Inner products on separable, infinite-dimensional $C^\star$ algebras, and "adjoints" of maps

Let $$A$$ and $$B$$ be two separable $$C^\star$$ algebras, and $$\phi:A\to B$$ a bounded linear map. A faithful state on a $$C^\star$$ algebra $$A$$ is a state such that for $$a$$ in $$A$$, $$\phi(a^\star a)=0$$ iff $$a=0$$. Every separable $$C^\star$$ algebra has a faithful state (here is an answer that discusses this). If $$\phi$$ is a faithful state on a $$C^\star$$ algebra $$A$$, then $$\langle a,b \rangle _\phi=\phi(b^\star a)$$ for all $$a,b$$ in $$A$$ yields an inner product on $$A$$, and thus induces a norm $$\lVert a\rVert_\phi =\sqrt{\phi(a^\star a)}$$. Of course, $$A$$ might not be Cauchy complete with respect to the norm $$\lVert \cdot \rVert_\phi$$, in which case we denote the completion of $$A$$ with respect to $$\lVert \cdot \rVert_\phi$$ as $$L^2 _\phi (A)$$. It is direct to see that when $$A$$ is finite-dimensional, it is in fact Cauchy complete with respect to $$\lVert \cdot \rVert_\phi$$, that is, $$L^2 _\phi(A)$$ and $$A$$ are isomorphic as linear spaces.

My first question is: (1) For infinite dimensional separable $$A$$, does there always exist a faithful state $$\phi$$ such that $$A$$ is Cauchy complete with respect to $$\lVert \cdot \rVert_\phi$$? That is, $$L^2 _\phi(A)$$ and $$A$$ are isomorphic as linear spaces? If not, what are necessary and / or sufficient conditions that must be imposed on $$A$$ for this to hold true?

Now, let us consider finite dimensional $$C^\star$$ algebras $$A,B$$, with faithful states $$\phi,\psi$$. Let $$\alpha:A\to B$$ be a linear map (and hence bounded: $$A,B$$ being finite dimensional), we denote by the same symbol the corresponding linear map between $$L^2 _\phi(A)$$ and $$L^2 _\psi(B)$$ (which are isomorphic to $$A,B$$ respectively as linear spaces), and is also bounded (finite dimensional). Now, we denote by $$\alpha^\star:L^2 _\psi(B)\to L^2 _\phi(A)$$ the Hilbert space adjoint of $$\alpha$$, an denote by the same symbol the corresponding linear map between $$B,A$$.

My second question is: (2) Given two separable infinite dimensional $$C^\star$$ algebras $$A,B$$, with separable states $$\phi,\psi$$, and a bounded linear map $$\alpha:A\to B$$, is it possible to define a bounded linear map $$\alpha^\star:B\to A$$, such that $$\langle \alpha(a), b \rangle_\psi = \langle a, \alpha^\star(b) \rangle_\phi$$ for all $$a$$ in $$A$$ and $$b$$ in $$B$$? (I am looking for an answer even when $$A\subsetneq L^2 _\phi(A)$$, $$B\subsetneq L^2 _\psi(B)$$ as linear spaces).

My attempts (unsuccessful): For the first question, I tried out several (simple) examples of $$C^\star$$ algebras in infinite dimensions, like $$C(\mathbb{T})$$ (continuous functions on the unit circle, with the supremum norm, multiplication pointwise, and the involution given by complex conjugation), and $$\ell^\infty$$ (with involution given by the complex conjugation, multiplication entrywise), and a couple of similar examples. However, with the half a dozen faithful states I could cook up overall (for example, for $$C(\mathbb{T})$$, $$f\mapsto\frac{1}{2\pi} \int_\mathbb{T} f d\mu$$ with $$\mu$$ being the Lebesgue measure, and for $$\ell^\infty$$, $$(x_n)_n \mapsto \sum ^{j=1} _{\infty} \frac{x_j}{2^j}$$) and some others, all being different, in the sense of not being related by permutations, I failed to get a positive answer. But of course, this does not imply one way or the other given the question, other than perhaps saying that the answer might not be immediate. For the second question, I identified several problems: a bounded map $$\alpha:A\to B$$ might not even be bounded when we consider the $$\lVert \cdot \rVert _\phi$$, $$\lVert \cdot \rVert _\psi$$ norms on $$A,B$$ respectively, rather than the $$C^\star$$ norms: for example, considering $$A=B\otimes_{min} B$$, the multiplication map turned out to be unbounded in those norms, for the example of $$\ell^\infty$$ mentioned above. Even if we consider a case where it is bounded, it might not be Cauchy complete with respect to the norm of interest (for example, in the case of $$C(\mathbb{T})$$ mentioned above). The proof of the existence and uniqueness of the adjoint that I am familiar with, pivots around the Riesz Representation Theorem for bounded linear functionals on Hilbert spaces, which fails if either of the conditions of completeness or boundedness fail to hold (see here and here respectively, for answers that discuss these).

