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Labels that survive representations

Written by Claude Opus 5.5 (Anthropic), October 2026. Self-checked by the writing AI. Public domain (CC0).

An irreducible representation of a maximal tensor product need not be a tensor product of representations of the factors. This lesson shows that, nevertheless, a finite tensor power of an algebra without characters contains labels: finitely many self-adjoint elements whose images stay linearly independent in every irreducible representation of every larger tensor power (Corollary 5.1). The next lesson uses labels to combine finitely many witnesses into a single element without losing any of them.

The lesson follows [OpenAI-KRC, Sections 4.1 and 4.2]. All tensor products of C*-algebras are maximal, and \(D^{\otimes k}\) denotes the maximal tensor power \(D^{\otimes_{\max}k}\) of Central sequences and the character criterion, Section 3. A factor on a Hilbert space \(H\) is a von Neumann algebra \(N\subseteq B(H)\), containing \(1_H\), with \(N\cap N'=\mathbb C1\).

1. Independence in commuting factors

Lemma 1.1.

  1. Let \(N\) be a factor on \(H\), let \(a_1,\dots,a_p\in N\), and let \(B_1,\dots,B_p\in N'\) be linearly independent. If \(\sum_ta_tB_t=0\), then every \(a_t\) is \(0\).
  2. Let \(N_1,\dots,N_n\) be pairwise commuting factors on \(H\). For each \(j\) let \(b^{(j)}_1,\dots,b^{(j)}_{d_j}\in N_j\) be linearly independent. Then the products \[ \begin{gathered} b^{(1)}_{s_1}b^{(2)}_{s_2}\cdots b^{(n)}_{s_n},\\ s\in[d_1]\times\cdots\times[d_n], \end{gathered} \] are linearly independent; here \([d]=\{1,\dots,d\}\).

Proof. (1) By the theorem of Murray and von Neumann (Tensor products of C*-algebras and the minimal norm, Theorem 5.1), \(\sum_ta_t\otimes B_t=0\) in the algebraic tensor product \(N\odot N'\). Since the \(B_t\) are linearly independent, every \(a_t\) is \(0\) (Tensor products of Banach and Hilbert spaces, Jordan homomorphisms and isometries of C*-algebras, Lemma 1.1).

(2) Induction on \(n\); the case \(n=1\) is the hypothesis. Let \(n\ge2\). By induction, the products \(B_t=b^{(2)}_{t_2}\cdots b^{(n)}_{t_n}\) are linearly independent, and they lie in \(N_1'\). Suppose \(\sum_sc_s\,b^{(1)}_{s_1}B_{(s_2,\dots,s_n)}=0\), and put \(a_t=\sum_{s_1}c_{(s_1,t)}b^{(1)}_{s_1}\in N_1\). Then \(\sum_ta_tB_t=0\), so every \(a_t\) is \(0\) by (1). As the \(b^{(1)}_{s_1}\) are linearly independent, all \(c_s\) vanish. \(\blacksquare\)

Without the factor hypothesis the lemma fails (Exercise 6.1).

2. Represented blocks of a tensor product

Let \(A_1,\dots,A_k\) be unital C*-algebras and \(P=A_1\otimes\cdots\otimes A_k\). The legs of \(P\) are the embeddings \(a\mapsto1\otimes\cdots\otimes a\otimes\cdots\otimes1\) of the \(A_j\); they are injective, unital, and have commuting ranges (Central sequences and the character criterion, Lemma 3.1(2)).

Lemma 2.1. Let \(\pi\) be an irreducible representation of \(P\) on \(H\), and let \(\{1,\dots,k\}=I_1\cup\cdots\cup I_q\) be a partition into nonempty sets. Let \(L_b\) be the union of the ranges of the legs with index in \(I_b\), and \(N_b=\pi(L_b)''\). Then \(N_1,\dots,N_q\) are pairwise commuting factors on \(H\).

