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Symmetric-power SU2 representation

Statement

Fix an integer $n \geq 0$ , let $V : = C^{2}$ carry the standard inner product linear in the first argument, with orthonormal basis $(e_{+}, e_{-})$ where $e_{+} : = (1, 0)$ and $e_{-} : = (0, 1)$ (Real and complex inner product spaces, with the inner product linear in the first argument), and let $V^{\otimes n}$ be the $n$ -fold algebraic tensor power with its finite string basis (The elementary tensors of two bases form the product basis of the tensor product) made orthonormal. Use $S_{n} = Sym ({0, \dots, n - 1})$ as in The finite symmetric group $S_{n}$ , one-line notation, and cycle notation. For $n \geq 1$ define the bijection $b_{n} : {0, \dots, n - 1} \to {1, \dots, n}$ by $b_{n} (a) = a + 1$ , and for $π \in S_{n}$ put $\hat{π} : = b_{n} \circ π \circ b_{n}^{- 1}$ . Define $U_{π}$ to be the place action of $\hat{π}$ in Commuting symmetric-group and linear actions on a tensor power: explicitly, for $x_{1}, \dots, x_{n} \in V$ , $U_{π} (x_{1} \otimes \dots \otimes x_{n}) = x_{π^{- 1} (0) + 1} \otimes \dots \otimes x_{π^{- 1} (n - 1) + 1}$ , extended linearly. Use the diagonal action $W^{\otimes n}$ of $W \in M_{2} (C)$ and the operators $Δ (T) : = \sum_{j = 1}^{n} 1^{\otimes (j - 1)} \otimes T \otimes 1^{\otimes (n - j)}$ . Put

$P_{+} : = \frac{1}{n!} \sum_{π \in S_{n}} U_{π}, S_{n} : = ran P_{+} \subseteq V^{\otimes n}, S U (2) : = {W \in M_{2} (C) : W^{*} W = I_{2}, \det W = 1},$

where $W^{*} : = {\overline{W}}^{T}$ is the conjugate transpose of $W$ and $I_{2}$ the identity matrix (Rectangular matrix multiplication and the identity matrix $I_{n}$ , including zero-sized shapes, For $n \geq 1$ , the determinant over a commutative ring by the Leibniz formula, and $∣ \det A ∣$ for a real matrix). For $W \in S U (2)$ define $U_{n} (W) : = W^{\otimes n} ∣_{S_{n}}$ . For $k \in {0, \dots, n}$ let $s_{k}$ be the sum of the elementary tensors $e_{ε_{1}} \otimes \dots \otimes e_{ε_{n}}$ over all $2^{n}$ strings $(ε_{1}, \dots, ε_{n}) \in {+, -}^{n}$ containing exactly $k$ entries $-$ , and put $v_{k} : = {(\binom{n}{k})}^{- 1 / 2} s_{k}$ .

For $n = 0$ use $V^{\otimes 0} = C$ , the empty string $s_{0} = v_{0} = 1$ , $0! = (\binom{0}{0}) = 1$ , the trivial group $S_{0}$ , $U_{π} = P_{+} = I_{C}$ , $W^{\otimes 0} = I_{C}$ for every $W$ , and $Δ (T) = 0$ . These conventions define $S_{0} = C$ and the trivial representation $U_{0}$ , with $J_{x} = J_{y} = J_{z} = 0$ .

Then:

(a) $P_{+}$ is the orthogonal projection onto the closed subspace $S_{n}$ , $U_{π}$ and $W^{\otimes n}$ commute for every $π \in S_{n}$ and every $W$ , and $W^{\otimes n} (S_{n}) \subseteq S_{n}$ ; the assignment $W \mapsto U_{n} (W)$ is well defined on all of $M_{2} (C)$ , and $U_{n} (W^{'}) U_{n} (W) = U_{n} (W^{'} W)$ holds for all $W, W^{'}$ ;

