## Distances between Probability Measures

Let $G$ be a finite quantum group and $M_p(G)$ be the set of states on the $\mathrm{C}^\ast$-algebra $F(G)$.

The algebra $F(G)$ has an invariant state $\int_G\in\mathbb{C}G=F(G)^\ast$, the dual space of $F(G)$.

Define a (bijective) map $\mathcal{F}:F(G)\rightarrow \mathbb{C}G$, by

$\displaystyle \mathcal{F}(a)b=\int_G ba$,

for $a,b\in F(G)$.

Then, where $\|\cdot\|_1^{F(G)}=\int_G|\cdot|$ and $\|\cdot\|_\infty^{F(G)}=\|\cdot\|_{\text{op}}$, define the total variation distance between states $\nu,\mu\in M_p(G)$ by

$\displaystyle \|\nu-\mu\|=\frac12 \|\mathcal{F}^{-1}(\nu-\mu)\|_1^{F(G)}$.

(Quantum Total Variation Distance (QTVD))

Standard non-commutative $\mathcal{L}^p$ machinary shows that:

$\displaystyle \|\nu-\mu\|=\sup_{\phi\in F(G):\|\phi\|_\infty^{F(G)}\leq 1}\frac12|\nu(\phi)-\mu(\phi)|$.

(supremum presentation)

In the classical case, using the test function $\phi=2\mathbf{1}_S-\mathbf{1}_G$, where $S=\{\nu\geq \mu\}$, we have the probabilists’ preferred definition of total variation distance:

$\displaystyle \|\nu-\mu\|_{\text{TV}}=\sup_{S\subset G}|\nu(\mathbf{1}_S)-\mu(\mathbf{1}_S)|=\sup_{S\subset G}|\nu(S)-\mu(S)|$.

In the classical case the set of indicator functions on the subsets of the group exhaust the set of projections in $F(G)$, and therefore the classical total variation distance is equal to:

$\displaystyle \|\nu-\mu\|_P=\sup_{p\text{ a projection}}|\nu(p)-\mu(p)|$.

(Projection Distance)

In all cases the quantum total variation distance and the supremum presentation are equal. In the classical case they are equal also to the projection distance. Therefore, in the classical case, we are free to define the total variation distance by the projection distance.

## Quantum Projection Distance $\neq$ Quantum Variation Distance?

Perhaps, however, on truly quantum finite groups the projection distance could differ from the QTVD. In particular, a pair of states on a $M_n(\mathbb{C})$ factor of $F(G)$ might be different in QTVD vs in projection distance (this cannot occur in the classical case as all the factors are one dimensional).

The following are two solutions to this problem. The first used a bit of help on Math Stack Exchange, and the second a more elegant solution from Jean-Christophe Bourin (communicated by Uwe Franz). They are similar — the first constructing an explicit map, the second using a phase.

### The Supremum is Attained at An Involution

Embed the multi-matrix $F(G)$ into a single (block) $M_{|G|}(\mathbb{C})$.

Consider the supremum presentation. As everything is in finite dimensions, the supremum is actually a maximum. Suppose that the max is attained for a self-adjoint involutive $\phi\in F(G)$. Then

$\displaystyle p=\mathbf{1}_S=:\frac{\mathbf{1}_G+\phi}{2}$

is a projection:

$\displaystyle p^2=\frac{\phi^2+2\phi+\mathbf{1}_G}{4}=\frac{2\mathbf{1}_G+2\phi}{4}=p$,

and $p^\ast=p$ follows from self-adjointness of $\phi$. Using such a $p$, the projection distance is equal to the QTVD.

If, however, there was a pair of states such that the max is not attained at an involution, then the projection distance for these two states would be strictly smaller than the QTVD.

Consider, therefore, two states on $M_n(\mathbb{C})$ given by two density matrices $\rho_1$ and $\rho_2$.

$\displaystyle \|\nu-\mu\|=\max_{\|\phi\|_\infty\leq 1} |\nu(\phi)-\mu(\phi)|$

$\displaystyle = \max_{\|\phi\|_\infty\leq 1}|\text{Tr}(\rho_1\phi)-\text{Tr}(\rho_2\phi)|$

$\displaystyle =\max_{\|\phi\|_\infty \leq 1}|\text{Tr}((\rho_1-\rho_2)\phi)|$.

From here we use the argument of Hanno.

Define $\theta:=\rho_1-\rho_2$. Now use the following inequality:

$\displaystyle |\text{Tr}(\theta\phi)|\leq \|\theta\|_1\|\phi\|_{\infty}$,

where $\|\theta\|_1=\text{Tr}(|\theta|)$ and in this case $\|\phi\|_\infty$ can be taken to equal one so that

$\displaystyle \|\nu-\mu\|\leq \|\theta\|_1$.

Note that $\theta\in F(G)_{\text{sa}}$ so that $\theta$ is diagonalisable with real eigenvalues. Fix a basis in which $\theta$ is given by a diagonal matrix $\theta=\text{diag}(\lambda_1,\dots,\lambda_n)$. Therefore $\|\theta\|_1=\text{Tr}(|\theta|)=\sum_{i=1}^n|\lambda_i|$.

Now define the operator norm one diagonal matrix $\phi$ by:

$\displaystyle \phi_{ii}=\begin{cases} 1 & \text{ if }\lambda_i=0\\ \text{sgn}(\lambda_i) & \text{ else} \end{cases}$.

Note $\phi$ is a self-adjoint involution and furthermore:

$\theta \phi=\text{diag}(|\lambda_1|,\dots,|\lambda_n|)$,

has trace equal to $|\theta|$. Therefore the max of $\frac12|\nu(\phi)-\mu(\phi)|$ (over norm one elements) is attained at a self-adjoint and involutive $\phi$ and so via $\mathbf{1}_S=\frac{1+\phi}{2}$ is a projection such that

$\|\nu-\mu\|=|\nu(\mathbf{1}_S)-\mu(\mathbf{1}_S)|$.

### The Supremum is attained at a Phase

As above, define $\theta:=\rho_1-\rho_2$. Let $\phi=U$, the phase of $\theta$. Using the argument of Peter, we note that $U$ can be taken to Hermitian. Therefore

$\theta=U|\theta|\Rightarrow U^*\theta=|\theta|\Rightarrow U\theta=|\theta|$,

and so

$\text{Tr}(\theta U)=\text{Tr}(U\theta)=\text{Tr}(|\theta|)=\|\theta\|_1$.

Note now

$\displaystyle \|\nu-\mu\|\leq \|\theta\|_1$,

but this upper bound is attained at $U$. As a hermitian unitary, $U$ is involutive and the above arguments follow through.

## Conclusion

This means there is no counterexample to

Projection Distance = QTVD.

What this means is that we can take as a definition the quantum total variation distance to be the same as the classical total variation distance:

$\displaystyle \|\nu-\mu\|=\sup_{S\subset G}|\nu(S)-\mu(S)|$,

where $S\subset G$ in the sense that there is a unital $\mathrm{C}^*$-algebra, denoted $F(S)$, of $F(G)$; and $\nu(S):=\nu(\mathbf{1}_S)$.