# Convergence rate for homogenization of a nonlocal model with oscillating
coefficients

Li Lin111School of Mathematics and Statistics & Center for Mathematical
Sciences, Huazhong University of Science and Technology,Wuhan, 430074, China.
E-mail address<EMAIL_ADDRESS>& Jinqiao Duan222Corresponding author.
Department of Applied Mathematics, Illinois Institute of Technology, Chicago,
Illnois 60616, USA. E-mail address<EMAIL_ADDRESS>
###### Abstract

This letter deals with homogenization of a nonlocal model with Lévy-type
operator of rapidly oscillating coefficients. This nonlocal model describes
mean residence time and other escape phenomena for stochastic dynamical
systems with non-Gaussian Lévy noise. We derive an effective model with a
specific convergence rate. This enables efficient analysis and simulation of
escape phenomena under non-Gaussian fluctuations.

###### keywords:

Convergence rate, mean residence time, escape phenomena, Lévy noise, nonlocal
elliptic equations

###### MSC:

[2020] 34A08, 74Q10 , 41A25

††journal: Journal of Applied Mathematics Letters

##  1 Introduction

We consider the homogenization of the following nonlocal partial differential
equation

$\begin{cases}\mathcal{A}^{\epsilon}u_{\epsilon}(x)=f(x),\qquad x\in D,\\\
u_{\epsilon}|_{D^{c}}=g(x),\end{cases}$ (1)

arising in the study of escape phenomena of stochastic dynamical systems under
Lévy fluctuations [1]. Here $D$ is a bounded domain in $\mathbb{R}^{d}$. When
$f=-1$ and $g=0$, the solution of this equation is the mean residence time for
such a stochastic system in domain $D$. Specifically, the nonlocal operator
depends on a small positive scale parameter $\varepsilon$ and is defined by
$\mathcal{A}^{\epsilon}u=\frac{1}{2}\mathcal{D}(\Theta^{\epsilon}\mathcal{D}^{*}u),$
where the coefficient
$\Theta^{\epsilon}(x,z)=\Theta(\frac{x}{\epsilon},\frac{z}{\epsilon})$ is of
period $1$ in $(x,z)$ and for a certain positive constant $\lambda$,
$\frac{1}{\lambda}<\Theta^{\epsilon}(x,z)<\lambda.$

The nonlocal linear operator $\mathcal{D}$ and its adjoint operator
$\mathcal{D}^{*}$ are defined as follows[16]. Given the antisymmetric kernel
function $\gamma(x,z)=(z-x)\frac{1}{|z-x|^{\frac{2+d+\alpha}{2}}}$ (note that
$\gamma(z,x)=-\gamma(x,z)$), the nonlocal divergence $\mathcal{D}$ is defined
by
$\mathcal{D}(\beta)(x):=\int_{\mathbb{R}^{d}}(\beta(x,z)+\beta(z,x))\cdot\gamma(x,z)dz,$
$x\in D.$ The adjoint operator $\mathcal{D^{*}}$ of $\mathcal{D}$ is then
given by $\mathcal{D^{*}}(\phi)(x,z)=-(\phi(z)-\phi(x))\gamma(x,z)$ for
$x,z\in D.$

By the way, if $\Theta\equiv 1$, then we see a relation with the nonlocal
Laplace operator
$\frac{1}{2}\mathcal{D}\mathcal{D^{*}}=-(-\Delta)^{\alpha/2}.$ The nonlocal
Laplace operator $(-\Delta)^{\alpha/2}$ is defined by

$(-\Delta)^{\alpha/2}u(x)=\int_{\mathbb{R}^{d}\backslash\\{x\\}}\frac{u(z)-u(x)}{|z-x|^{d+\alpha}}dz,$

where the integral is in the sense of Cauchy principal value, and it is the
generator of a symmetric $\alpha$-stable Lévy motion [2]

##  2 Main result

###  2.1 Function spaces

In this subsection, we define the following two important spaces. One is the
space $\mathcal{V}^{D}$, which is the counterpart of the classical Sobolev
space $H^{1}(D)$. The other is the space $\mathcal{X}^{D}$, as an analogue of
the usual trace space $H^{1/2}(\partial D).$

We set $\nu(x,y)=\gamma^{2}(x,z)$. we introduce a quadratic form[3, 4, 5] for
$u:\mathbb{R}^{d}\rightarrow\mathbb{R}$

$\mathcal{E}_{D}(u,u)=\frac{1}{2}\int_{\mathbb{R}^{d}\times\mathbb{R}^{d}\backslash
D^{c}\times D^{c}}(u(x)-u(z))^{2}\nu(x,z)dxdz.$

Now we define the corresponding Sobolev space [7]:

$\mathcal{V}^{D}=\\{u:\mathbb{R}^{d}\rightarrow\mathbb{R}\;\text{such
that}\;\mathcal{E}_{D}(u,u)<\infty\\}.$

We also denote

$\mathcal{V}^{D}_{0}=\\{u\in\mathcal{V}^{D};u=0\;\;\;a.e.\;on\;D^{c}\\}.$

More precisely, $\mathcal{V}^{D}_{0}=H_{0}^{\alpha/2}(D)$ from [8, Theorem
3.33]. The space $H_{0}^{\alpha/2}(D)$ is the standard fractional Sobolev
space. For $s<0$, we denote $H^{s}(D)$ as the dual space of $H_{0}^{-s}(D)$.
It is clear that $\mathcal{V}^{D}$ is a Hilbert subspace of $L^{2}(D)$, with
the induced norm
$||u||^{2}_{\mathcal{V}^{D}}=||u||^{2}_{L^{2}(D)}+\mathcal{E}_{D}(u,u).$

We will use the notation $\mathbb{T}^{d}$ to denote the $d$-dimensional unit
torus. The space $H^{\alpha/2}_{\\#}(\mathbb{T}^{d})$ of $1$-periodic
functions $u\in H^{\alpha/2}$ such that $\int_{\mathbb{T}^{d}}u(y)dy=0$ will
be interest in this study. Provided with the norm,
$\|u\|_{H^{\alpha/2}_{\\#}(\mathbb{T}^{d})}=\left(\int_{\mathbb{T}^{d}}\int_{\mathbb{T}^{d}}\frac{|u(y)-u(\eta)|^{2}}{|y-\eta|^{d+\alpha}}dyd\eta\right)^{\frac{1}{2}}.$

We let

$\mathcal{E}_{D}(u,v)=\frac{1}{2}\int_{\mathbb{R}^{d}\times\mathbb{R}^{d}\backslash
D^{c}\times D^{c}}(u(x)-u(z))(v(x)-v(z))\nu(x,z)dxdz,$

and

$a^{\epsilon}(u,v)=\frac{1}{2}\int_{\mathbb{R}^{d}\times\mathbb{R}^{d}\backslash
D^{c}\times D^{c}}\Theta^{\epsilon}(x,z)(u(x)-u(z))(v(x)-v(z))\nu(x,z)dxdz,$

if the integrals are absolutely convergent, in particular for
$u,v\in\mathcal{V}^{D}.$

By a solution of (1) we mean a weak solution, which is defined as every
function $u_{\epsilon}\in\mathcal{V}^{D}$ equal to $g$ a.e. on $D^{c}$ such
that for every $\phi\in\mathcal{V}^{D}_{0},$
$a^{\epsilon}(u_{\epsilon},\phi)=\int f\phi dx,$ this integral is infinite,
e.g., if $D$ is bounded and $f\in L^{2}(D).$

######  Remark 1.

Without loss of generality, we take the coefficient $\Theta^{\epsilon}(x,z)$
to be a symmetric function. In fact, we can define the symmetric and anti-
symmetric parts of $\Theta^{\epsilon}$:

$\Theta^{\epsilon}_{s}(x,z)=\frac{1}{2}(\Theta^{\epsilon}(x,z)+\Theta^{\epsilon}(z,x))\quad\text{and}\quad\Theta^{\epsilon}_{a}(x,z)=\frac{1}{2}(\Theta^{\epsilon}(x,z)-\Theta^{\epsilon}(z,x)).$

By [6, Lemma 3.2],
$\left(\Theta^{\epsilon}_{a}(x,z)\mathcal{D}^{*}u,\mathcal{D}^{*}\phi\right)=0.$
Then
$\mathcal{D}(\Theta^{\epsilon}(x,z)\mathcal{D}^{*}u)=\mathcal{D}(\Theta^{\epsilon}_{s}(x,z)\mathcal{D}^{*}u).$

Let $G^{\epsilon}_{D}(x,y)$ and $P^{\epsilon}_{D}(x,y)$ be the Green function
and Poisson kernel of $D$ for $\mathcal{A^{\epsilon}}$ respectively. At the
same time we let $G_{D}(x,y)$ and $P_{D}(x,y)$ be the Green function and
Poisson kernel for $\mathcal{A^{\epsilon}}$ with $\Theta^{\epsilon}=1$ on $D$,
respectively. In this case
$\mathcal{A^{\epsilon}}u=-2(-\Delta)^{\frac{\alpha}{2}}u.$

For $g:D^{c}\rightarrow\mathbb{R}$, we let $P^{\epsilon}_{D}[g](x)=g(x)$ for
$x\in D^{c}$ and
$P^{\epsilon}_{D}[g](x)=\int_{D^{c}}g(y)P^{\epsilon}_{D}(x,y)dy$ for$x\in D.$
Furthermore, $u=P^{\epsilon}_{D}[g]$ is the unique solution of the following
homogeneous Dirchlet problem:

$\begin{cases}\mathcal{A}^{\epsilon}u_{\epsilon}=0,\qquad x\in D,\\\
u^{\epsilon}|_{D^{c}}=g.\end{cases}$ (2)

######  Remark 2 ([4]).

For $f\in H^{-\alpha/2}(D)\text{ and }g\in\mathcal{X}^{D}$, we have the
existence and uniqueness of equations (1) and (2) in $\mathcal{V}^{D}$.

Next, for $\omega,\xi\in D^{c},$ we let
$\gamma_{D}(\omega,\xi)=\int_{D}\int_{D}\nu(\omega,x)G_{D}(x,y)\nu(y,\xi)dxdy=\int_{D}\nu(\xi,x)P_{D}(x,\omega)dx.$
For $g:D^{c}\rightarrow\mathbb{R}$ we introduce a quadratic form
$\mathcal{H}_{D}(g,g)=\frac{1}{2}\int_{D^{c}\times
D^{c}}(g(\omega)-g(\xi))^{2}\gamma_{D}(\xi,\omega)d\omega d\xi.$
Then we define as in [7] a new space
$\mathcal{X}^{D}=\\{g:D^{c}\rightarrow\mathbb{R}\;\text{such
that}\;\mathcal{H}_{D}(g,g)<\infty\\}.$

If $g\in\mathcal{X}^{D}$ and $x\in D,$ we obtain
$\int_{D^{c}}g(z)^{2}P_{D}(x,z)dz<\infty.$ We fix an arbitrary (reference)
point $x_{0}\in D.$ For $g\in\mathcal{X}^{D},$ we let
$|g|^{2}_{D^{c}}=\int_{D^{c}}g(z)^{2}P_{D}(x_{0},z)dz$ (we omit $x_{0}$ from
the notation). Then $\mathcal{X}^{D}$ is a Hilbert space with the induced norm
$||g||^{2}_{\mathcal{X}^{D}}=|g|^{2}_{D^{c}}+\mathcal{H}_{D}(g,g).$

###  2.2 Effective equation and convergence rate

Our main result is that the heterogeneous model (1) is approximated by a
homogenized effective model (3) below, with convergence rate $\frac{1}{2}$.
This is stated in the following theorem.

######  Theorem 1.

For scale parameter $\varepsilon$ sufficiently small, the solution
$u_{\epsilon}$ of heterogeneous model (1) is approximated by the solution
$u_{0}$ of the following homogenized equation

$\displaystyle\begin{cases}-a_{1}(-\Delta)^{\alpha/2}u_{0}-a_{2}\mathcal{F}u_{0}(x)=f(x),\qquad
x\in D,\\\ u_{0}|_{D^{c}}=g(x).\end{cases}$ (3)

with coefficients

$\displaystyle
a_{1}=\int_{\mathbb{T}^{d}\times\mathbb{T}^{d}}\Theta(y,\eta)dyd\eta,\qquad
a_{2}=\frac{1}{2}\int_{\mathbb{T}^{d}\times\mathbb{T}^{d}}\Theta(y,\eta)\mathcal{D}_{y}^{*}\chi(y)dyd\eta,$
$\displaystyle\zeta(u_{0})(x)=\frac{1}{|D|}\int_{D}(D^{*}u_{0})(x,z)dz,\qquad\mathcal{F}u_{0}(x)=\mathcal{D}|_{D}\zeta(u_{0})(x)=\int_{D}\left[\zeta(u_{0})(x)+\zeta(u_{0})(z)\right]\gamma(x,z)dz,$

where the function $\chi(y)$ is the unique solution of the following
variational problem

$\begin{cases}$${\hat{a}}(\chi,v)=\int_{\mathbb{T}^{d}\times\mathbb{T}^{d}}\Theta(y,\eta)D^{*}_{y}vdyd\eta,\\\
\chi\in H_{\\#}^{\alpha/2}(\mathbb{T}^{d}).\\\ \end{cases}$ (4)

Moreover, if $f\in C^{\infty}(\overline{D})$ and $\chi\in
L^{\infty}(\mathbb{T}^{d})$, then $u_{\epsilon}$ has the following asymptotic
expansion

$u_{\epsilon}=u_{0}-\varepsilon^{\frac{1+\alpha}{2}}\frac{1}{|D|}\int_{D}(D^{*}u_{0})(x,z)dz\cdot\chi(\frac{x}{\epsilon})+R(\varepsilon),$

and there exists a constant $C$ (independent of $\varepsilon$) such that the
remainder is estimated as

$\left\|R(\varepsilon)\right\|_{\mathcal{V}^{D}}=\left\|u_{\epsilon}-(u_{0}-\varepsilon^{\frac{1+\alpha}{2}}\frac{1}{|D|}\int_{D}(D^{*}u_{0})(x,z)dz\cdot\chi(\frac{x}{\epsilon}))\right\|_{\mathcal{V}^{D}}\leq
C\varepsilon^{1/2}.$

This says that $u_{\epsilon}\rightarrow u_{0},$ in Sobolev space
$\mathcal{V}^{D},$ with convergence rate $\frac{1}{2}.$

In order to prove this theorem, we recall some lemmas. The next result is due
to [7].

######  Lemma 1.

Let $D\subset\mathbb{R}^{d}$ be bounded, open and Lipschitz, $|\partial D|=0.$

  * 1.

