done. Rows 3 and 4 will give rise to the opposite faces of the first four rows. In all, the first four rows are all the faces that contain $\lfloor s \rfloor$, either as $\lfloor r \rfloor$ or $\lceil r \rceil$. The next 8 rows describe the faces that contain a vertex at distance 1 from $\lfloor s \rfloor$. Again we have that the first rows, rows 5 to 8, are positively oriented while the second half, rows 9 to 12 correspond to theirs opposite faces. Finally, rows 13 to 16 are symmetric to the the first 4 rows but its faces contain $\lceil s \rceil$ instead of $\lfloor s \rfloor$. $\square$