4 The inner product for the multi-matrix algebra

4.1 On $M_{n}$

Definition 4.1.1

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Given a linear functional $ϕ : M_{n} \to C$ , we define $ϕ_{Q} = \sum_{i, j} ϕ (e_{i j}) e_{j i}$ .

Lemma 4.1.2

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Given a linear functional $ϕ : M_{n} \to C$ , we get $ϕ$ is given by $x \mapsto Tr (ϕ_{Q} x)$ .

Proof ▶

Definition 4.1.3

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Given C $^{*}$ -algebras $A, C$ , we say a linear map $f : A \to C$ is positive when $0 \leq f (a)$ for all $0 \leq a$ . In other words, $f$ maps positive elements in $A$ to positive elements in $C$ .

As any matrix $x \in M_{n}$ is positive if and only if there exists some $y \in M_{n}$ such that $x = y^{*} y$ , we get that a linear functional $f$ on $M_{n}$ is a positive map when $0 \leq f (x^{*} x)$ for any matrix $x \in M_{n}$ .

Lemma 4.1.4

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Given a linear functional $ϕ$ on $M_{n}$ , we have,
$ϕ$ is positive $\Leftrightarrow$ $0 \leq ϕ_{Q}$ .
Here, $ϕ_{Q}$ is the matrix of $ϕ$ defined in Definition 4.1.1.

Proof ▶

By Lemma 4.1.2 we have $ϕ (x) = Tr (ϕ_{Q} x)$ for all $x \in M_{n}$ .

$(\Rightarrow)$: Suppose $ϕ$ is positive. By Lemma 1.2.10, we get $0 \leq \sum_{i} ϕ (x_{i} x_{i}^{*})$ for any tuple $(x_{i})$ in $C^{n}$ . So then for any $x \in C^{n}$ , we have $0 \leq ϕ (x x^{*}) = Tr (ϕ_{Q} x x^{*}) = x^{*} ϕ_{Q} x$ , which means $ϕ_{Q}$ is positive semi-definite.
$(\Leftarrow)$: Suppose $ϕ_{Q}$ is positive semi-definite. Then since $ϕ_{Q}$ is positive semi-definite, we have $ϕ_{Q} = ϕ_{Q}^{1 / 2} ϕ_{Q}^{1 / 2}$ , where $ϕ_{Q}^{1 / 2}$ is also positive semi-definite (and so is self-adjoint). So for any $x \in M_{n}$ , we have $ϕ (x^{*} x) = Tr (ϕ_{Q} x^{*} x) = Tr ({(x ϕ_{Q}^{1 / 2})}^{*} (x ϕ_{Q}^{1 / 2})) \geq 0$ .

Thus $ϕ$ is a positive map if and only if our matrix $ϕ_{Q}$ is positive semi-definite.

Proposition 4.1.5

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Given a linear functional $ϕ : M_{n} \to C$ , we get $ϕ$ is real (star-preserving) if and only if $ϕ_{Q}$ is self-adjoint.
Here, $ϕ_{Q}$ is the matrix given by $ϕ$ defined in Definition 4.1.1.

Proof ▶

$(\Rightarrow)$

Suppose $ϕ$ is star-preserving, i.e., $ϕ (x^{*}) = \overset{―}{ϕ (x)}$ for all $x \in M_{n}$ . We let $x \in M_{n}$ , and compute using Lemma 4.1.2,

Tr (ϕ_{Q} x^{*}) = ϕ (x^{*}) = \overset{―}{ϕ (x)} = \overset{―}{Tr (ϕ_{Q} x)} = Tr (x^{*} ϕ_{Q}^{*}),

And so $ϕ_{Q} = ϕ_{Q}^{*}$ .

$(\Leftarrow)$

Suppose $ϕ_{Q}$ is self-adjoint. Using Lemma 4.1.2, $ϕ$ is given by $x \mapsto Tr (ϕ_{Q} x)$ . And so for any $x \in M_{n}$ , we get $ϕ (x^{*}) = Tr (ϕ_{Q} x^{*}) = Tr ({(x ϕ_{Q})}^{*}) = \overset{―}{Tr (x ϕ_{Q})} = \overset{―}{ϕ (x)}$ . Thus, $ϕ$ is real.

Corollary 4.1.6

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If $A \in M_{n}$ . Then
$0 \leq A$ and is invertible $\Leftrightarrow$ $A$ is positive-definite.

