The Fell-Doran problem is a problem in functional analysis. It goes as follows: Let $A$ be a complex unital algebra, $X$ a locally convex space, and $L(X)$ the algebra of all continuous endomorphisms of $X$. Suppose that we have a representation of $A$ on $X$, by which we simply mean an algebra homomorphism $$ T : A \rightarrow L(X) $$ which is irreducible (no proper closed invariant subspace) and has trivial commutant (any bounded operator commuting with all the $T_a$ must be a multiple of the identity). The Fell-Doran problem is: Is $T(A)$ dense in $L(X)$ in the strong operator topology?

My question is: Is this a problem having to do with the fact that we didn't require a topology on our algebra? In other words, what can be said about the case when $A$ is actually a 'topological algebra' and the map $T$ is required to be continuous in some sense? Does that make the problem trivial, i.e. is the answer then automatically yes?

By the way, I have heard that so far there is almost no progress on the Fell-Doran problem in general; not even for Hilbert spaces! The only thing that is known is that there exists a certain concrete space where the answer is affirmative.

  • $\begingroup$ Do you mean "locally convex topological vector space"? $\endgroup$ Nov 16, 2009 at 19:40

3 Answers 3


I don't agree with Andrew: more specifically, if $A$ is the algebra of compact operators on a Hilbert space $H$, then let $T\in L(H)$. If, say, $H$ is separable, then let $(e_n)$ be an orthonormal sequence, and let $P_n$ be the orthogonal projection on the span of $e_1,\cdots,e_n$. Then $P_n$ is compact, and $P_n(x)\rightarrow x$ in norm for each $x\in H$. In particular, $P_nT\in A$ for each $A$ and $P_nT\rightarrow T$ is the strong operator topology.

More generally, the Kaplansky Density Theorem (see your favourite book on Operator Algebras, or Wikipedia) shows that if $A\subseteq L(H)$ is a $*$-closed algebra, then the unit ball of $A$ is strong operator dense in the unit ball of the von Neumann algebra which $A$ generates. So if $A$ has trivial commutant, then the von Neumann bicommutant theorem shows that $A$ generates all of $L(H)$. In particular, $A$ is strong operator dense in $L(H)$.

Of course, the question is: what happens if $A$ isn't $*$-closed...

  • $\begingroup$ Then I've misunderstood what "strong operator topology" means. I interpreted it as norm convergence, now you're saying that it's pointwise convergence (but pointwise with respect to the norm convergence on the space), is that correct? $\endgroup$ Nov 16, 2009 at 19:39
  • $\begingroup$ Yes! It's just the topology given by the seminorms $T\mapsto \|T(x)\|$ as $x$ varies over $H$. $\endgroup$ Nov 16, 2009 at 19:42
  • $\begingroup$ Bleugh. Now you say it, I remember getting all confused over the vast array of names for all of these different topologies! I have no problem with the actual topologies, just their names. Okay, now this sounds a lot like the approximation question which is known to be false. $\endgroup$ Nov 16, 2009 at 20:01
  • $\begingroup$ Do I understand this correctly? Mathhew is saying above that in the case of a $*$-structure preserving representation $T$ of a $*$-algebra on a seperable Hilbert space $H$, the Fell-Doran problem indeed trivializes in the sense that it follows from the "Kaplansky Density Theorem" and the "von Neumann bicommutant theorem" that the image of $T$ is then operator dense inside $L(H)$. That was spirit of my quest: I wanted to know to what extent the Fell-Doran is a "pathological" problem (having to do with the fact that one doesn't require the rep to preserve relevant structure) or a "natural" one. $\endgroup$ Nov 17, 2009 at 10:16
  • $\begingroup$ This is going to sound a bit whingey, but that's not what you asked, Bruce! You specifically asked about topology on the algebra. Also, you asked about a locally convex (topological vector?) space and this is for a Hilbert space where everything is almost always simpler than for other spaces. $\endgroup$ Nov 17, 2009 at 10:27

Actually, digging a bit deeper, I've found this paper by Zelazko: Colloquium Mathematicum which states the problem as: If $A\subseteq L(X)$ is topologically irreducible (that is, the orbit under $A$ of any non-zero vector is dense) and the commutant of $A$ is trivial, is $A$ totally irreducible. This means that for each $n$, if $x=(x_1,\cdots,x_n)$ is a vector in $X^n$ with linearly independent co-ordinates, then $x$ should be cyclic for $A^n$ (where $A^n$ acts on $X^n$ is the obvious way).

Off hand, I don't see what, if anything, this has to do with $A$ being strong operator topology dense in $L(X)$. Or is there more than one conjecture??

  • 1
    $\begingroup$ Sorry, got than wrong: for the definition of $A$ being totally irreducible, we should have that $A$ (not $A^n$) acts on $X^n$. This then does seem to be equivalent to $A$ being strong operator dense in $L(X)$ (at least if $X$ is a Banach space). For if $T\in L(X)$ and $x_1,\cdots,x_n\in X$, we have that $\{ (a(x_1),\cdots,a(x_n)) : a\in A\}$ is dense in $X^n$, in particular, it gets very close to $(T(x_1),\cdots,T(x_n))$. This shows that $A$ is strongly dense in $L(X)$. The converse is similar (approx the identity). $\endgroup$ Nov 16, 2009 at 20:11
  • $\begingroup$ You can edit your own answers, by the way. It's a lot easier on the maths if it's in an answer than in a comment! (And something's going a little wrong with those less-than-or-equal-to signs. If I get another 70 rep then I'll be able to take a look) $\endgroup$ Nov 16, 2009 at 20:18
  • $\begingroup$ Thanks, yes indeed I asked this question after attending a seminar yesterday by Wieslaw Zelasko (the same one from the Colloquium Mathematicum paper) where he asked it. $\endgroup$ Nov 17, 2009 at 10:01

I don't know, but I would be very surprised if a topology made the problem any easier since I would imagine that you could always use the induced topology from $L(X)$ on $A$ so if you can answer it in the topological setting then you can answer it in the discrete setting. More specifically, the question is about the image of $T$ but imposing a topology on $A$ only messes around with the domain so I'd be surprised if it had a significant effect.

However I suspect that I haven't understood the problem very well since it seems as though a simple example would be where $A$ was the compact operators together with the unit. That seems to fit the conditions but it certainly isn't dense in the strong operator topology (though it is in the weak topology).

Edit: As I suspected, I hadn't. As Matthew points out, the topology is not the strong operator topology but the weak operator topology with respect to the strong topology on $X$. That is, $T_\gamma \to T$ if $T_\gamma x \to Tx$ for all $x$.

This, then, sounds a lot like the approximation question which is known to be false for arbitrary Banach spaces (paper of Eno, though I forget the exact reference) since in a space where the approximation property fails one could take the subalgebra of finite rank operators (plus the identity to make it unital).

However, my first point still seems valid.

  • $\begingroup$ You mean Enflo ;-> But that's a great idea! Yes, if $X$ fails to have the approximation property, then "morally" (this area is full of technicalities) your algebra $A$ (being the finite ranks with unity) cannot be strongly dense, but it does have the other properties... The problem is that "morally" does not equal "proof". $\endgroup$ Nov 17, 2009 at 9:25
  • $\begingroup$ (Yes, Enflo. Was half-asleep at the time. Thanks.) Surely the people who've thought about Fell-Doran know about the approximation problem, so there must be something extremely non-trivial in the technicalities. $\endgroup$ Nov 17, 2009 at 10:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.