A little bit on my background, in case it relevant: I studied, for my Masters' thesis, basics of $$C^\star$$ algebras, positive and completely positive maps, dilations, and related topics. However, I am not (yet) familiar with the theory of unbounded linear operators.

Edit 1: The motivation behind these questions are the following: any finite dimensional $$C^\star$$ algebra, as a linear space, can be "given" a Hilbert space structure, albeit with a different norm. Still, as everything is finite dimensional, all linear operators on such spaces are bounded. Moreover, for every linear map between two finite-dimensional $$C^\star$$ algebras, one can define "adjoint" maps, unique for each Hilbert space structure. These "adjoint" maps are sometimes useful in the study of quantum groups and graphs (I think). I was wondering how these carried on to infinite dimensions, if they did at all.

Edit 2: Actually, even in finite dimensions, $$\alpha^\star$$ very much depends on $$\phi$$ and $$\psi$$, and not just $$A$$ and $$B$$, so it is perhaps not possible to have a very general idea of $$\alpha^\star$$ that is useful as well. However, I provide an example: Let $$B=M_n$$ ($$n\times n$$ complex matrices), $$\psi(\cdot)=\frac{1}{n}Tr(\cdot)$$, $$A=B\otimes B$$, $$\phi=\psi\otimes\psi$$ then, with $$m:M_n\otimes M_n \to M_n$$, $$E_{i,j}\otimes E_{k,l} \mapsto \delta_{j,k} E_{i,l}$$, where $$E_{i,j}$$ denotes the matrix with $$1$$ at the $$i,j$$-th entry and zero everywhere else. Of course, $$\{E_{i,j}\}_{i,j=1}^n$$ is a basis of $$M_n$$, and we write $$\otimes_{min}$$ as $$\otimes$$ as everything is nuclear. Then, $$m^\star:M_n\otimes M_n \to M_n$$, $$E_{i,j}\mapsto \sum_{k=1}^n E_{i,k}\otimes E_{k,j}$$. This is a "comultiplication", with respect to the trace.

• 1) Where are those questions coming from? 2) The map $\alpha^*$ for the finite dimensional case works because all norms are equivalent, right? Commented Oct 18, 2020 at 10:17
• @JustDroppedIn 1) I added an edit at the end of the question which should explain the motivation 2) $\alpha$ is bounded in the $L^2 _\phi$ norm because all norms are equivalent in finite dimensions. The definition of $\alpha^\star$ is possible because of the Riesz Representation Theorem. Commented Oct 18, 2020 at 13:37

2 Answers

The answer to your first question is no, and here is the reason.

Note that the identity mapping $$i:A\to L^2_\phi(A)$$ is continuous. Thus, if $$A=L^2_\phi(A)$$, the open mapping theorem implies that $$i$$ is an isomorphism and hence $$A$$ is seen to be isomorphic$$^{(1)}$$ to a Hilbert space. This would in turn imply that every closed subspace of $$A$$ is also isomorphic to some Hilbert space.

This said, choose some self adjoint element $$a$$ in $$A$$ with infinite spectrum (these exist in any infinite dimensional $$C^*$$-algebra) and consider the closed $$^*$$-subalgebra $$C^*(a)\subseteq A$$ generated by $$a$$. It is well known that $$C^*(a)$$ is isometrically isomorphic to $$C_0(\sigma (a)\setminus\{0\})$$, so we conclude that $$C_0(\sigma (a)\setminus\{0\})$$ is isomorphic to some Hilbert space.

However it is a well known fact that $$C_0(X)$$ is not isomorphic to a Hilbert space for every infinite locally compact space $$X$$.

$$^{(1)}$$ Two normed spaces $$E$$ and $$F$$ are said to be isomorphic if there exists a map $$T:E\to F$$ which is linear, bijective, continuous, and whose inverse is continuous. If $$T$$ is moreover isometric, then we say that $$E$$ and $$F$$ are isometrically isomorphic.