(b) $S U (2)$ is a subgroup of the multiplicative monoid of $M_{2} (C)$ with identity $I_{2}$ ; for $W \in S U (2)$ the operator $U_{n} (W)$ is unitary with inverse $U_{n} (W^{- 1}) = U_{n} (W)^{*}$ , so $U_{n} : S U (2) \to U (S_{n})$ is a group homomorphism (Strongly continuous unitary representations, invariant linear subspaces and intertwiners, Linear isometries, and orthogonal or unitary operators on finite-dimensional inner product spaces);

(c) the vectors $v_{0}, \dots, v_{n}$ form an orthonormal basis of $S_{n}$ ; in particular $S_{n}$ is $(n + 1)$ -dimensional;

(d) the operators $J_{i} : = \frac{1}{2} Δ (σ_{i}) ∣_{S_{n}}$ , $i = x, y, z$ , where $σ_{x}, σ_{y}, σ_{z}$ are the Pauli matrices of Finite spin matrices, are self-adjoint and satisfy, with $J_{\pm} : = J_{x} \pm i J_{y}$ , $J^{2} : = J_{x}^{2} + J_{y}^{2} + J_{z}^{2}$ ,

$J_{z} v_{k} = (\frac{n}{2} - k) v_{k}, J_{+} v_{k} = \sqrt{k (n - k + 1)} v_{k - 1}, J_{-} v_{k} = \sqrt{(k + 1) (n - k)} v_{k + 1},$

with $v_{- 1} : = 0$ and $v_{n + 1} : = 0$ ; moreover $[J_{z}, J_{\pm}] = \pm J_{\pm}$ , $[J_{+}, J_{-}] = 2 J_{z}$ , $[J_{i}, J_{j}] = i \sum_{k} ε_{i j k} J_{k}$ and $J^{2} = \frac{n}{2} (\frac{n}{2} + 1) I_{S_{n}}$ ; the map $v_{k} \mapsto e_{n / 2 - k}$ is a unitary equivalence of $(J_{x}, J_{y}, J_{z})$ with the spin- $\frac{n}{2}$ matrices of Finite spin matrices;

(e) for each $i$ the curve $A_{i} (t) : = \cos (t / 2) I_{2} - i \sin (t / 2) σ_{i}$ ( $t \in R$ ) lies in $S U (2)$ , satisfies $A_{i} (s) A_{i} (t) = A_{i} (s + t)$ and $A_{i}^{'} (0) = - \frac{i}{2} σ_{i}$ , and $t \mapsto V_{i} (t) : = U_{n} (A_{i} (t))$ is a strongly continuous one-parameter unitary group on $S_{n}$ (Strongly continuous one-parameter unitary group) whose infinitesimal generator $G_{i}$ (Infinitesimal generator of a unitary group) is $G_{i} = - i J_{i}$ on all of $S_{n}$ ; equivalently $\frac{1}{t} (V_{i} (t) v - v) \to - i J_{i} v$ for every $v \in S_{n}$ ;

(f) $U_{n}$ is strongly continuous on $S U (2)$ : if $W_{m} \to W_{0}$ in $M_{2} (C)$ with all $W_{m} \in S U (2)$ and $W_{0} \in S U (2)$ , then $∥ U_{n} (W_{m}) - U_{n} (W_{0}) ∥ \to 0$ in the operator norm on $S_{n}$ .

This is finite-dimensional linear algebra: the theorem constructs a unitary representation of $S U (2)$ and its generators. It does not assert that $S U (2)$ or any spin observable describes a physical particle, and it makes no measurement or statistics claim.

Facts & Assumptions

Given: $n \geq 0$ ; the space $V = C^{2}$ with its orthonormal basis $(e_{+}, e_{-})$ ; the tensor power $V^{\otimes n}$ ; the place permutations $U_{π}$ and diagonal actions $W^{\otimes n}$ ; the projection $P_{+}$ ; the subspace $S_{n}$ ; the strings $s_{k}$ and vectors $v_{k}$ ; the group $S U (2)$ ; the Pauli matrices $σ_{x}, σ_{y}, σ_{z}$ ; the operators $Δ (T)$ and $J_{i}$ .

