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A Cartesian tensor invariant under all proper rotations fixing a unit vector $n$ is axisymmetric. Without assuming a symmetric tensor, its general second-order form is the displayed combination of the transverse identity, axial projector and axial antisymmetric part. For $n=e_3$ this isIndeed, imposing $RT{R}^T=T$ for a quarter-turn about $e_3$ forces $T_{11}=T_{22}$, $T_{12}=-T_{21}$ and all mixed axial-transverse entries to vanish. The planar block is a linear combination of the identity and the planar rotation generator, so it commutes with every rotation about the axis. Thus quarter-turn invariance already suffices for a second-order tensor. A symmetric tensor has $\omega =0$, recovering the axisymmetric symmetric rank-two tensor form. The antisymmetric term is allowed because only proper rotations about the fixed axis are imposed.

A symmetric tensor invariant under rotations about a fixed nonzero vector $y$ has the form $\alpha I+\beta y{y}^T$. Choose the third coordinate axis parallel to $y$: rotation by a half-turn makes the mixed axial-transverse entries vanish, and all transverse rotations make the transverse block a scalar multiple of the identity. Its two transverse eigenvalues are $\alpha$ and its axial eigenvalue is $\alpha +\beta |y{|}^2$. The trace and the quadratic contraction $y^{T}Ty$ determine these two coefficients. Full isotropy holds exactly when $\beta =0$.

A dummy index is summed over its full specified range. Its name can be changed consistently without changing the expression, as in $A_{ij}{B}_{ij}=A_{ji}{B}_{ji}$. A free index specifies a component and cannot be renamed in only one term of an equation. Relabelling dummy indices is useful for tensor contraction arguments involving a symmetric tensor and an antisymmetric tensor.

The commutant of the permutation action on $V^{\otimes n}$ identifies with the symmetric tensors in $\mathrm{End}(V{)}^{\otimes n}$. Polarization spanning of symmetric tensors spans it by $a^{\otimes n}$. Interpolating $(a+tI{)}^{\otimes n}$ at $n+1$ values where $a+tI$ is invertible expresses each such power using invertible ones. Hence the Schur algebra is the linear span of the diagonal general linear action.

The six smallest complex irreducible representations of SU(3) have highest weights $(0,0),(1,0),(0,1),(2,0),(0,2),(1,1)$ and dimensions $1,3,3,6,6,8$. Their Young diagram row lengths are $(0,0),(1,0),(1,1),(2,0),(2,2),(2,1)$. They are realized by scalars, defining vectors, covectors (equivalently antisymmetric tensors of degree two), upper and lower symmetric tensors of degree two, and mixed tensors $T^i{}_j$ with zero trace, respectively. The Weyl dimension formula is $\dim (p,q)=(p+1)(q+1)(p+q+2)/2$. For $p+q\ge 3$ this is at least ten, proving that the list exhausts the six smallest representations.

A free massive spin-two field is a symmetric tensor obeying a massive Klein-Gordon equation, vanishing divergence, and vanishing trace. Its rest-frame polarizations form the symmetric traceless square of the $SO(D-1)$ vector, with $(D-2)(D+1)/2$ degrees of freedom.

Using the given linear momentum balance, multiply by the velocity and integrate over the fixed volume:This is integration by parts component by component, or the divergence theorem. The first term is boundary stress power. With component-first velocity gradient $K_{ij}={\partial }_j{u}_i$, the second is the double contraction $\Phi =\sigma :K$.

For a Newtonian fluid, $\sigma =-pI+2\mu e$, where $e=(K+K^T)/2$ is the rate-of-strain tensor and $\mu \ge 0$ is dynamic viscosity. Incompressibility removes the pressure contribution, and the symmetric tensor $e$ is orthogonal to the antisymmetric part of $K$. ThereforeThe displayed energy identity follows for the stated linear balance. A full advective momentum equation in a fixed control volume would also contribute an advective kinetic energy flux. The nonnegativity statement here is for a purely Newtonian viscous stress; a viscoelastic stress can exchange stored free energy with the flow.

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.