If $g\in\mathcal{X}^{D},$ then $P^{\epsilon}_{D}[g]\in\mathcal{V}^{D}$ and
$\mathcal{E}_{D}(P^{\epsilon}_{D}[g],P^{\epsilon}_{D}[g])=\mathcal{H}_{D}(g,g).$

  * 2.

If $u\in\mathcal{V}^{D},$ then $g=u|_{D^{c}}\in\mathcal{X}^{D}$ and
$\mathcal{E}_{D}(u,u)\geq\mathcal{H}_{D}(g,g).$

Let $X=\\{X_{t}\\}_{t\geq 0}$ be a Lévy process with $(0,\nu,0)$ as the Lévy
triplet on a probability space $(\Omega,\mathcal{F},\mathbb{P}).$ We introduce
the time of the first exit of $X$ from $D,$ $\tau_{D}=\tau_{D}(X)=\inf\\{t\geq
0:X_{t}\notin D\\}.$

######  Lemma 2.

The assumptions are the same as in Lemma 1. Then for every
$g\in\mathcal{X}^{D},$ there exists a positive constant $C(D)$ such that
$||P^{\epsilon}_{D}[g]||_{\mathcal{V}^{D}}\leq C(D)||g||_{\mathcal{X}^{D}}.$

###### Proof.

We write $U\subset\subset D$ if U is an open set, its closure $\overline{U}$
is bounded, and $\overline{U}\subset D.$ Let $\tilde{u}^{\epsilon}$ be the
unique solution of the homogeneous Dirchlet problem (2) and $\mathbb{E}^{x}$
be the expectation for $X_{t}$ start at $x\in D$. We have
$\mathbb{E}^{x}\tilde{u}_{\epsilon}(X_{\tau_{D}})=\int_{D^{c}}\tilde{u}_{\epsilon}(y)P_{D}^{\epsilon}(x,y)dy.$
Then we obtain

$||P^{\epsilon}_{D}[g]||^{2}_{L^{2}(D)}=\int_{D}(\int_{D^{c}}g(y)P_{D}^{\epsilon}(x,y)dy)^{2}dx\leq\int_{D}\int_{D^{c}}g(y)^{2}P_{D}^{\epsilon}(x,y)dydx,$

due to the fact that $\int_{D^{c}}P_{D}^{\epsilon}(x,y)dy=1,$ for $x\in D.$
That is to say
$||P^{\epsilon}_{D}[g]||^{2}_{L^{2}(D)}\leq\mathbb{E}^{x}\tilde{u}^{2}_{\epsilon}(X_{\tau_{D}}).$

Note that $\mathbb{E}^{x}\tilde{u}_{\epsilon}(X_{\tau_{D}})$ is a closed
martingale[7, Remark 4.4] and for $x\in U,$
$x\rightarrow\int_{U^{c}}g(y)^{2}P_{U}^{\epsilon}(x,y)dy$ satisfies the
Harnack inequality[9]. Thus

$\begin{split}||P^{\epsilon}_{D}[g]||^{2}_{L^{2}(D)}&\leq\lim_{x\in
U\subset\subset
D}\int_{U}\int_{U^{c}}g(y)^{2}P_{U}^{\epsilon}(x,y)dydx\leq\lim_{x\in
U\subset\subset
D}\int_{U}C\int_{U^{c}}g(y)^{2}P_{U}^{\epsilon}(x_{0},y)dydx=C(D)|g|_{D^{c}}^{2}.\end{split}$

By Lemma 1, we obtain

$\begin{split}||P^{\epsilon}_{D}[g]||^{2}_{\mathcal{V}^{D}}&=||P^{\epsilon}_{D}[g]||^{2}_{L^{2}(D)}+\mathcal{E}_{D}(P^{\epsilon}_{D}[g],P^{\epsilon}_{D}[g])\leq
C(D)|g|_{D^{c}}^{2}+\mathcal{H}_{D}(g,g)\leq
C(D)||g||^{2}_{\mathcal{X}^{D}}.\end{split}$

Hence Lemma 2 follows. ∎

Next, we obtain a uniform estimate concerning the solution $u_{\epsilon}$ for
the original heterogeneous equation (1).

######  Lemma 3.

Let $f$ in $H^{-\alpha/2}(D)$, $g$ in $\mathcal{X}^{D}$ and $u_{\epsilon}$ be
the unique solution of the original heterogenous equation (1). Then there
exist two positive constants $C_{1},C_{2}$ such that

$||u_{\epsilon}||_{\mathcal{V}^{D}}\leq
C_{1}||f||_{H^{-\alpha/2}(D)}+C_{2}||g||_{\mathcal{X}^{D}}.$

###### Proof.

From Lemma 2, we can obtain the following conclusion. For every
$g\in\mathcal{X}^{D},$ there exists $G\in\mathcal{V}^{D}$ and a linear
operator $\rho$ such that $\rho(G)=G|_{D^{c}}=g$ and
$||G||_{\mathcal{V}^{D}}\leq C(D)||g||_{\mathcal{X}^{D}}.$ For every
$v\in\mathcal{V}_{0}^{D},$ we have

$\begin{split}|(\mathcal{A}^{\epsilon}G,v)|&=\frac{1}{2}|(\mathcal{D}(\Theta^{\epsilon}\mathcal{D}^{*}G),v)|\leq\lambda
C||\mathcal{D}^{*}G||_{L^{2}(D\times\mathbb{R}^{d})}||\mathcal{D}^{*}v||_{L^{2}(D\times\mathbb{R}^{d})}\leq
4\lambda C||G||_{\mathcal{V}^{D}}||v||_{\mathcal{V}_{0}^{D}}.\end{split}$

Then we obtain
$\mathcal{D}(\Theta^{\epsilon}\mathcal{D}^{*}G)\in(\mathcal{V}_{0}^{D})^{*}.$
That is to say, for $x\in D,$ we have
$f-\mathcal{D}(\Theta^{\epsilon}\mathcal{D}^{*}G)\in H^{-\alpha/2}.$ Recall
that

$a^{\epsilon}(u,v)=\frac{1}{2}\int_{\mathbb{R}^{d}\times\mathbb{R}^{d}\backslash
D^{c}\times D^{c}}\Theta^{\epsilon}(x,z)(u(x)-u(z))(v(x)-v(z))\nu(x,z)dxdz,$

for every $v\in\mathcal{V}_{0}^{D}.$ We can find a unique
$v_{\epsilon}\in\mathcal{V}_{0}^{D}$ such that
$a^{\epsilon}(v_{\epsilon},v)=<f-\frac{1}{2}\mathcal{D}(\Theta^{\epsilon}\mathcal{D}^{*}G),v>_{(\mathcal{V}_{0}^{D})^{*},\mathcal{V}_{0}^{D}}.$
Due to the Poincaré inequality [4], there exists a constant $C\geq 1,$ for
every $u\in\mathcal{V}_{0}^{D},$

$||u||_{L^{2}(D)}^{2}\leq C\int_{\mathbb{R}^{d}\times\mathbb{R}^{d}\backslash
D^{c}\times D^{c}}(u(x)-u(z))^{2}\nu(x,z)dxdz.$

We thus obtain
$||u||_{\mathcal{V}_{0}^{D}}\leq(2C+1)||\mathcal{D}^{*}u||_{L^{2}(D\times\mathbb{R}^{d})}.$
In other words, the space $\mathcal{V}_{0}^{D}$ can be equipped by the norm
$||\mathcal{D}^{*}u||_{L^{2}(D\times\mathbb{R}^{d})}.$ Then
$a^{\epsilon}(v,v)\geq C||v||^{2}_{\mathcal{V}_{0}^{D}}$ and
$|a^{\epsilon}(u,v)|\leq
C||u||_{\mathcal{V}_{0}^{D}}||v||_{\mathcal{V}_{0}^{D}}.$ From the Lax-Milgram
theorem,

$\begin{split}\frac{1}{\lambda}||v_{\epsilon}||_{\mathcal{V}_{0}^{D}}^{2}\leq
a^{\epsilon}(v_{\epsilon},v_{\epsilon})=<f-\frac{1}{2}\mathcal{D}(\Theta^{\epsilon}\mathcal{D}^{*}G),v_{\epsilon}>_{(\mathcal{V}_{0}^{D})^{*},\mathcal{V}_{0}^{D}}\leq||f-\frac{1}{2}\mathcal{D}(\Theta^{\epsilon}\mathcal{D}^{*}G)||_{H^{-\alpha/2}(D)}||v_{\epsilon}||_{\mathcal{V}_{0}^{D}}.\end{split}$

We can see that

$||v_{\epsilon}||_{\mathcal{V}_{0}^{D}}\leq\lambda||f-\frac{1}{2}\mathcal{D}(\Theta^{\epsilon}\mathcal{D}^{*}G)||_{H^{-\alpha/2}(D)}.$

Set $u_{\epsilon}=v_{\epsilon}+G.$ By the linearity of $\rho$, we have
$\rho(u_{\epsilon})=\rho(G)=g.$ Furthermore,

$a^{\epsilon}(u_{\epsilon},v)=a^{\epsilon}(v_{\epsilon},v)+a^{\epsilon}(G,v)=(f,v),$

which means that $u_{\epsilon}$ is the unique solution of the original
heterogenous equation (1). Then

$\begin{split}||u_{\epsilon}||_{\mathcal{V}^{D}}&\leq||u_{\epsilon}-G||_{\mathcal{V}^{D}}+||G||_{\mathcal{V}^{D}}\leq
C||v_{\epsilon}||_{\mathcal{V}_{0}^{D}}+\sqrt{C(D)}||g||_{\mathcal{X}^{D}}\\\
&\leq\lambda
C||f-\frac{1}{2}\mathcal{D}(\Theta^{\epsilon}\mathcal{D}^{*}G)||_{H^{-\alpha/2}(D)}+\sqrt{C(D)}||g||_{\mathcal{X}^{D}}.\end{split}$

On the other hand,

$\begin{split}<\mathcal{D}(\Theta^{\epsilon}\mathcal{D}^{*}G),v>_{(\mathcal{V}_{0}^{D})^{*},\mathcal{V}_{0}^{D}}&=\int_{\mathbb{R}^{d}\times\mathbb{R}^{d}\backslash
D^{c}\times
D^{c}}\Theta^{\epsilon}(x,z)\mathcal{D}^{*}G\mathcal{D}^{*}vdxdz\leq\lambda
C||\mathcal{D}^{*}G||_{L^{2}(D\times\mathbb{R}^{d})}||\mathcal{D}^{*}v||_{L^{2}(D\times\mathbb{R}^{d})}\\\
&\leq
C||G||_{\mathcal{V}^{D}}||\mathcal{D}^{*}v||_{L^{2}(D\times\mathbb{R}^{d})}\leq
C||g||_{\mathcal{X}^{D}}||v||_{\mathcal{V}_{0}^{D}}.\end{split}$

That implies $||\mathcal{D}(\Theta^{\epsilon}\mathcal{D}^{*}G)||\leq
C||g||_{\mathcal{X}^{D}}$. Hence $||u_{\epsilon}||_{\mathcal{V}^{D}}\leq
C_{1}||f||_{H^{-\alpha/2}(D)}+C_{2}||g||_{\mathcal{X}^{D}}.$ This completes
the proof. ∎

###  2.3 Proof of Theorem 1

We are now ready to prove our main result in Theorem 1.

Step 1: First, we will derive the homogenized equation for $\varepsilon$
sufficiently small. For a function $\upsilon(x,y)$, we define

$(\mathcal{D}^{*}_{x}\upsilon)(x,z,y)=-(\upsilon(z,y)-\upsilon(x,y))\gamma(x,z)$

and

$\begin{split}(\mathcal{D}_{x}\mathcal{D}^{*}_{x}\upsilon)(x,y)&=2\int_{\mathbb{R}^{d}}-(\upsilon(z,y)-\upsilon(x,y))\gamma^{2}(x,z)dz\\\
&=-2(-\Delta)^{\alpha/2}_{x}\upsilon(x,y).\end{split}$

Denote $\eta=\frac{z}{\epsilon}$ a variable on the period:
$\eta\in\mathbb{T}^{d}.$ We look for a formal asymptotic expansion:

$u_{\epsilon}=u_{0}(x,\frac{x}{\epsilon})+\varepsilon^{\frac{1+\alpha}{2}}u_{1}(x,\frac{x}{\epsilon})+o(\varepsilon^{\frac{1+\alpha}{2}}),$

with $u_{i}(x,y)$, for $i=1,2$, such that

$\begin{cases}u_{i}(x,y)\quad\text{is defined for}\quad
x\in{D}\quad\text{and}\quad y\in{\mathbb{T}^{d}},\\\
u_{i}(\cdot,y)\quad\text{is $1$-periodic}.\\\ \end{cases}$

For every function $h,$ we denote $h^{\epsilon}(x)=h(\frac{x}{\epsilon})$.
Thus

$\mathcal{D}^{*}u_{\epsilon}=\mathcal{A}_{1}u_{0}+\varepsilon^{-\frac{1+\alpha}{2}}(\mathcal{A}_{0}u_{0})^{\epsilon}+\varepsilon^{\frac{1+\alpha}{2}}\mathcal{A}_{1}u_{1}+(\mathcal{A}_{0}u_{1})^{\epsilon}+o(\varepsilon^{\frac{1+\alpha}{2}}),$

where

$\mathcal{A}_{0}v(x,y):=(\mathcal{D}^{*}_{y}v)(x,y,\eta),\mathcal{A}_{1}v(x,y):=(\mathcal{D}^{*}_{x}v)(x,z,y).$

Then we have $\mathcal{A}_{0}u_{0}=0$. Furthermore, $u_{0}(x,y)=u_{0}(x).$

Let $C_{per}(\mathbb{T}^{d}))$ be the subspace of $C(\mathbb{R}^{n})$ of
$1$-periodic functions. For every $v\in\mathcal{M}(D,C_{per}(\mathbb{T}^{d}))$
($\mathcal{M}(D)$ is the space of functions in $C^{\infty}$ with compact
support), we denote $v_{\epsilon}=v(x,\frac{x}{\epsilon}).$ We conclude

$\begin{split}(f,v_{\epsilon})&=\frac{1}{2}(\Theta^{\epsilon}(\mathcal{A}_{1}u_{0}+(\mathcal{A}_{0}u_{1})^{\epsilon}),D^{*}v_{\epsilon})=\frac{1}{2}(\Theta^{\epsilon}(\mathcal{A}_{1}u_{0}+(\mathcal{A}_{0}u_{1})^{\epsilon}),D^{*}v_{\epsilon})_{L^{2}(D\times
D)}\\\ &+\int_{D\times
D^{c}}\Theta^{\epsilon}(u_{0}(x)-g(z))v_{\epsilon}(x)\gamma^{2}(x,z)dzdx:=I_{1}^{\epsilon}+I_{2}^{\epsilon}.\end{split}$

