A Symmetry-Based Decomposition Approach to Eigenvalue Problems

Abstract

In this paper, we propose a decomposition approach to differential eigenvalue problems with Abelian or non-Abelian symmetries. In the approach, we divide the original differential problem into eigenvalue subproblems which require less eigenpairs and can be solved independently. Our approach can be seamlessly incorporated with grid-based discretizations such as finite difference, finite element, or finite volume methods. We place the approach into a two-level parallelization setting, which saves the CPU time remarkably. For illustration and application, we implement our approach with finite elements and carry out electronic structure calculations of some symmetric cluster systems, in which we solve thousands of eigenpairs with millions of degrees of freedom and demonstrate the effectiveness of the approach.

Appendices

Appendix 1: Basic Concept of Group Theory

In this appendix, we include some basic concepts of group theory for a more self-contained exposition. They could be found in standard textbooks like [8, 9, 34].

A group $G$ is a set of elements $\{R\}$ with a well-defined multiplication operation which satisfy the following requirements:

• 1. The set is closed under the multiplication.

• 2. The associative law holds.

• 3. There exists a unit element $E$ such that $ER=RE=R$ for any $R\in G$.

• 4. There is an inverse $R^{-1}$ in $G$ to each element $R$ such that $RR^{-1} = R^{-1}R = E$.

If the commutative law of multiplication also holds, $G$ is called an Abelian group. Group $G$ is called a finite group if it contains a finite number of elements. And this number, denoted by $g$, is said to be the order of the group. The rearrangement theorem tells that the elements of $G$ are only rearranged by multiplying each by any $R\in G$, i.e., $RG=G$ for any $R\in G$.

An element $R_1\in G$ is called to be conjugate to $R_2$ if $R_2=SR_1S^{-1}$, where $S$ is some element in the group. All the mutually conjugate elements form a class of elements. It can be proved that group $G$ can be divided into distinct classes. Denote the number of classes as $n_c$. In an Abelian group, any two elements are commutative, so each element forms a class by itself, and $n_c$ equals the order of the group.

Groups $G=\{R\}$ and $G^{\prime }=\{R^{\prime }\}$ are called to be homomorphic, if there exists a correspondence between the elements of $G$ and $G^{\prime }$ as $R\leftrightarrow R^{\prime }_1,R^{\prime }_2,\ldots$, which means that if $RS=T$ then the product of any $R^{\prime }_i$ with any $S^{\prime }_j$ will belong to the set $\{T^{\prime }_1,T^{\prime }_2,\ldots \}$. This is a many-to-one correspondence in general. If the correspondence specializes to be one-to-one, the two groups are called to be isomorphic.

A matrix representation of group $G$ means a group of matrices which is homomorphic to $G$. Two representations are said to be equivalent if they are associated by a similarity transformation. If a representation can not be equivalent to representations of lower dimensionality, it is called irreducible. Any matrix representation with nonzero determinants is equivalent to a unitary representation, i.e., a representation by unitary matrices.

The number of all the inequivalent, irreducible, unitary representations is equal to $n_c$, which is the number of classes in $G$. The Celebrated Theorem tells that

$$\begin{aligned} \sum _{\nu =1}^{n_c}d_{\nu }^2=g, \end{aligned}$$

where $d_{\nu }$ denotes the dimensionality of the $\nu$th representation. Since the number of classes of an Abelian group equals the number of elements, an Abelian group of order $g$ has $g$ one-dimensional irreducible representations.

The groups used in this paper are all crystallographic point groups. Groups $D_2,\;D_{2h},\;D_{2d}$ and $D_4$ are four dihedral groups; the first two groups are Abelian and the other two are non-Abelian. In Tables 2 and 7, $C_{nj}$ denotes a rotation about axis $Oj$ by $2\pi /n$ in the right-hand screw sense and $I$ is the inversion operation [8]. The $Oj$ axes are illustrated in Fig. 6. We refer to textbooks like [8, 9, 34] for more details about crystallographic point groups.