[F1]

For $n \geq 1$ , iterating The elementary tensors of two bases form the product basis of the tensor product gives the finite basis of all $2^{n}$ strings $e_{ε_{1}} \otimes \dots \otimes e_{ε_{n}}$ ; for $n = 0$ the basis is $1 \in C$ . Define the inner product by $⟨ \sum c_{ε} e_{ε}, \sum d_{ε} e_{ε} ⟩ = \sum c_{ε} \overline{d_{ε}}$ . Expansion on elementary tensors gives the product inner product. The finite tensor product is thus the standard finite-dimensional coordinate inner product space; no completion or countable choice is used. The universal multilinear property is Finite iterated tensor products represent multilinear maps independently of parenthesization.

[F2]

For $n \geq 1$ , conjugation by $b_{n}$ is a bijection from the zero-indexed $S_{n}$ of The finite symmetric group $S_{n}$ , one-line notation, and cycle notation to the group on tensor positions ${1, \dots, n}$ of Commuting symmetric-group and linear actions on a tensor power; it preserves composition because $\hat{π} \hat{σ} = b_{n} π σ b_{n}^{- 1} = \hat{π σ}$ , and ${\hat{π}}^{- 1} (j) = π^{- 1} (j - 1) + 1$ for $1 \leq j \leq n$ . Thus the stated $U_{π}$ are well-defined linear maps by [F1], permute the orthonormal string basis, are unitary, satisfy $U_{π} U_{σ} = U_{π σ}$ and have $U_{π}^{*} = U_{π^{- 1}}$ . There are $n!$ permutations: prescribing the images of $0, \dots, n - 1$ successively leaves $n, n - 1, \dots, 1$ possibilities. In the finite average $P_{+} = (1 / n!) \sum_{π} U_{π}$ , inversion and left or right multiplication by a permutation reindex the summands. Thus $P_{+}^{*} = P_{+}$ , $U_{π} P_{+} = P_{+} U_{π} = P_{+}$ and $P_{+}^{2} = P_{+}$ ; its range consists exactly of the invariant vectors, since every invariant vector $v$ satisfies $P_{+} v = v$ . For $n = 1$ , $\hat{π} (1) = 1$ and $U_{π} = I_{V}$ ; for $n = 0$ , all identities follow from the separate empty-tensor conventions.

[F3]

On the spanning string basis, expansion of matrix entries gives $(S \otimes T) (S^{'} \otimes T^{'}) = (S S^{'}) \otimes (T T^{'})$ , and evaluation of the inner products of [F1] gives $(S \otimes T)^{*} = S^{*} \otimes T^{*}$ . Thus a tensor product of unitary matrices is unitary. Applying the same $W$ to every factor commutes with any place permutation, since both rules send $v_{1} \otimes \dots \otimes v_{n}$ to the tensor of the same permuted list $W v_{j}$ ; this identity is also valid for $n = 0$ (Commuting symmetric-group and linear actions on a tensor power).

[F4]

For a bounded operator $L$ on a finite-dimensional Hilbert space, the operator norm is $∥ L ∥ = \sup_{∥ z ∥ \leq 1} ∥ L z ∥$ , and $∥ L ∥ \leq \sum_{j} ∥ L e_{j} ∥$ for every orthonormal basis $(e_{j})$ ; a linear map with finite-dimensional domain is bounded, and $∥ L z ∥ \leq ∥ L ∥ ∥ z ∥$ (The operator norm as the least bound and as the unit-sphere or unit-ball supremum, A bounded linear operator between normed spaces). In orthonormal product coordinates, writing $z = \sum_{b} z_{b} \otimes e_{b}$ gives $∥ (S \otimes I) z ∥^{2} = \sum_{b} ∥ S z_{b} ∥^{2} \leq ∥ S ∥^{2} \sum_{b} ∥ z_{b} ∥^{2}$ ; similarly $∥ I \otimes T ∥ \leq ∥ T ∥$ . Since $S \otimes T = (S \otimes I) (I \otimes T)$ and $∥ L M z ∥ \leq ∥ L ∥ ∥ M ∥ ∥ z ∥$ , this proves $∥ S \otimes T ∥ \leq ∥ S ∥ ∥ T ∥$ , and iteration supplies the tensor estimates below.