Proof ▶

$(\Rightarrow)$

Suppose $0 \leq A$ and $A = A^{1 / 2} A^{1 / 2}$ is invertible. Then $A^{1 / 2}$ is also invertible and positive semi-definite. Let $v \in C^{n}$ be non-zero. Then, we compute,

⟨ v ∣ A v ⟩ = ⟨ v ∣ A^{1 / 2} A^{1 / 2} v ⟩ = ⟨ A^{1 / 2} v ∣ A^{1 / 2} v ⟩ > 0,

since $A^{1 / 2} v \neq 0$ (as $A^{1 / 2}$ is invertible). Note that, in the second equality, we use the self-adjointed-ness of $A^{1 / 2}$ since it is positive semi-definite. So we are done.

$(\Leftarrow)$

Suppose $0 < A$ . Then obviously $0 \leq A$ , so we only need to check if it is invertible. Suppose the contrary, i.e., $A$ is not invertible. Then there exists a non-zero $v \in C^{n}$ such that $A v = 0$ . But then, by the hypothesis, we get $0 < ⟨ v ∣ A v ⟩ = 0$ , which is a contradiction. Thus $A$ is invertible.

Lemma 4.1.7

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Given a positive definite matrix $Q \in M_{n}$ , we have $Tr (Q x^{*} x) = 0$ if and only if $x = 0$ for any $x \in M_{n}$ .

Proof ▶

We have $Tr ({(x Q^{1 / 2})}^{*} (x Q^{1 / 2})) = Tr (Q x^{*} x) = 0$ if and only if $x Q^{1 / 2} = 0$ . As $Q^{1 / 2}$ is positive definite, we get it is also invertible by Corollary 4.1.6, and so $x Q^{1 / 2} = 0$ if and only if $x = 0$ . And so we are done.

Definition 4.1.8

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A linear functional $f$ on $A$ is said to be faithful if $f (x^{*} x) = 0$ if and only if $x = 0$ for all $x \in A$ .

Proposition 4.1.9

✓

Given a linear functional $ϕ : M_{n} \to C$ , we have
$ϕ$ is a positive and faithful map $\Leftrightarrow$ $ϕ_{Q}$ is positive-definite.
Again, $ϕ_{Q}$ is the matrix given by $ϕ$ defined in Definition 4.1.1.

Proof ▶

By Lemma 4.1.2, we have $ϕ$ is given by $x \mapsto Tr (ϕ_{Q} x)$ , and by Lemma 4.1.4, we know $ϕ$ is positive if and only if $0 \leq ϕ_{Q}$ . So we need to show that faithfulness of a positive linear functional is equivalent to $ϕ_{Q}$ being positive definite.

$(\Rightarrow)$: Suppose $ϕ$ is faithful and positive. So we have $ϕ (x) = 0$ if and only if $x = 0$ for any positive semi-definite matrix $x \in M_{n}$ . Let $0 \neq x \in C^{n}$ . Then we have $x x^{*} \neq 0$ as $x \neq 0$ . This means, by faithfulness and positivity of $ϕ$ we get $ϕ (x x^{*}) \neq 0$ as $x x^{*}$ is a non-zero positive semi-definite matrix. And so $0 < ϕ (x x^{*}) = Tr (ϕ_{Q} x x^{*}) = x^{*} ϕ_{Q} x$ , which means $ϕ_{Q}$ is positive definite.
$(\Leftarrow)$: Suppose $ϕ_{Q}$ is positive definite. Then for any $x \in M_{n}$ , we get $ϕ (x^{*} x) = Tr (ϕ_{Q} x x^{*}) = 0$ if and only if $x = 0$ using Lemma 4.1.7.

Thus $ϕ$ is a faithful and positive linear functional if and only if the matrix $ϕ_{Q}$ is positive definite.

Theorem 4.1.10

✓

Given a linear functional $ϕ : M_{n} \to C$ , then the following are equivalent,

$ϕ$ is positive and faithful,
$ϕ_{Q}$ is pos-def and $\forall x \in M_{n} : ϕ (x) = Tr (Q x)$ ,
$M_{n} \times M_{n} \to C : (x, y) \mapsto ϕ (x^{*} y)$ defines an inner product on $M_{n}$ .

Proof ▶

The equivalence of the first and second parts is from Proposition 4.1.9.

For any $x, y \in M_{n}$ , let ${⟨ x ∣ y ⟩}_{ϕ} = ϕ (x^{*} y)$ . Then, clearly,

{⟨ x ∣ α y + β z ⟩}_{ϕ} = ϕ (x^{*} (α y + β z)) = α ϕ (x^{*} y) + β ϕ (x^{*} z) = α {⟨ x ∣ y ⟩}_{ϕ} + β {⟨ x ∣ z ⟩}_{ϕ},

by linearity of $ϕ$ , for any $x, y, z \in M_{n}$ and $α, β \in C$ .