By [15, Lemma 2.34], for every $v\in\mathcal{M}(D,C_{per}(\mathbb{T}^{d})),$
$I_{1}^{\epsilon},I_{2}^{\epsilon}$ converges, as $\varepsilon$ goes to $0.$
Now take
$v_{\epsilon}(x)=v_{0}(x)+\varepsilon^{\frac{1+\alpha}{2}}v_{1}(x,\frac{x}{\epsilon}),$
where $v_{0}(x)\in L^{2}(D)$ and $v_{1}\in L^{2}(D,C_{per}(\mathbb{T}^{d})).$
As $\varepsilon$ goes to $0,$ we have

$I_{1}^{\epsilon}\rightarrow\frac{1}{2}\int_{\mathbb{T}^{d}\times\mathbb{T}^{d}}\int_{D\times
D}\Theta(y,\eta)(\mathcal{D}^{*}u_{0}+\mathcal{D}_{y}^{*}u_{1})(\mathcal{D}^{*}v_{0}+\mathcal{D}_{y}^{*}v_{1})dxdzdyd\eta.$
(5)

On one hand, let $v_{0}=0,$ we have $I_{2}^{\epsilon}\rightarrow 0$ and

$\begin{split}\int_{D\times
D}(\Theta(y,\eta)\mathcal{D}_{y}^{*}u_{1},\mathcal{D}_{y}^{*}v_{1})_{L^{2}({\mathbb{T}^{d}\times\mathbb{T}^{d}})}dxdz&=-\int_{D\times
D}(\Theta(y,\eta)\mathcal{D}^{*}u_{0},\mathcal{D}_{y}^{*}v_{1})_{L^{2}{(\mathbb{T}^{d}\times\mathbb{T}^{d}})}dxdz.\end{split}$
(6)

For all $w,v\in H_{\\#}^{\alpha/2}(\mathbb{T}^{d})$, we introduce two
quadratic forms:
${\hat{a}}(w,v)=\frac{1}{2}\int_{\mathbb{T}^{d}\times\mathbb{T}^{d}}\Theta(y,\eta)D^{*}_{y}wD^{*}_{y}vdyd\eta.$
From equation (6), we have
$|D|\int_{D}{\hat{a}}(u_{1},v)dx=-\int_{D}(\int_{D}\mathcal{D}^{*}u_{0}(x,z)dz){\hat{a}}(\chi,v)dx,$
where $\chi(y)$ is the unique solution of the following variational problem

$\begin{cases}{\hat{a}}(\chi,v)=\int_{\mathbb{T}^{d}\times\mathbb{T}^{d}}\Theta(y,\eta)D^{*}_{y}vdyd\eta,\\\
\chi\in H_{\\#}^{\alpha/2}(\mathbb{T}^{d}),\\\ \end{cases}$ (7)

for all $v\in H_{\\#}^{\alpha/2}(\mathbb{T}^{d}).$ So

$u_{1}(x,y)=-\frac{1}{|D|}\int_{D}(D^{*}u_{0})(x,z)dz\cdot\chi(y)\in
L^{2}(\mathbb{R}^{d},H_{\\#}^{\alpha/2}(\mathbb{T}^{d})).$ (8)

Moreover, let $v_{1}=0,$ we conclude that

$I_{2}\rightarrow\frac{1}{2}\int_{\mathbb{T}^{d}\times\mathbb{T}^{d}}\Theta(y,\eta)dyd\eta\int_{D\times
D^{c}}(u_{0}(x)-g(z))v_{0}(x)\nu(x,z)dxdz,$

as $\varepsilon$ goes to $0.$ Substituting the representation of $u_{1}$ in
(8) into the equation (5), we have

$\displaystyle\begin{cases}-a_{1}(-\Delta)^{\alpha/2}u_{0}-a_{2}\mathcal{F}u_{0}(x)=f(x),\qquad
x\in D,\\\ u_{0}|_{D^{c}}=g(x).\end{cases}$ (9)

Here

$\displaystyle
a_{1}=\int_{\mathbb{T}^{d}\times\mathbb{T}^{d}}\Theta(y,\eta)dydn,\qquad
a_{2}=\frac{1}{2}\int_{\mathbb{T}^{d}\times\mathbb{T}^{d}}\Theta(y,\eta)\mathcal{D}_{y}^{*}\chi
dydn,$
$\displaystyle\zeta(u_{0})(x)=\frac{1}{|D|}\int_{D}(D^{*}u_{0})(x,z)dz,\quad\mathcal{F}u_{0}(x)=\mathcal{D}|_{D}\zeta(u_{0})(x)=\int_{D}\left[\zeta(u_{0})(x)+\zeta(u_{0})(z)\right]\gamma(x,z)dz.$

Step 2: In this step, we use the letter $C$ for a constant independent of
$\varepsilon$. We will prove the convergence rate to be $\frac{1}{2}$, in the
Sobolev space $\mathcal{V}^{D}.$

Setting

$Z_{\epsilon}(x)=u_{\epsilon}(x)-(u_{0}+\varepsilon^{\frac{1+\alpha}{2}}u_{1})(x,\frac{x}{\epsilon}),$

we have

$\begin{cases}\mathcal{A}_{\epsilon}Z_{\epsilon}=-\frac{1}{2}\varepsilon^{\frac{1+\alpha}{2}}\mathcal{D}(\Theta\mathcal{D}_{x}^{*}u_{1})^{\epsilon}:=\frac{1}{2}\varepsilon^{\frac{1+\alpha}{2}}F_{\epsilon}(x)\quad\text{in}\;D,\\\
Z_{\epsilon}|_{D^{c}}=-\varepsilon^{\frac{1+\alpha}{2}}u_{1}(x,\frac{x}{\epsilon}):=\varepsilon^{\frac{1+\alpha}{2}}K_{\epsilon}(x):=\varepsilon^{\frac{1+\alpha}{2}}K(x,\frac{x}{\epsilon})\quad\text{on}\;D^{c}.\end{cases}$
(10)

We can easily check that $F_{\epsilon}\in H^{-\alpha/2}(D).$ Let us now look
at the function $K_{\epsilon}.$ We prove the following estimate:
$||K_{\epsilon}||_{\mathcal{X}^{D}}\leq C\varepsilon^{-\alpha/2}.$

For a large enough constant $N,$ we set $M^{c}=\\{x,|x|>N\\}.$ Introduce the
function $m_{\epsilon}$ defined as follows:

$\begin{cases}m_{\epsilon}=1\quad\text{if}\quad dist(x,\partial
D)\leq\varepsilon,x\in D\;\text{or}\;x\in D^{c}/M^{c}\\\
m_{\epsilon}=0\quad\text{if}\quad dist(x,\partial D)\geq 2\varepsilon,x\in
D\;\text{or}\;x\in M^{c}\\\ ||\nabla
m_{\epsilon}||_{L^{\infty}(\mathbb{R}^{d})}\leq\varepsilon^{-1}C,\end{cases}$

Moreover, we set $m^{\epsilon}\in C^{\infty}(\mathbb{R}^{d}/\partial M),$ and
the derivative of the function $m^{\epsilon}$ at $\\{x\in D:dist(x,\partial
D)=2\varepsilon\\}$ is $0.$ Set $\psi_{\epsilon}=m_{\epsilon}K_{\epsilon}.$
The support of $\psi_{\epsilon}$ in the domain $D$ is a neighbourhood of
thickness $2\varepsilon$ which we denote by $U^{\epsilon}.$

First of all, we prove the estimates
$\left\|\psi_{\epsilon}\right\|_{\mathcal{V}^{U^{\epsilon}}}\leq
C\varepsilon^{-\alpha/2}.$

Clearly, from the definition of $m_{\epsilon}$ and the regularity properties
of $u_{0},$ we have $\left\|\psi_{\epsilon}\right\|_{L^{2}(U^{\epsilon})}\leq
C.$ Moreover, we have

$\begin{split}\mathcal{D}^{*}\psi_{\epsilon}&=\mathcal{D}^{*}(m_{\epsilon}K_{\epsilon})=((m_{\epsilon}K_{\epsilon})(x)-(m_{\epsilon}K_{\epsilon})(z))\gamma(x,z)\\\
&=m_{\epsilon}(x)\left((\mathcal{D}^{*}_{x}K)(x,z,\frac{x}{\epsilon})+\epsilon^{-\frac{1+\alpha}{2}}(\mathcal{D}^{*}_{y}K)^{\epsilon}|_{x=z}(z,\frac{x}{\epsilon},\frac{z}{\epsilon})\right)+(\mathcal{D}^{*}m_{\epsilon})(x,z)\cdot
K_{\epsilon}(z).\end{split}$

That is to say [6, Lemma 3.2],

$\begin{split}\mathcal{E}_{U^{\epsilon}}(\psi_{\epsilon},\psi_{\epsilon})&\leq
C\int_{U^{\epsilon}\times\mathbb{R}^{d}}(\mathcal{D}^{*}\psi_{\epsilon})^{2}dzdx\leq\int_{U^{\epsilon}\times\mathbb{R}^{d}}(m_{\epsilon})^{2}(x)(\mathcal{D}^{*}_{x}K)^{2}(x,z,\frac{x}{\epsilon})dzdx\\\
&+\varepsilon^{-1-\alpha}\int_{U^{\epsilon}\times\mathbb{R}^{d}}(m_{\epsilon})^{2}(x)\left((\mathcal{D}^{*}_{y}K|_{x=z})^{\epsilon}(z,\frac{x}{\epsilon},\frac{z}{\epsilon})\right)^{2}dzdx\\\
&+\int_{U^{\epsilon}\times\mathbb{R}^{d}}(\mathcal{D}^{*}m_{\epsilon})^{2}(x,z)\cdot(K_{\epsilon})^{2}(z)dxdz:=J_{1}+J_{2}+J_{3}.\end{split}$

We set $K(x,\frac{x}{\epsilon})=h(x)\chi(\frac{x}{\epsilon}),$ here
$h(x)=\frac{1}{|D|}\int_{D}(D^{*}u_{0})(x,z)dz.$ Clearly, from the regularity
of properties of $u_{0}$, $\chi$, we have

$\displaystyle J_{1}$
$\displaystyle=\int_{U^{\epsilon}\times\mathbb{R}^{d}}(m_{\epsilon})^{2}(x)(\mathcal{D}^{*}_{x}h)^{2}(x,z)\chi^{2}(\frac{x}{\epsilon})dzdx\leq
C\int_{U^{\epsilon}\times\mathbb{R}^{d}}(\mathcal{D}^{*}_{x}h)^{2}(x,z)dzdx\leq
C,$ $\displaystyle J_{2}$
$\displaystyle=\varepsilon^{-1-\alpha}\int_{U^{\epsilon}\times\mathbb{R}^{d}}(m_{\epsilon})^{2}(x)\left((\mathcal{D}^{*}_{y}\chi)^{\epsilon}\right)^{2}h^{2}(z)dzdx=\int_{U^{\epsilon}\times\\{\mathbb{R}^{d}\cap|x-z|\geq
1\\}}(m_{\epsilon})^{2}(x)\frac{(\chi^{\epsilon}(x)-\chi^{\epsilon}(z))^{2}}{|x-z|^{d+\alpha}}h^{2}(z)dzdx$
$\displaystyle+\varepsilon^{-1-\alpha}\int_{U^{\epsilon}\times\\{\mathbb{R}^{d}\cap|x-z|\leq
1\\}}(m_{\epsilon})^{2}(x)\left((\mathcal{D}^{*}_{y}\chi)^{\epsilon}\right)^{2}h^{2}(z)dzdx\leq
C\int_{U^{\epsilon}\times\\{\mathbb{R}^{d}\cap|x-z|\geq
1\\}}(m_{\epsilon})^{2}(x)h^{2}(z)dzdx$
$\displaystyle+\varepsilon^{-1-\alpha}\int_{U^{\epsilon}\times\\{\mathbb{R}^{d}\cap|x-z|\leq
1\\}}(m_{\epsilon})^{2}(x)\left((\mathcal{D}^{*}_{y}\chi)^{\epsilon}\right)^{2}h^{2}(z)dzdx\leq
C\varepsilon^{-1-\alpha}\int_{U^{\epsilon}\times
D}(\mathcal{D}^{*}u_{0})^{2}dzdx,$ $\displaystyle J_{3}$
$\displaystyle=\int_{U^{\epsilon}\times\mathbb{R}^{d}}(\mathcal{D}^{*}m_{\epsilon})^{2}(x,z)\cdot(K_{\epsilon})^{2}(z)dzdx=\int_{U^{\epsilon}\times\\{\mathbb{R}^{d}\cap|x-z|\leq
1\\}}\frac{[(m^{\epsilon})^{\prime}(\xi)]^{2}(x-z)^{2}}{|x-z|^{d+\alpha}}h^{2}(z)\chi^{2}(\frac{z}{\epsilon})dzdx$
$\displaystyle+C\int_{U^{\epsilon}\times\\{\mathbb{R}^{d}\cap|x-z|\geq
1\\}}\frac{1}{|x-z|^{d+\alpha}}h^{2}(z)\chi^{2}(\frac{z}{\epsilon})dzdx\leq
C\varepsilon^{-2}\int_{U^{\epsilon}\times D}(\mathcal{D}^{*}u_{0})^{2}dzdx.$

That is to say
$\mathcal{E}_{U^{\epsilon}}(\psi_{\epsilon},\psi_{\epsilon})\leq
J_{1}+J_{2}+J_{3}\leq C\varepsilon^{-1-\alpha}\int_{U^{\epsilon}\times
D}(\mathcal{D}^{*}u_{0})^{2}dzdx.$
We can use a result from [10](Chapter $1$, Lemma $1.5$), which states that
there exists positive constants $C$, independent of $\varepsilon,$ such that
$\left\|u_{0}\right\|_{L^{2}(U^{\epsilon})}\leq
C\varepsilon^{\frac{1}{2}}\left\|u_{0}\right\|_{H^{1}(D)}.$ We can conclude

$\int_{U^{\epsilon}}\left(\int_{D}(\mathcal{D}^{*}u_{0})^{2}dx\right)dz\leq
C\varepsilon\left\|u_{0}\right\|^{2}_{H^{1}(D)}.$

Then
$\left\|\psi_{\epsilon}\right\|^{2}_{\mathcal{V}^{U^{\epsilon}}}=\left\|\psi_{\epsilon}\right\|^{2}_{L^{2}(U^{\epsilon})}+\mathcal{E}_{U^{\epsilon}}(\psi_{\epsilon},\psi_{\epsilon})\leq
C\varepsilon^{-\alpha}.$ That is to say

$\left\|\psi_{\epsilon}\right\|_{\mathcal{V}^{U^{\epsilon}}}\leq
C\varepsilon^{-\alpha/2}.$ (11)