[F5]

$W^{*} W = I_{2}$ means that $W$ is a unitary operator on $C^{2}$ ; in orthonormal bases the matrix of the adjoint is the conjugate transpose, and $\det (W^{*}) = \overline{\det W}$ , $\det (W^{'} W) = \det W^{'} \det W$ (In orthonormal bases, the matrix of the adjoint is the conjugate transpose of the matrix, Coordinate columns $[v]_{B}$ and matrices $[T]_{B}^{C}$ of linear maps relative to ordered bases, For same-sized finite square matrices over a commutative ring, $\det (A B) = \det (A) \det (B)$ , Linear isometries, and orthogonal or unitary operators on finite-dimensional inner product spaces, For $n \geq 1$ , the determinant over a commutative ring by the Leibniz formula, and $∣ \det A ∣$ for a real matrix).

[F6]

The Pauli matrices satisfy $σ_{i}^{*} = σ_{i}$ , $σ_{x}^{2} = σ_{y}^{2} = σ_{z}^{2} = I_{2}$ , $σ_{x} σ_{y} = i σ_{z}$ , $σ_{y} σ_{z} = i σ_{x}$ , $σ_{z} σ_{x} = i σ_{y}$ ; also $σ_{+} + σ_{-} = 2 σ_{x}$ and $σ_{+} - σ_{-} = 2 i σ_{y}$ for $σ_{+} : = σ_{x} + i σ_{y}$ , $σ_{-} : = σ_{x} - i σ_{y}$ , and $σ_{+} e_{-} = 2 e_{+}$ , $σ_{+} e_{+} = 0$ , $σ_{-} e_{+} = 2 e_{-}$ , $σ_{-} e_{-} = 0$ , $σ_{z} e_{\pm} = \pm e_{\pm}$ (Finite spin matrices).

[F7]

The sine and cosine addition identities hold (The addition formulas for sine and cosine). Sine and cosine are differentiable with $\sin^{'} = \cos$ , $\cos^{'} = - \sin$ , and $\sin^{2} + \cos^{2} = 1$ ; derivatives of sums, scalar multiples and compositions are computed by the algebra of derivatives and the chain rule (The derivatives of sine and cosine are cosine and minus sine, A function differentiable at $c$ is continuous at $c$ , Parity and the Pythagorean identity for sine and cosine, Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ , The chain rule, in one line from Carathéodory: if $g$ is differentiable at $c$ and $f$ is differentiable at $g (c)$ , then $f \circ g$ is differentiable at $c$ with $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$ , The derivative $f^{'} (c) = \lim_{x \to c} \frac{f (x) - f (c)}{x - c}$ of $f : A \to R$ at a point $c \in A$ that is a limit point of $A$ , and differentiability on a set).

[F8]

For integers $0 \leq k \leq n$ , $(\binom{n}{k}) = n! / (k! (n - k)!) > 0$ and $(\binom{n}{k}) = (\binom{n}{n - k})$ and $(\binom{n}{k - 1}) / (\binom{n}{k}) = k / (n - k + 1)$ for $1 \leq k \leq n$ ; the ratio follows by cancelling the factorials using $k! = k (k - 1)!$ and $(n - k + 1)! = (n - k + 1) (n - k)!$ . Each string with $k$ minus signs corresponds bijectively to its set of minus positions, so there are $(\binom{n}{k})$ such strings and the string sums split accordingly ( $(\binom{n}{k}) k! (n - k)! = n!$ for $k \leq n$ ; hence $(\binom{n}{k}) k! = n^{\underline{k}}$ , the quotient $n! / (k! (n - k)!)$ is a natural number, and $(\binom{n}{k}) = (\binom{n}{n - k})$ , The factorial $n!$ and the falling factorial $n^{\underline{k}}$ , defined by recursion in $N$ , The set $[A]^{k}$ of $k$ -element subsets and the binomial coefficient $(\binom{n}{k}) : = ∣ [n]^{k} ∣$ , A finite sum in a commutative monoid indexed by an arbitrary finite set).

