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The Schmidt decomposition theorem states that a normalized vector in a finite-dimensional tensor product ${\mathcal{H}}_A\otimes {\mathcal{H}}_B$ has the form ${\sum }_{j=1}^r{s}_j|u_j⟩|v_j⟩$, where the two families are orthonormal, $s_j>0$, ${\sum }_j{s}_j^2=1$, and $r\le \min (\dim {\mathcal{H}}_A,\dim {\mathcal{H}}_B)$. Extending each family to an orthonormal basis for two qubits givesThe nonzero Schmidt coefficients can be made positive by absorbing phases into the basis vectors. Both are positive exactly when the pure state is entangled. A product state has Schmidt rank one, so one coefficient is zero; the source's assertion of two positive coefficients for every pure state needs this exception.

Choose each local orthonormal basis independently as the computational basis. In that basis ${\sigma }_z=|0⟩⟨0|-|1⟩⟨1|$. This is a local change of coordinates, implemented by separate unitary matrices, rather than a physical restriction on the original quantum state. The associated Pauli operators supply the other two local axes. Up to an irrelevant common phase, each unitary change of qubit basis corresponds to a rotation of its Bloch sphere.

Put $s=2{\lambda }_0{\lambda }_1$. The Schmidt-basis Pauli correlation tensor is diagonal. The Pauli operators ${\sigma }_x\otimes {\sigma }_x$ exchange $|00⟩$ and $|11⟩$, while ${\sigma }_y\otimes {\sigma }_y$ do so with minus signs, and ${\sigma }_z\otimes {\sigma }_z$ leaves both fixed. HenceMixed components vanish: those containing one $z$ and one transverse Pauli operator map the occupied basis vectors outside their span, while the $xy$ and $yx$ matrix elements are purely imaginary and cancel for real Schmidt coefficients. By bilinearity, for arbitrary real vectors,For a CHSH inequality test take unit quantum measurement axesThese CHSH axes for an entangled pure two-qubit state giveThe local bound is two: for each hidden state, $A(B+B^{\prime })+A^{\prime }(B-B^{\prime })$ is $\pm 2$ when all four outcomes are $\pm 1$, and averaging cannot increase its absolute value. Thus every entangled pure two-qubit state violates a CHSH inequality, the content of Gisin's theorem. A product state has $s=0$ and does not violate it, so the unqualified final claim in the question is false for that case. The maximum $2\sqrt{2}$ occurs for equal Schmidt coefficients. This excludes local hidden-variable theories satisfying measurement independence, but does not permit faster-than-light signalling: the local reduced density matrix and its quantum measurement probabilities are unchanged by the remote choice of axis.

In an ontological model of a quantum system, a preparation of $|\psi ⟩$ gives a probability distribution ${\mu }_{\psi }$ over a physical state $\lambda$. A fixed quantum measurement has response probabilities ${\xi }_j(\lambda )\ge 0$ with ${\sum }_j{\xi }_j=1$, reproducing the Born rule after averaging over ${\mu }_{\psi }$. The PBR theorem says that, assuming preparation independence and the quantum predictions, distributions for distinct pure states cannot overlap with positive probability. Thus the quantum state is determined by the physical state in the psi-ontic model sense.

Preparation independence says that separately prepared systems have independent physical states: a product preparation is represented by the product of their individual ontic distributions. This is an assumption about the underlying physical states, not merely a statement that experimental preparation choices are independent. The excluded psi-epistemic model hypothesis is that the same physical state can occur with positive probability for two different pure-state preparations. The theorem neither excludes additional hidden variables nor claims that every interpretation which speaks of information is ruled out without these assumptions.

For the two given qubit preparations, consider the following PBR exclusion measurement for zero and plus, written in the ordered basis $00,01,10,11$:Direct inner products show that these four vectors are an orthonormal basis. Each labelled vector is orthogonal to the correspondingly labelled product preparation, so the associated projective measurement satisfiesThis is an example of antidistinguishable quantum states: every outcome excludes one possible preparation, although the preparations cannot be perfectly distinguished.

To prove the contradiction without requiring deterministic quantum measurement responses, choose a common dominating measure for ${\mu }_0,{\mu }_{+}$ and write their densities $m_0,m_{+}$. If they overlap, $\nu (\lambda )=\min (m_0,m_{+})$ has mass $\epsilon >0$. By preparation independence, every one of the four product densities dominates $\nu ({\lambda }_1)\nu ({\lambda }_2)$. Its total mass is ${\epsilon }^2$. The zero Born rule probability for outcome $xy$ implies that its nonnegative response ${\xi }_{xy}({\lambda }_1,{\lambda }_2)$ vanishes almost everywhere for its matching preparation, hence also under this common product measure. All four responses would then vanish on a set of positive measure, contradicting their sum being one. Therefore ${\mu }_0$ and ${\mu }_{+}$ are mutually singular.

