Is there a canonical inner product for finite dimensional $C^*$-algebras

Question

So this might be a fairly basic question, but I can't answer it myself.

A $C^*$-algebra $A$ is a Banach space with additional properties: so a vector space, but without an inner product, only a norm.

If, however, we are only considering the finite-dimensional case, then this algebra $A$ will also be canonically isomorphic (as a vector space) to $\mathbb{C}^n$ for some $n$, via say $\phi:A\xrightarrow{\sim} \mathbb{C}^n$. But on this space, we do have an inner product.

Is it therefore sensible (in the following way) to define $\langle x,y\rangle_A\equiv \langle \phi(x),\phi(y) \rangle_{\mathbb{C}^n}$?

Sensible as in: For an operator $T$ on $A$, we can now define the adjoint as usual, $\langle T^*x ,y\rangle = \langle x,Ty\rangle$, and this definition is 'good'

Answer 1

These are comments that are unfortunately too long to write as comments. I assume all algebras are unital, but this is not a problem because passing to a unionisation will preserve finite dimensionality.

If your algebra is commutative and finite dimensional then there is a canonical identification of the algebra with $\ell^2(X)=\{f:X\to\Bbb C\}$ where $X$ is a finite set. This is called the Gelfand transform. On $\ell^2(X)$ you have a canonical inner product: $$(f,g) := \frac1{|X|}\sum_{x\in X}\overline{f(x)}g(x) $$

Now unless $X$ is a single point, ie the algebra is $\Bbb C$, you will not have that the $C^*$ norm is induced by this inner product.

Looking at the non-commutative case, let us be inspired by matrices. Here you have an inner product defined on $M_{n\times n}$ via: $$(A,B):=\frac{\mathrm{Tr}(A^*B)}{n}$$ This inner product does not induce the same norm as the $C^*$ norm! Now the trace (or better said the normalised trace) is something called a state. This is a linear functional $\omega: \mathcal A\to\Bbb C$ so that $\omega(\Bbb1)=1$ and $\omega(P)≥0$ whenever $P$ is positive. Further the normalised trace is faithful, this means that if $A>0$ you have $\omega(A)>0$.

If you have a faithful state $\omega$ you automatically get an inner product in the same way, namely via $(a,b):=\omega(a^*b)$. The commutative finite dimensional case is a special case of this, where we have a canonical faithful state via the Gelfand-transform.

The inner products considered here are all compatible with the $C^*$ structure in a certain way. We know that $\mathcal A$ acts linearly on itself via multiplication. So we have an algebra morphism $L:\mathcal A\to B(\mathcal A)$ via $a\mapsto L_a = (b\mapsto a\cdot b)$. Further: $$(a,L_b c) = \omega(a^*bc)=\omega((b^*a)^*c)=(L_{b^*}a,c) $$ And we verify that this is a $*$-morphism. A non-trivial result is that injective $*$-morphisms are isometries, so here $\mathcal A$ embeds as a sub-algebra of $B(\mathcal A)$. I think this is as good as you can get.

The reason why this is not an answer is that I don't know if there are canonical faithful states on non-commutative finite dimensional C* algebras.

Answer 2

Ok, so I came back to this question and I think I have an answer. I'm posting this separately to my "long comment". In the comment I motivated that one needs to canonically construct a faithful state that is invariant under inner automorphisms.

The unitaries $U(\mathcal A)\subset \mathcal A$ are a compact topological group and thus you have a canonical positive measure of mass $1$ on them. Lets call this measure $du$. If we have any state $\omega$ of $\mathcal A$, we can make this state invariant under inner automorphisms by taking the mean over $U(\mathcal A)$: $$\omega'(a):=\int_{U(\mathcal A)}\omega(uau^*)\ du$$ This averaging will preserve the faithfullness of $\omega$. Where can we find a canonical state over which we take the mean? Well we can just take the mean over all states. Faithful states are a bounded convex subset of the finite dimensional $\mathcal A'$, such sets also have probability measures on them (take the Lebesgue measure on the real vector space generated by the set and normalise so that the set has volume $1$). So we can just take the mean over all states: $$\omega(a):=\int_{\text{Faithful states of }\mathcal A}\nu(a)\,d\nu$$

As described in the comment a faithful state that is invariant under inner automorphisms induces a Hilbert space structure on $\mathcal A$, if we call this Hilbert space $H_\mathcal A$ then the map $$L: \mathcal A\to B(H_\mathcal A), a\mapsto (x\mapsto a\cdot x)$$ is an isometric $*$-morphism.