Representation theory of SU(2)

Finite-dimensional complex representations of $S{U}_2$ are completely reducible, and their irreducibles are indexed by nonnegative integers.

For the defining SU(2) action ${\eta }^{\prime }=A\eta$, the conjugate row ${\bar{\eta }}_{\alpha }=({\eta }^{\alpha }{)}^{\ast }$ transforms by $A^{†}=A^{-1}$. Thus ${\bar{\eta }}_{\alpha }{\eta }^{\alpha }$ is invariant. The alternating invariant tensor ${ϵ}_{12}=1$ satisfies $A^TϵA=ϵ$. Choose its inverse ${ϵ}^{12}=-1$; then ${ϵ}^{\alpha \gamma }{ϵ}_{\gamma \beta }={\delta }^{\alpha }{}_{\beta }$. Raising the conjugate index, ${\tilde{\eta }}^{\alpha }={ϵ}^{\alpha \beta }{\bar{\eta }}_{\beta }$, gives another defining spinor. The resulting antilinear intertwiner squares to $-I$, so the defining representation is a pseudoreal representation.

The irreducible representation $V_n={\mathrm{Sym}}^n({\mathbb{C}}^2)$ consists of homogeneous degree-$n$ polynomials in two variables, with the action induced from the standard two-dimensional representation. It has dimension $n+1$.

Let $V_n={\mathrm{Sym}}^n({\mathbb{C}}^2)$ be the homogeneous polynomial representation of SU2. An element $g$ of SU(2) has eigenvalues $z,z^{-1}$ with $|z|=1$. Its character isTo see this, choose an eigenbasis in the defining two-dimensional representation. The degree-$n$ monomials form a basis of the symmetric power; their eigenvalues are the products of $n-k$ copies of $z$ and $k$ copies of $z^{-1}$. Adding them computes the trace.

For $z\ne \pm 1$, the geometric series givesWith $z=e^{i\theta }$ this becomes $\sin ((n+1)\theta )/\sin \theta$. The original finite sum also handles the apparent singularities: ${\chi }_{V_n}(I)=n+1$ and ${\chi }_{V_n}(-I)=(-1{)}^n(n+1)$.

Every finite-dimensional complex $S{U}_2$-representation is a direct sum of the pairwise nonisomorphic irreducibles $V_n={\mathrm{Sym}}^n({\mathbb{C}}^2)$.

The standard invariant alternating form identifies $V_1$ with its dual. Its symmetric powers identify every $V_n$ with $V_n^{\ast }$, and complete reducibility then gives $V\cong V^{\ast }$ for every finite-dimensional complex $S{U}_2$-representation.

On $V_n$, the central element $-I$ acts as $(-1{)}^n$. It therefore acts trivially on $V_n\otimes V_n$, although it need not act trivially on the tensor square of a representation containing irreducibles of both parities.

Clebsch-Gordan coefficients are the change-of-basis coefficients between an uncoupled tensor-product basis and a basis adapted to the irreducible decomposition of the tensor product.

Apply a total ladder operator to a coupled state and evaluate against an uncoupled product state. The total operator is the sum of the two factor operators, so its one coupled ladder amplitude times the corresponding one of the Clebsch-Gordan coefficients equals the sum of two factor ladder amplitudes times shifted coefficients. An overall common rescaling of the ladder operators cancels from the recurrence. Nonzero coefficients obey $m_1+m_2=m$, and the positive ladder amplitudes require a consistent phase convention within each irreducible summand.

// Target: lie-theory.bigb

A symmetrized form of the Clebsch-Gordan coefficients used for coupling three angular momenta. The relation is $⟨j_1{m}_1,j_2{m}_2|j_3{m}_3⟩=(-1{)}^{j_1-j_2+m_3}\sqrt{2j_3+1}\left(\begin{array}{ccc}{j}_1& j_2& j_3\\ m_1& m_2& -m_3\end{array}\right)$. The symbol vanishes unless the magnetic indices sum to zero and the angular momenta obey the triangle and integrality conditions. Products of two such symbols express the Gaunt integral; the zero-magnetic-index symbol enforces even total multipole for integer spherical harmonics.

For $m,n\ge 0$,The identity follows by multiplying the weight characters and comparing their nested weight strings.

Realize $V_n={\mathrm{Sym}}^n{\mathbb{C}}^2$. Contract $r$ pairs of indices, one from each symmetric tensor, with the alternating invariant tensor $ϵ$, and symmetrize the remaining indices. This is a nonzero intertwining operator into $V_{m+n-2r}$. In the bihomogeneous polynomial realization, $(z_1{w}_2-z_2{w}_1{)}^r{z}_1^{m-r}{w}_1^{n-r}$ is a highest-weight vector of weight $m+n-2r$. Its lowering string has length $m+n-2r+1$. These inequivalent irreducible summands exhaust the tensor product, since their dimensions sum to $(m+1)(n+1)$.

For $Q^{\alpha \beta \gamma }$, put $S=Q^{(\alpha \beta \gamma )}$, $d_1^{\alpha }={ϵ}_{\beta \gamma }{Q}^{\alpha \beta \gamma }$, $d_2^{\alpha }={ϵ}_{\beta \gamma }{Q}^{\beta \alpha \gamma }$ and $d_3^{\alpha }={ϵ}_{\beta \gamma }{Q}^{\beta \gamma \alpha }$. With ${ϵ}_{12}=1$ and inverse ${ϵ}^{12}=-1$, the two-dimensional alternating identity gives $d_1-d_2+d_3=0$. The inverse decomposition is $Q^{\alpha \beta \gamma }=S^{\alpha \beta \gamma }+{ϵ}^{\alpha \beta }(2d_1^{\gamma }-d_2^{\gamma })/3-{ϵ}^{\alpha \gamma }(d_1^{\beta }+d_2^{\beta })/3$. The symmetric part is the spin-$3/2$ SU(2) representation, and the two independent contractions are spin-$1/2$ representations. This is the spin coupling underlying ground-state three-quark baryons, before the Pauli constraint on three-quark flavour multiplets is imposed.

// Target: quantum-field-theory.bigb

On the multiplicity-one summand $V_{2n-2j}\subset V_n\otimes V_n$, interchange of tensor factors acts by $(-1{)}^j$. A highest-weight vector exhibiting the sign is

For the irreducible representation $V_n={\mathrm{Sym}}^n({\mathbb{C}}^2)$ of SU(2), the exterior square decomposes asFor $n=0$, the sum is empty and the exterior square is zero.

The Clebsch-Gordan decomposition for SU2 first gives $V_n\otimes V_n\cong {⨁}_{j=0}^n{V}_{2n-2j}$. By flip parity in the SU2 tensor square, exchanging the two tensor factors acts on the $j$th summand by $(-1{)}^j$. The exterior square is the minus-one eigenspace of this exchange, so it selects exactly the odd indices $j=2k+1$. Substitution gives the displayed formula.

For example, ${\bigwedge }^2{V}_2\cong V_2$ has dimension three, while ${\bigwedge }^2{V}_3\cong V_4\oplus V_0$ has dimension $5+1=6$, as required for the exterior square of a four-dimensional vector space.

The alternating summands in the tensor square give ${\bigwedge }^2{V}_4\cong V_6\oplus V_2$. Since $V_4$ is self-dual and has trivial determinant,