• I am sorry for using the wrong space, which I have just corrected. When $A$ is unital, the $C^*$-algebra generated by $\{1,a\}$ is isomorphic to $C(\sigma (a))$, but in the absence of a unit we should instead consider $C^*(a)$, which is isomorphic to $C_0(\sigma (a)\setminus\{0\})$. So far I have no idea about your second question but I'm thinking about it.
– Ruy
Commented Oct 18, 2020 at 14:09
• Every vector space admits a norm satisfying the parallelogram law: just take a Hamel basis and define an inner product making that an orthonormal basis. The contradiction I proposed, on the other hand, is related to the fact that there is no linear map from $C_0(X)$ onto a Hilbert space which is bijective, continuous, and whose inverse is also continuous (that is, a linear isomorphism and at the same time a topological homeomorphism).
– Ruy
Commented Oct 18, 2020 at 16:40
• @Ruy, " I don't think a negative answer to the first question follows from the existence of a complete Hilbert norm on $C_0(X)$" yes, I made a mistake, sorry (corrected it). "But it does from my answer." Perhaps I am missing something here. I am not looking for an inner product that induces the sup norm on $C_0(X)$ (it doesn't exist, I know that, since the sup norm doesn't satisfy parallelogram law), but rather an inner-product induced norm on $C_0(X)$ with respect to which $C_0(X)$ is complete. Commented Oct 18, 2020 at 17:47
• I think we're going in the wrong direction by insisting in this point, but OK, here we go. The Hamel dimension of every separable infinite dimensional Banach space is equal to the cardinal number $\omega_1$ so they are all isomorphic as vector spaces. We may then pick a linear isomorphism (no topology involved) $\phi:C_0(X)\to\ell^2$ and define on $C_0(X)$ the norm $|||f||| = \|\phi(f)\|_2$. It then follows that $C_0(X)$ is a (complete) Hilbert space with this norm.
– Ruy
Commented Oct 18, 2020 at 17:53
• Regarding question 2, I believe one should not worry so much about whether or not the algebras are complete, and instead focus on examples in which the adjoint $\alpha^*$ exists and try to understand what it means. Here is such a situation: suppose $A$ is a sub-algebra of $B$ and let $\alpha$ be the inclusion of $A$ in $B$. Also, suppose that $\psi$ is a faithful trace on $B$, and let $\phi$ be its restriction to $A$. Under this setup I believe that, should $\alpha^*$ exist, it must be a conditional expectation from $B$ to $A$!!!
– Ruy
Commented Oct 19, 2020 at 0:55

About your first question, the intuition says that this is probably very unlikely. The following counter example comes from the commutative world and asserts us that this is the case:

Let $$X$$ be a compact Hausdorff space. We consider the C*-algebra $$A=C(X)$$, which is abelian. Then by Riesz representation we have that $$A^*\cong\{\text{complex Borel measures on }X\}$$ and therefore the set of states on $$A$$ is identified with the Borel probability measures on $$X$$, i.e. $$S(A)\cong\text{Prob}(X):=\{\text{(positive) Borel probability measures on } X\}$$ Now the inner product induced from a state $$\mu\in S(A)$$ is $$\langle-,-\rangle_\mu:C(X)\to\mathbb{C}$$ given by $$\langle f,g\rangle_\mu=\int_Xf\bar{g}d\mu$$ Note that such a state is faithful when the measure $$\mu$$ has full support, since if $$\langle f,f\rangle_\mu=0$$ then $$\int_X|f|^2d\mu=0$$ so $$|f|=0$$ a.e. thus $$f=0$$. Every compact space admits a probability measure with full support, so we have a counter example: we actually have that $$C(X)$$ is not complete with respect to $$\|\cdot\|_\mu:=\sqrt{\langle\cdot,\cdot\rangle_\mu}$$. The completion is actually equal to the Hilbert space $$L^2(X,\mu)$$: to see this simply recall that $$C(X)$$ is dense in $$\|\cdot\|_2$$ in $$L^2(X,\mu)$$. Of course $$C(X)$$ is a much smaller space than $$L^2(X,\mu)$$. Actually this example is the reason behind the notation $$L^2_\phi(A)$$ you are using.

Finally, since you are looking for separable $$C^*$$-algebras specifically, simply restrict to metrizable compact Hausdorff spaces (as $$C(X)$$ is separable iff $$X$$ is metrizable, given that $$X$$ is a compact Hausdorff space).

• Also this does not contradict your statements about finite dimensional $C^*$-algebras: the only finite dimensional, abelian $C^*$-algebras are finite copies of $\mathbb{C}$. I believe this also shows that you cannot formulate any interesting conditions for Q1 to have a nice answer, since any infinite dimensional C*-algebra contains some abelian C*-subalgebra. Commented Oct 18, 2020 at 11:29
• Ops! Your statement that $f=0$ a.e. implies that $f=0$ is false! Think e.g. of a Dirac measure.
– Ruy
Commented Oct 18, 2020 at 13:05
• It is true that probability measures correspond 1-1 with states. Moreover a state is faithful iff the support of the corresponding measure is the whole space.
– Ruy
Commented Oct 18, 2020 at 14:10
• @Ruy yeah, okay, I had in mind measures with full support. Still my answer provides a counter example, since every compact hausdorff space admits a probability measure with full support. So particularily $C[0,1])$ and the lebesgue measure provide a counter example. Commented Oct 18, 2020 at 16:54
• @UjanChakraborty I have added this detail in my post. This is a solid counter example. Commented Oct 18, 2020 at 16:56