Fig. 6

Rotation axes in Tables 2 and 7

Appendix 2: Proof of Proposition 2

Proof

(a) Since $\{P_R\}$ are unitary operators, we have

$$\begin{aligned} {{\fancyscript{P}}^{(\nu )}_{ml}}^* = \frac{d_{\nu }}{g} \sum _{R\in G} \varGamma ^{(\nu )}(R)_{ml} P_{R^{-1}} = \frac{d_{\nu }}{g} \sum _{S\in G} \varGamma ^{(\nu )}(S^{-1})_{ml} P_S, \end{aligned}$$

which together with the fact that $\varGamma ^{(\nu )}$ is a unitary representation derives

$$\begin{aligned} {{\fancyscript{P}}^{(\nu )}_{ml}}^* = \frac{d_{\nu }}{g} \sum _{S\in G} {\varGamma ^{(\nu )}(S)}^*_{lm} P_S = {\fancyscript{P}}^{(\nu )}_{lm}. \end{aligned}$$

(b) It follows from the definition that

$$\begin{aligned} {\fancyscript{P}}^{(\nu )}_{ml} {\fancyscript{P}}^{(\nu ^{\prime })}_{m^{\prime }l^{\prime }}&= \left( \frac{d_{\nu }}{g}\sum _{R\in G} {\varGamma ^{(\nu )}(R)}^*_{ml}P_R\right) \left( \frac{d_{\nu ^{\prime }}}{g} \sum _{S\in G} {\varGamma ^{(\nu ^{\prime })}(S)}^*_{m^{\prime }l^{\prime }}P_S\right) \\&= \frac{d_{\nu }d_{\nu ^{\prime }}}{g^2} \sum _{R\in G}{\varGamma ^{(\nu )}(R)}^*_{ml} \left( \sum _{S\in G} {\varGamma ^{(\nu ^{\prime })}(S)}^*_{m^{\prime }l^{\prime }}P_{RS} \right) . \end{aligned}$$

Note that the rearrangement theorem implies that, when $S$ runs over all the group elements, $S^{\prime }=RS$ for any $R$ also runs over all the elements. Hence we get

$$\begin{aligned} {\fancyscript{P}}^{(\nu )}_{ml} {\fancyscript{P}}^{(\nu ^{\prime })}_{m^{\prime }l^{\prime }}&= \frac{d_{\nu }d_{\nu ^{\prime }}}{g^2} \sum _{R\in G}{\varGamma ^{(\nu )}(R)}^*_{ml} \left( \sum _{S^{\prime }\in G} {\varGamma ^{(\nu ^{\prime })}(R^{-1}S^{\prime })}^*_{m^{\prime }l^{\prime }}P_{S^{\prime }} \right) \\&= \frac{d_{\nu }d_{\nu ^{\prime }}}{g^2} \sum _{S^{\prime }\in G} \left( \sum _{R\in G} {\varGamma ^{(\nu )}(R)}^*_{ml}~{\varGamma ^{(\nu ^{\prime })}(R^{-1}S^{\prime })}^*_{m^{\prime }l^{\prime }} \right) P_{S^{\prime }}. \end{aligned}$$