Secondly, we will show
$\left\|K_{\epsilon}\right\|_{\mathcal{X}^{D}}=\left\|\psi_{\epsilon}\right\|_{\mathcal{X}^{D}}+C.$
Observe now that $\psi_{\epsilon}=K_{\epsilon}$ on $D^{c}/M^{c}$ and
$\psi_{\epsilon}=0$ on $M^{c}.$ Then

$\displaystyle\left\|K_{\epsilon}\right\|^{2}_{\mathcal{X}^{D}}$
$\displaystyle=C\int_{D^{c}/M^{c}\times
M^{c}}(K_{\epsilon}(x)-K_{\epsilon}(z))^{2}\gamma_{D}(x,z)dzdx+\int_{D^{c}/M^{c}\times
D^{c}/M^{c}}(K_{\epsilon}(x)-K_{\epsilon}(z))^{2}\gamma_{D}(x,z)dzdx$
$\displaystyle+\int_{M^{c}\times
M^{c}}(K_{\epsilon}(x)-K_{\epsilon}(z))^{2}\gamma_{D}(x,z)dzdx,$
$\displaystyle\left\|\psi_{\epsilon}\right\|^{2}_{\mathcal{X}^{D}}$
$\displaystyle=C\int_{D^{c}/M^{c}\times
M^{c}}(\psi_{\epsilon}(x)-\psi_{\epsilon}(z))^{2}\gamma_{D}(x,z)dzdx+\int_{D^{c}/M^{c}\times
D^{c}/M^{c}}(K_{\epsilon}(x)-K_{\epsilon}(z))^{2}\gamma_{D}(x,z)dzdx.$

Recall that [7, Theorem 2.6]

$\gamma_{D}(x,z)\approx\left\\{\begin{aligned}
&\nu(\delta_{D}(z))\nu(\delta_{D}(x))\quad\text{if}\quad
diam(D)\leq\delta_{D}(x),\delta_{D}(z),\\\
&\nu(\delta_{D}(z))/V(\delta_{D}(x))\quad\text{if}\quad\delta_{D}(x)\leq
diam(D)\leq\delta_{D}(z),\\\
&\frac{\nu(r(x,z))V^{2}(r(x,z))}{V(\delta_{D}(x))V(\delta_{D}(z))}\quad\text{if}\quad\delta_{D}(x),\delta_{D}(z)\leq
diam(D),\end{aligned}\right.$

here $\delta_{D}(x)=dist(x,\partial
D),r(x,y)=\delta_{D}(x)+|x-y|+\delta_{D}(x)$ and $V(r)=Cr^{\alpha/2}.$ Then,
we have

$\begin{split}&\int_{M^{c}\times
M^{c}}(K_{\epsilon}(x)-K_{\epsilon}(z))^{2}\gamma_{D}(x,z)dzdx\leq
C\int_{M^{c}\times
M^{c}}(K_{\epsilon}(x)-K_{\epsilon}(z))^{2}\left(\delta_{D}(x)\right)^{-1-\alpha}\left(\delta_{D}(z)\right)^{-1-\alpha}dzdx\leq
C,\end{split}$

and

$\begin{split}&\int_{D^{c}/M^{c}}\int_{\times
M^{c}}(K_{\epsilon}(x)-K_{\epsilon}(z))^{2}\gamma_{D}(x,z)dzdx\\\
&=C\int_{D^{c}/M^{c}\cap\\{x:\delta_{D}(x)\leq
diam(D)\\}}\int_{M^{c}}(K_{\epsilon}(x)-K_{\epsilon}(z))^{2}\left(\delta_{D}(x)\right)^{-\alpha/2}\left(\delta_{D}(z)\right)^{-1-\alpha}dzdx\\\
&+C\int_{D^{c}/M^{c}\cap\\{x:\delta_{D}(x)\geq
diam(D)\\}}\int_{M^{c}}(K_{\epsilon}(x)-K_{\epsilon}(z))^{2}\left(\delta_{D}(x)\right)^{-1-\alpha}\left(\delta_{D}(z)\right)^{-1-\alpha}dzdx\\\
&\leq C.\end{split}$

Then, we get the conclusion

$\left\|K_{\epsilon}\right\|^{2}_{\mathcal{X}^{D}}=\left\|\psi_{\epsilon}\right\|^{2}_{\mathcal{X}^{D}}+C.$
(12)

Next, we will show $\left\|\psi_{\epsilon}\right\|^{2}_{\mathcal{V}^{D}}\leq
max\\{C,1\\}\left\|\psi_{\epsilon}\right\|^{2}_{\mathcal{V}^{U^{\epsilon}}}+C.$
In fact,

$\begin{split}\left\|\psi_{\epsilon}\right\|^{2}_{\mathcal{V}^{D}}&=\left\|\psi_{\epsilon}\right\|^{2}_{L^{2}(D)}+C\int_{D\times\mathbb{R}^{d}}(\mathcal{D}^{*}\psi_{\epsilon})^{2}dxdz=\left\|\psi_{\epsilon}\right\|^{2}_{L^{2}(U^{\epsilon})}+C\int_{U^{\epsilon}\times\mathbb{R}^{d}}(\mathcal{D}^{*}\psi_{\epsilon})^{2}dxdz\\\
&+C\int_{D/U^{\epsilon}}\int_{\left(D/U^{\epsilon}\right)^{c}}(m^{\epsilon}(x))^{2}K^{2}(x,\frac{x}{\epsilon})\nu(x,z)dzdx\\\
&\leq
max\\{C,1\\}\left\|\psi_{\epsilon}\right\|^{2}_{\mathcal{V}^{U^{\epsilon}}}+C\int_{D/U^{\epsilon}}\int_{\left(D/U^{\epsilon}\right)^{c}}(m^{\epsilon}(x))^{2}K^{2}(x,\frac{x}{\epsilon})\nu(x,z)dzdx.\end{split}$

From the fact that the derivative of the function $m^{\epsilon}$ at $\\{x\in
D:dist(x,\partial D)=2\varepsilon\\}$ is $0,$ we have

$\left\|\psi_{\epsilon}\right\|^{2}_{\mathcal{V}^{D}}\leq
max\\{C,1\\}\left\|\psi_{\epsilon}\right\|^{2}_{\mathcal{V}^{U^{\epsilon}}}+C.$
(13)

Finally, we will get the convergence rate. Combining (11), (12), (13) and [7,
Corollary 5.1] we conclude

$\begin{split}\left\|K_{\epsilon}\right\|^{2}_{\mathcal{X}^{D}}&=\left\|\psi_{\epsilon}\right\|^{2}_{\mathcal{X}^{D}}+C\leq
C\left\|\psi_{\epsilon}\right\|^{2}_{\mathcal{V}^{D}}+C\leq
C\left\|\psi_{\epsilon}\right\|^{2}_{\mathcal{V}^{U^{\epsilon}}}+C\leq
C\varepsilon^{-\alpha}.\end{split}$

We thus estimate from Lemma 3

$\begin{split}\left\|Z_{\epsilon}\right\|_{\mathcal{V}^{D}}&\leq
C\varepsilon^{\frac{1+\alpha}{2}}\left\|F_{\epsilon}\right\|_{H^{-\alpha/2}(D)}+C\varepsilon^{\frac{1+\alpha}{2}}\left\|K_{\epsilon}\right\|_{\mathcal{X}^{D}}\leq
C\varepsilon^{\frac{1+\alpha}{2}}+\varepsilon^{\frac{1+\alpha}{2}}\varepsilon^{-\frac{\alpha}{2}}c_{11}\leq
C\varepsilon^{\frac{1}{2}}.\end{split}$

This completes the proof of Theorem 1.

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The K-homology of 2-dimensional Crystallographic Groups

Hang Wang Research Center for Operator Algebras (RCOA) of East China Normal University (ECNU), Shanghai 200241, China. Email<EMAIL_ADDRESS>Xiufeng Yao Research Center for Operator Algebras (RCOA) of East China Normal University (ECNU), Shanghai 200241, China. Email<EMAIL_ADDRESS>
In this paper we compute the topological K-homology of 2-dimensional crystal groups. Our method focuses on the fixed points of group action and simplifies the calculation of the K-homology of universal space. The result also verifies the Baum-Connes Conjecture of 2-dimensional crystal groups.

§ INTRODUCTION

A crystal could be thought as a structure consisting of small balls arranged periodically. The balls are constantly arranged in the same way to form the basic pattern (Cell) and spread evenly throughout the space. For example, the crystal structures made of molecules or atoms that are common in physics and chemistry such as diamonds.

The initial motivation for the study of crystal groups came from Hilbert's question 18. Our understanding of the algebraic structure of crystal groups also evolved around Hilbert's 18th question [1]. Bieberbach defined the crystal group as a subgroup of the Euclidean group, consisting of the semi-direct product of lattice and point group. He also defined the equivalence of crystal groups by using the conjugate of projective transformations [2], and proved that there are only finite equivalence classes of crystal groups in n-dimensional space [3]. This is the first fundamental result of Hilbert's 18th question. And then, Zassenhaus proposed the Jordan-Zassenhaus theorem [4], which proves that a geometric equivalence class of crystals contains at most a finite number of arithmetic equivalence classes. This is a fundamental result of integral representation theory and the key to Hilbert's 18th problem.

From the point of view of operator algebra, Zassenhaus's work allows us to take one crystal group as an ordinary finite-dimensional amenable group, that is, a finite locally compact group with Haar measure. B.D. Evans showed that the C*-algebra of a crystal group is a section on vector bundle [5]. The base space of this vector bundle is the equivalence class of the Pontryagin duality of the lattice about the group actions of the point group , and the fibres are matrices. K. F. Taylor found the way to compute the explicit matrix formulation of the group C*-algebra of any locally compact group with finite-index commutative normal subgroups, and then we could use matrix to describe the C*-algebra of a crystal group [6]. J. M. G. Fell and R. S. Doran gave the method to compute the irreducible representations of the group C*-algebra for a crystal group [7]. It should be noted that, in Evans's description, the base space can be regarded as the fundamental domain of the orbit space. The group action of point group on a boundary point determines whether its isotropy subgroup is trivial. Eric M. Pohorecky used this method to calculate some examples of 2-dimensional crystal groups [8].

Due to the development of string theory and solid state physics, the topological K-theory of crystal groups has received much attention [9]. The difficulty here stems from the special semi-direct product structure of crystal groups, and we usually have to use some cohomology tools to discuss the cases on the boundary of the fundamental domain. Mingze Yang studied the topology of the group C* -algebras of two-dimensional crystal groups and calculated their K- theory [10]. He also found it very difficult in higher dimensional cases.

To solve this problem, we associate to each crystal group $G$ with a universal space $\underline{E}G$ and the assembly map

\begin{align}\label{bc1}\
K^{G}_{\bullet}(\underline{E}G) \overset{\mu}{\rightarrow} K_{\bullet}(C_{r}^{*}(G)) .
\end{align}

The left-hand-side is the equivariant K-homology of $\underline{E}G$, and the right-hand-side is the K-theory of the reduced group C*-algebra of $G$. This is what we are familiar with in the formulation of the Baum-Connes Conjecture.

Using E-theory and KK-theory, Higson and Kasparov proved the Baum-Connes Conjecture for countable discrete groups, which act isometrically and metrically on an infinite-dimensional Euclidean space and include crystal groups [11]. Thus we could identify the difficult topolocical K-theory of the group C*-algebra to the better accessible equavariant homology theory given by equivatiant topological K-theory. Davis and Lück calculated the results of both sides of Eq.<ref> in the case of $Z^{n} \rtimes_{\rho} Z_{p}$( p is a prime). Some physicists have applied similar methods to calculations in crystalline materials [12]. Recently, Gomi and Thiang introduced T-duality to the study of the Baum-Connes conjecture of crystal groups by using twisted K-theory [13].

The main structure and motivation of this paper are as follows. Section 2 contains necessary preliminaries about crystal groups and equivariant K-homology. In section 2.1 we review the algebraic structure of crystal groups and relevant cohomology. Section 2.1 includes some theorems about the topological K-homology of finite groups, especially the equations what we need in the computation. Section 3 is the core of this paper. We propose a method that greatly simplifies the calculation of the conjugacy classes of elements of finite order. We propose a method that greatly simplifies the calculation. Basically, this method just needs to find the center of transformation of the plane patterns. We list the detailed calculation results of K-homology for all 17 2-dimensional crystal groups. These results are consistent with the known result about K-theory[10] and verify the Baum-Connes Conjecture. Our original motivation is from the both side of the assembly map <ref> having the similar form of homology groups. We hope that the calculation will inspire us to explore an explicit description of the assembly map <ref>.

§ PRELIMINARIES

In this sections we recall some relevant definitions and theorems.

§.§ Crystal Group and Cohomology

In the previous, the objects arranged periodically include not only the "balls" but also the basic patterns composed of balls, which are generally called "cells". The cells should have special properties that allow them to be arranged in geometric symmetry and periodicity. As the mathematical abstraction of crystal, a crystal group describes the transformations of cells in n-dimensional Euclidean space $\mathbb{R}^{n}$. Furthermore, all these transformations are rigid motions, i.e. isometries in $\mathbb{R}^{n}$. In Euclidean geometry, we have the following assertion [14]:

Let $O(n)$ denote the orthogonal group in $\mathbb{R}^{n}$, $V=\{t(v)| v\in V\}$ denote the vector space of translations to distinguish it from the $\mathbb{R}^{n}$. Every isometry can be written in a unique way as a composition $t(v) \circ \phi$, where $t(v)\in V$ and $\phi\in O(n)$. In addition, all isometries (or rigid motions) of $\mathbb{R}^{n}$ form a group, which we call the Euclidean group $Isom(\mathbb{R}^n)$ and

\begin{align*}
(t, \phi)\cdot(t', \phi')&=(t+\phi(t'), \phi \phi'), \\ Isom(\mathbb{R}^{n})&=V \rtimes O(n).
\end{align*}

A n-dimensional crystal group is the subgroup of $Isom(\mathbb{R}^n)$, Zassenhaus gave the most common definition of crystal group:

An abstract group G is isomorphic to an n-dimensional crystallographic group if and only if G contains a finite index, normal, free abelian subgroup of rank n, that is also maximal abelian.

Now let us analyze the algebric structure of crystal groups under the framework of Euclidean groups. According to Def.<ref>, a crystal group fits into an exact sequence

\begin{align}
M \stackrel{i}{\rightarrow} G \stackrel{p}{\rightarrow} D,
\end{align}

where $i$ and $p$ are natural map.

The translation part of crystal group generates the free abelian subgroup $M$. This group is often called the lattice. $D$ denotes finite quotient group $G/M$ which is also called the point group, is also a subgroup of $O(n)$. The point group represents rotation and symmetry and generates cells, while the lattice spreads cells all over the space.