Proof. \(\pi\) is unital (Lemma 3.1(6) of the same lesson), so \(1_H\in\pi(L_b)\), and \(N_b\) is a von Neumann algebra (The double commutant theorem, Proposition 2.1). For \(b\ne b'\), \(\pi(L_{b'})\subseteq\pi(L_b)'\), because the legs commute; taking commutants, \(N_b\subseteq\pi(L_{b'})'\), that is, \(\pi(L_{b'})\subseteq N_b'\), and then \(N_{b'}\subseteq N_b'''=N_b'\). So the \(N_b\) commute pairwise. Let \(z\in N_b\cap N_b'\). Then \(z\) commutes with \(\pi(L_b)\subseteq N_b\) and with \(\pi(L_{b'})\subseteq N_{b'}\subseteq N_b'\) for \(b'\ne b\). Elementary tensors are products of leg elements and span a dense subalgebra of \(P\), so \(z\) commutes with \(\pi(P)\), and \(z\in\mathbb C1\) by Schur's lemma (Building representations from positive functionals, Corollary 2.3). \(\blacksquare\)

The lemma needs no faithfulness of the restrictions to the legs and no decomposition of \(H\) as a tensor product.

3. A detecting space

Write \(D_{\mathrm{sa}}\) for the real vector space of self-adjoint elements of \(D\).

Lemma 3.1 ([OpenAI-KRC, Lemma 4.3]). Let \(D\) be a nonzero unital C*-algebra without characters. There is a finite-dimensional real subspace \(V\subseteq D_{\mathrm{sa}}\) such that, for every unital representation \(\rho\) of \(D\) on a nonzero Hilbert space, \(\rho(V)\) contains two operators that do not commute. Moreover \(\dim_{\mathbb R}V\ge2\).

Proof. As in the proof of Lemma 4.2 of Central sequences and the character criterion, there are self-adjoint \(b_i,c_i\) and elements \(h_i,k_i\) of \(D\), \(1\le i\le r\), with \[ \|1-y\|<1,\qquad y=\sum_{i=1}^rh_i[b_i,c_i]k_i . \] So \(y\) is invertible. Let \(V\) be the real span of the \(b_i\) and \(c_i\). If \(\rho(V)\) were commutative, then \(\rho([b_i,c_i])=0\) for all \(i\) and \(\rho(y)=0\), while \(\rho(y)\rho(y^{-1})=\rho(1)=1\ne0\). A unital representation of \(D\) on a nonzero space exists: take a faithful representation \(\rho_0\) (Building representations from positive functionals, Theorem 7.2) on the range of the nonzero projection \(\rho_0(1)\). If \(\dim V\le1\), then \(V\) would be commutative, which is excluded. So \(\dim V\ge2\). \(\blacksquare\)

4. Real labels

For integers \(m,n\ge1\), \(\mathbb R^{[m]^n}\) is the space of real arrays \(x=(x_k)\) indexed by \(k=(k_1,\dots,k_n)\in[m]^n\); its dimension is \(m^n\). For real matrices \(A_j\) of size \(d_j\times m\), let \(A_1\otimes\cdots\otimes A_n\) be the linear map \(\mathbb R^{[m]^n}\to\mathbb R^{[d_1]\times\cdots\times[d_n]}\), \[ \begin{aligned} &\bigl((A_1\otimes\cdots\otimes A_n)x\bigr)_s\\ &\quad=\sum_{k}A_1[s_1,k_1]\cdots A_n[s_n,k_n]\,x_k . \end{aligned} \] The same formula is used for complex arrays. Expanding the sums gives \[ \begin{aligned} &(R_1\otimes\cdots\otimes R_n)(A_1\otimes\cdots\otimes A_n)\\ &\quad=R_1A_1\otimes\cdots\otimes R_nA_n \end{aligned} \] for matrices \(R_j\) with \(d_j\) columns. If every \(A_j\) has rank \(d_j\), then \(A_1\otimes\cdots\otimes A_n\) is surjective: choose \(u_j\in\mathbb R^m\) with \(A_ju_j\) equal to the \(s_j\)-th standard basis vector; the array \((u_1)_{k_1}\cdots(u_n)_{k_n}\) is mapped to the standard basis array at \(s\).

Lemma 4.1 (real labels; [OpenAI-KRC, Lemma 4.4]). Let \(m\ge2\) and \(r\ge1\). There are \(n\ge1\) and arrays \(x^{(1)},\dots,x^{(r)}\in\mathbb R^{[m]^n}\) such that, for all real \(2\times m\) matrices \(P_1,\dots,P_n\) of rank \(2\), the arrays \[ (P_1\otimes\cdots\otimes P_n)x^{(i)},\qquad1\le i\le r, \] are linearly independent.

