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The Kronecker sum of square matrices combines their Kronecker products with identity matrices. If $Ax=\lambda x$ and $By=\mu y$, then $x\otimes y$ is an eigenvector with eigenvalue $\lambda +\mu$. Tensor-product spatial grids therefore turn separated coordinate Laplacians into a Kronecker sum. Two Hermitian matrices give a Hermitian Kronecker sum, allowing an energy method without computing every eigenvector.

Use the basis matrices $M_{ab}=E_{ab}-E_{ba}$ from solution. Define the Pauli matrices and the two-by-two identity byThe following five Hermitian four-by-four matrices, written as Kronecker products, realize a Euclidean Clifford algebra:Define the Lie algebra representation on ${\mathbb{C}}^4$ byThese are anti-Hermitian matrices, appropriate to the real compact Lie algebra. For complete explicitness, all ten representation matrices areEach table entry is the image under $\rho$ of the basis matrix immediately to its left. Since the two-by-two factors are specified above, these Kronecker products specify every entry of each four-by-four matrix.

To verify the representation law, the Clifford algebra relations implyApply this identity to $\rho (M_{cd})=\frac{1}{4}[{\gamma }_c,{\gamma }_d]$ and use the commutator product rule. The result isexactly the Lie bracket relation in solution. Thus $\rho$ is a Lie algebra representation.

It remains to identify its highest weight, rather than just its dimension. The commuting Hermitian Cartan subalgebra operators areTheir simultaneous eigenvalues are $(x,y)=(\pm \frac{1}{2},\pm \frac{1}{2})$. Converting with $(a,b)=(x-y,2y)$ gives exactly the four weights of $V(0,1)$ in solution. A state with $(x,y)=(\frac{1}{2},\frac{1}{2})$ has highest weight $(0,1)$, so this Clifford realization of the so5 spin representation is the requested $r$.

Finally, $\rho (M_{12}{)}^2=-I_4/4$, soThis explicitly exhibits the distinction between the spinor representation of $\mathfrak{so}(5)$, or of $\mathrm{Spin}(5)$, and a single-valued group representation of $SO(5)$.

All three vector spaces now have the same dimension $n$. Each of the nine blocks of $P$ commutes with the common irreducible representation, so the Schur lemma givesConversely $R(g)=I_3\otimes R_1(g)$, so the Kronecker product identity immediately verifies commutation for every $A$. The commutant of a semisimple representation acts arbitrarily on the three-dimensional multiplicity vector space while acting as a scalar on each irreducible factor. Its complex dimension is nine.

The componentwise product is closed in $G_1\times G_2$. Associativity follows componentwise from associativity in each group:The identity is $(e_1,e_2)$, and $(g_1,g_2{)}^{-1}=(g_1^{-1},g_2^{-1})$. These verify all the group axioms, so this is the direct product of groups.

Write the paper's Kronecker product $\times$ as $\otimes$. Its mixed-product identity and the representation laws of $R_1,R_2$ giveAlso $R(e_1,e_2)=I_{n_1}\otimes I_{n_2}=I_{n_1{n}_2}$, andThus the images are invertible and this is a group representation of dimension $n_1{n}_2$, called the external tensor product of group representations. No irreducibility assumption is needed for this construction.

Put $h=1/(m+1)$ and impose homogeneous Dirichlet boundary conditions, as required by the stated continuum eigenvalues. Let $T=\mathrm{tridiag}(1,-2,1)$ be the unscaled one-dimensional Dirichlet discrete Laplacian. For $1\le k\le m$, the vectors $v_j^{(k)}=\sin (jk\pi h)$ satisfyThe identity follows by adding the two neighboring sines; the values at $j=0,m+1$ vanish. These $m$ mutually orthogonal vectors form the discrete sine transform basis. Consequently their tensor products give all $m^2$ two-dimensional modes, not merely a few candidate eigenvectors.

In a compatible grid ordering the axial operator is $A_5=h^{-2}(T\otimes I+I\otimes T)$. The nine-point operator isIndeed, the added Kronecker product supplies the diagonal neighbors with weight $1/6$, subtracts $1/3$ from each axial neighbor and adds $2/3$ to the central coefficient. Thus it has exactly the coefficients of the second discretization. On $v^{(k)}\otimes v^{(l)}$ the eigenvalues areandEquivalently, $h^2{\lambda }_{9,kl}=-10/3+4[\cos (k\pi h)+\cos (l\pi h)]/3+2\cos (k\pi h)\cos (l\pi h)/3$.

For fixed mode numbers $k,l$, expansion of the sines giveswhereas the mixed Kronecker product correction givesThus both eigenvalue approximations are second order for fixed modes. The leading nine-point error includes an additional positive term ${\pi }^4{h}^2{k}^2{l}^2/6$.

There is also an exact comparison, not just an asymptotic one. Since $\sin x<x$ for $0<x<\pi /2$, and both sine factors in the correction are positive,Both estimates lie on the less-negative side of the continuum Laplacian eigenvalue, so the five-point formula is the better eigenvalue estimate for every corresponding interior mode. The five-point versus nine-point Laplacian eigenvalue accuracy comparison does not contradict the useful isotropy or special Poisson accuracy properties of a nine-point stencil. Those properties concern a different error criterion. The fixed-mode expansion is not uniform for modes with $k$ or $l$ comparable to $h^{-1}$; wavelengths close to the mesh scale need not have small relative spectral error.