Similarly to Euclidean groups, the point group $D$ acts on lattice $M$ by pulling its elements back to G and conjugating. More precisely, if $d\in D$,$a\in M$, we write the action of $d$ on $a$ as

\begin{align}
d.a =\gamma(d)i(a)\gamma(d)^{-1}=(\varphi(d),d)(a,1)(\varphi(d),d)^{-1}=(d(a),1),
\end{align}

where $\gamma(d)=(\varphi(d),d)$ is a pullback. However, there is a problem we need to point out here. $\gamma$ is given by restricting a pullback from $O(n)$ to $V \rtimes O(n)$ to subgroup $D$. And every pullback is determined by a map $\varphi: D\rightarrow V$. So it may not be a group homomorphism and is not unique.

The solution to this problem is composing with the exact sequence

\begin{align}\label{eqq1}
M\rightarrow V \rightarrow V/M.
\end{align}

For an n-dimensional crystal group $G$, considering lattice $M \cong \mathbb{Z}^{n}$ and V being $\mathbb{R}^{n}$, we could write Eq.<ref> as

\begin{align}
\label{eqq2}
0\rightarrow M\cong \mathbb{Z}^{n} \rightarrow \mathbb{R}^{n}\rightarrow \mathbb{R}^{n} / \mathbb{Z}^{n}\cong \mathbb{T}^{n}\rightarrow 0,
\end{align}

where $\mathbb{T}^{n}$ denotes n-torus. And then we could have a well-defined map $s: D\rightarrow \mathbb{T}^{n}$ associating with $\varphi$. $s$ satifies the condition

\begin{align}
\label{1cocycle}
s(d_{1}d_{2})=s(d_{1})+d_{1} (s(d_{2})).
\end{align}

Thus s is a group 1-cocycle. Meanwhile, let $\gamma(d)\gamma(c)=\gamma(dc)\alpha(d,c)$, we could define a 2-cocycle

\begin{align}
\alpha(d,c):=\gamma(d)\gamma(c) \gamma(dc)^{-1}&=(\varphi(d),d)(\varphi(c),c)(\varphi(dc),dc)^{-1} \\
\end{align}

It is easy to construct cohomology group $H^{\bullet -1}(D,\mathbb{T}^{n})\cong H^{\bullet}(D,\mathbb{Z}^{n}) $ of $G$ with coefficients in $\mathbb{T}^{n}\cong \mathbb{R}^{n}/\mathbb{Z}^{n} \cong V/M$. For a finite group, we often care about its cohomology and homology which are key to K-theory and many other questions. For a crystal group, its cohomology group distinguishes it from other crystal groups with the same lattice and point group. In other words, there exists one-to-one correspondence between crystal groups and the orbits their normalizer acting on the cohomology group $H^{1}(D,\mathbb{T}^{n})$ [15]. We could classify different crystal groups by computing $H^{1}(D,\mathbb{T}^{n})$. Pólya [16] and Niggli [17] showed that there are 17 different 2-dimensional crystal groups.

§.§ Equivarian K-homology of the Crystallographic Groups

In this part, we will introduce the main theorem we will be using to compute the K-homology of crystal groups. Most definitions and results can be found in [18] and [19].

For a discrete group $G$, the Baum-Connes Conjecture allows us to obtain the K-theory of the reduced group C*-algebra $C^{*}_{r}(G)$ of $G$ by computing the equivariant K-homology of the universal space $\underline{E}G$ for proper $G$-actions. First, the definition of proper action is as follows:

The $G$-space $X$ is a topolocical space with a given continuous action of $G$ on $X$

$$ G \times X \rightarrow X.$$

The action of $G$ on $X$ is proper if for every $p\in X$, there exists a tripe $(U,H,\rho)$ such that

(i) $U$ is an open neighborhood of $p$, with $gu\in U$for all $(g,u)\in G\times U$,

(ii)$H$ is a compact subgroup of $G$,

(iii)$\rho:U\rightarrow G/H$is a $G$-map from $U$ to the homogeneous space $G/H$.

The topolocical spaces with group action also have the notion of properness.

A $G$-space $X$ is proper if for every $x\in X $, there is a  $G$-invariant open subset $U\subseteq X$ containing $x$, a finite subgroup $H$ of $G$, and a $G$-equivariant map from $U$to $G/H$.

Now we can use proper property to define universal spaces. It is important to notice that the universal spaces associated with a group are not unique, and we have group-quivariantly homotopy equivalence. But now we just introduce the notation $\underline{E}G$. Some papers also use universal example to represent it.

The universal space $\underline{E}G$ is a proper $G$-space for proper actions of $G$, if for any proper $G$-space $X$, there exists a $G$-map $f: X\rightarrow\underline{E}G$, and any two $G$-maps from $X$ to $\underline{E}G$ are $G$-homotopic.

On one hand, [20] shows that an n-dimensional crystal group acts properly on $\mathbb{R}^{n}$ and $\mathbb{R}^{n}$ is a model of $\underline{E}G$ for it. Considering lattice $M\cong \mathbb{Z}^{n}$ acts freely on $\mathbb{R}^{n}\cong\underline{E}G$, the group action of $G$ on $\mathbb{R}^{n}$ could be regarded as the group action of the point group $D$ on torus $\mathbb{T}^{n}\cong \mathbb{R}^{n}/\mathbb{Z}^{n}$ by Eq.<ref>. Then we obtain the first main equation in the computation of $K^{G}_{\bullet}(\underline{E}G)$:

\begin{align}\label{Khomology}
K^{\mathbb{Z}^{n}\rtimes D}_{\bullet}(\mathbb{R}^{n})\cong
\end{align}

We will use one of the last three items in computation depending on different crystal groups in the next section.

On the other hand, as a G-CW-complexes, the skeleton filtration of $\underline{E}G$ could be used to construct Bredon homology $H_{i}^{\mathfrak{F i n}}\left(G ; K_{j}^{G}(\cdot)\right)$ which is associated with Chern character. So we could use Chern character and associated equivariant homology group to reduce the computation of $K^{G}_{\bullet}(\underline{E}G)$. The following theorem is originally from [21] and summarized by Valette[18].

Let $(X,A)$ be a pair of proper G-CW-complexes. Then there is a decomposition

\begin{align}\label{1}
C h_{\bullet}^{G}: \bigoplus_{i \in \mathbb{Z}} H_{\bullet+2 i}^{\mathfrak{F i n}}\left(X, A ; \mathbb{Q} \otimes R_{\mathbb{C}}\right) \stackrel{\cong}{\longrightarrow} K_{\bullet}^{G}(X, A) \otimes \mathbb{Q} .
\end{align}

For the groups on the left hand side one has isomorphisms

\begin{align}\label{2}
H_{k}^{\mathfrak{F} \operatorname{in}}\left(X, A ; \mathbb{Q} \otimes R_{\mathbb{C}}\right) \cong \bigoplus_{[g] \in \operatorname{FC}(G)} H_{k}\left(\left(X^{g}, A^{g}\right) / C_{G}(g) ; \mathbb{Q}\right), \quad k \in \mathbb{N}.
\end{align}

$FC(G)$ is the set of conjugacy classes of elements of finite order in $G$. $(\cdot)^{g}$ denotes stabilizer, the set of fixed points of some element $g$. And $C_{G}(g)$ is centralizer of $g$.

For a 2-dimensional crystal group $G=\mathbb{Z}^{2}\rtimes D$, the group action of $G$ on $\underline{E}G\cong \mathbb{R}^{2}$ could decent to $\mathbb{T}^{2}$ as Eq.<ref>. So we can rewrite using Eq.<ref> and Eq.<ref> as

\begin{align}
\label{cp}
K^{G}_{\bullet}(\underline{E}G)\cong \left\{\begin{matrix}
K^{\mathbb{Z}^{n}\rtimes D}_{\bullet}(\mathbb{R}^{n})\cong \bigoplus_{i \in \mathbb{Z}} \bigoplus_{[g] \in \operatorname{FC}(G)} H_{i+2k}\left(\left(\mathbb{R}^{2}\right)^{g} / C_{G}(g)\right), \\
K^{D}_{\bullet}(\mathbb{T}^{n})\cong \bigoplus_{i \in \mathbb{Z}} \bigoplus_{[g] \in \operatorname{FC}(D)} H_{i+2k}\left(\left(\mathbb{T}^{2}\right)^{g} / C_{D}(g)\right),
\end{matrix}\right. \quad k \in \mathbb{N}.
\end{align}

§ THE K-HOMOLOGY OF 2D CRYSTAL GROUPS

From the end of the previous section, we know that to compute the K-homology of a crystal group, we need to compute the related homogy group. This requires first to find the conjugacy classes of all elements of finite order in group $G$, and then find the corresponding fixed-point sets and centralizers. The key is to find all conjugacy classes $FC(G)$ or $FC(D)$ in $\mathbb{R}^{2}$ or $\mathbb{T}^{2}$.

§.§ The Computation of Conjugacy Classes

For any crystal group $G=A\rtimes D$, let $g=(a,d), a\in A, d\in D$ be an arbitrary element, which means the group action of $G$ is composed by the group action of lattice $A$ and point group $D$. If $G$ is 2-dimensional, the lattice could be writen as $A=<t,s>\cong \mathbb{Z}^{2}$, and $t$, $s$ are linearly independent vectors in $\mathbb{R}^{2}$. Then we could use $\{t,s\}$ as a group of coordinate basis and write an element $ a=mt+ns\in A, (m,n\in\mathbb{Z})$ as $a=(m,n)$ for convenience. Forthermore, $a=(m,n)=((m,n),id_{D})$ denotes the embedding of $a$ in $G$.

For an element $d$ in the point group $D$ with its inverse $d^{-1}$, $\gamma(d)=(u_{d},M_{d})$ denotes the pullback of $d$ from $D$ to $G$. And $u_{d}=(u_{d}^{1},u_{d}^{2})\in \mathbb{R}^{2}$is the translation part, while $M_{d}$ is the transformation matrix in the orthogonal group $O(2)$. From now on, we can write

\begin{align}
G=A\rtimes D=\{a\cdot d|a\in A,d\in D\}=\{(a,id_{D})\cdot (u_{d},M_{d})=(a+u_{d},M_{d})|a\in A, d\in D \}.
\end{align}

The group action of D on A is

\begin{align}
\end{align}

let's think for a moment in terms of affine groups. If $y=(y^{1},y^{2}),x=(x^{1},x^{2}),b=(b^{1},b^{2})\in\mathbb{R}^{2}$ is three translation vectors and $(0,0)$ is the coordinate origin, for any martrix $A\in O(2)$, we can define an affine transformation similar to the previous form

\begin{align}
y=Ax+b=((b^{1},b^{2}),E_{2\times 2})\cdot((0,0),A),
\end{align}

where $E_{2\times 2}$ is $2\times 2$ unitary matrix and could also denotes $id_{D}$. Picking $(m,n)$ as new coordinate origin, we could obtain the new affine translation under new coordidate

\begin{align}\label{1.1}
y'&=((m,n),E_{2\times 2})((b^{1},b^{2}),E_{2\times 2})((0,0),A)((-m,-n),E_{2\times 2})x \\
&=((b^{1},b^{2})+(m,n)+ -A(m,n),A)x.
\end{align}

Just like before, using $d$ denotes an orthogonal transformation in a 2-dimensional crystal group $G$, the pullback of $d$ is $\gamma(d)=(u_{d},M_{d})=((u_{d}^{1},u_{d}^{2}),M_{d})$. The group action of lattice $A$ on $d$ gives conjugacy class in the following form:

\begin{align}
\end{align}

where $a=(m,n)=((m,n),id_{D})\in A$. Thus, in analogy to Eq.<ref>, the group action of a lattice on a point group can be thought of as a corresponding orthogonal transformation at the origin of a point in the lattice.

There are three classes of finite ordered elements of pullback of $D$ in 2-dimensional crystal group $G$, we write $\sigma$, $\rho$ and $d$ respectively. $\sigma$ means rotation, $\rho$ means mirror reflection and $d$ means symmetry about a point. $t=(m,n)\in\mathbb{Z}^{2}$is an arbitrary point in the lattice $A$. Since the order of point group is finite, we can compute conjugacy classes just by conjugating by $t$.

For rotation $\sigma$, $\gamma(\sigma)=((0,0),M_{\sigma})$, we could obtain the conjugacy classes of $\sigma$ in $G$:

\begin{align}
\begin{aligned}
t^{-1} \sigma t &=\left((-m,-n), id_{D}\right)\left((0,0), M_{\sigma}\right)\left((m, n), id_{D}\right) \\
&=\left((-m,-n), id_{D}\right)\left((m, n) M_{\sigma}, M_{\sigma}\right) \\
&=\left((-m, n)+(m, n)M_{\sigma}, M_{\sigma}\right).
\end{aligned}
\end{align}

We can easily find the the element of conjugacy class of rotation $\sigma$ could be obtained by changing the rotation origin to $(-m,-n)$.

For mirror reflection $\rho=\left((0,0), M_{\rho}\right)$,

\begin{align}
t^{-1} \sigma t=\left((-m,-n)+(m, n) M_{\rho}, M_{\rho}\right)=\left((-m, 0)+(m, 0) M_{\rho}, M_{\rho}\right).
\end{align}

This is equivalent to shifting the reflection axis $l_{\rho}$ by $(-m,-n)$.

Finally, for symmetry about the origin point, let
$$\gamma(d)=((u,v),M_{d}),\quad M_{d}=\begin{bmatrix}
-1&0 \\
\end{bmatrix}.$$

\begin{align}
\begin{aligned}
t^{-1} \rho t &=\left((-m,-n), id_{D}\right)\left((u, v), M_{d}\right)\left((m, n), id_{D}\right) \\
&=\left((-m,-n), id_{D}\right)\left((u, v)+(-m,-n), M_{d}\right) \\
&=\left(\left(\frac{1}{2} u-m, \frac{1}{2} v-n\right)+\left(m-\frac{1}{2} u, n-\frac{-1}{2} v\right) M_{d}, M_{d}\right).
\end{aligned}
\end{align}

This is the same as symmetry about $\frac{1}{2}\gamma(d)+(-m,-n)$.

The above calculation about conjugacy classes shows that we only need to find all these three classes of group actions of finite order in one cell. More precisely, if $g=(v,\phi)$ is an element of finite order in $G$( i.e. $(v,\phi)^{k}=(0,id_{D})),k\in\mathbb{N}$), it must be a pullback from $D$ because the orders of translation vectors in lattice are infinite. Since $\phi^{k}=id_{D}$, it satisfies equation

\begin{align}\label{eq}
\end{align}

So finding all elements of finite order in $G$ just amounts to solve the equation Eq.<ref> for every $\phi \in D$. Considering $\phi(v+\phi(v)+\cdots+\phi^{k-1}(v))=0$, we only need to find all the axes and centers of reflections and centers of rotations in one cell.

