Let $V$ be a locally convex topological vector space over $\mathbb C$, and let $A=\mathrm{End}(V)$ be its algebra of continuous linear endomorphisms (viewed just as a $\mathbb{C}$-algebra, not as a topological algebra).

Does the algebra $A$ remember $V$?

Namely, is it true that every representation of $A$ on a locally convex topological vector space $W$ for which the associated map $A\to \mathrm{End}(W)$ is an isomorphism is isomorphic to $V$ as an $A$-representation, and the isomorphism $V\cong W$ is a homeomorphism?

  • 2
    $\begingroup$ In the case of Banach spaces I think this follows from an old result of Eidelheit, but since the conclusion of that result is usually stated as the weaker conclusion "$V\cong W$ as TVS" I'd need to look at the proof again in more detail $\endgroup$ – Yemon Choi Jan 21 '18 at 23:36
  • $\begingroup$ @AndreaHenriues I am sorry if my question is elementary. For a compact space $X$ can we recover the topology of $X$ from the algebra $A=End(C(X))$ where $C(X)$ is considered as a banach space?What is the answer of your question in the particular case $V=C(X)$? $\endgroup$ – Ali Taghavi Jan 22 '18 at 8:11
  • $\begingroup$ @AliTaghavi This should really be a separate question, but it is not elementary! The answer is no since C(X) and C([0,1]) are isomorphic as Banach spaces whenever X is an uncountable compact metric space; this result is known as Milutin's theorem $\endgroup$ – Yemon Choi Jan 23 '18 at 2:21
  • $\begingroup$ Clarification/correction to my earlier comment, in light of the answer below: Eidelheit's theorem says that if E and F are Banach spaces and End(E) and End(F) are isomorphic as algebras, then E and F are isomorphic as TVSes. $\endgroup$ – Yemon Choi Jan 23 '18 at 2:22

If $V$ is a Mackey space (for example, a Banach space), $\text{End} (V)$ coincides with $\text{End} (V_\sigma)$, where $V_\sigma$ denotes $V$ with the weak topology.


Yes, you can recover the $A$-module $V$. There is a minimal ideal in $End(V)$, namely the ideal $F$ of finite-rank-operators. It is in fact the unique minimal two-sided ideal, because for every $f\neq 0$ one can find a finite-rank operator $g$ with $gf\neq 0$.

$F$ is isomorphic to the algebraic tensor product $V \otimes_{alg} V^\ast$ as a $A$-$A$-bimodule via the canonical map $v\otimes\phi \mapsto \phi(-)v$. This also proves that $A$ acts transitively on $V\setminus\{0\}$. One can readily verify that the left ideals contained in $F$ are in 1-to-1 correspondence to finite-dimensional quotients of $V$ via $(V\xrightarrow{q}\mathbb{C}^n) \leftrightarrow \{f \mid f\text{ factors through }q\}$ and this left ideal is isomorphic to $\sum_{i=1}^n V\otimes q_i \leq V\otimes V^\ast$ as $A$-modules. In particular, all the minimal left ideals contained in $F$ are isomorphic to $V$ as $A$-modules.

Now if you want $End(V)$ to also recover the topology of $V$, you need more information, I think. Of course you can recover $V^\ast$ as a vector space and the duality $V^\ast\times V \to \mathbb{C}$ with the same ideas as above by looking at right ideals and the product of minimal right and left ideals, but that only gives upper and lower bounds on the topology of $V$ (this is the Mackey-Arens theorem).

This also answers the second question in the affirmative: If $A\cong End(W)$, then $V\cong W$ because both are isomorphic to the minimal left ideals contained in the minimal two-sided ideal.

  • $\begingroup$ Concerning the second: I didn't use the word "irreducible", but I did say "for which the associated map A→End(W) is an isomorphism". $\endgroup$ – André Henriques Jan 22 '18 at 1:16
  • $\begingroup$ Sorry, I really should learn not to write these things late at night... This is the third or fourth time I misread something like that. Anyway, I have corrected my answer. $\endgroup$ – Johannes Hahn Jan 22 '18 at 1:18
  • $\begingroup$ Sorry: I edited the last paragraph of my question and added "and the isomorphism V≅W is a homeomorphism", because that's the thing I'm most interested in. With this modification, the last paragraph of your answer only answers half of my question. $\endgroup$ – André Henriques Jan 22 '18 at 1:25
  • $\begingroup$ You're right. I do not know how to recover the topology of $V$ and I'd guess that it isn't possible in general. $\endgroup$ – Johannes Hahn Jan 22 '18 at 1:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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