For the second pair, group the $2n$ independent preparations into two blocks of $n$. Define $|{\Psi }_i⟩=|{\psi }_i{⟩}^{\otimes n}$. The tensor-power reduction of PBR overlap givesEach block therefore has an effective two-dimensional Hilbert space. Explicitly,are orthonormal and $|{\Psi }_2⟩=(|e_0⟩+|e_1⟩)/\sqrt{2}$. Embed the four-vector exclusion basis above into the tensor product of these two block spaces. To obtain a complete quantum measurement on all $2n$ qubits, add the orthogonal complement of that four-dimensional subspace to one of its four projectors. All four allowed preparations lie in the subspace, so the four forbidden probabilities remain zero.

If the single-copy preparation distributions overlap with common mass $\epsilon >0$, their $2n$-fold products all dominate the common measure ${\nu }^{\otimes 2n}$ of mass ${\epsilon }^{2n}>0$. The same zero-response contradiction now applies to the four block preparations. ThusThe argument is exact for the ideal devices specified in the question; no finite experimental resolution or noisy overlap bound is assumed.

The tensor differential. Use homological grading, so each differential lowers degree by one. The tensor product of chain complexes hasfor $x\in C_p$. This Koszul sign rule givesThus the graded tensor product is a chain complex.

The Hom differential. Write $M_j={\prod }_p{\mathrm{Hom}}_R(C_p,C_{p+j}^{\prime })$. For a degree-$j$ element of this graded Hom complex of chain complexes, use the prescribed differentialThe next application uses degree $j-1$, soIn degree zero, $d_{M}f=f{d}_C-d_{C^{\prime }}f$, so its kernel consists exactly of chain maps. A degree-one element $h$ has $d_{M}h=h{d}_C+d_{C^{\prime }}h$, which is exactly the change between two chain maps related by a chain homotopy. ConsequentlyThis is natural: precomposition and postcomposition by chain maps preserve both degree-zero cycles and degree-zero boundaries.

The dual complex and the sign adjustment. Define the reversed dual chain complexHere $d_X\varphi$ is a functional on $C_{1-i}$, so it belongs to $X_{i-1}$, and $d_X^2=0$. The absence of an additional sign in $d_X$ is intentional.

Interpret finite generation of the free chain complexes as finiteness of their total graded free modules. Then only finitely many degrees occur, and the finite-free evaluation isomorphisms assemble into a graded isomorphismwhere $\varphi \in X_i$, $y\in C_j^{\prime }$, and this component is zero outside $C_{-i}$. The dual module construction and evaluation make $\psi$ natural. Finite rank is needed for evaluation to be an isomorphism; finite total support also makes the sums on the tensor side agree with the products on the Hom functor side.

Under $\psi$, the Hom functor differential has the formThe usual tensor product of chain complexes instead has second coefficient $(-1{)}^i$. SetThis is integer-valued for every $j\in \mathbb{Z}$, including negative $j$, and satisfies $\rho (j)-\rho (j-1)=j-1$. Conjugating the usual tensor differential by $A$ leaves the first term unchanged and changes its second coefficient to $(-1{)}^{i+j-1}$. Hence $d_M=\psi AD{A}^{-1}{\psi }^{-1}$, giving the sign conjugation for the tensor-Hom identificationThis is an isomorphism of chain complexes, not merely of their homology. If one instead assumes only degreewise finite rank with unbounded grading, the ordinary tensor need not identify with the product defining $M$; the finite-total convention is essential to this last conclusion.

Interpret the paired indices $(i,k)$ as incoming and $(j,l)$ as outgoing. For self-conjugate particles in the real vector representation of the orthogonal group, crossing the second incoming particle with the second outgoing one givesThe continuation $\theta \mapsto i\pi -\theta$ exchanges the trace and permutation tensor structures and leaves the identity structure fixed. Therefore crossing symmetry imposesThese are analytic relations; they do not determine the three functions completely.