§.§ K-homology of 2-dimensional Crystal Groups

In the following, we list what we need to compute the K-homology of 17 2-dimensional crystal groups. We will introduce their point groups and related pullback, and write down the conjugacy classes of elements of finite order. We use the same notation as before for convenience.

According to Eq.<ref>, we choose $\mathbb{R}^{2}$ or $\mathbb{T}^{2}$ for convenience. Next, We design a table to list all representative of conjugacy classes, the related stabilizer $X^{g}$, the centralizer $C_{\bullet}(g)$ and their quotient $X^{g}/C_{\bullet}(g)$, and the homology groups of the quotients:
$$\bigoplus_{k \in even} \bigoplus_{[g] \in FC[\bullet]} H_{k}\left(X^{g}/C_{\bullet}(g)\right)\quad and\quad \bigoplus_{k \in odd} \bigoplus_{[g] \in FC[\bullet]} H_{k}\left(X^{g}/C_{\bullet}(g)\right).$$

K-homology group $K_{0}(\underline{E}G)$ and $K_{1}(\underline{E}G)$ are given by these direct sums of homology groups.

In addition, we share some figures to show the patterns of crystal groups with the translation vectors and axes and centre points of rotations and reflections. Let $I^{2}=[-1,1]\times[-1,1]\cong \mathbb{R}^{2}/\mathbb{Z}^{2}\subseteq \mathbb{R}^{2}$. This article also includes the images of $I^{2}$ containing the fundamental domains to help the readers understanding the calculations of homology groups. Unless otherwise specified, all of the points we've given in tables are in $I^{2}$. $I^{2}$ brings us another benefit that the group actions of $D$ on $\mathbb{T}^{2}$ coulb be showed on $I^{2}$ by defining map

\begin{align}
\begin{aligned}
\mathbb{T}^{2} &\rightarrow I^{2} \\
(z,w)=(e^{i\pi t},e^{i\pi s})&\mapsto (t,s).
\end{aligned}
\end{align}

§.§ p1

Fundamental Domain of p1

We compute $K^{G}_{\bullet}(\underline{E}G)\cong K^{D}_{\bullet}(\mathbb{T}^{2})$.

1. The group $p1\cong \mathbb{Z}^{2}\cong A=<t,s>$, and its point group $D$ is trivial.

2. p1 has no untrivial elements of finite order.

3. See Table <ref>.

to 0.95X[c]X[c]
representative             identity
$X^{g}$                    $\mathbb{T}^{2}$

$C_{D}(g)$           $D$

$X^{g}/C_{D}(g)$     $\mathbb{T}^{2}$

$\bigoplus_{k \in even} H_{k}\left(X^{g}/C_{D}(g)\right)$     $\mathbb{Z}\oplus \mathbb{Z}=\mathbb{Z}^{2}$

$\bigoplus_{k \in odd} H_{k}\left(X^{g}/C_{D}(g)\right)$      $\mathbb{Z}^{2}$

The computation about p1

4. $K_{0}(p1)=\mathbb{Z}^{2},K_{1}(p1)=\mathbb{Z}^{2}$.

§.§ p2

Fundamental Domain of p2

We compute $K^{G}_{\bullet}(\underline{E}G)\cong K^{D}_{\bullet}(\mathbb{T}^{2})$.

1. Group $p2\cong \mathbb{Z}^{2}\rtimes D$. Let $t=(t_{1},0)$ and $s=(s_{1},s_{2})$ be a basis for $A$ ($s_{1},t_{1},t_{2}\neq0$ ).

Point group $D=\{id_{D},\sigma\}$ and the pullback:

$$ \gamma(\sigma)=((0,0),M_{\sigma}),\quad M_{\sigma}=\begin{bmatrix}
-1 & 0\\
0 & -1
\end{bmatrix}.$$

2. $FC(G)=FC(D)=\{[id_{D}],[\sigma]\}$.

3. See Table <ref>.

to 0.95X[c]X[c]X[c]
representative             identity          $\sigma$
$X^{g}$                    $\mathbb{T}^{2}$     $(\pm 1,\pm 1)\in \mathbb{T}^{2} $

$C_{D}(g)$           $D$        $D$

$X^{g}/C_{D}(g)$     Fundamental Domain     $(1,1),(0,1)$, $(1,0),(0,0) $

$\bigoplus \limits_{k \in even} H_{k}(X^{g}/C_{D}(g))$     $\mathbb{Z}^{2}$     $\mathbb{Z}^{4}$

$\bigoplus \limits_{k \in odd} H_{k}(X^{g}/C_{D}(g))$      0     0

The computation about p2

4. $K_{0}(p2)=\mathbb{Z}^{6}, K_{1}(p2)=0$.

§.§ p3

Fundamental Domain of p3

We compute $K^{G}_{\bullet}(\underline{E}G)\cong K^{D}_{\bullet}(\mathbb{T}^{2})$.

1. Let $t=(t_{1},0)$ and $s=(-\frac{1}{2}t_{1},\frac{\sqrt{3}}{2}t_{1})$ form a basis for $A$ ($t_{1}\neq0$ ).

Point group $D=\{id_{D},\sigma,\sigma^{2}\}$ and the pullback:

\begin{align*}
\gamma(\sigma)=((0,0),M_{\sigma}),\quad \gamma(\sigma^{2})=((0,0),M_{\sigma^{2}}) \\ M_{\sigma}=\begin{bmatrix}
0 & 1\\
-1 & -1
\end{bmatrix}, \quad
-1 & -1\\
1 & 0
\end{bmatrix}.
\end{align*}

2. $FC(G)=FC(D)=\{[id_{D}],[\sigma],[\sigma^{2}]\}$.

3. See Table <ref>.

to 0.95X[c]X[c]X[c]X[c]
representative             identity          $\sigma$     $\sigma^{2}$
$X^{g}$                    $\mathbb{T}^{2}$     $(\frac{2}{3}, \frac{2}{3})$,$(-\frac{2}{3},-\frac{2}{3})$,$(0,0)$      same as for $\sigma$

$C_{D}(g)$           $D$        $D$

$X^{g}/C_{D}(g)$     Fundamental Domain     $(\frac{2}{3}, \frac{2}{3})$,$(-\frac{2}{3},-\frac{2}{3})$,$(0,0)$      same as for $\sigma$

$\bigoplus \limits_{k \in even} H_{k}(X^{g}/C_{D}(g))$     $\mathbb{Z}^{2}$     $\mathbb{Z}^{3}$     $\mathbb{Z}^{3}$

$\bigoplus \limits_{k \in odd} H_{k}(X^{g}/C_{D}(g))$      0    0     0

The computation about p3

4. $K_{0}(p3)=\mathbb{Z}^{8}, K_{1}(p3)=0$.

§.§ p4

Fundamental Domain of p4

We compute $K^{G}_{\bullet}(\underline{E}G)\cong K^{D}_{\bullet}(\mathbb{T}^{2})$.

1. Let $t=(t_{1},0)$ and $s=(0,t_{1})$ form a basis for $A$ ($t_{1}\neq0$ ).

Point group $D=\{id_{D},\sigma,\sigma^{2},\sigma^{3}\}$ and the pullback:

\begin{align*}
\gamma(\sigma)=((0,0),M_{\sigma}),\quad \gamma(\sigma^{2})=((0,0),M_{\sigma^{2}}),\quad \gamma(\sigma^{3})=((0,0),M_{\sigma^{3}}) \\
0 & 1\\
-1 & 0
\end{bmatrix}, \quad
-1 & 0\\
0 & -1
\end{bmatrix}, \quad
0 & -1\\
1 & 0
\end{bmatrix}.
\end{align*}

2. $FC(G)=FC(D)=\{[id_{D}],[\sigma],[\sigma^{2}],[\sigma^{3}]\}$.

3. See Table <ref>.

to 0.95X[c]X[c]X[c]X[c]X[c]
representative             identity          $\sigma$     $\sigma^{2}$    $\sigma^{3}$
$X^{g}$                    $\mathbb{T}^{2}$     $(0,0)$, $(1,1)$     $(1,1),(0,1)$, $(1,0),(0,0)$    $(0,0)$, $(-1,-1)$

$C_{D}(g)$           $D$        $D$    $D$     $D$

$X^{g}/C_{D}(g)$     Fundamental Domain     $(0,0)$, $(1,1)$     $(1,1),(0,0)$, $(1,0)$    $(0,0)$, $(-1,-1)$

$\bigoplus \limits_{k \in even} H_{k}(X^{g}/C_{D}(g))$     $\mathbb{Z}^{2}$     $\mathbb{Z}^{2}$     $\mathbb{Z}^{3}$    $\mathbb{Z}^{2}$

$\bigoplus \limits_{k \in odd} H_{k}(X^{g}/C_{D}(g))$      0    0     0       0

The computation about p4

4. $K_{0}(p4)=\mathbb{Z}^{9}, K_{1}(p4)=0$.

§.§ p6

Fundamental Domain of p6

We compute $K^{G}_{\bullet}(\underline{E}G)\cong K^{\mathbb{Z}^{2}\rtimes D}_{\bullet}(\mathbb{R}^{2})$.

1. Let $t=(t_{1},0)$ and  $s=(\frac{1}{2}t_{1},\frac{\sqrt{3}}{2}t_{1})$ form a basis for $A$ ($t_{1}\neq0$ ).

Point group $D=\{id_{D},\sigma,\sigma^{2},\sigma^{3},\sigma^{4},\sigma^{5}\}$ and the pullback:

\begin{align*}
\gamma(\sigma)=((0,0),M_{\sigma}),\quad \gamma(\sigma^{2})=((0,0),M_{\sigma^{2}}),\quad \gamma(\sigma^{3})=((0,0),M_{\sigma^{3}}) \\
\gamma(\sigma^{4})=((0,0),M_{\sigma^{4}}),\quad \gamma(\sigma^{5})=((0,0),M_{\sigma^{5}})\\
0 & 1\\
-1 & 1
\end{bmatrix}, \quad
-1 & 1\\
-1 & 0
\end{bmatrix}, \quad
-1 & 0\\
1 & -1
\end{bmatrix} \\
0 & -1\\
1 & -1
\end{bmatrix}, \quad
1 & -1\\
1 & 0
\end{bmatrix} .
\end{align*}

2. $FC(G)=\{[id_{D}],[\sigma],[\sigma^{2}],[\sigma^{3}],[\sigma^{4}],[\sigma^{5}],[t\circ \sigma^{3}],[s\circ \sigma^{3}],[s\circ t\circ \sigma^{3}]\}$, where

\begin{aligned}
t \circ \sigma^{3} &=\left((1,0), M_{\sigma^{3}}\right)=\left(\left(\frac{1}{2}, 0\right)+\left(-\frac{1}{2}, 0\right) M_{\sigma^{3}}, M_{\sigma^{3}}\right) ,\\
s \circ \sigma^{3} &=\left((0,1), M_{\sigma^{3}}\right)=\left(\left(0, \frac{1}{2}\right)+\left(0,-\frac{1}{2}\right) M_{\sigma^{3}}, M_{\sigma^{3}}\right) ,\\
s\circ t\circ \sigma^{3} &=\left((1,1), M_{\sigma^{3}}\right)=\left(\left(\frac{1}{2}, \frac{1}{2}\right)+\left(-\frac{1}{2}, \frac{1}{2}\right) M_{\sigma^{3}}, M_{\sigma^{3}}\right).
\end{aligned}

3. See Table <ref>.

to 0.95X[c]X[c]X[c]X[c]X[c]
representative             identity          $\sigma$     $\sigma^{2}$     $\sigma^{3}$
$X^{g}$                    $\mathbb{R}^{2}$     $(0,0)$   $(0,0)$   $(0,0)$

$C_{D}(g)$           $G$        $D$    $D$     $D$

$X^{g}/C_{D}(g)$     Fundamental Domain     $(0,0)$   $(0,0)$   $(0,0)$

$\bigoplus \limits_{k \in even} H_{k}(X^{g}/C_{D}(g))$     $\mathbb{Z}^{2}$     $\mathbb{Z}$     $\mathbb{Z}$    $\mathbb{Z}$

$\bigoplus \limits_{k \in odd} H_{k}(X^{g}/C_{D}(g))$      0    0     0       0
$\sigma^{4}$   $\sigma^{5}$   $t\circ \sigma^{3}$   $s\circ \sigma^{3}$   $s\circ t\circ \sigma^{3}$

$(0,0)$   $(0,0)$   $(\frac{1}{2},0)$   $(0,\frac{1}{2})$   $(\frac{1}{2},\frac{1}{2})$

$D$   $D$   $id_{D},t\circ \sigma^{3}$   $id_{D},s\circ \sigma^{3}$   $id_{D},s\circ t\circ \sigma^{3}$

$(0,0)$   $(0,0)$   $(\frac{1}{2},0)$   $(0,\frac{1}{2})$   $(\frac{1}{2},\frac{1}{2})$

$\mathbb{Z}$     $\mathbb{Z}$     $\mathbb{Z}$    $\mathbb{Z}$   $\mathbb{Z}$

0   0   0   0   0

The computation about p6

4. $K_{0}(p6)=\mathbb{Z}^{10}, K_{1}(p6)=0$.

§.§ pm

Fundamental Domain of pm

We compute $K^{G}_{\bullet}(\underline{E}G)\cong K^{\mathbb{Z}^{2}\rtimes D}_{\bullet}(\mathbb{R}^{2})$.

1. Let $t=(t_{1},0)$ and $s=(0,s_{2})$ form a basis for $A$ ($t_{1},s_{2}\neq0$ ).

Point group $D=\{id_{D},\rho\}$ and the pullback:

\begin{align*}
\gamma(\rho)=((0,0),M_{\rho}),\quad
1 & 0\\
0 & -1
\end{bmatrix}.
\end{align*}

2. $FC(G)=\{[id_{D}],[\rho],[s\circ \rho]\}$.

3. See Table <ref>.

to 0.95X[c]X[c]X[c]X[c]
representative             identity          $\rho$     $s\circ \rho$
$X^{g}$                    $\mathbb{R}^{2}$     $(x,0)$    $(x,-\frac{1}{2})$

$C_{D}(g)$           $G$        $t,id_{D},\rho$    $t,\rho$

$X^{g}/C_{D}(g)$     Fundamental Domain     $S^{1}$     $S^{1}$

$\bigoplus \limits_{k \in even} H_{k}(X^{g}/C_{D}(g))$      $\mathbb{Z}$     $\mathbb{Z}$    $\mathbb{Z}$

$\bigoplus \limits_{k \in odd} H_{k}(X^{g}/C_{D}(g))$      $\mathbb{Z}$     $\mathbb{Z}$    $\mathbb{Z}$

The computation about pm

4. $K_{0}(pm)=\mathbb{Z}^{3}, K_{1}(pm)=\mathbb{Z}^{3}$.