We may calculate as follows

$$\begin{aligned} \sum _{R\in G} {\varGamma ^{(\nu )}(R)}^*_{ml} {\varGamma ^{(\nu ^{\prime })}(R^{-1}S^{\prime })}^*_{m^{\prime }l^{\prime }}&= \sum _{R\in G} {\varGamma ^{(\nu )}(R)}^*_{ml} \left( \sum _{n=1}^{d_{\nu ^{\prime }}} {\varGamma ^{(\nu ^{\prime })}(R^{-1})}^*_{m^{\prime }n}~{\varGamma ^{(\nu ^{\prime })}(S^{\prime })}^*_{nl^{\prime }} \right) \\&= \sum _{R\in G} {\varGamma ^{(\nu )}(R)}^*_{ml} \left( \sum _{n=1}^{d_{\nu ^{\prime }}} {\varGamma ^{(\nu ^{\prime })}(R)}_{nm^{\prime }}~{\varGamma ^{(\nu ^{\prime })}(S^{\prime })}^*_{nl^{\prime }} \right) \\&= \sum _{n=1}^{d_{\nu ^{\prime }}} {\varGamma ^{(\nu ^{\prime })}(S^{\prime })}^*_{nl^{\prime }} \left( \sum _{R\in G} {\varGamma ^{(\nu )}(R)}^*_{ml}~{\varGamma ^{(\nu ^{\prime })}(R)}_{nm^{\prime }} \right) , \end{aligned}$$

which together with the great orthogonality theorem yields

$$\begin{aligned} \sum _{R\in G} {\varGamma ^{(\nu )}(R)}^*_{ml} {\varGamma ^{(\nu ^{\prime })}(R^{-1}S^{\prime })}^*_{m^{\prime }l^{\prime }} = \delta _{\nu \nu ^{\prime }} \delta _{lm^{\prime }}\frac{g}{d_{\nu ^{\prime }}} {\varGamma ^{(\nu )}(S^{\prime })}^*_{ml^{\prime }}. \end{aligned}$$

Thus we arrive at

$$\begin{aligned} {\fancyscript{P}}^{(\nu )}_{ml} {\fancyscript{P}}^{(\nu ^{\prime })}_{m^{\prime }l^{\prime }} = \delta _{\nu \nu ^{\prime }} \delta _{lm^{\prime }} \frac{d_{\nu }}{g} \sum _{S^{\prime }\in G} {\varGamma ^{(\nu )}(S^{\prime })}^*_{ml^{\prime }} P_{S^{\prime }} = \delta _{\nu \nu ^{\prime }} \delta _{lm^{\prime }} {\fancyscript{P}}^{(\nu )}_{ml^{\prime }}. \end{aligned}$$

$\square$

Appendix 3: An Example

We take the Laplacian in square $(-1,1)^2$ as an example to illustrate the subproblem formulation in Corollary 2. Namely, we consider the decomposition of the following eigenvalue problem

$$\begin{aligned} \left\{ \begin{array}{rcll} -\Delta u &{} = &{}\lambda u &{}\quad \text{ in }~\varOmega =(-1,1)^2,\\ u &{} = &{}0 &{}\quad \text{ on }~\partial \varOmega . \end{array} \right. \end{aligned}$$

(27)

Note that $G=\{E,\sigma _x,\sigma _y,I\}$ is a symmetry group associated with (27), where $E$ represents the identity operation, $\sigma _x$ a reflection about $x$-axis, $\sigma _y$ a reflection about $y$-axis, and $I$ the inversion operation. We see that $G$ is an Abelian group of order 4, and has 4 one-dimensional irreducible representations as shown in Table 9.

Table 9 Representation matrices of example group $G$

According to Theorem 1 and Corollary 2, eigenvalue problem (27) can be decomposed into 4 subproblems (due to $\sum _{\nu =1}^{n_c}d_{\nu }=4$). And the symmetry characteristic conditions, the third equation in (13), for the 4 subproblems are

$$\begin{aligned} \{u^{(1)}(R_1x),u^{(1)}(R_2x),u^{(1)}(R_3x),u^{(1)}(R_4x)\}&= \{1, 1, 1, 1\} ~u^{(1)}(x), \end{aligned}$$

(28)

$$\begin{aligned} \{u^{(2)}(R_1x),u^{(2)}(R_2x),u^{(2)}(R_3x),u^{(2)}(R_4x)\}&= \{1, 1,-1,-1\} ~u^{(2)}(x), \end{aligned}$$