For unitarity, act on the two-species tensor product with the O(N)-invariant S-matrix $S=S_{2}I+S_{3}P+S_{1}K$. The invariant operators obey $P^2=I$, $PK=KP=K$ and $K^2=NK$. Writing $S_a^{-}=S_a(-\theta )$ and multiplying $S(\theta )S(-\theta )=I$ givesThe three invariant operators are independent for $N\ge 2$. Equivalently, the orthogonal-invariant scattering channels have amplitudesand each satisfies $s_a(\theta )s_a(-\theta )=1$. The trace singlet, symmetric traceless and antisymmetric channels have dimensions $1$, $N(N+1)/2-1$ and $N(N-1)/2$. For $N=2$ these are $1,2,1$. With Hermitian analyticity of a two-particle S-matrix, $S_a(-\theta )=S_a(\theta {)}^{\ast }$ on the real axis, so each channel has unit modulus. This expresses conservation of scattering probability.

The Faddeev-Zamolodchikov algebra orders particle operators by rapidity. Its associativity requires that a product of three operators be independent of parentheses and, in particular, that both sequences of adjacent exchanges give the same final ordered species word with the same coefficient. This is the Faddeev-Zamolodchikov associativity constraint, or spectral Yang-Baxter equation. The exchanges do not use the ordinary creation/annihilation normal ordering convention.

Take ${\theta }_1>{\theta }_2>{\theta }_3$, with $\theta ={\theta }_1-{\theta }_2$, ${\theta }^{\prime }={\theta }_2-{\theta }_3$, and abbreviateFor $N=2$, start with $A_1({\theta }_1)A_1({\theta }_2)A_2({\theta }_3)$ and compare the coefficient of $A_1({\theta }_3)A_2({\theta }_2)A_1({\theta }_1)$. First exchange positions $1,2$, then $2,3$, then $1,2$. The initial equal-index exchange givesFrom the first term, the specified final word is reached through $b_2{c}_3$; from the second term, the intermediate equal-index exchange uses $b_1$ to produce species $1$, followed by $c_2$. Thus this exchange route has coefficientFor the other route, first exchange positions $2,3$, then $1,2$, then $2,3$. The first exchange is between different indices and gives $c_2{A}_1({\theta }_1)A_2({\theta }_3)A_1({\theta }_2)+c_3{A}_1({\theta }_1)A_1({\theta }_3)A_2({\theta }_2)$. The first term reaches the target through $b_3{a}_3$, and the second reaches it through $(b_1+b_2+b_3)a_2$. ThereforeEquate these two coefficients, cancel the common term $a_2{b}_2{c}_3$, and rearrange:Restoring the three arguments gives the required scalar Yang-Baxter equation. The derivation starts in the ordered physical region $\theta ,{\theta }^{\prime }>0$; the identity extends to other values by the same scattering analytic continuation, wherever its factors are defined. Associativity must hold coefficient by coefficient for every species word; this displayed relation is one necessary component.

Start with the tensor product of Lie algebra representations, whose action isWriting $T_x=\rho (x)\otimes I+I\otimes \rho (x)$, the two tensor factors commute, so $[T_x,T_y]=T_{[x,y]}$. This verifies the Lie algebra representation identity over every field.

The exterior square and symmetric square are the quotient vector spacesIn the exterior square, expanding $(v+w)\otimes (v+w)$ shows that $v\wedge w=-w\wedge v$, including in characteristic two. Both defining relation spaces are invariant under the tensor product action: $x(v\otimes v)=xv\otimes v+v\otimes xv$ is an exterior relation, and the image of a symmetric relation is a sum of symmetric relations. Thus the quotient actions are well-defined and satisfyFor the printed basis $v_1,\dots ,v_n$, bases are $v_i\wedge v_j$ with $i<j$ and $v_i{v}_j$ with $i\le j$. Their dimensions are $n(n-1)/2$ and $n(n+1)/2$ respectively.

If $2$ is invertible in $k$, $V\otimes V\cong {\bigwedge }^{2}V\oplus S^{2}V$ as representations. Define the flip $\tau (v\otimes w)=w\otimes v$. It commutes with the Lie algebra action and satisfies ${\tau }^2=I$. Thereforeare complementary invariant linear projections. The mapsidentify $S^{2}V$ with $\mathrm{im}{P}_{+}$ and ${\bigwedge }^{2}V$ with $\mathrm{im}{P}_{-}$. Their inverses are the corresponding quotient maps restricted to these subspaces. This proves the assertion for every field of odd characteristic, and also for characteristic zero.