[F9]

Complex conjugation is additive and multiplicative, $i^{2} = - 1$ , $\overline{i} = - i$ , and $\overline{z^{- 1}} = {\overline{z}}^{- 1}$ for $z \neq 0$ (Conjugation is an involutive real-field automorphism, $z \overline{z} = ∣ z ∣^{2}$ , and modulus is definite, multiplicative, and subadditive, The complex numbers as $R [x] / (x^{2} + 1)$ , with the real embedding and imaginary unit $i$ ).

[F10]

For $s = \frac{n}{2}$ the operators $J_{z}, J_{\pm}, J_{x}, J_{y}, J^{2}$ of Finite spin matrices act on their orthonormal basis $e_{m}$ , $- s \leq m \leq s$ , by $J_{z} e_{m} = m e_{m}$ , $J_{+} e_{m} = \sqrt{(s - m) (s + m + 1)} e_{m + 1}$ , $J_{-} e_{m} = \sqrt{(s + m) (s - m + 1)} e_{m - 1}$ , with endpoint conventions $e_{s + 1} = e_{- s - 1} = 0$ , have self-adjoint $J_{x}, J_{y}, J_{z}$ and $J_{-} = J_{+}^{*}$ , satisfy $[J_{z}, J_{\pm}] = \pm J_{\pm}$ , $[J_{+}, J_{-}] = 2 J_{z}$ and $J^{2} = s (s + 1) I$ , and have $∥ J_{+} e_{m} ∥^{2} = (s - m) (s + m + 1)$ .

Proof

1.1givenF10algebra

[The zero tensor power] For $n = 0$ , the conventions give $P_{+} = I$ , $S_{0} = C v_{0}$ , $U_{0} (W) = I$ , and $J_{i} = 0$ . Clauses (a)-(c) follow immediately; all ladder and Casimir formulas in (d) vanish, and $v_{0} \mapsto e_{0}$ identifies the zero operators with the spin-zero model [F10]. Each $V_{i} (t) = I$ has derivative zero and is strongly continuous, and $U_{0}$ is constant, proving (e)-(f); the subgroup and $A_{i}$ assertions, independent of $n$ , are proved below. For the tensor and basis calculations in the remaining steps assume $n \geq 1$ .

1.2F2F3algebra

[The symmetric subspace and invariance] By [F2] $P_{+}$ is an orthogonal projection and $U_{π} P_{+} = P_{+}$ ; hence $P_{+} v = v$ exactly for $v \in S_{n}$ , and $S_{n} = \ker (I - P_{+})$ is closed because the finite matrix $I - P_{+}$ is continuous. For $W \in M_{2} (C)$ and $π \in S_{n}$ , [F3] gives $W^{\otimes n} P_{+} = \frac{1}{n!} \sum_{π} W^{\otimes n} U_{π} = \frac{1}{n!} \sum_{π} U_{π} W^{\otimes n} = P_{+} W^{\otimes n}$ , so $v \in S_{n}$ implies $P_{+} (W^{\otimes n} v) = W^{\otimes n} P_{+} v = W^{\otimes n} v$ , i.e. $W^{\otimes n} (S_{n}) \subseteq S_{n}$ ; therefore $U_{n} (W) : = W^{\otimes n} ∣_{S_{n}}$ is a well-defined linear map on $S_{n}$ for every $W$ .

1.3F6F7F9algebra

[The curves $A_{i}$ ] Using $I_{2}^{*} = I_{2}$ , $σ_{i}^{*} = σ_{i}$ , $σ_{i}^{2} = I_{2}$ , [F7] and [F9]: $A_{i} (t)^{*} A_{i} (t) = \cos (t / 2)^{2} I_{2} + \sin (t / 2)^{2} σ_{i}^{2} - i \cos (t / 2) \sin (t / 2) (σ_{i} - σ_{i}^{*}) = (\cos^{2} + \sin^{2}) (t / 2) I_{2} = I_{2}$ , and direct computation of the three $2 \times 2$ determinants from the displayed matrices gives $\det A_{i} (t) = \cos^{2} (t / 2) + \sin^{2} (t / 2) = 1$ ; so $A_{i} (t) \in S U (2)$ . The identities $\cos (s + t) = \cos s \cos t - \sin s \sin t$ and $\sin (s + t) = \sin s \cos t + \cos s \sin t$ , applied with $σ_{i}^{2} = I_{2}$ , give $A_{i} (s) A_{i} (t) = A_{i} (s + t)$ . Differentiating the displayed formula at $t = 0$ with [F7] and [F9] gives $A_{i}^{'} (0) = - \frac{i}{2} σ_{i}$ .

