Exchange symmetry, Schur–Weyl and Temperley–Lieb spin algebra

On a word $w=u_{i_1}\cdots u_{i_k}$ define $w^{\ast }=u_{i_k}\cdots u_{i_1}$, extend by $({\sum }_w{c}_{w}w{)}^{\ast }={\sum }_w\overline{c_w}{w}^{\ast }$, and set $1^{\ast }=1$. Reversal of words and [F2] give $(xy{)}^{\ast }=y^{\ast }{x}^{\ast }$ and $(x^{\ast }{)}^{\ast }=x$ in the free algebra. Thus this is a conjugate-linear anti-involution before taking the quotient.

The star fixes each relation $u_i^2-2u_i$ and each palindromic relation $u_i{u}_{i+1}{u}_i-u_i$ and $u_{i+1}{u}_i{u}_{i+1}-u_{i+1}$. It sends a distant commutator $u_i{u}_j-u_j{u}_i$ to its negative. Hence the star sends every generator of the defining ideal into that ideal.

Every element of the ideal is a finite sum of terms $xry$ with $r$ a displayed defining relation; its star is $y^{\ast }{r}^{\ast }{x}^{\ast }$, which is again in the ideal by step 2.1. The ideal is therefore star-invariant, so the operation descends to the quotient and remains conjugate-linear, anti-multiplicative, and involutive there. If $N\le 1$ there are no generators or relations, and the same construction is just coefficient conjugation.

On elementary tensors, applying $\rho (\pi )$ after ${\mathrm{\Delta }}_N(U)$ gives $U{v}_{{\pi }^{-1}(1)}\otimes \cdots \otimes U{v}_{{\pi }^{-1}(N)}$, which is also the result of applying ${\mathrm{\Delta }}_N(U)$ after $\rho (\pi )$. The product formula and [F6] make $\rho (\pi )$ unitary because it permutes the factors, and make $U^{\otimes N}$ unitary because $⟨Uv,Uw{⟩}_V=⟨v,w{⟩}_V$ for every factor. For $N=0$ both actions are the identity on $\mathbb{C}$.

The normalized trace formula is an inner product on $M_{\lambda }$. Choose an orthonormal basis $(s_k{)}_{k=1}^{d_{\lambda }}$ of $S^{\lambda }$ using [F9]. The adjoint identity [F7], conjugate symmetry, and the diagonal trace formula [F12] give $d_{\lambda }^{-1}\mathrm{Tr}(g^{\ast }f)=d_{\lambda }^{-1}{\sum }_k⟨f(s_k),g(s_k){⟩}_{H_N}\mathrm{.}$ This expression is linear in $f$, conjugate-linear in $g$, conjugate symmetric, and for $g=f$ is a sum of nonnegative squared norms. If it is zero, every $f(s_k)=0$, hence $f=0$; so it is positive definite. The product inner product on $S^{\lambda }\otimes M_{\lambda }$ is well-defined by its finite product basis from [F6] and the tensor basis theorem, and is positive definite by the same coordinate formula.

By [F2] and [F3], the irreducible $S_N$-types occurring in $H_N$ are exactly the modules $S^{\lambda }$ with $\lambda ⊢N$ and $\mathrm{\ell }(\lambda )\le d$. Their multiplicity spaces are $M_{\lambda }={\mathrm{Hom}}_{S_N}(S^{\lambda },H_N)$, and the algebraic Schur–Weyl decomposition places $S_N$ on the Specht factor and $GL(V)$ on the multiplicity factor. This gives the algebraic partition and factor- placement inputs for the local evaluation argument.

By [F6], $H_N$ is finite-dimensional. In any nonzero invariant subspace of the current remainder, choose an invariant subspace $W$ of least positive dimension; it is irreducible. Its orthogonal complement in that remainder is invariant: for $x\perp W$ and $w\in W$, $⟨\rho (\pi )x,w⟩=⟨x,\rho (\pi {)}^{-1}w⟩=0$ because $W$ is invariant under $\rho (\pi {)}^{-1}$ and the action is unitary. [F8] gives the orthogonal splitting at each stage. Repeating in the complement, whose dimension strictly decreases, yields a finite orthogonal decomposition of all $H_N$ into irreducible copies. By [F2], each copy has a unique Specht type. The orthogonal projection onto one copy commutes with $S_N$, since that copy and the sum of the other invariant copies are invariant. Its restriction to an inequivalent irreducible copy is an intertwiner, so [F4] forces it to vanish. Hence copies of distinct Specht types are orthogonal.

