Embeddings and branching: why the map matters

A branching rule is the decomposition of a representation after pulling it back along a specified homomorphism. The two group names do not determine the answer. Different maps from $SU(2)$ to $SU(3)$ give different multiplets, and a block map into $SU(5)$ carries a finite kernel that constrains every abelian charge in its output.

Throughout, representations are finite-dimensional, continuous and complex, in characteristic zero; the groups are compact. Thus restrictions are completely reducible. Familiarity with highest weights and representation descent is useful. The map atlas records maps separately from their domain and codomain groups.

Restriction is composition

Given $\iota:H\to G$ and $\rho:G\to GL(V)$, the restricted action is

$$\rho|_\iota=\rho\circ\iota, \qquad V|_\iota\cong\bigoplus_\mu m_\mu W_\mu.$$

Here $W_\mu$ runs over irreducible representations of the domain $H$. Every element of $\ker\iota$ acts trivially on every summand. Consequently these summands descend to the image $\iota(H)\cong H/\ker\iota$. A map with a nontrivial kernel is a homomorphism from $H$, and an embedding of this quotient into $G$.

Conjugating $\iota$ by an element of $G$ gives an equivalent restriction, with the intertwiner supplied by $\rho$ of that element. More generally, there can be inequivalent maps with the same domain and codomain. Their differentials describe the Lie-algebra maps, but the differential alone does not record a discrete kernel. Always retain the global map as well.

For connected compact groups, one effective calculation is to restrict the full character to a maximal torus of $H$, then decompose it into irreducible characters. This requires the actual torus map; it does not mean retaining only the parent’s highest weight. The opening “Branching rules” section of the official SageMath tutorial explains this character-restriction method.

Two maps from SU(2) to SU(3)

Let $V_n=\operatorname{Sym}^n\mathbb C^2$ denote the $SU(2)$ irrep with highest weight $n=2j$, dimension $n+1$, and central action $(-1)^n$. Use $SU(3)$ Dynkin labels $(p,q)$ with $\mathbf3=(1,0)$ and $\overline{\mathbf3}=(0,1)$.

The regular map is $\iota_{\mathrm r}(U)=\operatorname{diag}(U,1)$. The principal map is $\iota_{\mathrm p}(U)=\operatorname{Sym}^2U$, written in an orthonormal basis of $V_2$. It is unitary and has determinant one: the eigenvalues of $U$ are $z,z^{-1}$, so those of its symmetric square are $z^2,1,z^{-2}$. These maps give

Map	Kernel	Image in SU(3)	Defining representation pulled back
Regular	$\{I\}$	A block $SU(2)$	$V_1\oplus V_0$
Principal	$\{I,-I\}$	A conjugate of the real $SO(3)$ subgroup	$V_2$

For the principal kernel, $\operatorname{Sym}^2U=I$ forces $z^2=1$; a unitary $U$ with both eigenvalues $+1$ or both $-1$ is $I$ or $-I$. The spin-one representation is the complexification of the real rotation representation. This identifies the image with $SO(3)$, not a second embedded copy of $SU(2)$. For an explicit generator comparison, see Haber, Physics 251 Solution Set 4, Problem 5, equations (44)–(46), pp. 14–15. The regular and principal atlas records preserve the distinction.

Now pull back the same two parent representations. Functoriality of tensor and symmetric powers lets us compute without a branching table:

$$\begin{aligned} \mathbf6|_{\mathrm r} &=\operatorname{Sym}^2(V_1\oplus V_0) =V_2\oplus V_1\oplus V_0,\\ \mathbf6|_{\mathrm p} &=\operatorname{Sym}^2 V_2 =V_4\oplus V_0. \end{aligned}$$

The last line separates symmetric traceless tensors of the rotation triplet from their trace. Both dimensions are six, but only the principal restriction is forced to contain exclusively even $n$. For the adjoint, use $\mathbf3\otimes\overline{\mathbf3}=\mathbf1\oplus\mathbf8$ and $V_a\otimes V_b=\bigoplus_{k=0}^{\min(a,b)}V_{a+b-2k}$:

$$\begin{aligned} \mathbf8|_{\mathrm r}&=V_2\oplus2V_1\oplus V_0, & 8&=3+2(2)+1,\\ \mathbf8|_{\mathrm p}&=V_4\oplus V_2, & 8&=5+3. \end{aligned}$$

The doublet multiplicity in the first line matters. A list of distinct dimensions would lose two dimensions and the actual representation.

Read multiplicities from restricted weights

For $\operatorname{diag}(z,z^{-1})\in SU(2)$, the two torus maps are

$$\iota_{\mathrm r}:z\longmapsto\operatorname{diag}(z,z^{-1},1), \qquad \iota_{\mathrm p}:z\longmapsto\operatorname{diag}(z^2,1,z^{-2}).$$

The adjoint character is the defining character times its dual minus one. In these examples the defining restricted character is self-dual, giving

$$\begin{aligned} \chi_{\mathbf8}|_{\mathrm r} &=(z+1+z^{-1})^2-1 =z^2+2z+2+2z^{-1}+z^{-2},\\ \chi_{\mathbf8}|_{\mathrm p} &=(z^2+1+z^{-2})^2-1 =z^4+2z^2+2+2z^{-2}+z^{-4}. \end{aligned}$$

Write a restricted character as $\sum_m c_m z^m$. Since $V_n$ contributes one copy of each weight $n,n-2,\ldots,-n$, for $n\geq0$ its multiplicity is

$$\boxed{m_n=c_n-c_{n+2}.}$$

For the regular adjoint this gives $m_2=1$, $m_1=2$, $m_0=2-1=1$. For the principal adjoint it gives $m_4=1$, $m_2=2-1=1$, $m_0=2-2=0$. This calculation both reconstructs the character and checks the singlet count. The SU(3) branching reference extends it to a complete weight-enumeration method. Compare the regular adjoint and principal adjoint in the calculator.

