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For the block embedding $SU(m)\times SU(n)\subset SU(m+n)$ with $m,n\ge 2$, the centralizer is the block-scalar subgroup $C=\{\mathrm{diag}(z{I}_m,w{I}_n):|z|=|w|=1,z^m{w}^n=1\}$. Schur lemma makes the diagonal blocks scalar and kills the off-diagonal intertwiners, since the two blocks are inequivalent representations of the product group.

Put $d=\gcd (m,n)$, $a=m/d$, $b=n/d$. The map $(z,w)\mapsto z^a{w}^b$ has image the $d$th roots of unity and connected kernel $(z,w)=(e^{ib\theta },e^{-ia\theta })$. A Bezout identity completes $(a,b)$ to an invertible integer coordinate change on the two-dimensional torus, proving $C\cong U(1)\times {\mathbb{Z}}_d$. Thus the full centralizer is a circle group exactly when $m,n$ are coprime. For $m=n=2$, $(zw{)}^2=1$ gives two circles, a useful counterexample to the unrestricted connectedness claim.

For a finite-dimensional complex semisimple representation $V={⨁}_{\lambda }({\mathbb{C}}^{m_{\lambda }}\otimes V_{\lambda })$ with pairwise inequivalent irreducible representations $V_{\lambda }$, its commuting endomorphisms are exactly ${⨁}_{\lambda }(A_{\lambda }\otimes I_{V_{\lambda }})$. To prove it, write each block as an intertwiner and apply Schur's lemma: blocks between inequivalent constituents vanish, and blocks between fixed copies of the same constituent are scalar. The dimension is ${\sum }_{\lambda }{m}_{\lambda }^2$.

Let $S=s=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$. Entrywise complex conjugation interchanges the diagonal entries of $r$ and fixes $s$, soSince $r,s$ generate the group, these identities hold for every element. Thus $S$ is an invertible intertwiner, provingThe character of a tensor product representation is the product of the factor characters. Hence ${\chi }_{R_2\otimes R_2}=(4,1,0)$. Its character inner products with the three rows of solution give multiplicities $(1,1,1)$, soFor explicit Clebsch-Gordan decomposition of group representations bases, let $e_{+},e_{-}$ be the coordinate basis. The symmetric tensor $e_{+}\otimes e_{-}+e_{-}\otimes e_{+}$ spans $1$; the antisymmetric tensor $e_{+}\otimes e_{-}-e_{-}\otimes e_{+}$ spans $\epsilon$; and the ordered pair $e_{-}\otimes e_{-},e_{+}\otimes e_{+}$ carries $R_2$, since $r$ acts by $\mathrm{diag}(\alpha ,{\alpha }^2)$ and $s$ exchanges them. In particular $S^2{R}_2=1\oplus R_2$ and ${\Lambda }^2{R}_2=\epsilon$.

Multiplying by the sign representation preserves the character of $R_2$, because its character is zero on all three reflections. Thus $\epsilon \otimes R_2\cong R_2$, and associativity givesThereforeThe dimensions give $8=1+1+3\cdot 2$. Its two one-dimensional summands can also be seen directly: $e_{+}^{\otimes 3}+e_{-}^{\otimes 3}$ and $e_{+}^{\otimes 3}-e_{-}^{\otimes 3}$ transform as $1$ and $\epsilon$, respectively.

A group representation on a vector space $V$ over a field $K$ is a group homomorphism $R:G\to \mathrm{GL}(V)$. In particular, $R(gh)=R(g)R(h)$, $R(e)=I$ and $R(g^{-1})=R(g{)}^{-1}$. Choosing a basis identifies it with a matrix representation.

Two group representations $R$ on $V$ and $S$ on $W$ are equivalent representations when an invertible linear map $T:V\to W$ satisfies $TR(g)=S(g)T$ for every $g$. Thus $S(g)=TR(g)T^{-1}$; they differ only by a choice of basis.