Since $S_N$ is finite by [F11], choose a basis $(v_j)$ of the finite- dimensional $S^{\lambda }$ and define its coordinate product $⟨\sum a_j{v}_j,\sum b_j{v}_j{⟩}_0:=\sum a_j\overline{b_j}$. Average this product over $S_N$ by $⟨v,w⟩:=\mathrm{\mid }{S}_N{\mathrm{\mid }}^{-1}{\sum }_{\pi \in S_N}⟨\rho (\pi )v,\rho (\pi )w{⟩}_0$. Each summand on $(v,v)$ is nonnegative and the identity term is positive for $v\mathrm{\ne }0$, while right multiplication reindexes the finite sum; hence this is a positive definite invariant Hermitian product. This proves existence; now use the invariant product fixed in the Statement. For each irreducible copy $K_a$ of type $S^{\lambda }$, choose an equivariant isomorphism $j_a:S^{\lambda }\to K_a$. Its adjoint exists by [F10], and $j_a^{\ast }{j}_a$ commutes with $S_N$: the adjoint identity [F7], equivariance of $j_a$, and unitarity of both representations give $\rho (\pi {)}^{-1}{j}_a^{\ast }{j}_a\rho (\pi )=j_a^{\ast }{j}_a$. By [F5] this is $c_{a}I$; injectivity gives $c_a>0$, so replacing $j_a$ by $c_a^{-1\mathrm{/}2}{j}_a$ makes it unitary. For any $f\in M_{\lambda }$, decompose $f={\sum }_b{P}_{b}f$ using the orthogonal projections onto all irreducible copies in $H_N$; each $P_{b}f$ is an intertwiner because $P_b$ commutes with $S_N$ by step 2.1. If $K_b$ has type different from $S^{\lambda }$, then $P_{b}f=0$ by [F4] and [F2]. If $K_b$ has type $S^{\lambda }$, then $j_b^{-1}{P}_{b}f:S^{\lambda }\to S^{\lambda }$ is an intertwiner, hence is a scalar by [F5]. Thus the maps $j_a$ span $M_{\lambda }$. Their ranges are orthogonal, so $d_{\lambda }^{-1}\mathrm{Tr}(j_b^{\ast }{j}_a)={\delta }_{ab}$; they are an orthonormal basis in the stated normalized Hilbert-Schmidt inner product. By [F9], take an orthonormal basis of each $S^{\lambda }$; evaluation sends the tensor product basis to the union of the corresponding orthonormal bases of all copies $K_a$. This union is an orthonormal basis of $H_N$, so $\mathrm{\Phi }$ is unitary.

For $\pi \in S_N$, evaluation gives $\mathrm{\Phi }({\rho }_{\lambda }(\pi )s\otimes f)=f({\rho }_{\lambda }(\pi )s)=\rho (\pi )f(s)=\rho (\pi )\mathrm{\Phi }(s\otimes f)$ because $f$ is an intertwiner. For $U\in U(d)$, ${\mathrm{\Delta }}_N(U)\circ f$ is again an intertwiner by [F1], and evaluation gives $\mathrm{\Phi }(s\otimes ({\mathrm{\Delta }}_N(U)\circ f))={\mathrm{\Delta }}_N(U)f(s)={\mathrm{\Delta }}_N(U)\mathrm{\Phi }(s\otimes f)$. Thus the actions have the asserted factor placement. Since ${\mathrm{\Delta }}_N(U)$ is unitary and $\mathrm{\Phi }$ is unitary, the postcomposition action is unitary on each multiplicity factor: on the orthogonal summand $S^{\lambda }\otimes M_{\lambda }$ the operator is $I\otimes T_U$, and evaluating on any unit vector of the nonzero $S^{\lambda }$ factor shows that $T_U$ preserves inner products; $T_{U^{-1}}$ is its inverse.