Over every field, $S^2(V\oplus W)\cong S^{2}V\oplus S^{2}W\oplus (V\otimes W)$. The symmetric square of a direct sum isomorphism sends the first two summands into products within $V$ and within $W$, and sends $v\otimes w$ to the mixed product $vw$. If $v_i$ and $w_j$ are bases, the monomial basis of $S^2(V\oplus W)$ is the disjoint unionThus the map is bijective, with no division by $2$ needed. The Leibniz rule for the action preserves each of these three summands and agrees with its usual Lie algebra representation action, proving equivariance.

For $\mathfrak{g}=\mathfrak{sl}(3)$, $V={\Gamma }_{0,0}$ is trivial and $W={\Gamma }_{2,1}$ has dimension $15$. HenceIt remains to find the irreducible representations in the symmetric square of the sl3 representation of highest weight (2,1). We give the formal character calculation explicitly.

Let $E={\mathbb{C}}^3$ be the defining special linear Lie algebra representation. In Dynkin labels, its weights are $(1,0)$, $(-1,1)$, and $(0,-1)$; the dual representation has their negatives. The equivariant contractionis surjective. Its kernel has dimension $18-3=15$. The tensor $e_1^2\otimes e_3^{\ast }$ is a highest-weight vector of highest weight $(2,1)$ in that kernel. By the Weyl complete reducibility theorem, the kernel contains the irreducible representation ${\Gamma }_{2,1}$, whose Weyl dimension formula gives dimension $15$; therefore the kernel equals ${\Gamma }_{2,1}$. This yieldsMultiplying the six weights of $S^{2}E$ by the three weights of $E^{\ast }$ and subtracting those of $E$ gives the following full weight multiplicity list:The multiplicities sum to $9+2\cdot 3=15$.

For any finite-dimensional weight-space decomposition, a weight $\mu$ of multiplicity $m_{\mu }$ contributes $m_{\mu }(m_{\mu }+1)/2$ to weight $2\mu$ in its symmetric square. Distinct weights $\mu ,\nu$ contribute $m_{\mu }{m}_{\nu }$ to $\mu +\nu$. Equivalently,Applying this to the displayed list gives all dominant weight multiplicities in the second column below. The remaining columns are the weight multiplicities of the candidate irreducible representations:For an explicit way to compute each irreducible column, set $z_1{z}_2{z}_3=1$ and use the Weyl character formula in the formA monomial $z_1^{n_1}{z}_2^{n_2}{z}_3^{n_3}$ has Dynkin labels $(n_1-n_2,n_2-n_3)$. Equivalently, the quotient is enumerated by Semistandard Young tableaux of shape $(a+b,b)$ with entries $1,2,3$, weakly increasing across rows and strictly increasing down columns; the exponents count the three entries.

The five irreducible columns sum to the $S^{2}W$ column. These are all its dominant weights, and all five candidate characters have no other dominant weights. Every Weyl group orbit meets the dominant chamber, and weight multiplicities are constant on Weyl group orbits. Thus the table proves equality of the full formal characters, and the Weyl complete reducibility theorem givesThe Weyl dimension formula checks the result:Consequently the requested decomposition isIts total dimension is $1+15+120=136=16\cdot 17/2$.

A smooth vector field is a smooth section of a vector bundle $X:M\to TM$ of the tangent bundle, so $X_p\in T_{p}M$. In a manifold chart, $X=X^i{\partial }_i$ with smooth coefficients. It acts on a smooth function by $Xf=X^i{\partial }_{i}f$. For a diffeomorphism $\alpha :M\to N$, the pushforward of a vector field isThe Lie bracket of vector fields is the commutator of their actions on smooth functions:The second derivatives of $f$ cancel, so this is again a vector field. The corresponding Jacobi identity is $[X,[Y,Z]]+[Y,[Z,X]]+[Z,[X,Y]]=0$.

A local flow of $X$ is a smooth map $(t,p)\mapsto {\varphi }_t(p)$ on an open neighbourhood of $\{0\}\times M$ satisfying ${\varphi }_0(p)=p$ and ${\partial }_t{\varphi }_t(p)=X_{{\varphi }_t(p)}$. On a sufficiently small neighbourhood and time interval, each ${\varphi }_t$ is a diffeomorphism onto its image, with inverse ${\varphi }_{-t}$. Uniqueness of integral curves of a vector field gives ${\varphi }_{s+t}={\varphi }_s\circ {\varphi }_t$ wherever both sides are defined. These are local statements; no assumption of a complete vector field is required.