A nonzero group representation is a reducible representation if there is a vector subspace $U$ with $0\ne U\ne V$ and $R(g)U\subseteq U$ for every $g$. It is an irreducible representation if it has no such invariant subspace. Reducibility alone does not assert the existence of an invariant complementary vector subspace; that stronger conclusion says that the representation is a semisimple representation.

The Schur lemma states that an intertwiner $T$ between irreducible representations is either zero or an isomorphism. Consequently an intertwiner between inequivalent irreducible representations is zero. For a finite-dimensional irreducible representation over the complex numbers, every commuting endomorphism is scalar:For the first assertion, the kernel and image of $T$ are invariant subspaces, so a nonzero $T$ is injective and surjective. For the scalar conclusion, choose an eigenvalue $\lambda$ of $T$. The intertwiner $T-\lambda I$ has a nonzero kernel, so it must vanish. The finite-dimensional complex hypothesis matters: the scalar conclusion need not hold over the real numbers, as a real planar rotation through $\pi /2$ is an irreducible representation of the cyclic group of order four, while the rotation itself is a commuting nonscalar endomorphism.

Write $P=(P_{ij}{)}_{i,j=1}^3$, where $P_{ij}:{\mathbb{C}}^{m_j}\to {\mathbb{C}}^{m_i}$. Block multiplication makes the commuting condition equivalent toThus every block is an intertwiner. For $i\ne j$, the irreducible representations are inequivalent, so the Schur lemma gives $P_{ij}=0$. On each diagonal block, the same Schur lemma gives $P_{ii}={\lambda }_i{I}_{m_i}$. ThereforeConversely, every displayed matrix commutes with $R(g)$, proving both necessity and sufficiency. This commutant of a semisimple representation has complex dimension three. The parameters may vanish: $P$ was not required to be invertible.

Now $m_1=m_2=n$ and the first two matrix representations agree in the chosen bases. Each of $P_{11},P_{12},P_{21},P_{22}$ is therefore a commuting endomorphism of that common irreducible representation, so each is scalar by the Schur lemma. All blocks connecting either of these copies with the third vanish, since the third irreducible representation is inequivalent. HenceEvery such matrix commutes with $R(g)$. In particular the off-diagonal scalars $b,c$ can mix the two identical copies. The commutant of a semisimple representation is $M_2(\mathbb{C})\oplus \mathbb{C}$, of complex dimension five. If the first two representations were merely equivalent, rather than literally equal, the mixed blocks would instead be scalar multiples of a fixed invertible intertwiner and its inverse.

First choose the same orthonormal basis for the left and right copies of the two-dimensional unitary representation $D$. Then the family representation is $F=D\oplus \epsilon$. Because the barred fermion transforms with $F(g{)}^{†}$ and the scalar field $H$ is a trivial representation, invariance of the Yukawa interaction requiresWrite $\lambda$ in a $2+1$ block form. The off-diagonal blocks would be intertwiners between inequivalent irreducible representations $D$ and $\epsilon$, so the Schur lemma makes them zero. The doublet block must be a scalar matrix, while the singlet block is arbitrary. ThereforeEquivalently, the allowed Yukawa interaction is $H[a({\bar{\psi }}_L^1{\psi }_R^1+{\bar{\psi }}_L^2{\psi }_R^2)+b{\bar{\psi }}_L^3{\psi }_R^3]$, together with its Hermitian conjugate when constructing a real Lagrangian. The tensor product rules explain the two free parameters: both $D^{\ast }\otimes D$ and ${\epsilon }^{\ast }\otimes \epsilon$ contain one trivial representation, while the mixed products contain none.

There is a useful basis qualification. The basis used in solution is not orthonormal: its Gram matrix is $K=\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)$, and $D(g{)}^{†}KD(g)=K$. If the fields are expressed in exactly those coordinates, with the barred components defined by ordinary conjugate transpose, the invariant doublet coefficient is $aK$, rather than $a{I}_2$. Canonically normalizing the kinetic term returns the boxed form. Independently chosen left and right orthonormal bases replace $a{I}_2$ by $aS$, where $S$ is their fixed unitary intertwiner; the physical degeneracy is unchanged.