A fixed vector does not determine its stabilizer

A singlet under $H$ says that $H$ fixes a vector. Its full stabilizer in $G$ may be larger. For the regular adjoint singlet, choose the traceless Hermitian matrix $v=\operatorname{diag}(1,1,-2)$, with adjoint action $v\mapsto UvU^{-1}$. A matrix fixes $v$ precisely when it preserves its two eigenspaces. Therefore

$$\operatorname{Stab}_{SU(3)}(v)=S(U(2)\times U(1)),$$

which contains the regular $SU(2)$ and an extra circle. The branching singlet alone would miss that circle.

Conversely, realize $\mathbf6$ as complex symmetric matrices with action $S\mapsto USU^T$. The vector $S=I_3$ is fixed exactly when $UU^T=I_3$. Together with unitarity this implies $U=\overline U$, so its stabilizer inside $SU(3)$ is precisely the real $SO(3)$. This explicitly realizes the principal singlet. Whether a field theory chooses either vector as a stable vacuum still depends on its potential and other fields; the physics guide develops that distinction.

Keep the circle and its kernel in the SU(5) block map

For $A\in SU(3)$, $B\in SU(2)$ and $z=e^{i\theta}\in U(1)$ with $\theta$ of period $2\pi$, define

$$\begin{aligned} \iota:SU(3)\times SU(2)\times U(1)&\longrightarrow SU(5),\\ (A,B,z)&\longmapsto\operatorname{diag}(z^2A,z^{-3}B). \end{aligned}$$

The determinant is $z^{6-6}=1$. The generator $X=\operatorname{diag}(2,2,2,-3,-3)$ is primitive because $2$ and $3$ are coprime. The image is $H=S(U(3)\times U(2))$: for any such pair of blocks $(C,D)$ choose $z$ with $z^6=\det C$ and set $A=z^{-2}C$, $B=z^3D$. Both determinants then equal one. Setting the image equal to the identity gives

$$\ker\iota=\{(z^{-2}I_3,z^3I_2,z):z^6=1\}\cong\mathbb Z_6, \qquad H\cong\frac{SU(3)\times SU(2)\times U(1)}{\mathbb Z_6}.$$

Baez–Huerta, The Algebra of Grand Unified Theories, §3.1, p. 34 gives this block homomorphism and its kernel with the weak block first and parameter $\alpha=z^{-1}$. The map record uses the color-first convention above.

A subscript below denotes the integer $X$ charge, so the defining representation restricts as

$$\mathbf5\longmapsto(\mathbf3,\mathbf1)_2 \oplus(\mathbf1,\mathbf2)_{-3}.$$

Apply $\Lambda^2(V\oplus W)=\Lambda^2V\oplus(V\otimes W)\oplus \Lambda^2W$. Charges add, $\Lambda^2\mathbf3=\overline{\mathbf3}$, and $\Lambda^2\mathbf2=\mathbf1$. Hence

$$\boxed{\mathbf{10}\longmapsto (\overline{\mathbf3},\mathbf1)_4 \oplus(\mathbf3,\mathbf2)_{-1} \oplus(\mathbf1,\mathbf1)_{-6}.}$$

This is the exterior-square construction in Baez–Huerta, §3.1, pp. 38–39, expressed in the integer charge convention just defined. Check it in three ways. Dimensions give $10=3+6+1$. The trace of $X$ is $3(4)+6(-1)+1(-6)=0$. Finally, if $t_3$ is the $SU(3)$ triality and $t_2=n\bmod2$ the $SU(2)$ parity, the kernel generator acts trivially iff

$$-2t_3+3t_2+X\equiv0\pmod6.$$

The three summands yield respectively $-4+4=0$, $-2+3-1=0$ and $-6\equiv0$. This last check uses the global subgroup, not just its Lie algebra. A hypothetical $(\mathbf1,\mathbf1)_1$ would be a perfectly valid product-group representation but could not occur in this restriction.

For the Standard Model convention $Q_{\mathrm{electric}}=T_3+Y$, set $Y=-X/6$. The three hypercharges are then $-2/3$, $1/6$, and $1$. Baez–Huerta use twice this hypercharge, so their numbers must be divided by two before comparison. A different normalization is legitimate only when its generator and period are translated with it. Continue to SU(5) and Standard Model multiplets for the particle interpretation and to the charged branching calculator to reproduce the decomposition.

Scope and next calculation

These methods compute representation multiplicities. They do not supply normalized Clebsch–Gordan bases, coupling constants, mass eigenvalues, or a proof of a vacuum’s dynamical stability. The local SU(3) tool accepts $0\leq p,q\leq6$ for the two maps above. The charged block tool supports $3\leq N\leq6$ and canonical Young diagrams with at most twelve boxes; the block reference states its full input contract and general formula. These are implementation bounds, not mathematical bounds on restriction.

As a final check, conjugate the ten-dimensional result. All $X$ charges change sign and $\mathbf3$ exchanges with $\overline{\mathbf3}$, while the $SU(2)$ doublet is self-dual. Thus $\overline{\mathbf{10}}$ gives $(\mathbf3,\mathbf1)_{-4}\oplus (\overline{\mathbf3},\mathbf2)_1\oplus (\mathbf1,\mathbf1)_6$. Each summand still satisfies the kernel condition. A dimension-only branching list would fail to distinguish these two parent representations. Retain the parent labels, chosen map and charge convention when saving a calculation.