Does there exist a nontrivial (i.e. other than $\mathbb{C}$) example of a $C^*$-algebra which is also a Hilbert space (in the same norm, of course)?

For $\mathbb{C}^n$ with $n > 1$ the answer is no by uniqueness of norm in $C^*$-algebras, since $\mathbb{C}^n$ is a $C^*$-algebra in the $\ell^\infty$ norm, which is not given by an inner product. What about more generally?


I will first assume that the $C^*$-algebra is unital. If it is not one-dimensional, then because every element of the algebra has the form $a+ib$ with $a$ and $b$ self-adjoint, there is a self-adjoint element $a$ that is not a scalar multiple of the identity. Then the spectrum of $a$ contains at least $2$ distinct elements. Otherwise, if say $\lambda$ were the only element of the spectrum of $a$, then $a-\lambda 1$ would be a self-adjoint element with spectral radius $0$, hence $a=\lambda 1$.

Since the spectrum of $a$ has at least $2$ elements, there exist continuous ($\mathbb C$-valued or even positive real-valued) functions $f$ and $g$ on the spectrum of $a$ such that $\|f(a)\|=\|g(a)\|=1$ and $f(a)g(a)=0$. It follows that $\|f(a)+g(a)\|=\|f(a)-g(a)\|=1$. Hence

$$2\|f(a)\|^2+2\|g(a)\|^2=4>2=\|f(a)+g(a)\|^2+\|f(a)-g(a)\|^2,$$ in violation of the parallelogram law. Alternatively, $\{f(a)+tg(a):t\in[-1,1]\}$ is a closed convex set with infinitely many elements of minimal norm.

If the algebra is nonunital, you can still find $a$, $f$, and $g$ as above, but you need $f(0)=g(0)=0$, so the spectrum of $a$ should have $2$ distinct nonzero elements, which means slightly more is needed to see that such $a$ exists. This feels like overkill, but in this case the algebra is infinite dimensional, which implies that it has a self-adjoint element with infinite spectrum. Note that a self-adjoint element whose spectrum has only one nonzero element is a scalar multiple of a projection, and there should be an easier way to see that there are self-adjoint elements that are not multiples of projections, but none comes to mind.

  • $\begingroup$ I asked a question about a simpler way to get $a$ in the nonunital case: math.stackexchange.com/questions/536444/… $\endgroup$ – Jonas Meyer Oct 26 '13 at 0:38
  • $\begingroup$ Did I get you right that requiring $f(0)=0$ is to ensure that $f(A)\in\mathcal{A}$? Btw nice answer!! :D $\endgroup$ – C-Star-W-Star Oct 18 '15 at 12:59
  • $\begingroup$ @AlexanderFrei: Yes. Thanks. $\endgroup$ – Jonas Meyer Jun 15 '16 at 19:49

No. Every infinite dimensional C*-algebra is a strict subset of its second dual.

Edited to add: Perhaps an easier way to see it is to use the fact that the extreme points of the unit ball of a C*-algebra are precisely the partial isometries. It is not too hard to see that any C*-algebra of dimension greater than 1 contains elements of norm one that are not partial isometries.

Edit the second: Just can't resist giving one more reason. First, the unit ball of a non-unital C*-algebra has no extreme points at all. Otherwise, call the unit $e$ and the cone of positive elements $P$, then the unit ball in the self-adjoint part of the C*-algebra is $B_{\text{sa}}=(e-P)\cap(-e+P)$. In particular, $B_{\text{sa}}\subset(e-P)$. It is easy to that if this holds for a unit vector $e$ in a Hilbert space, the cone $P$ must be the entire half space $\{v\colon \langle v,e\rangle\ge0\}$. In geometric terms, the unit ball of the self-adjoint part of a C*-algebra has a sharp point at the unit, while the unit ball of a Hilbert space is smooth.

  • $\begingroup$ Harald, do you have a reference for the first statement? I have a vague recollection that "reflexive $C^\ast$-algebras are finite-dimensional" follows from some results by Ghahramani, but don't remember if there were earlier and more direct proofs $\endgroup$ – user16299 Feb 26 '12 at 11:51
  • $\begingroup$ @YemonChoi: Sorry, it's one of those things I seem to have known “forever”, and I can't dig up a good reference. It's been years and years since I did any work in this field, and memory goes rusty. But if my subconscious dregs up a reference, I'll be sure to note it here. $\endgroup$ – Harald Hanche-Olsen Feb 26 '12 at 16:21
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    $\begingroup$ All extreme points are partial isometries, but not all partial isometries are extreme points in general. This doesn't affect the rest of your answer. One characterization of the extreme points in the unit ball of a unital C* algebra $A$ is as the set of all partial isometries $v$ such that $(1-v^*v) A(1-vv^*)=\{0\}$. (In particular, isometries and coisometries are extreme points.) This is due to Kadison, seen in jstor.org/stable/1969534 $\endgroup$ – Jonas Meyer Feb 28 '12 at 2:44
  • $\begingroup$ @JonasMeyer: You're right, I was too quick there. (Another reference: Sakai, C*-algebras and W*-algebras, 1.6.4 and 1.6.5.) $\endgroup$ – Harald Hanche-Olsen Feb 28 '12 at 8:24
  • $\begingroup$ Wow, now I understand how much stronger your answer is: There is even no infinite dimensional TVS that is a Hilbert space and a C*-algebra. (There the norms may even differ up to they induce the TVS topology). Contrary every finite dimensional (Hausdorff) TVS is a Hilbert space and a C*-algebra. (There the norms will even differ up to they induce the TVS topology.) This completely disproves the even more general problem about equivalence up to TVS.. Wwoooow!!! :OOO (+1) $\endgroup$ – C-Star-W-Star Jan 5 '17 at 13:20

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