Use upper indices for the defining representation of SU(6) and lower indices for its dual representation. The Young tableaux below display tensor-slot labels; their shapes determine the Young diagrams. Full columns of height six are determinant representations and can be removed for SU(6).Here all indices run from one to six. The highest weight Dynkin labels follow from the differences of successive row lengths. In the same order the row lengths are $(1)$, $(1,1,1,1,1)$, $(2)$, $(1,1)$ and $(2,1,1,1,1)$.

The defining vector transforms as $v^a\mapsto U^a{}_b{v}^b$. For a unitary representation, the conjugate representation is equivalent to the dual representation, so $\overline{6}$ can be written as a covector with $w_a\mapsto w_b(U^{-1}{)}^b{}_a$. Its five-box column is equivalently a totally antisymmetric tensor $A^{a_1\cdots a_5}$, related by the invariant alternating tensor $w_a={ϵ}_{a{a}_1\cdots a_5}{A}^{a_1\cdots a_5}/5!$.

The symmetric tensor transforms as $S^{ab}\mapsto U^a{}_c{U}^b{}_d{S}^{cd}$, with $6(6+1)/2=21$ independent components. The antisymmetric tensor has the same transformation law but obeys $A^{ab}=-A^{ba}$, giving $6(6-1)/2=15$ independent components. For the Adjoint representation, $T\mapsto UT{U}^{-1}$; the trace constraint removes one component from the $36$ entries of a mixed tensor, leaving $35$. Its six-box Young diagram is obtained by converting the five-index column to a lower index and selecting the traceless mixed tensor.

Thus the requested dimensions, in the order above, are

Use upper indices for the defining representation of SU(3) and lower indices for its dual representation. A Young diagram with row lengths $({\ell }_1,{\ell }_2)$ has highest weight Dynkin labels $(p,q)=({\ell }_1-{\ell }_2,{\ell }_2)$; a full column of height three is a determinant representation and can be removed for SU(3). The Weyl dimension formula reduces toThe six irreducible representations of smallest dimensions are $1,3,\overline{3},6,\overline{6},8$. The following box arrays draw their Young diagrams; filling boxes by tensor-slot labels gives representative Young tableaux:Their associated tensors, in the same order, can be taken as follows:

These low-dimensional SU3 tensor representations have dimensions $1,3,3,6,6,8$ by direct component counting. To check that none were missed, if $p+q\ge 3$ then $(p+1)(q+1)\ge p+q+1$, so the displayed dimension is at least $(p+q+1)(p+q+2)/2\ge 10$. The pairs with $p+q\le 2$ are precisely the six listed pairs.

The squares of $I,-I,\pm A,\pm B$ are $I$, whereas $(\pm C{)}^2=-I$. Therefore ${\chi }_{\rho }(g^2)=(2,2,2,2,-2)$. The symmetric square and exterior square formulas from solution giveComparison with the character table gives the Clebsch-Gordan decomposition of group representationsFor explicit bases, put $e_1=(1,0{)}^T,e_2=(0,1{)}^T$. The symmetric tensors $e_1\otimes e_1+e_2\otimes e_2$, $e_1\otimes e_2+e_2\otimes e_1$, and $e_1\otimes e_1-e_2\otimes e_2$ transform as $L_{++},L_{+-},L_{-+}$ respectively. The antisymmetric tensor $e_1\otimes e_2-e_2\otimes e_1$ transforms as $L_{--}$, also the determinant representation of $\rho$. Adding these direct sums recovers $\rho \otimes \rho ={⨁}_{s,t}{L}_{st}$.

Write $[r_1,r_2,\dots ]$ for a Young diagram with row lengths $r_i$, and depict an empty box by $\square$. Appending the box of the fundamental representation to another single box gives a row or a column:The corresponding symmetric tensors and antisymmetric tensors have dimensions $4\cdot 5/2=10$ and $4\cdot 3/2=6$, soFor the conjugate representation, the invariant volume tensor identifies $\bar{4}\cong {\Lambda }^{3}4$, represented by a column of three boxes. Adding one box givesA column of four boxes is the determinant representation, which is trivial on $SU(4)$. The remaining Young diagram has Dynkin labels $(1,0,1)$ and dimension fifteen. For example the hook-content formula givesEquivalently, a mixed tensor $M^i{}_j$ transforms by $M\mapsto UM{U}^{†}$ and splits into its trace and traceless matrix parts. Thereforewhere $15$ is the Adjoint representation.