If $ϕ$ is faithful, we have ${⟨ x ∣ x ⟩}_{ϕ} = ϕ (x^{*} x) = 0$ if and only if $x = 0$ for any $x \in M_{n}$ . And $ϕ$ is positive if and only if $0 \leq {⟨ x ∣ x ⟩}_{ϕ}$ for any $x \in M_{n}$ . So if ${⟨ \cdot ∣ \cdot ⟩}_{ϕ}$ defines an inner product on $M_{n}$ , then we get $ϕ$ is faithful and positive. So it remains to show that, given $ϕ$ is a faithful and positive linear functional, we get $\overset{―}{{⟨ x ∣ y ⟩}_{ϕ}} = {⟨ y ∣ x ⟩}_{ϕ}$ for any $x, y \in M_{n}$ .

Suppose $ϕ$ is faithful and positive. By Proposition 4.1.9, we get the matrix $ϕ_{Q}$ is positive definite (and so is self-adjoint). Using Proposition 4.1.5, we get $ϕ$ is real, and so, for any $x, y \in M_{n}$ , we get, $\overset{―}{{⟨ x ∣ y ⟩}_{ϕ}} = \overset{―}{ϕ (x^{*} y)} = ϕ (y^{*} x) = {⟨ y ∣ x ⟩}_{ϕ}$ . Therefore, ${⟨ \cdot ∣ \cdot ⟩}_{ϕ} : (x, y) \mapsto ϕ (x^{*} y)$ is a well-defined inner product on $M_{n}$ .

4.2 On $⨁_{i} M_{n_{i}}$

Definition 4.2.1

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For each $i$ , we fix a faithful and positive linear functional $ψ_{i}$ on $M_{n_{i}}$ , and we let $Q_{i} \in M_{n_{i}}$ denote the positive definite matrix such that $ψ_{i} : x \mapsto Tr (Q_{i} x)$ (so each $Q_{i} = \sum_{j, k} ψ_{i} (e_{j k}) e_{k j}$ ) – see Proposition 4.1.1.

Let $ψ$ be a faithful positive linear functional on $⨁_{i} M_{n_{i}}$ given by $ψ = \sum_{i} ψ_{i} \circ p_{i}$ , where each $p_{i}$ is the projection map $⨁_{j} M_{n_{j}} \to M_{n_{i}}$ , and we let $Q = ⨁_{i} Q_{i}$ . So then, given $x \in ⨁_{i} M_{n_{i}}$ , we get $ψ (x) = Tr (Q x)$ , where $Tr$ here is defined by the sum of the diagonals in each matrix block.

By Theorem 4.1.10, we define the inner product on each $M_{n_{i}}$ by

{⟨ x ∣ y ⟩}_{ψ_{i}} = ψ_{i} (x^{*} y) = Tr (Q_{i} x^{*} y),

for all $x, y \in M_{n_{i}}$ . We denote $(M_{n_{i}}, ψ_{i})$ to be the Hilbert space given by this inner product.
We define the inner product on $⨁_{i} M_{n_{i}}$ by

{⟨ x ∣ y ⟩}_{ψ} = ψ (x^{*} y) = Tr (Q x^{*} y),

for all $x, y \in ⨁_{i} M_{n_{i}}$ , where $Tr$ here is defined as the sum of the diagonals in each matrix block.

Proposition 4.2.2

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The adjoint of $p s i$ on $(⨁_{i} M_{n_{i}}, ψ)$ and $C$ is given by $C \to B : x \mapsto x 1$ . In other words, $ψ^{*} = | 1 ⟩$ .

Proof ▶

For any $x \in C$ and $y \in ⨁_{i} M_{n_{i}}$ , we have,

{⟨ ψ^{*} (x) ∣ y ⟩}_{ψ} = {⟨ x ∣ ψ (y) ⟩}_{C} = \overset{―}{x} ψ (y) = \overset{―}{x} {⟨ 1 ∣ y ⟩}_{ψ} = {⟨ x 1 ∣ y ⟩}_{ψ} .

Thus $ψ^{*} (x) = x 1$ for any $x \in C$ .

Proposition 4.2.3

We get ${[ι_{s} (e_{i j} Q_{s}^{- 1 / 2})_{i, j = 1}^{n_{s}}]}_{s = 1}^{K}$ is an orthonormal basis of $(⨁_{i} M_{n_{i}}, ψ)$ .
Here, $Q = ψ_{Q}$ from Proposition 4.1.1.

Proof ▶

Recall that, given an orthonormal basis $f = (f_{i})$ on a Hilbert space $H$ , we let $R_{f}$ be the linear isomorphism $H ≅ C^{\dim H}$ given by $R_{f} (x)_{i} = ⟨ f_{i} ∣ x ⟩$ .