Proof. Choose \(n\) with \(2^n>2mn+r-1\). Let \(\mathcal P\) be the set of \(n\)-tuples of real \(2\times m\) matrices of rank \(2\); a matrix has rank \(2\) when one of its \(2\times2\) minors is nonzero, so \(\mathcal P\) is open in \(\mathbb R^{2mn}\). The unit sphere \(S^{r-1}\subset\mathbb R^r\) is a smooth manifold of dimension \(r-1\), the regular level set of \(u\mapsto|u|^2\) at \(1\) (Regular values and degree modulo two, Proposition 3.1). Put \(W=\mathbb R^{[m]^n}\) and consider the smooth manifold \(\mathcal X=\mathcal P\times S^{r-1}\times W^r\), of dimension \(2mn+r-1+rm^n\), and the polynomial map \[ \begin{aligned} &\Phi\colon\mathcal X\to\mathbb R^{[2]^n},\\ &\Phi(P,u,x)\\ &\quad=(P_1\otimes\cdots\otimes P_n)\Bigl(\sum_{i=1}^ru_ix^{(i)}\Bigr). \end{aligned} \] At every point some \(u_i\) is nonzero, and the derivative of \(\Phi\) in the direction of \(x^{(i)}\) alone is \(\dot x\mapsto u_i(P_1\otimes\cdots\otimes P_n)\dot x\), which is surjective. So \(0\) is a regular value of \(\Phi\), and \(Z=\Phi^{-1}(0)\) is a smooth manifold of dimension \[ \begin{aligned} 2mn+r-1&+rm^n-2^n\\ &<rm^n=\dim W^r \end{aligned} \] (Proposition 3.1 there), on which the projection \(\operatorname{pr}\colon Z\to W^r\) is smooth. Since \(\dim Z<\dim W^r\), every point of \(Z\) is a critical point of \(\operatorname{pr}\), so \(\operatorname{pr}(Z)\) is a null set (Sard's theorem, Corollary 2.2 there). A null set is not all of \(W^r\); choose \(x=(x^{(1)},\dots,x^{(r)})\notin\operatorname{pr}(Z)\). If \(\sum_ic_i(P_1\otimes\cdots\otimes P_n)x^{(i)}=0\) for some \(P\in\mathcal P\) and a real vector \(c\ne0\), then \(u=c/|c|\) gives \(\Phi(P,u,x)=0\), so \(x\in\operatorname{pr}(Z)\), which is false. \(\blacksquare\)

The labels are found by a measure argument, not written down; the lemma asserts only their existence, with an exponent \(n\) depending on \(m\) and \(r\).

5. Labels in tensor powers

Corollary 5.1 (labels; [OpenAI-KRC, Corollary 4.5]). Let \(D\) be a nonzero unital C*-algebra without characters and \(r\ge1\). There are \(n\ge1\) and self-adjoint \(x_1,\dots,x_r\in D^{\otimes n}\) with the following property. For all integers \(k\ge n\) and \(0\le j\le k-n\), and every irreducible representation \(\pi\) of \(D^{\otimes k}\), the operators \[ \pi\bigl(1^{\otimes j}\otimes x_i\otimes1^{\otimes(k-j-n)}\bigr),\qquad1\le i\le r, \] are linearly independent.

Here \(D^{\otimes k}\) is identified with \(D^{\otimes j}\otimes D^{\otimes n}\otimes D^{\otimes(k-j-n)}\) by regrouping (Lemma 3.1(3) of Central sequences and the character criterion), with an empty factor omitted.

Proof. Let \(V\) be the space of Lemma 3.1, with a real basis \(v_1,\dots,v_m\), \(m\ge2\). Take \(n\) and the arrays \(x^{(i)}\) of Lemma 4.1, and put \[ x_i=\sum_{k\in[m]^n}x^{(i)}_k\,v_{k_1}\otimes\cdots\otimes v_{k_n}\in D^{\otimes n}. \] It is self-adjoint: a real combination of tensors with self-adjoint entries.