Use the standard chain ordering of the five simple roots of $SU(6)$. A Young diagram of row lengths $r_1\ge \cdots \ge r_6=0$ has highest weight Dynkin labels $a_i=r_i-r_{i+1}$. A full column of six boxes can be removed because it is the trivial determinant representation. The required data areIn the same order, the Young tableaux areEach upper index transforms by $U$, and each lower index by $U^{-1}$. In particular $M\mapsto UM{U}^{-1}$ is the complex Adjoint representation. The column of five boxes can be represented either by $w_i$ or by the totally antisymmetric tensor $B^{i_1\dots i_5}$, using the invariant six-index alternating tensor:Thus it really is the conjugate representation, rather than a second independent six-dimensional choice. Similarly converting the lower index of $M^i{}_j$ into five alternating upper indices gives the displayed hook Young diagram; the removed trace would be its full six-box column component.

The dimension counts areThe first two count independent entries of symmetric tensors and antisymmetric tensors; the last removes one trace from a mixed tensor. The hook-content formula independently gives these dimensions from the Young diagrams.

Let index $0$ denote the first coordinate, which the embedded $SU(5)$ fixes, and let $a,b=1,\dots ,5$ denote the remaining coordinates. The fundamental representation and its conjugate representation immediately giveFor a symmetric tensor, $S^{00}$ is a trivial representation, $S^{0a}=S^{a0}$ is a $5$, and $S^{ab}$ is the fifteen-dimensional symmetric square of a group representation of $5$. For an antisymmetric tensor, $A^{00}=0$, $A^{0a}=-A^{a0}$ is a $5$, and $A^{ab}$ is the ten-dimensional exterior power ${\Lambda }^{2}5$. HenceHere the left-hand $15$ is the antisymmetric tensor of $SU(6)$, while the right-hand $15$ in the first equation is the symmetric tensor of $SU(5)$; equal dimensions alone do not identify representations of different groups.

Finally decompose the traceless block matrixThe traceless part of $D$ is the Adjoint representation $24$, the column $c$ is $5$, the row $b$ is $\bar{5}$, and the one remaining diagonal trivial representation is generated by $\mathrm{diag}(-5,1,1,1,1,1)$. ThusAll dimensions match: $6=1+5$, $21=15+5+1$, $15=10+5$ and $35=24+5+5+1$.

The unbroken group is the stabilizer subgroup of the vacuum expectation value. For a nonzero vector vacuum expectation value, an orthogonal transformation puts $⟨\varphi ⟩=v{e}_N$. A special orthogonal transformation fixing it has the block form $\mathrm{diag}(A,1)$ with $A\in \mathrm{SO}(N-1)$. ConsequentlyThe zero vacuum expectation value leaves the entire special orthogonal group unbroken.

For a real symmetric tensor write its vacuum expectation value as a real symmetric matrix $Q$. The transformation is $Q\mapsto OQ{O}^T$. The spectral theorem for real symmetric matrices gives an orthonormal basis in whichBecause $OQ{O}^T=Q$ is equivalent to $OQ=QO$, an unbroken transformation preserves each eigenspace. The complete unbroken group is thereforeIts identity component is $H^0={\prod }_i\mathrm{SO}(m_i)$. This symmetric tensor vacuum stabilizer shows why the representation alone does not determine a unique breaking pattern:

For example a rank-one tensor $Q=v{e}_N{e}_N^T$ preserves $O(N-1)$ embedded by $A\mapsto \mathrm{diag}(A,\det A)$. Its continuous unbroken group is $\mathrm{SO}(N-1)$, but it also allows reversal of the distinguished axis. Restricting to a traceless symmetric tensor imposes ${\sum }_i{m}_i{q}_i=0$ and otherwise leaves the same classification.