2.1step 1.2F3F5algebra

[Representation law and unitarity] For $W, W^{'} \in M_{2} (C)$ , [F3] gives $(W^{'})^{\otimes n} W^{\otimes n} = (W^{'} W)^{\otimes n}$ on $V^{\otimes n}$ , hence on $S_{n}$ by step 1.2 and restriction: $U_{n} (W^{'}) U_{n} (W) = U_{n} (W^{'} W)$ . If $W \in S U (2)$ , then $W$ is unitary by [F5], [F6]; by [F3] $W^{\otimes n}$ is unitary on $V^{\otimes n}$ , and for $v \in S_{n}$ , $∥ U_{n} (W) v ∥ = ∥ W^{\otimes n} v ∥ = ∥ v ∥$ with $W^{\otimes n}$ mapping $S_{n}$ bijectively onto $S_{n}$ (its inverse is $(W^{- 1})^{\otimes n}$ restricted, where $W^{- 1} = W^{*} \in S U (2)$ ); hence $U_{n} (W)$ is unitary with $(U_{n} (W))^{*} = U_{n} (W^{- 1})$ . This gives the operator statements of (a) and (b).

2.2step 1.2F1F2F8algebra

[The basis $v_{k}$ and the dimension] Fix $k$ and let $w = e_{ε_{1}} \otimes \dots \otimes e_{ε_{n}}$ have exactly $k$ minus entries. In $n! P_{+} w = \sum_{π} U_{π} w$ every string with exactly $k$ minus entries occurs once for each of the $k! (n - k)!$ permutations carrying the pattern of $w$ to it, so $n! P_{+} w = k! (n - k)! s_{k}$ and $P_{+} w = {(\binom{n}{k})}^{- 1} s_{k}$ by [F8]; since the $2^{n}$ elementary tensors span $V^{\otimes n}$ [F1], this shows $S_{n} = ran P_{+} \subseteq span {s_{0}, \dots, s_{n}}$ , while $s_{k} \in S_{n}$ because each $U_{π}$ permutes the strings with $k$ minus entries, so $P_{+} s_{k} = s_{k}$ . Distinct elementary tensors are orthonormal [F1], and two summands from $s_{k}$ and $s_{l}$ with $k \neq l$ are distinct elementary tensors, so $⟨ s_{k}, s_{l} ⟩ = δ_{k l} (\binom{n}{k})$ by [F8]; hence $v_{0}, \dots, v_{n}$ are orthonormal and span $S_{n}$ , an orthonormal basis of the $(n + 1)$ -dimensional space $S_{n}$ .

2.3step 1.2F2F3F6algebra

[The operators $Δ (σ_{i})$ and their commutation] For $T, T^{'} \in M_{2} (C)$ the place sums satisfy $[Δ (T), Δ (T^{'})] = Δ ([T, T^{'}])$ and $Δ (T)^{*} = Δ (T^{*})$ : expanding the definitions, the summands at different places commute and the summand at a common place contributes $[T, T^{'}]$ , while the adjoint of a place sum is the place sum of the adjoints [F3]. Consequently $[Δ (σ_{i}), Δ (σ_{j})] = Δ ([σ_{i}, σ_{j}])$ ; and $Δ (σ_{i})$ commutes with every $U_{π}$ because permuting places preserves the multiset of places, so by [F2] each $Δ (σ_{i})$ preserves $S_{n}$ , and its restriction $J_{i} = \frac{1}{2} Δ (σ_{i}) ∣_{S_{n}}$ is self-adjoint.