§.§ cm

Fundamental Domain of cm

We compute $K^{G}_{\bullet}(\underline{E}G)\cong K^{\mathbb{Z}^{2}\rtimes D}_{\bullet}(\mathbb{R}^{2})$.

1. Let $t=(t_{1},0)$ and $s=(\frac{1}{2}t_{1},\frac{1}{2}s_{2})$ form a basis for $A$ ($t_{1},s_{2}\neq0$ ).

Point group $D=\{id_{D},\rho\}$ and the pullback:

\begin{align*}
\gamma(\rho)=((0,0),M_{\rho}),\quad
1 & 0\\
1 & -1
\end{bmatrix}.
\end{align*}

2. $FC(G)=\{[id_{D}],[\rho]\}$.

3. See Table <ref>.

to 0.95X[c]X[c]X[c]
representative             identity          $\rho$
$X^{g}$                    $\mathbb{R}^{2}$     $(x,\frac{x}{2})$

$C_{D}(g)$           $G$        $2t+s,id_{D},\rho$

$X^{g}/C_{D}(g)$     Fundamental Domain     $S^{1}$

$\bigoplus \limits_{k \in even} H_{k}(X^{g}/C_{D}(g))$      $\mathbb{Z}$     $\mathbb{Z}$

$\bigoplus \limits_{k \in odd} H_{k}(X^{g}/C_{D}(g))$      $\mathbb{Z}$     $\mathbb{Z}$

The computation about cm

4. $K_{0}(cm)=\mathbb{Z}^{2}, K_{1}(cm)=\mathbb{Z}^{3}$.

§.§ pg

Fundamental Domain of pg

We compute $K^{G}_{\bullet}(\underline{E}G)\cong K^{\mathbb{Z}^{2}\rtimes D}_{\bullet}(\mathbb{R}^{2})$.

1. Let $t=(t_{1},0)$ and $s=(0,s_{2})$ form a basis for $A$ ($t_{1},s_{2}\neq0$ ).

Point group $D=\{id_{D},\rho\}$ and the pullback:

\begin{align*}
\gamma(\rho)=((\frac{1}{2},0),M_{\rho}),\quad
1 & 0\\
o & -1
\end{bmatrix}.
\end{align*}

2. $FC(G)=\{[id_{D}]\}$.

3. See Table <ref>.

to 0.95X[c]X[c]
representative             identity
$X^{g}$                    $\mathbb{R}^{2}$

$C_{D}(g)$           $G$

$X^{g}/C_{D}(g)$     Cylindar

$\bigoplus \limits_{k \in even} H_{k}(X^{g}/C_{D}(g))$      $\mathbb{Z}$

$\bigoplus \limits_{k \in odd} H_{k}(X^{g}/C_{D}(g))$      $\mathbb{Z}\oplus \mathbb{Z}_{2} $

The computation about pg

4. $K_{0}(pg)=\mathbb{Z}, K_{pg}(cm)=\mathbb{Z}\oplus \mathbb{Z}_{2}$.

§.§ cmm2

Fundamental Domain of cmm2

We compute $K^{G}_{\bullet}(\underline{E}G)\cong K^{\mathbb{Z}^{2}\rtimes D}_{\bullet}(\mathbb{R}^{2})$.

1. Let $t=(t_{1},0)$ and  $s=(\frac{1}{2}t_{1},\frac{1}{2}s_{2})$ form a basis for $A$ ($t_{1},s_{2}\neq0$ ).

Point group $D=\{id_{D},\sigma,\rho_{1},\rho_{2}\}$ and the pullback:

\begin{align*}
\gamma(\sigma)=((0,0),M_{\sigma}),\quad \gamma(\rho_{1})=((0,0),M_{\rho_{1}}),\quad \gamma(\rho_{2})=((0,0),M_{\rho_{2}}) \\
-1 & 0\\
0 & -1
\end{bmatrix}, \quad
1 & 0\\
1 & -1
\end{bmatrix}, \quad
-1 & 0\\
-1 & 1
\end{bmatrix} .
\end{align*}

2. $FC(G)=\{[id_{D}],[\sigma],[\rho_{1}],[\rho_{2}],[t\circ \sigma]\}$, where

\begin{aligned}
t \circ \sigma =((1,0),M_{\sigma})=((\frac{1}{2},0)+(-\frac{1}{2},0)M_{\sigma},M_{\sigma}).
\end{aligned}

3. See Table <ref>.

to 0.95X[c]X[c]X[c]X[c]X[c]X[c]
representative             identity          $\sigma$     $\rho_{1}$     $\rho_{2}$    $t\circ\sigma$
$X^{g}$                    $\mathbb{R}^{2}$     $(0,0)$   $(0,0)$   $(0,0)$   $(\frac{1}{2},0)$

$C_{D}(g)$           $G$        $D$    $D$     $D$    $t\circ\sigma$

$X^{g}/C_{D}(g)$     Fundamental Domain     $(0,0)$   $(0,0)$   $(0,0)$   $(\frac{1}{2},0)$

$\bigoplus \limits_{k \in even} H_{k}(X^{g}/C_{D}(g))$     $\mathbb{Z}$     $\mathbb{Z}$     $\mathbb{Z}$    $\mathbb{Z}$   $\mathbb{Z}$

$\bigoplus \limits_{k \in odd} H_{k}(X^{g}/C_{D}(g))$
0   0   0   0   0

The computation about cmm2

4. $K_{0}(cmm2)=\mathbb{Z}^{5}, K_{1}(cmm2)=0$.

§.§ pmm2

Fundamental Domain of pmm2

We compute $K^{G}_{\bullet}(\underline{E}G)\cong K^{\mathbb{Z}^{2}\rtimes D}_{\bullet}(\mathbb{R}^{2})$.

1. Let $t=(t_{1},0)$ and  $s=(0,s_{2})$ form a basis for $A$ ($t_{1},s_{2}\neq0$ ).

Point group $D=\{id_{D},\sigma,\rho_{1},\rho_{2}\}$ and the pullback:

\begin{align*}
\gamma(\sigma)=((0,0),M_{\sigma}),\quad \gamma(\rho_{1})=((0,0),M_{\rho_{1}}),\quad \gamma(\rho_{2})=((0,0),M_{\rho_{2}}) \\
-1 & 0\\
0 & -1
\end{bmatrix}, \quad
1 & 0\\
0 & -1
\end{bmatrix}, \quad
-1 & 0\\
0 & 1
\end{bmatrix} .
\end{align*}

2. $FC(G)=\{[id_{D}],[\sigma],[\rho_{1}],[\rho_{2}],[t\circ \sigma],[s\circ\sigma],[t\circ s\circ\sigma],[t\circ\rho_{2}],[s\circ\rho_{1}]\}$, where

\begin{aligned}
t \circ \sigma =((1,0),M_{\sigma})=((\frac{1}{2},0)+(-\frac{1}{2},0)M_{\sigma},M_{\sigma}), \\
s \circ \sigma =((0,1),M_{\sigma})=((0,\frac{1}{2})+(0,-\frac{1}{2})M_{\sigma},M_{\sigma}),\\
s\circ t \circ \sigma =((1,1),M_{\sigma})=((\frac{1}{2},\frac{1}{2})+(-\frac{1}{2},-\frac{1}{2})M_{\sigma},M_{\sigma}),\\
t \circ \rho_{2} =((1,0),M_{\rho_{2}})=((\frac{1}{2},0)+(-\frac{1}{2},0)M_{\rho_{2}},M_{\rho_{2}}),\\
s \circ \rho_{1} =((0,1),M_{\rho_{1}})=((0,\frac{1}{2})+(0,-\frac{1}{2})M_{\rho_{1}},M_{\rho_{1}}).
\end{aligned}

3. See Table <ref>.

to 0.95X[c]X[c]X[c]X[c]X[c]
representative             identity          $\rho_{1}$     $\rho_{2}$     $\sigma$
$X^{g}$                    $\mathbb{R}^{2}$     $(x,0)$   $(0,y)$   $(0,0)$

$C_{D}(g)$           $G$        $t,\sigma,\rho_{1},\rho_{2}$    $s,\sigma,\rho_{1},\rho_{2}$     $D$

$X^{g}/C_{D}(g)$     Fundamental Domain     $\{(x,0)|0\leqslant x\leqslant 1\}$   $\{(0,y)|0\leqslant y\leqslant 1\}$   $(0,0)$

$\bigoplus \limits_{k \in even} H_{k}(X^{g}/C_{D}(g))$     $\mathbb{Z}$     $\mathbb{Z}$     $\mathbb{Z}$    $\mathbb{Z}$

$\bigoplus \limits_{k \in odd} H_{k}(X^{g}/C_{D}(g))$
0   0   0   0
$t\circ\sigma$     $s\circ\sigma$   $s\circ t\circ\sigma$   $t\circ\rho_{2}$   $s\circ\rho_{1}$
$(0,0)$   $(0,\frac{1}{2})$   $(\frac{1}{2},\frac{1}{2})$   $(\frac{1}{2},y)$   $(x,\frac{1}{2})$

$T\circ\sigma$   $\rho_{1},\rho_{2},\sigma$   $\rho_{1},\rho_{2},\sigma$   $t,\rho_{1},\rho_{2},\sigma$   $s,\rho_{1},\rho_{2},\sigma$

$(0,0)$   $(0,\frac{1}{2})$   $(\frac{1}{2},\frac{1}{2})$   $\{(\frac{1}{2},y)|0\leqslant y\leqslant 1\}$   $\{(x,\frac{1}{2})|0\leqslant x\leqslant 1\}$

$\mathbb{Z}$     $\mathbb{Z}$     $\mathbb{Z}$    $\mathbb{Z}$    $\mathbb{Z}$

0   0   0   0   0

The computation about pmm2

4. $K_{0}(pmm2)=\mathbb{Z}^{9}, K_{1}(pmm2)=0$.

§.§ pmg

Fundamental Domain of pmg

We compute $K^{G}_{\bullet}(\underline{E}G)\cong K^{\mathbb{Z}^{2}\rtimes D}_{\bullet}(\mathbb{R}^{2})$.

1. Let $t=(t_{1},0)$ and  $s=(0,s_{2})$ form a basis for $A$ ($t_{1},s_{2}\neq0$ ).

Point group $D=\{id_{D},\sigma,\rho_{1},\rho_{2}\}$ and the pullback:

\begin{align*}
\gamma(\sigma)=((\frac{1}{2},0),M_{\sigma}),\quad \gamma(\rho_{1})=((\frac{1}{2},0),M_{\rho_{1}}),\quad \gamma(\rho_{2})=((0,0),M_{\rho_{2}}) \\
-1 & 0\\
0 & -1
\end{bmatrix}, \quad
1 & 0\\
0 & -1
\end{bmatrix}, \quad
-1 & 0\\
0 & 1
\end{bmatrix} .
\end{align*}

2. $FC(G)=\{[id_{D}],[\sigma],[\rho_{2}],[s\circ\sigma]$, where

\begin{aligned}
s \circ \sigma =((\frac{1}{2},1),M_{\sigma})=((\frac{1}{4},\frac{1}{2})+(-\frac{1}{4},-\frac{1}{2})M_{\sigma},M_{\sigma}).
\end{aligned}

3. See Table <ref>.

to 0.95X[c]X[c]X[c]X[c]X[c]
representative             identity          $\rho_{2}$     $\sigma$     $s\circ\sigma$
$X^{g}$                    $\mathbb{R}^{2}$     $(x,0)$   $(\frac{1}{4},0)$   $(\frac{1}{4},\frac{1}{2})$

$C_{D}(g)$           $G$        $t,id_{D},\rho_{2}$    $\sigma,id_{D}$       $id_{D},s\circ\sigma$

$X^{g}/C_{D}(g)$     Fundamental Domain     $\{(x,0)|-1\leqslant x\leqslant 1\}$   $(\frac{1}{4},0)$   $(\frac{1}{4},\frac{1}{2})$

$\bigoplus \limits_{k \in even} H_{k}(X^{g}/C_{D}(g))$     $\mathbb{Z}$     $\mathbb{Z}$     $\mathbb{Z}$    $\mathbb{Z}$

$\bigoplus \limits_{k \in odd} H_{k}(X^{g}/C_{D}(g))$
0   $\mathbb{Z}$   0   0

The computation about pmg

4. $K_{0}(pmg)=\mathbb{Z}^{4}, K_{1}(pmg)=\mathbb{Z}$.

§.§ pgg2

Fundamental Domain of pgg2

We compute $K^{G}_{\bullet}(\underline{E}G)\cong K^{\mathbb{Z}^{2}\rtimes D}_{\bullet}(\mathbb{R}^{2})$.

1. Let $t=(t_{1},0)$ and  $s=(0,s_{2})$ form a basis for $A$ ($t_{1},s_{2}\neq0$ ).

Point group $D=\{id_{D},\sigma,\rho_{1},\rho_{2}\}$ and the pullback:

\begin{align*}
\gamma(\sigma)=((1,0),M_{\sigma}),\quad \gamma(\rho_{1})=((\frac{1}{2},\frac{1}{2}),M_{\rho_{1}}),\quad \gamma(\rho_{2})=((\frac{1}{2},\frac{1}{2}),M_{\rho_{2}}) \\
-1 & 0\\
0 & -1
\end{bmatrix}, \quad
1 & 0\\
0 & -1
\end{bmatrix}, \quad
-1 & 0\\
0 & 1
\end{bmatrix} .
\end{align*}

2. $FC(G)=\{[id_{D}],[\sigma],[s\circ\sigma]$, where

\begin{aligned}
s \circ \sigma =((1,1),M_{\sigma})=((\frac{1}{2},\frac{1}{2})+(-\frac{1}{2},-\frac{1}{2})M_{\sigma},M_{\sigma}).
\end{aligned}

3. See Table <ref>.

to 0.95X[c]X[c]X[c]X[c]
representative             identity           $\sigma$     $s\circ\sigma$
$X^{g}$                    $\mathbb{R}^{2}$    $(\frac{1}{2},0)$   $(\frac{1}{2},\frac{1}{2})$

$C_{D}(g)$           $G$        $\sigma,id_{D}$       $id_{D},s\circ\sigma$

$X^{g}/C_{D}(g)$     Fundamental Domain    $(\frac{1}{2},0)$   $(\frac{1}{2},\frac{1}{2})$

$\bigoplus \limits_{k \in even} H_{k}(X^{g}/C_{D}(g))$     $\mathbb{Z}$     $\mathbb{Z}$     $\mathbb{Z}$

$\bigoplus \limits_{k \in odd} H_{k}(X^{g}/C_{D}(g))$
$\mathbb{Z}_{2}$   0   0

The computation about pgg2

4. $K_{0}(pgg2)=\mathbb{Z}^{3}, K_{1}(pgg2)=\mathbb{Z}_{2}$.