For an ordered pair put $c=(x+y)\mathrm{/}2$ and $z=(x-y)\mathrm{/}2$. This is an invertible linear coordinate change, with inverse $(c,z)\mapsto (c+z,c-z)$; the collision-free condition is $z\mathrm{\ne }0$. In polar coordinates write $z=ru$ with $r=\mathrm{\parallel }z\mathrm{\parallel }>0$ and $u\in S^1$. These formulas and their inverses are continuous, so they give a homeomorphism $F_2({\mathbb{R}}^2)\cong {\mathbb{R}}^2\times (0,\mathrm{\infty })\times S^1$. Transposition fixes $c,r$ and sends $u$ to $-u$. If $A:=S^1\mathrm{/}(u\sim -u)$ has quotient topology, the map $(c,r,u)\mapsto (c,r,[u])$ is an open quotient map: the circle quotient is open because the saturation of an open set is its union with its antipodal image, and products of open rectangles show the same for the product map. Its fibers are exactly the transposition orbits. Thus it induces a homeomorphism $C_2({\mathbb{R}}^2)\cong X:={\mathbb{R}}^2\times (0,\mathrm{\infty })\times A\mathrm{.}$

Let $\eta :\mathbb{R}\mathrm{/}\mathbb{Z}\to S^1$ be the homeomorphism $\eta ([t])=(\cos 2\pi t,\sin 2\pi t)$ from [F5]. The antipodal map on $S^1$ corresponds under $\eta$ to $[t]\mapsto [t+1\mathrm{/}2]$. The continuous map $D_2:\mathbb{R}\mathrm{/}\mathbb{Z}\to \mathbb{R}\mathrm{/}\mathbb{Z}$, $D_2([t]):=[2t]$, is well-defined and continuous by [F6] applied to $t\mapsto [2t]$ on $\mathbb{R}$; its fibres are exactly the pairs $[t],[t+1\mathrm{/}2]$. Hence $\psi :A\to \mathbb{R}\mathrm{/}\mathbb{Z}$, $\psi ([\eta ([t])]):=[2t]$, is well-defined and continuous by the quotient topology. Its inverse is $[s]\mapsto [\eta ([s\mathrm{/}2])]$; changing $s$ by an integer changes $s\mathrm{/}2$ by a half-integer, which is the same class in $A$. This inverse is continuous by [F6] applied to the continuous map $s\mapsto [\eta ([s\mathrm{/}2])]$ on $\mathbb{R}$; the two formulas compose to the identity, so $\psi$ is a homeomorphism. Under the homeomorphism of step 1.1, $q_0$ is $(0,1,[(1,0)])$, which $\psi$ sends to $[0]$.

On $X$ define $r(c,\rho ,a):=\psi (a)$ and $i([t]):=(0,1,{\psi }^{-1}([t]))$. These maps are continuous, preserve the specified basepoints, and $r\circ i$ is the identity. The maps $H_s(c,\rho ,a):=((1-s)c,(1-s)\rho +s,a),0\le s\le 1,$ are a continuous based homotopy from ${\mathrm{id}}_X$ to $i\circ r$: the radius is positive since $(1-s)\rho +s>0$, and the basepoint $(0,1,[(1,0)])$ is fixed for every $s$.

By [F3], $(r\circ i{)}_{\ast }=r_{\ast }{i}_{\ast }=\mathrm{id}$ and the based homotopy in step 3.1 gives $(i\circ r{)}_{\ast }=i_{\ast }{r}_{\ast }=\mathrm{id}$ on the based-loop classes of [F2]. Thus $r_{\ast }$ and $i_{\ast }$ are inverse isomorphisms. The pointed homeomorphism of step 1.1 transports this isomorphism back to $C_2({\mathbb{R}}^2)$; applying [F4] gives ${\pi }_1(C_2({\mathbb{R}}^2),q_0)\cong \mathbb{Z}$.

In the coordinates of step 1.1, $\gamma$ keeps $c=0$ and $r=1$ and has $u(t)=(\cos \pi t,\sin \pi t)$. Its image under $r$ is $[t]={\omega }_1(t)$, which is the positive generator by [F4]. Replacing $\pi t$ by $2\pi t$ describes a full turn of the relative direction; its image is $[2t]={\omega }_2(t)$, so its class is twice $[\gamma ]$. This identifies exchange and winding orientation without using a diagram.