Write $X=\left(\begin{array}{cc}A& B\\ C& D\end{array}\right)$. Commuting with all $\mathrm{diag}(U_1,U_2)$ requiresFor $n_1,n_2\ge 2$, the defining complex representations of each factor are irreducible. Schur lemma gives $A=z{I}_{n_1}$ and $D=w{I}_{n_2}$. The two block vector spaces are inequivalent representations of the product group, even when $n_1=n_2$: the first factor acts nontrivially on just the first block. Thus their off-diagonal intertwiners vanish, $B=C=0$. Unitarity and the unit determinant give

The assertion that this full group is always a circle needs correction. Put $d=\gcd (n_1,n_2)$, $a=n_1/d$, $b=n_2/d$. The continuous group homomorphismmaps this centralizer onto the $d$th roots of unity. Its kernel is the identity component of a Lie group, parametrized injectively by $e^{i\theta }\mapsto (e^{ib\theta },e^{-ia\theta })$. Coprimality of $a,b$ proves injectivity; a Bezout identity proves that this parametrization exhausts the kernel. Extending the primitive integer row $(a,b)$ to a determinant-one integer matrix gives coordinates in which the constraint is simply $t^d=1$. ConsequentlyFor a concrete counterexample, take $n_1=n_2=2$. The constraint is $(zw{)}^2=1$, giving two separate circles $zw=1$ and $zw=-1$. The element $\mathrm{diag}(I_2,-I_2)$ lies in the second component, so this centralizer cannot be isomorphic as a Lie group to the connected circle group.

For the intended specialization of this centralizer of two special unitary blocks, $d=\gcd (2,3)=1$ and therefore the full centralizer really is a circle group:We use this primitive integer U(1) charge normalization below. If exactly one $n_i$ equals $1$ and the other is at least $2$, the same block-scalar formula still holds. If $n_1=n_2=1$, both factors are trivial and the centralizer is all of SU(2), another exception to the printed claim.

Schur's lemma states that an intertwiner $T:V\to W$ between irreducible representations is either zero or an isomorphism. In particular, inequivalent irreducible representations have no nonzero intertwiners. For a finite-dimensional irreducible representation over the complex numbers, every commuting endomorphism is scalar:Indeed, the kernel and image of an intertwiner are invariant subspaces, proving the first statement. For the second, choose an eigenvalue $\lambda$ of $T$. The intertwiner $T-\lambda I$ has nonzero kernel, so it must vanish. The scalar conclusion uses an algebraically closed coefficient field: over the real numbers, planar rotation through $\pi /2$ gives an irreducible real representation of ${\mathbb{Z}}_4$ and commutes with itself without being a real scalar matrix.

Partition $M$ into the same five-by-five blocks as $R$. The equation for block $(i,j)$ is $M_{ij}{R}_j(g)=R_i(g)M_{ij}$, so each block is an intertwiner. By Schur's lemma, blocks between the two inequivalent irreducible representations vanish, whereas blocks between identical copies are arbitrary scalar multiples of the relevant identity matrix. Thus the commutant of a semisimple representation isEquivalently, $M=(A\otimes I_{n_1})\oplus (B\otimes I_{n_2})$, where $A\in M_3(\mathbb{C})$ and $B\in M_2(\mathbb{C})$. Substitution proves every such matrix commutes with every $R(g)$. There is no invertibility requirement on $M$, and its solution space has complex dimension $3^2+2^2=13$, even if $n_1=n_2$.

Use the contraction splitting of a defining module tensor a dual symmetric square. For a quadratic polynomial $p$ in the dual variables, defineBoth maps are intertwiners. The first is the natural tensor contraction; the second uses the invariant tensor ${\sum }_i{e}_i\otimes y_i$. Directly, $C(Q_j)=4y_j$, so $C\iota =4I$ andThe vector $T_{1;33}=e_1\otimes y_3^2$ lies in $\ker C$ and is killed by both positive simple-root operators $E_{12},E_{23}$. It is a highest-weight vector of weight $L_1-2L_3={\omega }_1+2{\omega }_2$. By Weyl complete reducibility theorem, it generates the corresponding irreducible summand. The Weyl dimension formula for ${\mathfrak{sl}}_3$ gives $\dim {\Gamma }_{a,b}=(a+1)(b+1)(a+b+2)/2$, hence $\dim {\Gamma }_{1,2}=15$, exhausting the kernel. ThereforeThe kernel has the nine outer weights of multiplicity one and each inner weight $-L_j$ of multiplicity two; the dual summand has just the three weights $-L_j$, each once.