Proposition 4.2.4

Let $f$ be the orthonormal basis from Proposition 4.2.3. Then for $x \in ⨁_{i} M_{n_{i}}$ , we get $R_{f} (x)_{s, i j} = (x Q^{1 / 2})_{s, i j}$ .
Here, $Q = ψ_{Q}$ from Proposition 4.1.1.

Proof ▶

4.3 The modular automorphism

Let $B = ⨁_{i} M_{n_{i}}$ in this section.

Definition 4.3.1

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Given $t \in R$ , we define the algebra automorphism $σ_{t} : B ≅ B$ to be given by $a \mapsto Q^{- t} a Q^{t}$ with inverse $a \mapsto Q^{t} a Q^{- t}$ (so $σ_{t}^{- 1} = σ_{- t}$ ).

Proposition 4.3.2

✓

For any $x \in B$ , we get ${σ_{t} (x)}^{*} = σ_{- t} (x^{*})$ .
In other words, $σ_{t}^{r} = σ_{- t}$ .

Proof ▶

We compute,

{σ_{t} (x)}^{*} = {Q^{- t} x Q^{t}}^{*} = Q^{t} x^{*} Q^{- t} = σ_{- t} (x^{*}) .

Proposition 4.3.3

✓

For any $t \in R$ , $σ_{t}$ is self-adjoint.

Proof ▶

Let $x, y \in B$ . Then we compute

\begin{aligned} ⟨ x ∣ σ_{t}^{*} (y) ⟩ & = ⟨ σ_{t} (x) ∣ y ⟩ = ψ ({σ_{t} (x)}^{*} y) = ψ (σ_{- t} (x^{*}) y) \\ = Tr (Q σ_{- t} (x^{*}) y) = Tr (Q Q^{t} x^{*} Q^{- t} y) \\ = Tr (Q x^{*} Q^{- t} y Q^{t}) = ⟨ x ∣ σ_{t} (y) ⟩ . \end{aligned}

Thus $σ_{t}$ is self-adjoint.

Corollary 4.3.4

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For any $t \in R$ , we get $σ_{t}$ is also a co-algebra homomorphism.

Proof ▶

Using Proposition 3.1.3, it is enough to show that $σ_{t}^{*}$ is an algebra homomorphism. And this is true since $σ_{t}$ is self-adjoint by Proposition 4.3.3.

Lemma 4.3.5

✓

For any $x, y, z \in B$ , we get $⟨ x y ∣ z ⟩ = ⟨ y ∣ x^{*} z ⟩$ .

Proof ▶

Clearly, $⟨ x y ∣ z ⟩ = ψ (y^{*} x^{*} z) = ⟨ y ∣ x^{*} z ⟩$ .

Lemma 4.3.6

✓

For any $x, y, z \in B$ , we get $⟨ x y ∣ z ⟩ = ⟨ x ∣ z σ_{- 1} (y^{*}) ⟩$ .

Proof ▶

⟨ x y ∣ z ⟩ = Tr (Q y^{*} x^{*} z) = Tr (Q x^{*} z (Q y^{*} Q^{- 1})) = ⟨ x ∣ z σ_{- 1} (y^{*}) ⟩ .

Lemma 4.3.7

✓

We clearly get $σ_{0} = id$ .

Proof ▶

For any $x \in B$ , we clearly get $σ_{0} (x) = Q^{0} x Q^{0} = x$ .

Lemma 4.3.8

✓

For any $s, t \in R$ , we get $σ_{s} \circ σ_{t} = σ_{s + t}$ .

Proof ▶

For any $x \in B$ , we get □

σ_{s} σ_{t} (x) = Q^{- s} (Q^{- t} x Q^{t}) Q^{s} = Q^{- (s + t)} x Q^{s + t} = σ_{s + t} (x) .

From Proposition 3.1.4, we see that our Hilbert space on $B$ is a Frobenius algebra.

Proposition 4.3.9

For any $t \in R$ , we get $σ_{t} = id$ if and only if $t = 0$ or $ψ$ is tracial.

Proof ▶

4.4 Multiplication composed with co-multiplication

Proposition 4.4.1

Given $α \in C$ , we get $m m^{*} = α id$ if and only if $Tr (Q_{i}^{- 1}) = α$ for all $i \in [K]$ .

Proof ▶

4 The inner product for the multi-matrix algebra

4.1 On Mn

4.2 On ⨁iMni

4.3 The modular automorphism

4.4 Multiplication composed with co-multiplication

4.1 On $M_{n}$

4.2 On $⨁_{i} M_{n_{i}}$