A smooth tensor field of type $(r,s)$ is a smooth section of $(TM{)}^{\otimes r}\otimes (T^{\ast }M{)}^{\otimes s}$, where the factors are the tangent bundle and cotangent bundle. The flow definition of the Lie derivative of a tensor field isHere pullback applies $d{\varphi }_t^{-1}$ to each vector factor and the dual of $d{\varphi }_t$ to each covector factor, evaluating $T$ at ${\varphi }_t(p)$. Thus all tensors being differentiated lie in the same fibre over $p$.

For a smooth function, ${\varphi }_t^{\ast }f=f\circ {\varphi }_t$, and the chain rule givesFor a vector field $Y$, in local coordinates the expansions ${\varphi }_t(x)=x+tX(x)+O(t^2)$ and $d{\varphi }_t(x)=I+td{X}_x+O(t^2)$ giveConsequentlyPullback preserves tensor products and tensor contractions, so differentiation makes this definition a tensor derivation. It therefore agrees on every tensor field with the Lie derivative of a tensor field determined by these two formulas.

Now put ${\psi }_t=\alpha \circ {\varphi }_t\circ {\alpha }^{-1}$. The chain rule givesThus ${\psi }_t$ is the local flow of ${\alpha }_{\ast }X$. In the case $M=N$, if ${\alpha }_{\ast }X=X$, uniqueness of integral curves of a vector field makes ${\psi }_t={\varphi }_t$ on their common domains, or $\alpha \circ {\varphi }_t={\varphi }_t\circ \alpha$. Conversely, differentiating this commuting identity at zero gives $d{\alpha }_p{X}_p=X_{\alpha (p)}$, hence ${\alpha }_{\ast }X=X$. This proves that diffeomorphism invariance of a vector field is equivalent to commuting with its local flow, with every identity understood on the domain where its compositions exist.

For each point $x$, choose a global section $s$ whose germ $s_x$ generates the rank-one free ${\mathcal{O}}_{X,x}$-module ${\mathcal{L}}_x$. The section $s\otimes s$ is global, and tensor products commute with taking stalks. Thereforegenerates $(\mathcal{L}{\otimes }_{{\mathcal{O}}_X}\mathcal{L}{)}_x$. Thus the tensor square is globally generated. More generally, the tensor product of globally generated line bundles is globally generated: choose a generator section for each factor at the point and tensor them. This works on every scheme, including nonreduced schemes.

Write a local frame as $e_a$, its dual basis as ${ϵ}^a$, and set $D{e}_a=A^b{}_a{e}_b$. The one-form definition gives $D{ϵ}^a=-A^a{}_b{ϵ}^b$. Repeated indices are summed. Every tensor field has a unique local expansionDefine $DT$ by applying the scalar operator to its coefficient and applying $D$ to one frame factor at a time, adding all these terms. In components this isIn each term only the indicated slot is replaced. This formula is real-linear and satisfies the Leibniz rule for a tensor product: differentiating a coefficient product uses the scalar product rule, and the list of differentiated frame factors splits into the two factors' lists.

To check that it is intrinsic, write $e$ as the row of frame fields and $ϵ$ as the column of dual fields. Change frame by $e^{\prime }=eP$, where $P$ is an invertible smooth matrix. ThenDifferentiating the inverse matrix gives $Z(P^{-1})=-P^{-1}(ZP)P^{-1}$; it cancels the extra frame-change terms in the dual factors. The same cancellation in each tensor slot makes the two local formulas agree. Thus they glue to a global tensor derivation.

In particular, for arbitrary tensor types, not necessarily the same type,The resulting type is the sum of their two covariant counts and their two contravariant counts. This also supplies the required real-linearity for every type, including functions as type $(0,0)$.

Put $A^e=A{\otimes }_k{A}^{\mathrm{op}}$, so an $A$-bimodule is a left $A^e$-module through $(a\otimes b^{\mathrm{op}})m=amb$. The Hochschild cohomology isEquivalently, the Hochschild cochain complex has $C^n(A,M)={\mathrm{Hom}}_k(A^{\otimes n},M)$ and coboundary mapIts cohomology agrees with the displayed Ext functor because the bar resolution of an associative algebra is free over $A^e$ when $k$ is a field. The Hochschild cohomological dimension isThe supremum uses all bimodules, not merely the finitely generated one supplied in the question; an unbounded projective dimension is infinity.