On ${\mathbb{C}}^2\otimes {\mathbb{C}}^2$, direct application to the four basis vectors gives $s=\frac{1}{2}(I+{\sigma }_x\otimes {\sigma }_x+{\sigma }_y\otimes {\sigma }_y+{\sigma }_z\otimes {\sigma }_z)\mathrm{.}$ For $\mathrm{\mid }00⟩$ and $\mathrm{\mid }11⟩$, the sum of the three Pauli products is the vector itself; for $\mathrm{\mid }01⟩$ it is $2\mathrm{\mid }10⟩-\mathrm{\mid }01⟩$, and for $\mathrm{\mid }10⟩$ it is $2\mathrm{\mid }01⟩-\mathrm{\mid }10⟩$, so the displayed operator is the factor swap in all four cases. The symmetric three-dimensional subspace and antisymmetric line are an orthogonal decomposition; $s$ is $+I$ on the former and $-I$ on the latter, whose unit vector is $\omega$. Therefore $I-s=2P_{\omega }$, and $(I-s{)}^2=4P_{\omega }^2=4P_{\omega }=2(I-s)$. Tensoring with identities gives $e_i=2P_i$, self-adjointness, and $e_i^2=2e_i$ for every bond.

On three adjacent factors, $P_{12}$ has image spanned by $\mathrm{\mid }\omega {⟩}_{12}\otimes \mathrm{\mid }0⟩$ and $\mathrm{\mid }\omega {⟩}_{12}\otimes \mathrm{\mid }1⟩$. Applying $P_{23}$ and then $P_{12}$ to these two vectors gives, respectively, $P_{12}{P}_{23}{P}_{12}(\mathrm{\mid }\omega {⟩}_{12}\mathrm{\mid }0{⟩}_3)=\frac{1}{4}\mathrm{\mid }\omega {⟩}_{12}\mathrm{\mid }0{⟩}_3,P_{12}{P}_{23}{P}_{12}(\mathrm{\mid }\omega {⟩}_{12}\mathrm{\mid }1{⟩}_3)=\frac{1}{4}\mathrm{\mid }\omega {⟩}_{12}\mathrm{\mid }1{⟩}_3\mathrm{.}$ For example, the first intermediate vector is $P_{23}(\mathrm{\mid }\omega {⟩}_{12}\mathrm{\mid }0{⟩}_3)=(-\mathrm{\mid }001⟩+\mathrm{\mid }010⟩)\mathrm{/}(2\sqrt{2})$; $P_{12}$ kills $\mathrm{\mid }001⟩$ and sends $\mathrm{\mid }010⟩$ to $\mathrm{\mid }\omega {⟩}_{12}\mathrm{\mid }0{⟩}_3\mathrm{/}\sqrt{2}$. The second computation is the explicit calculation $P_{23}(\mathrm{\mid }\omega {⟩}_{12}\mathrm{\mid }1{⟩}_3)=(-\mathrm{\mid }101⟩+\mathrm{\mid }110⟩)\mathrm{/}(2\sqrt{2})$; $P_{12}$ kills $\mathrm{\mid }110⟩$ and sends $\mathrm{\mid }101⟩$ to $-\mathrm{\mid }\omega {⟩}_{12}\mathrm{\mid }1{⟩}_3\mathrm{/}\sqrt{2}$, again giving the coefficient $1\mathrm{/}4$. Since $P_{12}{P}_{23}{P}_{12}$ first projects into the displayed image, it follows on all eight basis vectors that $P_{12}{P}_{23}{P}_{12}=P_{12}\mathrm{/}4$. Reversing the three tensor factors exchanges $P_{12}$ and $P_{23}$ (the singlet line is unchanged by reversing its two entries), so also $P_{23}{P}_{12}{P}_{23}=P_{23}\mathrm{/}4$. Hence $e_i{e}_{i+1}{e}_i=8P_i{P}_{i+1}{P}_i=2P_i=e_i$ and the reversed relation holds. Operators on disjoint tensor factors commute, giving the distant relation.

The relations proved in steps 1.1 and 2.1 and [F3] give an algebra homomorphism from the presented quotient, since every defining relation has zero image. Each $e_i$ is self-adjoint by step 1.1, so the operator adjoint of a word reverses its factors and fixes its generator images; conjugating coefficients agrees with the algebra involution [F4]. Thus the representation preserves $\ast$.

The adjacent swap is the place action of the transposition $(i,i+1)$, since that transposition is its own inverse in the left-place convention. Therefore $e_i=I-\rho ((i,i+1))$. Applying the factor placement in [F5] shows that on each summand $S^{\lambda }\otimes M_{\lambda }$ it is $(I-{\rho }_{\lambda }((i,i+1)))\otimes I_{M_{\lambda }}$, as claimed.