The printed phrase “not reducible” is a misprint: the requested nonzero complementary submodules establish that this tensor square is reducible. The natural reason is that the flip $\tau (v\otimes w)=w\otimes v$ commutes with the Lie algebra representation. Its two eigenspaces are the symmetric square and the exterior square, givingIn the symmetric square, the four weights $\pm 2L_i$ and four weights $\pm L_1\pm L_2$ each occur once, while zero occurs twice. In the exterior square, the four weights $\pm L_1\pm L_2$ each occur once and zero occurs twice; the $\pm 2L_i$ are absent because repeated wedge factors vanish. These are the two requested complementary weight diagrams.

More precisely, the tensor-square decomposition of the defining symplectic representation identifies ${\mathrm{Sym}}^{2}V$ with the irreducible adjoint module $L(2{\omega }_1)$. The identification is an intertwiner sending a symmetric tensor $vw$ to the element of the symplectic Lie algebra $x\mapsto \omega (v,x)w+\omega (w,x)v$. For the exterior square, symplectic contraction of an exterior square has a five-dimensional kernel, the primitive exterior square, and the invariant tensor $\Omega =e_1\wedge e_3+e_2\wedge e_4$ supplies a complementary line. Thus the full irreducible decomposition isHere ${\omega }_1=L_1$ and ${\omega }_2=L_1+L_2$. The two zero weights in the exterior square split into one primitive zero weight and the invariant line.

Within each fixed unitary irreducible representation, choose the selected right singular vectors to be orthonormal. Their corresponding left singular vectors are also orthonormal, even when a singular value is repeated: choose an orthonormal basis in each eigenspace of $T_{\rho }^{\ast }{T}_{\rho }$ and put $B_p=T_{\rho }{A}_p/{\lambda }_p$. Therefore, when ${\rho }_p={\rho }_q$,This is (iv), with the unnormalized Hilbert-Schmidt inner product and the Kronecker delta. No orthogonality of differently shaped matrices is being asserted.

For the unheaded continuation, partition $b\in {\mathbb{C}}^m$ into blocks $b_p\in {\mathbb{C}}^{n_p}$. ThenThe Schur averaging of rectangular matrices formula isfor any $n_p\times n_q$ matrix $M$. The average is an intertwiner; the Schur lemma makes it zero for inequivalent irreducible representations, and a scalar multiple of the identity for the same representative. The trace determines that scalar in the latter case.

Apply this with $M=U(p{)}^{\ast }U(q)$. The Hilbert-Schmidt inner product normalization just proved givesExpanding the squared Euclidean norm now yieldsThus the final identity follows from averaging, although $U^{\ast }U$ itself need not be the identity. The whole construction is the spectral inverse theorem for the matrix-valued U2 quantity.

For chosen inequivalent unitary irreducible representations $\rho ,\sigma$ of a finite group and a $d_{\rho }\times d_{\sigma }$ matrix $M$, the average ${\mathbb{E}}_x\rho (x)M\sigma (x{)}^{\ast }$ is zero when $\rho \ne \sigma$. For $\rho =\sigma$, it is $(\mathrm{tr}M/d_{\rho })I$. This follows from the Schur lemma, since the average intertwines the two group representations. Equivalent representations in different bases require the corresponding intertwiner; the identity-matrix formula assumes literally the same representative.

For inequivalent chosen unitary irreducible representations of a finite group, uniform expectation givesThe Kronecker deltas in the first case refer to entries in the same chosen basis. Averaging an intertwiner and applying the Schur lemma proves the formula; summing diagonal entries gives character orthogonality.