Let $V={\mathbb{C}}^3$ be the defining flavor SU(3) representation. Under $U\in SU(3)$, $q^{\alpha }\mapsto U^{\alpha }{}_{\beta }{q}^{\beta }$. The antiquark transforms in the dual representation:Since $U$ is unitary, its lower-index column of components transforms by $U^{\ast }$, the complex conjugate numerical matrix. Thus $q^{\alpha }{\bar{q}}_{\alpha }$ is invariant. The defining triplet and its conjugate are inequivalent complex representations; conjugation reverses their weights. Write $3=V$ and $\overline{3}=V^{\ast }$, and distinguish a direct sum $\oplus$ from a tensor product $\otimes$.

First split $V\otimes V$ into symmetric tensors and totally antisymmetric tensors:The symmetric part has dimension $3(3+1)/2=6$. Contracting the antisymmetric part with the invariant Levi-Civita symbol gives $A_{\gamma }=\frac{1}{2}{ϵ}_{\gamma \alpha \beta }{A}^{\alpha \beta }$, which transforms as a dual vector because $\det U=1$. Its inverse is $A^{\alpha \beta }={ϵ}^{\alpha \beta \gamma }{A}_{\gamma }$. HenceThe symmetric part is the irreducible highest-weight representation with Dynkin labels $(2,0)$; its six monomials give the standard symmetric-square representation.

For three triplets, the antisymmetric first-pair sector becomes $V^{\ast }\otimes V$. Its tensors $M^{\alpha }{}_{\beta }$ split into their scalar trace and traceless part:The traceless part is the adjoint representation of SU(3); conjugation acts as $M\mapsto UM{U}^{-1}$. It is irreducible because an invariant subspace would be an ideal in the simple complex special linear Lie algebra ${\mathfrak{sl}}_3$.

The symmetric first-pair sector consists of $T^{abc}=T^{bac}$ and has dimension $6\cdot 3=18$. Complete symmetrization maps it onto ${\mathrm{Sym}}^{3}V$, whose dimension is $\left(\genfrac{}{}{0ex}{}{5}{3}\right)=10$. Its kernel is the mixed-symmetry eight-dimensional sector. More explicitly, the equivariant map $T^{abc}\mapsto M^a{}_e={ϵ}_{ebc}{T}^{abc}$ has trace zero because the first two indices of $T$ are symmetric; it annihilates the completely symmetric part and is not identically zero. Since the target adjoint module is irreducible, the map is onto its eight-dimensional space. Its kernel is consequently exactly the ten-dimensional completely symmetric part. Complete symmetrization supplies an invariant splitting, so this sector is $10\oplus 8$. Combining the sectors gives the tensor cube of the defining SU(3) representationHere the decuplet has Dynkin labels $(3,0)$, and the singlet is the fully antisymmetric contraction ${ϵ}_{abc}{T}^{abc}$. The two octets are distinct copies, not two different irreducible types.

For two conjugate SU(3) triplets with a triplet, begin with $T^a{}_{bc}$ and split its lower pair. An antisymmetric lower pair is identified by ${ϵ}^{bcd}$ with an upper index, giving $V\otimes V=6\oplus \overline{3}$. The symmetric lower-pair sector has dimension $3\cdot 6=18$, and its contraction is the equivariant mapIt is onto $V^{\ast }$, since the invariant embeddingsatisfies $C\iota (v)=v$. Every such tensor therefore splits uniquely as $\iota (C(T))+(T-\iota (C(T)))$. The latter tensor is symmetric in its lower indices and has zero contraction, with dimension $18-3=15$. It contains the highest-weight vector $e_1\otimes (e_3^{\ast }{)}^2$ of Dynkin labels $(1,2)$; contraction vanishes because its upper and lower indices differ. The Weyl dimension formula for an SU(3) module with labels $(p,q)$ is $\dim V_{(p,q)}=(p+1)(q+1)(p+q+2)/2$, which gives $15$ for $(1,2)$. These tensor representations are unitary, so invariant subspaces have invariant orthogonal complements; the irreducible highest-weight summand containing this vector exhausts the fifteen-dimensional kernel. ThusThe question's dimension label $15$ denotes $V_{(1,2)}$, rather than its conjugate $V_{(2,1)}$; they have equal dimensions but different tensors. Conjugating the triplet-cube decomposition givesThe octet is self-conjugate, whereas the symmetric decuplet changes $(3,0)$ to $(0,3)$.