3.1step 2.1F5F9algebra

[ $S U (2)$ is a subgroup] If $W, W^{'} \in S U (2)$ , then $(W W^{'})^{*} (W W^{'}) = W^{' *} W^{*} W W^{'} = W^{' *} W^{'} = I_{2}$ and $\det (W W^{'}) = \det W \det W^{'} = 1$ , so $W W^{'} \in S U (2)$ ; $I_{2}^{*} I_{2} = I_{2}$ and $\det I_{2} = 1$ , so $I_{2} \in S U (2)$ ; and for $W \in S U (2)$ one has $W^{*} W = W W^{*} = I_{2}$ (multiplying $W^{*} W = I_{2}$ by $W^{- 1}$ on the right gives $W^{*} = W^{- 1}$ , whence $W W^{*} = I_{2}$ ) and $\det W^{*} = \overline{\det W} = 1$ , so $W^{- 1} = W^{*} \in S U (2)$ . Hence $S U (2)$ is a subgroup, and $U_{n}$ is a group homomorphism by step 2.1.

3.2step 2.1step 2.3step 1.3F1F3F4F6F7algebra

[Generator of $t \mapsto U_{n} (A_{i} (t))$ ] By step 2.1, $V_{i} (t) : = U_{n} (A_{i} (t))$ satisfies $V_{i} (s) V_{i} (t) = V_{i} (s + t)$ and $V_{i} (0) = I$ , and each $V_{i} (t)$ is unitary. The telescoping identity $W^{\otimes k} - W_{0}^{\otimes k} = \sum_{j = 1}^{k} (W^{\otimes (j - 1)} \otimes (W - W_{0}) \otimes W_{0}^{\otimes (k - j)})$ , valid by [F3] with $W^{\otimes 0} : = I$ , gives $∥ W^{\otimes k} - W_{0}^{\otimes k} ∥ \leq k \max (∥ W ∥, ∥ W_{0} ∥)^{k - 1} ∥ W - W_{0} ∥$ by [F4]; with $W = A_{i} (t)$ , $W_{0} = A_{i} (t_{0})$ and $∥ A_{i} (\cdot) ∥ = 1$ , the explicit formula for $A_{i}$ and [F7] show that $∥ V_{i} (t) v - V_{i} (t_{0}) v ∥$ is bounded by $n ∥ A_{i} (t) - A_{i} (t_{0}) ∥ ∥ v ∥$ for all $v \in S_{n}$ by the operator-norm bound; so $V_{i}$ is a strongly continuous one-parameter unitary group. For the derivative, write $A_{i} (t) = I_{2} + t A_{i}^{'} (0) + r (t)$ with $∥ r (t) ∥ \leq ∣ t ∣ η (t)$ and $η (t) \to 0$ by [F6], [F7] and step 1.3; the same telescoping expansion with $W = A_{i} (t)$ , $W_{0} = I_{2}$ , using $∥ A_{i} (t) ∥ \leq 1$ and $∥ r (t) ∥ \leq ∣ t ∣ η (t)$ , gives $A_{i} (t)^{\otimes n} - I_{2}^{\otimes n} = t \sum_{j = 1}^{n} 1^{\otimes (j - 1)} \otimes A_{i}^{'} (0) \otimes 1^{\otimes (n - j)} + R (t)$ with $∥ R (t) ∥ \leq n ∣ t ∣ η^{'} (t)$ and $η^{'} (t) \to 0$ as $t \to 0$ ; restricting to $S_{n}$ and using step 2.3 and $A_{i}^{'} (0) = - \frac{i}{2} σ_{i}$ of step 1.3 gives $\frac{1}{t} (V_{i} (t) v - v) \to - \frac{i}{2} Δ (σ_{i}) v = - i J_{i} v$ for every $v \in S_{n}$ . Hence the infinitesimal generator is $G_{i} = - i J_{i}$ with domain $D (G_{i}) = S_{n}$ .