(29)

$$\begin{aligned} \{u^{(3)}(R_1x),u^{(3)}(R_2x),u^{(3)}(R_3x),u^{(3)}(R_4x)\}&= \{1,-1,-1, 1\} ~u^{(3)}(x), \end{aligned}$$

(30)

$$\begin{aligned} \{u^{(4)}(R_1x),u^{(4)}(R_2x),u^{(4)}(R_3x),u^{(4)}(R_4x)\}&= \{1,-1, 1,-1\} ~u^{(4)}(x), \end{aligned}$$

(31)

where $x\in \varOmega$ is an arbitrary point and subscripts of $\{u_1^{(\nu )}:\nu =1,2,3,4\}$ are omitted.

In Fig. 7, we illustrate four eigenfunctions of (27) belonging to different subproblems. We see that $u_2$ and $u_3$ are degenerate eigenfunctions corresponding to $\lambda =\frac{5}{4}\pi ^2$ with double degeneracy. In other words, a doubly-degenerate eigenvalue of the original problem becomes nondegenerate for subproblems. This implies a relation between symmetry and degeneracy [24, 29, 36]. Moreover, the first subproblem does not have this eigenvalue, which shows that the decomposition approach has improved the spectral separation.

Fig. 7

Four eigenfunctions of problem (27): $u_1$ keeps invariant under $\{E,\sigma _x,\sigma _y,I\}$ and satisfies Eq. (28), and $u_2, u_3$ and $u_4$ satisfy (29), (30) and (31), respectively

Appendix 4: Spatial Symmetry in Reciprocal Space

Plane wave method is widely used for solving the Kohn–Sham equations of crystals. Actually, plane waves may be regarded as grid-based discretizations in reciprocal space. We will show that the symmetry relation in real space is kept in reciprocal space. The solution domain $\varOmega$ of crystals can be spanned by three lattice vectors in real space. We denote them as $\mathbf{a}_i (i=1,2,3)$. If function $f$ is invariant with integer multiple translations of the lattice vectors, we then present the function in reciprocal space as like:

$$\begin{aligned} \hat{f}(\mathbf{q})=\frac{1}{N}\sum _{ \mathbf {r}}f(\mathbf{r}) e^{-\imath \mathbf{q}\cdot \mathbf{r}}, \end{aligned}$$

where $\mathbf{q}$ is any vector in reciprocal space satisfying $\mathbf{q}\cdot \mathbf{a}_i=\frac{2\pi n}{N_i}$ with $n$ an integer, $N_i$ the number of degrees of freedom along direction $\mathbf{a}_i~(i=1,2,3)$, and $N=N_1N_2N_3$ the total number of degrees of freedom. Assume that $f$ is kept invariant under coordinate transformation $R$ in $\varOmega$. We obtain from

$$\begin{aligned} \hat{f}(R\mathbf{q}) = \frac{1}{N}\sum _{ \mathbf{r}}f(\mathbf{r}) e^{-\imath (R\mathbf{q})\cdot \mathbf{r}} \end{aligned}$$

and the coordinate transformation $R$ can be represented as an orthogonal matrix that

$$\begin{aligned} \hat{f}(R\mathbf{q}) = \frac{1}{N}\sum _{ \mathbf{r}}f( \mathbf{r}) e^{-\imath \mathbf{q}\cdot \left( R^{-1}\mathbf{r}\right) }. \end{aligned}$$

Since

$$\begin{aligned} f(R^{-1}\mathbf{r})=f(\mathbf{r})\quad \forall \mathbf{r}\in \varOmega , \end{aligned}$$

we have

$$\begin{aligned} \hat{f}(R\mathbf{q}) = \frac{1}{N}\sum _{R^{-1}\mathbf{r}}f(R^{-1}\mathbf{r}) e^{-\imath \mathbf{q}\cdot \left( R^{-1}\mathbf{r}\right) } = \hat{f}(\mathbf{q}). \end{aligned}$$

Hence the decomposition approach is probably applicable to plane waves.