Now consider the flavor diquark $D_{\alpha }\propto {ϵ}_{\alpha \beta \gamma }{q}^{\beta }{q}^{\gamma }$. This notation describes an antisymmetric two-quark flavor state; color and spin factors are separate. In particular $D_3=(ud-du)/\sqrt{2}$ has isospin zero, baryon number $2/3$, strangeness zero and flavor hypercharge $2/3$. The strange antiquark has $I=0$, $B=-1/3$, $S=+1$ and $Y=2/3$. Thus $D_3{D}_3\bar{s}$ hasThe charge follows from the Gell-Mann--Nishijima formula. Each of the three factors has lower flavor index $3$, so their product is the completely symmetric tensor component $A_{333}$. It lies entirely in ${\mathrm{Sym}}^3{V}^{\ast }=\overline{10}$, the pentaquark antidecuplet. Equivalently, the symmetric diquark pair belongs to $\overline{6}$, and $\overline{6}\otimes \overline{3}=\overline{10}\oplus 8$, with this highest-hypercharge state in the antidecuplet.

It cannot be in an octet. The octet has the weights of a traceless $q^{\alpha }{\bar{q}}_{\beta }$ tensor, whose hypercharges are $Y_{\alpha }-Y_{\beta }$. For $(u,d,s)$ the quark values are $1/3,1/3,-2/3$, so these differences are only $0,+1,-1$; there is no $Y=2$ weight. The octet's top row instead has $Y=1$ and $I=1/2$. This is a flavor-representation assignment for a possible pentaquark, not a proof that the configuration binds or that its full color-spin-orbital wavefunction satisfies all Fermi statistics constraints.

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.

The integrand is symmetric in $i,j$, so $T$ is a symmetric tensor. The sphere and its area element are invariant under rotations, and the integrand transforms covariantly when $\mathbf{x},\mathbf{y}$ are rotated together. For fixed $\mathbf{y}$, rotations about its direction therefore leave $T$ unchanged. In axes with $\mathbf{y}$ along the third axis, the mixed transverse-axial entries vanish and the two transverse eigenvalues are equal. Thus it is an axisymmetric symmetric rank-two tensor,where $\alpha ,\beta$ depend on $a,n$ but not on the direction of $\mathbf{y}$. The restriction $n>-1$ makes the integral convergent near $\mathbf{x}=\mathbf{y}$: the integrand is of size distance${}^{2n}$ and the local area element has one further radial power.

Take the polar axis along $\mathbf{y}$. Because $|\mathbf{x}|=|\mathbf{y}|=a$,The area element is $a^2\sin \theta d\theta d\varphi$. Hence with $s=1-\cos \theta$,

For the second scalar integral, take the trace. It contracts the two chord-vector factors into their squared length:Put $K=\pi (2a{)}^{2n+2}$. The tensor form gives $3\alpha +\beta a^2=K/(n+1)$, while its axial contraction gives $\alpha +\beta a^2=K/(n+2)$. Solving yields the weighted chord tensor of a sphere coefficientsBecause $a>0$, this is an isotropic tensor exactly when $\beta =0$. Therefore $n=-1/2$ is the unique permitted isotropic exponent.

The Adjoint double cover from SU(2) to SO(3) is obtained by writing a traceless Hermitian matrix as $X=x_i{\sigma }_i$, where the ${\sigma }_i$ are the Pauli matrices, and defining $AX{A}^{†}=(R(A)x{)}_i{\sigma }_i$. Explicitly,Conjugation preserves $\frac{1}{2}\mathrm{tr}{X}^2=|x{|}^2$, so $R(A)$ is orthogonal; connectedness of SU(2) gives $\det R(A)=1$. Composition of conjugations makes $R$ a group homomorphism. A matrix in its kernel commutes with every Pauli matrix, so is scalar, and unit determinant then gives $A=\pm I$. Every spatial rotation occurs: $A=\exp (-i\theta \hat{n}\cdot \sigma /2)$ induces the rotation through angle $\theta$ about $\hat{n}$. ThusSince SU(2) is the simply connected three-sphere, this is the universal double cover; the two Lie algebras are isomorphic, although the groups are not.