3.3step 2.2step 2.3F1F6F8algebra

[Actions on the basis vectors] Fix $0 \leq k \leq n$ . On an elementary tensor with $k$ minus entries, $Δ (σ_{z})$ acts by $# + - # - = n - 2 k$ , and by linearity $Δ (σ_{z}) s_{k} = (n - 2 k) s_{k}$ , so $J_{z} v_{k} = (n / 2 - k) v_{k}$ . For $σ_{+} = σ_{x} + i σ_{y}$ one has $σ_{+} e_{+} = 0$ and $σ_{+} e_{-} = 2 e_{+}$ by [F6]. If $1 \leq k \leq n$ , each target string with $k - 1$ minus entries in $Δ (σ_{+}) s_{k}$ occurs with coefficient $2$ from each of its $n - k + 1$ plus positions, so $Δ (σ_{+}) s_{k} = 2 (n - k + 1) s_{k - 1}$ and $J_{+} v_{k} = (n - k + 1) \sqrt{(\binom{n}{k - 1}) / (\binom{n}{k})} v_{k - 1} = \sqrt{k (n - k + 1)} v_{k - 1}$ by [F8]. For $k = 0$ , every factor is $e_{+}$ , so $J_{+} v_{0} = 0$ , agreeing with the stated $v_{- 1} = 0$ convention. Similarly, if $0 \leq k < n$ , $σ_{-} e_{-} = 0$ and $σ_{-} e_{+} = 2 e_{-}$ give $Δ (σ_{-}) s_{k} = 2 (k + 1) s_{k + 1}$ , hence $J_{-} v_{k} = (k + 1) \sqrt{(\binom{n}{k + 1}) / (\binom{n}{k})} v_{k + 1} = \sqrt{(k + 1) (n - k)} v_{k + 1}$ by [F8]. For $k = n$ , every factor is $e_{-}$ , so $J_{-} v_{n} = 0$ , agreeing with $v_{n + 1} = 0$ .

4.1step 3.3F10algebra

[Commutators and identification with the spin- $n / 2$ model] By steps 2.3 and 3.3, $J_{z}, J_{\pm}$ satisfy $[J_{z}, J_{\pm}] = \pm J_{\pm}$ and $[J_{+}, J_{-}] = 2 J_{z}$ on the basis $(v_{k})$ ; substituting $J_{x} = \frac{1}{2} (J_{+} + J_{-})$ , $J_{y} = \frac{1}{2 i} (J_{+} - J_{-})$ and expanding gives $[J_{x}, J_{y}] = i J_{z}$ , $[J_{y}, J_{z}] = i J_{x}$ , $[J_{z}, J_{x}] = i J_{y}$ . Since $v_{k} \mapsto e_{n / 2 - k}$ maps the basis of $S_{n}$ onto the basis of the spin- $\frac{n}{2}$ space of Finite spin matrices, and by step 3.3 the three operators have there exactly the action prescribed in that proposition, the described map is a unitary equivalence intertwining $(J_{x}, J_{y}, J_{z})$ with the spin- $\frac{n}{2}$ matrices and carrying $J^{2}$ to $s (s + 1) I$ with $s = n / 2$ ; in particular $J^{2} = \frac{n}{2} (\frac{n}{2} + 1) I_{S_{n}}$ . This proves (a)-(d) except the strong-continuity clause.

5.1step 1.1step 1.2step 2.1step 3.1step 3.2F3F4algebra∎

[Strong continuity] Let $W_{m} \to W_{0}$ in $M_{2} (C)$ . By the telescoping identity of step 3.2 with [F3], $∥ W_{m}^{\otimes n} - W_{0}^{\otimes n} ∥ \leq n \max (∥ W_{m} ∥, ∥ W_{0} ∥)^{n - 1} ∥ W_{m} - W_{0} ∥$ , and $\max (∥ W_{m} ∥, ∥ W_{0} ∥)$ is bounded because $∥ W_{m} ∥ \to ∥ W_{0} ∥$ by [F4] and convergence of the entries; hence $∥ W_{m}^{\otimes n} - W_{0}^{\otimes n} ∥ \to 0$ . Restricting the operators to the invariant subspace $S_{n}$ can only decrease the operator norm, so $∥ U_{n} (W_{m}) - U_{n} (W_{0}) ∥ \to 0$ , proving (f); in particular $U_{n}$ is strongly continuous on $S U (2)$ , since norm continuity implies strong continuity by [F4]. This completes (a)-(f).

Symmetric-power SU2 representation

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources