3 The Single-Block Fundamental Theorem

An injective MPS tensor that generates the same matrix-product vectors as a second tensor of equal bond dimension is conjugate to it [ CPGSV21 , Corollary IV.5 ] : there is an invertible $X$ with $B^i = X A^i X^{-1}$ for every physical index $i$. The argument is purely algebraic. Equality of the matrix-product vectors gives $\operatorname{tr}(A^w)=\operatorname{tr}(B^w)$ for every word $w$; the length-two and length-three traces produce a linear map $T$ with $T(A^i)=B^i$ that is multiplicative and nonzero; a nonzero multiplicative endomorphism of the simple algebra $M_{D}(\mathbb {C})$ is a unital automorphism; and by Skolem–Noether every automorphism of $M_{D}(\mathbb {C})$ is conjugation by an invertible $X$. Evaluating at $A^i$ closes the proof. The same conjugation, one normal block at a time, is the central step of the multi-block theorem in Chapter 11.

3.1 The multiplicative linear extension

Theorem 3.1.1 Nondegeneracy of the trace pairing

✓

For $M \in M_{D}(\mathbb {C})$, $\operatorname{tr}(M N) = 0$ for all $N \in M_{D}(\mathbb {C})$ if and only if $M = 0$.

Proof ▶

Testing against the matrix unit $E_{ji}$ gives $\operatorname{tr}(M E_{ji}) = M_{ij}$, so $\operatorname{tr}(M N) = 0$ for all $N$ forces every entry of $M$ to vanish.

Lemma 3.1.2 Same matrix-product vectors give trace agreement

✓

If $A$ and $B$ generate the same matrix-product vectors, then $\operatorname{tr}(A^w) = \operatorname{tr}(B^w)$ for every word $w$.

Proof ▶

By (??), the configuration $\sigma = w$ gives $V^{(|w|)}(A)_w = \operatorname{tr}(A^w)$. Equality of the matrix-product vectors gives $V^{(N)}(A) = V^{(N)}(B)$ for all $N$, so $\operatorname{tr}(A^w) = \operatorname{tr}(B^w)$.

The trace conditions constrain every product of the matrices $A^i$. To bring the algebra structure of $M_{D}(\mathbb {C})$ to bear, record how a matrix pairs against the family $\{ A^i\} $ under the trace.

Definition 3.1.3 Trace pairing map

✓

For a tensor $A$, the trace pairing map $\Phi _A : M_{D}(\mathbb {C}) \to \mathbb {C}^d$ is defined by

\begin{align} \Phi _A(M)_i & = \operatorname{tr}(M A^i). \label{eq:single_trace_pairing} \end{align}

Theorem 3.1.4 The trace pairing map is injective for injective tensors

✓

If $A$ is injective, then $\ker \Phi _A = \{ 0\} $.

Proof ▶

If $\Phi _A(M) = 0$, then (??) gives $\operatorname{tr}(M A^i) = 0$ for every $i$. The $\{ A^i\} $ span $M_{D}(\mathbb {C})$, so $\operatorname{tr}(M N) = 0$ for all $N$, and $M = 0$ by Theorem 3.1.1.

Lemma 3.1.5 Injectivity transfers across a range inclusion

✓

Let $\Phi , \Psi : V \to W$ be $\mathbb {C}$-linear maps with $V$ finite-dimensional. If $\ker \Phi = \{ 0\} $ and $\operatorname{range}\Phi \subseteq \operatorname{range}\Psi $, then $\ker \Psi = \{ 0\} $.

Proof ▶

By rank–nullity, $\ker \Phi = \{ 0\} $ gives $\dim \operatorname{range}\Phi = \dim V$, and the range inclusion gives

\begin{align} \dim V & = \dim \operatorname{range}\Phi \le \dim \operatorname{range}\Psi \le \dim V. \notag \end{align}

so $\dim \operatorname{range}\Psi = \dim V$ and $\ker \Psi = \{ 0\} $.

Theorem 3.1.6 The linear extension exists and is unique

✓

If $A$ is injective and $A$, $B$ generate the same matrix-product vectors, then there is a unique linear map $T : M_{D}(\mathbb {C}) \to M_{D}(\mathbb {C})$ with $T(A^i) = B^i$ for all $i$.

Proof ▶

Length-two trace agreement gives $\Phi _A(A^i) = \Phi _B(B^i)$ for every $i$. The $\{ A^i\} $ span $M_{D}(\mathbb {C})$, so $\operatorname{range}\Phi _A \subseteq \operatorname{range}\Phi _B$. With $\ker \Phi _A = \{ 0\} $ (Theorem 3.1.4), Lemma 3.1.5 gives $\ker \Phi _B = \{ 0\} $. Let $g$ be a left inverse of $\Phi _B$ and set $T = g \circ \Phi _A$; then

\begin{align} T(A^i) & = g(\Phi _A(A^i)) = g(\Phi _B(B^i)) = B^i. \notag \end{align}

Uniqueness holds because the $\{ A^i\} $ span $M_{D}(\mathbb {C})$.

Theorem 3.1.7 The linear extension is multiplicative

✓

Under the hypotheses of Theorem 3.1.6, the map $T$ satisfies $T(MN) = T(M) T(N)$ for all $M, N \in M_{D}(\mathbb {C})$.

Proof ▶

Length-two trace agreement gives $\Phi _B(T(A^i)) = \Phi _B(B^i) = \Phi _A(A^i)$; since the $\{ A^i\} $ span $M_{D}(\mathbb {C})$, this extends to $\Phi _B \circ T = \Phi _A$, so $\operatorname{range}\Phi _A = \operatorname{range}(\Phi _B \circ T) \subseteq \operatorname{range}\Phi _B$. With $\ker \Phi _A = \{ 0\} $ (Theorem 3.1.4), Lemma 3.1.5 makes $\Phi _B$ injective. Length-three trace agreement gives, for all $i$, $j$, and $k$,

\begin{align} \Phi _B(T(A^i A^j))_k & = \Phi _A(A^i A^j)_k = \operatorname{tr}(A^i A^j A^k) = \operatorname{tr}(B^i B^j B^k) = \Phi _B(B^i B^j)_k. \notag \end{align}

Hence injectivity of $\Phi _B$ yields $T(A^i A^j) = B^i B^j$. Extend in two stages: for fixed $j$, the maps $M \mapsto T(M A^j)$ and $M \mapsto T(M) B^j$ agree on the spanning $\{ A^i\} $, hence $T(M A^j) = T(M) B^j$ for all $M$; then for fixed $M$, the maps $N \mapsto T(M N)$ and $N \mapsto T(M) T(N)$ agree on the spanning $\{ A^j\} $, hence $T(M N) = T(M) T(N)$ for all $N$.

3.2 Inner automorphism and the single-block theorem

The rigidity of the full matrix algebra now finishes the proof. A nonzero multiplicative map on $M_{D}(\mathbb {C})$ is forced to be bijective (simplicity), unit-preserving (surjectivity), and conjugation by a single invertible matrix (Skolem–Noether). The next four results establish these steps and the nonvanishing of $T$, which the final theorem combines into the gauge $X$.

Theorem 3.2.1 A nonzero multiplicative endomorphism of $M_{D}(\mathbb {C})$ is bijective

✓

A nonzero multiplicative $\mathbb {C}$-linear endomorphism of $M_{D}(\mathbb {C})$ is bijective.

Proof ▶

The kernel of a multiplicative linear map is a two-sided ideal. The algebra $M_{D}(\mathbb {C})$ is simple, so the kernel is $\{ 0\} $ or all of $M_{D}(\mathbb {C})$; the latter forces the map to be zero. Hence the map is injective, and surjective by finite-dimensionality.

Lemma 3.2.2 A surjective multiplicative map preserves the unit

✓

If $T : M_{D}(\mathbb {C}) \to M_{D}(\mathbb {C})$ is a surjective multiplicative $\mathbb {C}$-linear map, then $T(\mathbb {1}) = \mathbb {1}$, so $T$ is a $\mathbb {C}$-algebra homomorphism.

Proof ▶

Choose $X$ with $T(X) = \mathbb {1}$. Then

\begin{align} T(\mathbb {1}) & = T(X)T(\mathbb {1}) = T(X\mathbb {1}) = T(X) = \mathbb {1}. \notag \end{align}

Lemma 3.2.3 An injective tensor forces a nonzero partner

✓

Let $D \ge 1$. If $A$ is injective and $A$, $B$ generate the same matrix-product vectors, then the $B^i$ are not all zero.

Proof ▶

If $B^i = 0$ for every $i$, then Lemma 3.1.2 on length-one words gives $\operatorname{tr}(A^i) = 0$ for all $i$. The $\{ A^i\} $ span $M_{D}(\mathbb {C})$, so the trace functional vanishes identically, contradicting $\operatorname{tr}(\mathbb {1}) = D \neq 0$.

Theorem 3.2.4 Skolem–Noether

✓

Every $\mathbb {C}$-algebra automorphism $f$ of $M_{D}(\mathbb {C})$ is inner: there is $X \in \mathrm{GL}_D(\mathbb {C})$ with $f(M) = X M X^{-1}$ for all $M$.

Proof ▶

Under the identification $M_{D}(\mathbb {C}) \cong \operatorname{End}_{\mathbb {C}}(\mathbb {C}^D)$, every algebra automorphism of $\operatorname{End}_{\mathbb {C}}(\mathbb {C}^D)$ is conjugation by an invertible linear map. Translating back gives the invertible $X$.

Theorem 3.2.5 Single-block Fundamental Theorem

✓

Let $A$ be an injective MPS tensor and $B$ a tensor of the same bond dimension. If $A$ and $B$ generate the same matrix-product vectors, then there is $X \in \mathrm{GL}_D(\mathbb {C})$ with $B^i = X A^i X^{-1}$ for all $i$. In tensor-network notation the gauge conjugates the local tensor on its virtual legs, with the physical index $i$ on the upper leg:

$\begin{tenkz}[physical=up] \tnX{X} & \tn[up=$i$]{A} & \tnX{X^{-1}} \end{tenkz}$

Proof ▶

Theorem 3.1.6 gives the unique linear $T$ with $T(A^i) = B^i$, and Theorem 3.1.7 makes it multiplicative. It is nonzero: otherwise $B^i = 0$ for all $i$, against Lemma 3.2.3. By Theorem 3.2.1, $T$ is bijective, and by Lemma 3.2.2 it fixes $\mathbb {1}$, so $T$ is a $\mathbb {C}$-algebra automorphism. Theorem 3.2.4 gives $X \in \mathrm{GL}_D(\mathbb {C})$ with $T(M) = X M X^{-1}$; evaluating at $M = A^i$ gives $B^i = X A^i X^{-1}$.