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Let $A$ be a real/complex algebra (just a real/complex vector space with a multiplication; none PI's are required). Let $\mathcal{H}$ be a real/complex Hilbert space. Let $\operatorname{B}(\mathcal{H})$ be the algebra of bounded endomorphisms of $\mathcal{H}$ and let $\operatorname{Nil}(\mathcal{H})\subset\operatorname{B}(\mathcal{H})$ be the vector subspace of nilpotent bounded operators, i.e., such that $T^2=0$.

My question is:

There are known conditions on $A$ and on $\mathcal{H}$ to ensure the existence of representations $\rho:A\rightarrow \operatorname{B}(\mathcal{H})$ which factor trough the inclusion $\operatorname{Nil}(\mathcal{H})\hookrightarrow\operatorname{B}(\mathcal{H})$?

In other words: Under which conditions there exists an algebra homomorphism $\rho:A\rightarrow \operatorname{B}(\mathcal{H})$ whose image is contained in $\operatorname{Nil}(\mathcal{H})$?

Thank you for any help.

P.S: I'm specially interested when the Hilbert space is the Sobolev space of some nice bounded open set $U\subset \mathbb{R}^n$ and related structures.

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    $\begingroup$ All infinite-dimensional separable Hilbert spaces are isometric hence they have the same algebras of bounded operators. As you you look at algebraically defined entities, the question does not depend on the particular incarnation of $H$. $\endgroup$ Commented Mar 26, 2020 at 17:33
  • $\begingroup$ Thank you for this remark. $\endgroup$ Commented Mar 26, 2020 at 17:37
  • $\begingroup$ What are "PI's" in the first sentence, and can you clarify what you mean by "subalgebra of those nilpotent bounded operators which mutually commute"? $\endgroup$
    – Nik Weaver
    Commented Mar 26, 2020 at 19:40
  • $\begingroup$ PI = Polynomial Identity. $\endgroup$ Commented Mar 26, 2020 at 21:55
  • $\begingroup$ And I tried to clarify the meaning of "subalgebra of those nilpotent bounded operators which mutually commute" editing my question. Actually, it was bad written. Sorry for this and thanks for the comment. $\endgroup$ Commented Mar 26, 2020 at 22:02

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The problem is still ill-posed, because the set of nilpotent bounded operators is not a vector space. (The sum of two nilpotents need not be nilpotent.) Let's interpret the question as: which algebras admit faithful representations as algebras of nilpotent bounded operators?

Note that "nilpotent" usually means $T^n = 0$ for some $n$, not $T^2 = 0$. But whatever definition you take, it will automatically be shared by $A$ and any faithful representation of $A$. So the question is just: which nilpotent algebras can be faithfully represented as algebras of bounded operators?

In the finite dimensional case, a theorem of Cayley tells us that every algebra can be faithfully represented as an algebra of matrices. So in this case every nilpotent algebra has a faithful matrix representation. I think the same is true of any algebra whose dimension is countable. Sketch of proof: assume $A$ is unital and let $T_1, T_2, \ldots$ be a vector space basis for $A$. We regard $A$ as acting on itself by left multiplication and need to find an inner product with respect to which every left multiplication operator is bounded. We can do this by taking the $T_i$ to be mutually orthogonal, and inductively taking the Hilbert space norm of $T_i$ to be sufficiently large so as to ensure that the compression of $M_{T_j}$ (the left multiplication operator), for any $j < i$, to ${\rm span}(T_1, \ldots, T_i)$ has norm at most $(2-1/i)r_j$ where $r_j$ is the norm of its compression to ${\rm span}(T_1, \ldots, T_j)$.

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  • $\begingroup$ Uau! This is exactly what I need. Thank you very much! $\endgroup$ Commented Mar 27, 2020 at 13:17
  • $\begingroup$ You are welcome! $\endgroup$
    – Nik Weaver
    Commented Mar 27, 2020 at 13:25

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