An extension in this classification is a square-zero extension of an algebra: a short exact sequence $0\to M\to E\to A\to 0$, where $E$ and $A$ are unital algebras, $E\to A$ is unital, and $M$ is its two-sided ideal with $M^2=0$ and induced $A$-bimodule structure equal to the prescribed one. An equivalence of extensions is an algebra isomorphism of the middle terms commuting with the maps and inducing the identity on both $A$ and $M$. Arbitrary isomorphisms of middle algebras, or extensions without the square-zero ideal requirement, are not classified by this cohomology group.

Choose a $k$-linear map section $s:A\to E$ with $s(1)=1$. Its multiplication defectis a normalized Hochschild cocycle: $\mu (1,a)=\mu (a,1)=0$, and associativity in $E$ gives $\delta \mu =0$. Changing $s$ to $s+g$, where $g(1)=0$, changes the defect to $\mu +\delta g$, since $M^2=0$.

Conversely, for a normalized Hochschild cocycle $\mu$, put $E_{\mu }=A\oplus M$ as a vector space and defineThe Hochschild cocycle equation is exactly associativity, and $(1,0)$ is the identity. If ${\mu }^{\prime }=\mu +\delta g$, the map $E_{{\mu }^{\prime }}\to E_{\mu }$, $(a,m)\mapsto (a,m+g(a))$, is an equivalence. Conversely, every equivalence has this form after choosing sections. The normalized Hochschild cochain complex computes the same cohomology as the full complex: in the bar resolution of an associative algebra, the degenerate terms containing an inserted identity form a contractible subcomplex. Passing to the normalized bar resolution of an associative algebra, then applying the Hom functor, gives the same cohomology. Thus every class has a normalized representative. We obtain a bijection between $H{H}^2(A,M)$ and equivalence classes of square-zero extensions.

For a formal associative deformation, a completion convention is necessary. The usual star product lives on the formal power series moduleThis is the adic completion of a module applied to the ordinary tensor product, rather than literally the ordinary $A{\otimes }_{k}k[[t]]$ when $A$ is infinite-dimensional. For example, ${\sum }_{n\ge 0}{X}^n{t}^n$ lies in $k[X][[t]]$ but not in the ordinary tensor product, whose coefficient spaces have finite-dimensional span. The two agree when $A$ is finite-dimensional. We interpret the printed notation in this standard completed sense; the infinite iteration below requires that interpretation.

A star product is a $k[[t]]$-bilinear, unital product continuous for the adic topology satisfying associativity of the formIt is a trivial formal deformation if a $k[[t]]$-linear automorphism continuous for the adic topology $T=\mathrm{id}+{\sum }_{r\ge 1}{t}^r{T}_r$, with $T(1)=1$, satisfies $T(a\ast b)=T(a)T(b)$.

If $\mathrm{Dim}(A)\le 1$, then $H{H}^2(A,A)=0$. Suppose changes of coordinates have removed all coefficients below order $r$. The order-$r$ part of associativity then says $\delta {\mu }_r=0$. Hence ${\mu }_r=\delta g_r$ for a $k$-linear map $g_r:A\to A$. Its normalization gives $g_r(1)=0$, since $(\delta g_r)(1,1)=g_r(1)$. Transport the product by $T_r=\mathrm{id}+t^r{g}_r$:The order-$r$ coefficient becomes ${\mu }_r-\delta g_r=0$, and lower coefficients remain zero. Repeating constructs compatible changes of coordinates modulo every $t^N$. They converge in the adic topology to an invertible $T$ fixing $1$, with inverse obtained coefficient by coefficient. The limit product is ordinary multiplication. Therefore every star product is trivial under the completed formal-series convention. In fact, the argument only needs the vanishing of $H{H}^2(A,A)$, not all of Hochschild cohomological dimension at most one.

The Clebsch-Gordan decomposition for SU2 statesRealize the tensor product as polynomials of bidegree $(m,n)$ in $(x_1,y_1)$ and $(x_2,y_2)$. The determinant $D=x_1{y}_2-y_1{x}_2$ is invariant under the simultaneous SU2 action. For each $r$, the vectorhas torus weight $q=m+n-2r$ and is killed by $E=x_1{\partial }_{y_1}+x_2{\partial }_{y_2}$. Applying $F=y_1{\partial }_{x_1}+y_2{\partial }_{x_2}$ produces $q+1$ nonzero vectors of distinct weights, with $F^{q+1}{v}_r=0$. The relations $[E,F]=H$ and $H{v}_r=q{v}_r$ give $E{F}^j{v}_r=j(q-j+1)F^{j-1}{v}_r$, so their span is an irreducible copy of $V_q$.