A modulus of a number field is $\mathfrak{m}={\mathfrak{m}}_0{\mathfrak{m}}_{\infty }$, with ${\mathfrak{m}}_0$ a nonzero integral ideal and ${\mathfrak{m}}_{\infty }$ a subset of the real places. Let $I_{\mathfrak{m}}$ be the group of fractional ideals prime to ${\mathfrak{m}}_0$, and let $P_{\mathfrak{m}}$ consist of principal ideals $(a)$ with $a\equiv 1$ at each prime power in ${\mathfrak{m}}_0$ and $a>0$ at each selected real place. The ray class group isThe ray class field $K_{\mathfrak{m}}$ is the finite abelian extension for which the Artin map has kernel $P_{\mathfrak{m}}$ and induces an isomorphism ${\mathrm{Cl}}_{\mathfrak{m}}(K)\cong \mathrm{Gal}(K_{\mathfrak{m}}/K)$. Equivalently it is the maximal abelian extension whose conductor divides $\mathfrak{m}$.

To compute its order use the ray class exact sequence, including signatures:The middle map lifts a residue and sign tuple to an element of $K$ by weak approximation theorem, then takes its principal ideal as a ray class. Different lifts differ by a principal ray factor. Its kernel consists exactly of tuples represented by global units. Every ordinary ideal class has a representative prime to the finite modulus, again by weak approximation theorem; this gives surjectivity of the last map. If $U_{\mathfrak{m}}$ denotes the ray units, the kernel of the unit map is $U_{\mathfrak{m}}$. The Chinese remainder theorem and counting units in each prime-power quotient give the ray class number formula

First take $K=\mathbb{Q}(\sqrt{-3})$. Its ring of integers is $\mathbb{Z}[\omega ]$, where $\omega =(-1+\sqrt{-3})/2$ and ${\omega }^2+\omega +1=0$. The Minkowski bound for ideal classes is $(2/\pi )\sqrt{3}<2$, so every ideal class has a representative of norm one and $h_K=1$. Its units are the six roots of unity. Modulo seven, the polynomial $X^2+X+1$ has distinct roots $2,4$, and thereforeThe unit group of this quotient has order $36$. The unit $1+\omega$ maps to $(3,5)$, which has order six; consequently unit reduction is injective and $U_{(7)}=\{1\}$. The ray class number formula gives $36/6=6$.

We also identify the quotient explicitly. Conjugation interchanges the two factors, so the residue norm is $(x,y)\mapsto xy$. Its kernel has order six and equals the image of the six units, since $3\cdot 5=1(\mathrm{mod}7)$. HenceThe field $M=K({\zeta }_7)=\mathbb{Q}({\zeta }_{21})$ has degree six over $K$, since the degrees over $\mathbb{Q}$ are twelve and two. It is unramified outside seven by the cyclotomic computation and the base-change argument above. At primes over seven, $K$ has completion ${\mathbb{Q}}_7$, and $M/K$ is the tame degree-six cyclotomic extension; its conductor exponent is one. Its relative arithmetic Artin symbol at an ideal $\mathfrak{a}$ prime to seven acts byIf $\mathfrak{a}=(a)$ with $a\equiv 1(\mathrm{mod}7{\mathcal{O}}_K)$, then $N_{K/\mathbb{Q}}(a)\equiv 1(\mathrm{mod}7)$. Thus the Artin map kills the principal ray subgroup. It factors through the order-six ray class group; the residue norm quotient computed above shows that it is onto the degree-six Galois group, so the fields coincide. This proves the explicit ray class field of Q of square root minus three modulo seven:

Now take $K=\mathbb{Q}(\sqrt{-6})$, and write $s=\sqrt{-6}$. Its ring of integers is $\mathbb{Z}[s]$, its discriminant is $-24$, and its units are $\pm 1$. The quotient by $(s)$ iswhose two units are both images of $\pm 1$. The ray class exact sequence therefore gives ${\mathrm{Cl}}_{(s)}(K)\cong \mathrm{Cl}(K)$.

For its class number, the Minkowski bound for ideal classes is $(2/\pi )\sqrt{24}<4$, so each ideal class contains an integral ideal of norm at most three. The primes of norms two and three areThese are the only nontrivial possibilities within the bound, and their ideal classes coincide and have order at most two. The ideal ${\mathfrak{p}}_2$ is not principal: its being principal would give an integer solution to $x^2+6y^2=2$, which is impossible. Thus $h_K=2$, the ideal class group is $C_2$, and the ray field is the Hilbert class field.

We verify its actual identity rather than stopping at the degree. PutThe two quadratic fields $\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}(\sqrt{-3})$ are distinct, so $[H:\mathbb{Q}]=4$ and $[H:K]=2$. The integral elements $1,\sqrt{2},b,\sqrt{2}b$ are a rational basis. Their trace matrix is the Kronecker product of the two quadratic trace matrices, with discriminants $8$ and $-3$. Its determinant is therefore $8^2(-3{)}^2=576$. The discriminant-index formula for an integral lattice shows that the field discriminant divides $576$. On the other hand the relative discriminant ideal formula givesThe norm on the right is a positive integer, so equality is forced. Thus the relative discriminant ideal is the unit ideal, and $H/K$ is unramified at every finite prime. There are no real places of $K$. The extension is quadratic and therefore Abelian; its degree equals $h_K$, identifying it with the Hilbert class field of Q of square root minus six. Hence the ray class field of Q of square root minus six modulo square root minus six is

At an interior grid point, the five-point finite difference method givesLetwhere the zero Dirichlet boundary values are already incorporated. Stack the unknowns with the $y$ index varying fastest. Using the Kronecker product, defineThen the semidiscrete equation isMoreover,Thus the directional matrices commute, as in the five-point Dirichlet Laplacian as a Kronecker sum.