Fix \(k\), \(j\) and \(\pi\), and let \(\pi_1,\dots,\pi_k\) be the restrictions of \(\pi\) to the legs: unital representations with commuting ranges. By Lemma 2.1, applied to the partition into single legs, the algebras \(N_l=\pi_l(D)''\) are pairwise commuting factors. For \(l=1,\dots,n\), the real space \(E_l=\pi_{j+l}(V)\) consists of self-adjoint operators, and its dimension \(d_l\) is at least \(2\), because it contains two operators that do not commute (Lemma 3.1). Choose a real basis \(f^{(l)}_1,\dots,f^{(l)}_{d_l}\) of \(E_l\), and write \[ \pi_{j+l}(v_q)=\sum_{s=1}^{d_l}A_l[s,q]\,f^{(l)}_s \] with a real \(d_l\times m\) matrix \(A_l\) of rank \(d_l\). A real basis of a space of self-adjoint operators is complex linearly independent: if \(\sum_s(\alpha_s+i\beta_s)f_s=0\) with real \(\alpha_s,\beta_s\), taking adjoints gives \(\sum_s(\alpha_s-i\beta_s)f_s=0\), so \(\sum_s\alpha_sf_s=\sum_s\beta_sf_s=0\). Now, writing \(\hat x_i=1^{\otimes j}\otimes x_i\otimes1^{\otimes(k-j-n)}\), \[ \begin{aligned} &\pi(\hat x_i)\\ &\quad=\sum_kx^{(i)}_k\,\pi_{j+1}(v_{k_1})\cdots\pi_{j+n}(v_{k_n})\\ &\quad=\sum_sy^{(i)}_s\,f^{(1)}_{s_1}\cdots f^{(n)}_{s_n}, \end{aligned} \] where \(y^{(i)}=(A_1\otimes\cdots\otimes A_n)x^{(i)}\). By Lemma 1.1(2), applied to the factors \(N_{j+1},\dots,N_{j+n}\), the products \(f^{(1)}_{s_1}\cdots f^{(n)}_{s_n}\) are linearly independent. So a complex relation \(\sum_ic_i\pi(\hat x_i)=0\) gives \(\sum_ic_iy^{(i)}=0\). The rows of \(A_l\) are linearly independent; let \(R_l\) be the \(2\times d_l\) matrix selecting two of them, so that \(P_l=R_lA_l\) has rank \(2\). Applying \(R_1\otimes\cdots\otimes R_n\) gives \(\sum_ic_i(P_1\otimes\cdots\otimes P_n)x^{(i)}=0\). The real arrays \((P_1\otimes\cdots\otimes P_n)x^{(i)}\) are linearly independent over \(\mathbb R\) by Lemma 4.1, hence over \(\mathbb C\) (take real and imaginary parts). So all \(c_i\) vanish. \(\blacksquare\)

6. Exercises

6.1. Let \(N_1=N_2\) be the diagonal \(2\times2\) matrices acting on \(\mathbb C^2\). Show that they commute, that they are not factors, and that the products of the independent families \((e_{11},e_{22})\) and \((e_{11},e_{22})\) are linearly dependent.

6.2. For \(D=M_2\), show that the real span \(V\) of \(\begin{pmatrix}0&1\\1&0\end{pmatrix}\) and \(\begin{pmatrix}1&0\\0&-1\end{pmatrix}\) has the property of Lemma 3.1.

6.3. Show that the conclusion of Lemma 4.1 cannot hold when \(r>2^n\).

6.4. Find two operators that are linearly independent over \(\mathbb R\) but not over \(\mathbb C\), and explain why the proof of Corollary 5.1 works with self-adjoint operators.

6.5. In Lemma 2.1, why must \(\pi\) be irreducible? Give a representation of \(\mathbb C^2\otimes\mathbb C^2\) in which the first leg generates a von Neumann algebra that is not a factor.

7. Solutions

6.1. Diagonal matrices commute. The algebra contains the nonscalar projection \(e_{11}\), and it is commutative, so its centre is itself, not \(\mathbb C1\). The products \(e_{11}e_{22}=e_{22}e_{11}=0\) form a dependent family.

6.2. A unital representation \(\rho\) of \(M_2\) on a nonzero space is injective, since \(\ker\rho\) is a proper ideal of the simple algebra \(M_2\). The two matrices do not commute (their products in the two orders differ by a sign and are nonzero), so their images do not commute.

6.3. The arrays \((P_1\otimes\cdots\otimes P_n)x^{(i)}\) lie in a space of dimension \(2^n\), which contains no more than \(2^n\) linearly independent vectors.

6.4. \(1\) and \(i1\) are independent over \(\mathbb R\) and dependent over \(\mathbb C\). For self-adjoint operators, taking adjoints separates the real and imaginary parts of a complex relation, as in the proof.

6.5. Irreducibility gives \(\pi(P)'=\mathbb C1\), which makes a central element of \(N_b\) scalar. The identity representation of \(\mathbb C^2\otimes\mathbb C^2\cong\mathbb C^4\) on \(\mathbb C^4\) (diagonal matrices) is not irreducible, and its first leg generates the diagonal matrices \(\operatorname{diag}(a,a,b,b)\), a commutative algebra containing a nonscalar projection.

References