Define the conjugate fundamental spinor of SU(2) by ${\bar{\eta }}_{\alpha }=({\eta }^{\alpha }{)}^{\ast }$. Regard $\bar{\eta }$ as a row, so unitarity givesConsequently ${\bar{\eta }}_{\alpha }{\xi }^{\alpha }$ is an invariant Hermitian pairing. The basic invariant tensors are the Kronecker delta ${\delta }^{\alpha }{}_{\beta }$ and the alternating tensors ${ϵ}_{\alpha \beta }$ and ${ϵ}^{\alpha \beta }$. Choose ${ϵ}_{12}=1$, ${ϵ}^{12}=-1$, so that ${ϵ}^{\alpha \gamma }{ϵ}_{\gamma \beta }={\delta }^{\alpha }{}_{\beta }$. The unit-determinant identity $A^TϵA=ϵ$ makes $[\eta ,\xi ]={ϵ}_{\alpha \beta }{\eta }^{\alpha }{\xi }^{\beta }$ invariant; its inverse gives the corresponding upper-index invariant. These contractions generate the tensor invariants of the defining representation. In particular ${\tilde{\eta }}^{\alpha }={ϵ}^{\alpha \beta }{\bar{\eta }}_{\beta }$ transforms by $A$, showing that the conjugate representation is equivalent to the defining one. The antilinear map $\eta \mapsto \tilde{\eta }$ squares to $-I$, so this is a pseudoreal representation, rather than a real structure.

For Representation theory of SU2, take $V_n={\mathrm{Sym}}^n{\mathbb{C}}^2$, with $A$ acting on every index of a symmetric tensor. Its independent components are determined by the number $r$ of indices equal to $2$, so there are $n+1$ of them. Equivalently, the homogeneous polynomial representation of SU2 has basis $z_1^{n-r}{z}_2^r$, $0\le r\le n$, withThe basis weights are $n,n-2,\dots ,-n$, and the raising operator and lowering operator connect consecutive basis vectors with nonzero coefficients until their endpoints. Any nonzero invariant subspace contains a weight vector, by applying polynomial spectral projections in $H$. Raising it reaches $z_1^n$, and lowering then obtains the entire basis. Thus $V_n$ is an irreducible representation, of spin $j=n/2$. The classification of finite-dimensional representations of SU2 gives every finite-dimensional complex irreducible representation in this way. The central element $-I$ acts on a rank-$n$ tensor as $(-1{)}^n$, so descent of an SU(2) representation to SO(3) gives

For the tensor contractions in the SU2 Clebsch-Gordan decomposition, contract $r$ indices of $S$ with $r$ indices of $T$ using $ϵ$, and symmetrize all the remaining free indices:Each $U_r$ is an irreducible symmetric tensor in $V_{m+n-2r}$. To prove these are precisely the constituents, use two sets of formal variables $z,w$. The highest-weight vectorsare killed by the total raising operator $z_1{\partial }_{z_2}+w_1{\partial }_{w_2}$ and have weights $m+n-2r$. The determinant factor is invariant; lowering the remaining factor gives an irreducible string of dimension $m+n-2r+1$. Distinct highest weights give inequivalent summands. The unitary representation on this tensor product makes invariant complements available, and the dimension checkexhausts the space. Hence the contractions above are its irreducible projections, up to normalization, and the Clebsch-Gordan decomposition for SU2 is

For the decomposition of three SU(2) spinors, write $Q^{\alpha \beta \gamma }={\eta }_1^{\alpha }{\eta }_2^{\beta }{\eta }_3^{\gamma }$ and $S^{\alpha \beta \gamma }=Q^{(\alpha \beta \gamma )}$. The contractionstransform as defining spinors. Alternation in three indices in dimension two vanishes, giving $d_1-d_2+d_3=0$; there are exactly two independent contractions. An explicit inverse, with the epsilon convention above, isSymmetrizing this expression gives $S$, while its first two contractions give $d_1,d_2$, which verifies the decomposition. Thus