§.§ p31m

Fundamental Domain of p31m

We compute $K^{G}_{\bullet}(\underline{E}G)\cong K^{\mathbb{Z}^{2}\rtimes D}_{\bullet}(\mathbb{R}^{2})$.

1. Let $t=(t_{1},0)$ and  $s=(\frac{1}{2}t_{1},\frac{\sqrt{3}}{6}s_{2})$ form a basis for $A$ ($t_{1},s_{2}\neq0$ ).

Point group $D=\{id_{D},\sigma,\sigma^{2},\rho_{1},\rho_{2},\rho_{3}\}$ and the pullback of the genetors:

\begin{align*}
\gamma(\sigma)=((0,0),M_{\sigma}),\quad \gamma(\rho_{1})=((0,0),M_{\rho_{1}}), \\
-2 & 3\\
-1 & 1
\end{bmatrix}, \quad
1 & 0\\
1 & -1
\end{bmatrix} .
\end{align*}

2. $FC(G)=\{[id_{D}],[\sigma],[\rho_{1}],[-s\circ\sigma],[-s\circ t\circ \sigma]\}$, where

\begin{aligned}
s \circ \sigma =((0,1),M_{\sigma})=((-\frac{1}{3},1)+(\frac{1}{3},-1)M_{\sigma},M_{\sigma}),\\
s \circ t\circ \sigma =((1,1),M_{\sigma})=((\frac{1}{3},0)+(-\frac{1}{3},0)M_{\sigma},M_{\sigma}).
\end{aligned}

3. See Table <ref>.

to 0.95X[c]X[c]X[c]X[c]X[c]X[c]
representative             identity          $\rho_{1}$     $\sigma$     $s\circ\sigma$    $s\circ t\circ\sigma$
$X^{g}$                    $\mathbb{R}^{2}$     $(x,0)$   $(0,0)$   $(-\frac{1}{3},1)$   $(\frac{1}{3},0)$

$C_{D}(g)$           $G$        $t,\rho_{1}$    $id_{D},\sigma,\sigma^{2}$     $id_{D},s\circ\sigma$    $id_{D},s\circ t\circ\sigma$

$X^{g}/C_{D}(g)$     Fundamental Domain     $\{(x,0)|-1\leqslant x\leqslant 1\}$   $(0,0)$   $(-\frac{1}{3},0),0)$   $(\frac{1}{3},0)$

$\bigoplus \limits_{k \in even} H_{k}(X^{g}/C_{D}(g))$     $\mathbb{Z}$     $\mathbb{Z}$     $\mathbb{Z}$    $\mathbb{Z}$   $\mathbb{Z}$

$\bigoplus \limits_{k \in odd} H_{k}(X^{g}/C_{D}(g))$
0   $\mathbb{Z}$   0   0   0

The computation about p31m

4. $K_{0}(p31m)=\mathbb{Z}^{5}, K_{1}(p31m)=\mathbb{Z}$.

§.§ p3m1

Fundamental Domain of p3m1

We compute $K^{G}_{\bullet}(\underline{E}G)\cong K^{\mathbb{Z}^{2}\rtimes D}_{\bullet}(\mathbb{R}^{2})$.

1. Let $t=(t_{1},0)$ and  $s=(\frac{1}{2}t_{1},\frac{\sqrt{3}}{2}s_{2})$ form a basis for $A$ ($t_{1},s_{2}\neq0$ ).

Point group $D=\{id_{D},\sigma,\sigma^{2},\rho_{1},\rho_{2},\rho_{3}\}$ and the pullback of the genetors:

\begin{align*}
\gamma(\sigma)=((0,0),M_{\sigma}),\quad \gamma(\rho_{1})=((0,0),M_{\rho_{1}}), \\
-1 & 1\\
-1 & 0
\end{bmatrix}, \quad
1 & 0\\
1 & -1
\end{bmatrix} .
\end{align*}

2. $FC(G)=\{[id_{D}],[\sigma],[\rho_{1}],[t\circ\sigma],[2t\circ t\circ \sigma]\}$, where

\begin{aligned}
t \circ \sigma =((1,0),M_{\sigma})=((\frac{1}{3},\frac{1}{3})+(-\frac{1}{3},-\frac{1}{3})M_{\sigma},M_{\sigma}),\\
2t \circ t\circ \sigma =((2,0),M_{\sigma})=((\frac{2}{3},\frac{2}{3})+(-\frac{2}{3},-\frac{2}{3})M_{\sigma},M_{\sigma}).
\end{aligned}

3. See Table <ref>.

to 0.95X[c]X[c]X[c]X[c]X[c]X[c]
representative             identity          $\rho_{1}$     $\sigma$     $t\circ\sigma$    $2t\circ\sigma$
$X^{g}$                    $\mathbb{R}^{2}$     $(0,y)$   $(0,0)$   $(\frac{1}{3},\frac{1}{3})$   $(\frac{2}{3},\frac{2}{3})$

$C_{D}(g)$           $G$        $t,\rho_{1}$    $id_{D},\sigma,\sigma^{2}$     $id_{D},t\circ\sigma$    $id_{D}, 2t\circ\sigma$

$X^{g}/C_{D}(g)$     Fundamental Domain     $\{(0,y)|-1\leqslant y\leqslant 1\}$   $(0,0)$   $(\frac{1}{3},\frac{1}{3})$   $(\frac{2}{3},\frac{2}{3})$

$\bigoplus \limits_{k \in even} H_{k}(X^{g}/C_{D}(g))$     $\mathbb{Z}$     $\mathbb{Z}$     $\mathbb{Z}$    $\mathbb{Z}$   $\mathbb{Z}$

$\bigoplus \limits_{k \in odd} H_{k}(X^{g}/C_{D}(g))$
0   $\mathbb{Z}$   0   0   0

The computation about p3m1

4. $K_{0}(p3m1)=\mathbb{Z}^{5}, K_{1}(p3m1)=\mathbb{Z}$.

§.§ p4mm

Fundamental Domain of p4mm

We compute $K^{G}_{\bullet}(\underline{E}G)\cong K^{D}_{\bullet}(\mathbb{T}^{2})$.

1. Let $t=(t_{1},0)$ and  $s=(0,s_{2})$ form a basis for $A$ ($t_{1},s_{2}\neq0$ ).

Point group $D=\{id_{D},\sigma,\sigma^{2},\sigma^{3},\rho_{1},\rho_{2},\rho_{3},\rho_{4}\}$ and the pullback of the genetors:

\begin{align*}
\gamma(\sigma)=((0,0),M_{\sigma}),\quad \gamma(\rho_{1})=((0,0),M_{\rho_{1}}), \\
0 & 1\\
-1 & 0
\end{bmatrix}, \quad
1 & 0\\
0 & -1
\end{bmatrix} .
\end{align*}

2. $FC(G)=\{[id_{D}],[\sigma],[\sigma^{2}],[\rho_{1}],[\rho_{2}]\}$.

3. See Table <ref>.

to 0.95X[c]X[c]X[c]X[c]X[c]X[c]
representative             identity          $\sigma$     $\sigma^{2}$     $\rho_{1}$    $\rho_{2}$
$X^{g}$                    $\mathbb{T}^{2}$     $(\pm 1,\pm 1)$, $(0,0)$    $(\pm 1,\pm 1)$, $(0,0)$, $(0,\pm 1)$, $(\pm 1,0)$   $(x,0)$, $(x,1)$   $(x,x)$

$C_{D}(g)$           $D$        $D$    $D$     $D$    $D$

$X^{g}/C_{D}(g)$     Fundamental Domain     $(0,0),(1,1)$   $(0,0)$, $(1,1)$, $(0,1)$   $\{(x,0)|0\leqslant x\leqslant 1\}$, $\{(x,1)|0\leqslant x\leqslant 1\}$   $\{(x,x)|0\leqslant x\leqslant 1\}$

$\bigoplus \limits_{k \in even} H_{k}(X^{g}/C_{D}(g))$     $\mathbb{Z}$     $\mathbb{Z}^{2}$     $\mathbb{Z}^{3}$    $\mathbb{Z}^{2}$   $\mathbb{Z}$

$\bigoplus \limits_{k \in odd} H_{k}(X^{g}/C_{D}(g))$
0   0   0   0   0

The computation about p4mm

4. $K_{0}(p4mm)=\mathbb{Z}^{9}, K_{1}(p4mm)=0$.

§.§ p4mg

Fundamental Domain of p4mg

We compute $K^{G}_{\bullet}(\underline{E}G)\cong K^{D}_{\bullet}(\mathbb{T}^{2})$.

1. Let $t=(t_{1},0)$ and  $s=(0,s_{2})$ form a basis for $A$ ($t_{1},s_{2}\neq0$ ).

Point group $D=\{id_{D},\sigma,\sigma^{2},\sigma^{3},\rho_{1},\rho_{2},\rho_{3},\rho_{4}\}$ and the pullback of the genetors:

\begin{align*}
\gamma(\sigma)=((0,0),M_{\sigma}),\quad \gamma(\rho_{1})=((\frac{1}{2},\frac{1}{2}),M_{\rho_{1}}), \\
0 & 1\\
-1 & 0
\end{bmatrix}, \quad
1 & 0\\
0 & -1
\end{bmatrix} .
\end{align*}

2. $FC(G)=FC(D)=\{[id_{D}],[\sigma],[\sigma^{2}],[\rho_{2}],[s\circ\sigma],[s\circ \sigma^{2}]\}$, where

\begin{aligned}
s \circ \sigma =((0,1),M_{\sigma})=((-\frac{1}{2},\frac{1}{2})+(\frac{1}{2},-\frac{1}{2})M_{\sigma},M_{\sigma}),\\
s \circ \sigma^{2} =((0,1),M_{\sigma})=((0,\frac{1}{2})+(0,-\frac{1}{2})M_{\sigma^{2}},M_{\sigma^{2}}).
\end{aligned}

3. See Table <ref>.

to 0.95X[c]X[c]X[c]X[c]
representative             identity          $\rho_{2}$     $\sigma$
$X^{g}$                    $\mathbb{T}^{2}$     $(x,x)$    $(0,0)$

$C_{D}(g)$           $D$        $t+s$, $\rho_{2}$, $\rho_{4}$     $\sigma$, $\sigma^{2}$

$X^{g}/C_{D}(g)$     Fundamental Domain     $\{(x,x)|0\leqslant x\leqslant 1\}$   $(0,0)$

$\bigoplus \limits_{k \in even} H_{k}(X^{g}/C_{D}(g))$     $\mathbb{Z}$     $\mathbb{Z}$     $\mathbb{Z}$

$\bigoplus \limits_{k \in odd} H_{k}(X^{g}/C_{D}(g))$
0   0   0
$\sigma^{2}$     $s\circ\sigma$    $s\circ\sigma^{2}$

$(0,0)$     $(-\frac{1}{2},\frac{1}{2})$     $(0,\frac{1}{2})$

$\sigma$, $\sigma^{2}$     $s\circ\sigma$     $s\circ\sigma^{2}$

$(0,0)$     $(-\frac{1}{2},\frac{1}{2})$     $(0,\frac{1}{2})$

$\mathbb{Z}$     $\mathbb{Z}$     $\mathbb{Z}$

0   0   0

The computation about p4mg

4. $K_{0}(p4mg)=\mathbb{Z}^{6}, K_{1}(p4mg)=0$.

§.§ p6mm

Fundamental Domain of p6mm

We compute $K^{G}_{\bullet}(\underline{E}G)\cong K^{D}_{\bullet}(\mathbb{T}^{2})$.

1. Let $t=(t_{1},0)$ and  $s=(\frac{1}{2}t_{1},\frac{\sqrt{3}}{2}s_{2})$ form a basis for $A$ ($t_{1},s_{2}\neq0$ ).

Point group $D=\{id_{D},\sigma,\sigma^{2},\sigma^{3},\sigma^{4},\sigma^{5},\rho_{1},\rho_{2},\rho_{3},\rho_{4},\rho_{5},\rho_{6}\}$ and the pullback of the genetors:

\begin{align*}
\gamma(\sigma)=((0,0),M_{\sigma}),\quad \gamma(\rho_{1})=((0,0),M_{\rho_{1}}), \\
0 & 1\\
-1 & 1
\end{bmatrix}, \quad
1 & 0\\
1 & -1
\end{bmatrix} .
\end{align*}

2. $FC(G)=FC(D)=\{[id_{D}],[\sigma],[\sigma^{2}],[\sigma^{3}],[\rho_{1}],[\rho_{2}]\}$.

3. See Table <ref>.

to 0.95X[c]X[c]X[c]X[c]
representative             identity          $\sigma$     $\sigma^{2}$
$X^{g}$                    $\mathbb{T}^{2}$     $(0,0)$    $(0,0)$, $(\frac{2}{3},-\frac{2}{3})$, $(-\frac{2}{3},\frac{2}{3})$

$C_{D}(g)$           $D$        $id_{D}$, $\sigma$, $\sigma^{2}$, $\sigma^{3}$, $\sigma^{4}$, $\sigma^{5}$     $id_{D}$, $\sigma$, $\sigma^{2}$, $\sigma^{3}$, $\sigma^{4}$, $\sigma^{5}$

$X^{g}/C_{D}(g)$     Fundamental Domain     $(0,0)$   $(0,0)$, $(\frac{2}{3},\frac{2}{3})$

$\bigoplus \limits_{k \in even} H_{k}(X^{g}/C_{D}(g))$     $\mathbb{Z}$     $\mathbb{Z}$     $\mathbb{Z}^{2}$

$\bigoplus \limits_{k \in odd} H_{k}(X^{g}/C_{D}(g))$
0   0   0
$\sigma^{3}$     $\rho_{1}$    $\rho_{2}$

$(\pm 1,\pm 1)$, $(0,0)$, $(0,\pm 1)$, $(\pm 1,0)$     $(x,\frac{x}{2})$, $(x,\frac{x}{2}\pm 1)$      $(x,x)$

$id_{D}$, $\sigma$, $\sigma^{2}$, $\sigma^{3}$, $\sigma^{4}$, $\sigma^{5}$     $id_{D}$, $\rho_{1}$, $\sigma^{3}$     $id_{D}$, $\rho_{2}$, $\sigma^{3}$

$(0,0)$, $(1,0)$   $\{(x,\frac{x}{2})|0\leqslant x\leqslant 1\}$, $\{(x,\frac{x}{2}+1)|-1\leqslant x\leqslant 0\}$    $\{(x,x)|0\leqslant x\leqslant 1\}$

$\mathbb{Z}^{2}$     $\mathbb{Z}$     $\mathbb{Z}$

0   0   0

The computation about p6mm

4. $K_{0}(p6mm)=\mathbb{Z}^{8}, K_{1}(p6mm)=0$.

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