The discussion revolves around proving the existence of a homomorphism in a commutative unital subalgebra of bounded operators on a Hilbert space, specifically focusing on the relationship between the spectrum of an operator and homomorphisms from the algebra to the complex numbers. The scope includes theoretical aspects of functional analysis and operator theory.
Participants express differing views on the necessity of certain assumptions about the algebra A, such as self-adjointness and closure properties. The discussion remains unresolved regarding the implications of these assumptions on the existence of the homomorphism and the nature of the mappings involved.
Limitations include the dependence on the definitions of self-adjointness and closure in the operator norm topology, as well as unresolved questions about the nature of polynomial mappings and their properties in this context.
Hi Boromir, and welcome to MHB!Boromir said:Let A be a communitative unital subalgebra of B(H) (the set of bounded operators on a hilbert space). Let T be in A. Prove that for all x in the spectrum of T, there is a homomorphism h on A such that h(T)=x. Give the homomorphism explicity.
Opalg said:Hi Boromir, and welcome to MHB!
Can you give us some idea of what you know about such algebras? Is the algebra A supposed to be selfadjoint? If so, then the spectral theorem will apply. That might give you a handle on the problem.
Opalg said:Hi Boromir, and welcome to MHB!
Can you give us some idea of what you know about such algebras? Is the algebra A supposed to be selfadjoint? If so, then the spectral theorem will apply. That might give you a handle on the problem.
Opalg said:The key to this is the commutative Gelfand–Naimark theorem.
The problem is unchanged if we replace $A$ by its closure (in the operator norm topology). So we may as well assume that $A$ is closed. Now that you have added the crucial information that $A$ is selfadjoint (i.e. closed under the adjoint operation), it follows that $A$ is a commutative unital C*-algebra. The G–N theorem then tells you that there is an isometric isomorphism $\phi:A\to C(X)$ from $A$ to the algebra of continuous functions on a compact Hausdorff space $X$ (the structure space or spectrum of $A$). It follows fairly easily that the spectrum of $T$ is equal to the range of the function $\phi(T)$. Also, the homomorphisms from $A$ to the scalars are exactly the point evaluation functions on $X$, in other words functions of the form $S\mapsto (\phi(S))(x)$ for $x\in X$ and $S\in A$. That is all you need to deduce the result.
That also comes from the G–N theorem. If $T$ has an inverse then its Gelfand transform $\phi(T)$ never takes the value $0$ and therefore has an inverse, namely the function $f(x) = 1/((\phi(T))(x))$. The inverse Gelfand transform $\phi^{-1}(f)$ belongs to $A$ and is equal to $T^{-1}$.Boromir said:On a separate but related note, why is A closed under inverses i.e. if T has an inverse, then the inverse is in A
Opalg said:The key to this is the commutative Gelfand–Naimark theorem.
The problem is unchanged if we replace $A$ by its closure (in the operator norm topology). So we may as well assume that $A$ is closed. Now that you have added the crucial information that $A$ is selfadjoint (i.e. closed under the adjoint operation), it follows that $A$ is a commutative unital C*-algebra. The G–N theorem then tells you that there is an isometric isomorphism $\phi:A\to C(X)$ from $A$ to the algebra of continuous functions on a compact Hausdorff space $X$ (the structure space or spectrum of $A$). It follows fairly easily that the spectrum of $T$ is equal to the range of the function $\phi(T)$. Also, the homomorphisms from $A$ to the scalars are exactly the point evaluation functions on $X$, in other words functions of the form $S\mapsto (\phi(S))(x)$ for $x\in X$ and $S\in A$. That is all you need to deduce the result.
The product of the polynomials $p(T,T^*)$ and $g(T,T^*)$ is $p(T,T^*)g(T,T^*)$, not $p(g(T,T^*))$.Boromir said:Taking as a concrete example the set of polynomials generated by T and it's adjoint, you are saying that the map $p(T,T^*)$->$p(x,x*)$ where x is any fixed member of the spectrum is a homomorphism? Does it preserve the 'multiplication? Well, given polynomials p,g, we have $p(g(T,T^*))$->$p(g(x,x*))$ but that does not equal $p(x,x*)$$g(x,x*)$ which is the multiplication in C. Or am I missing something?
Yes of course the operation is not composition of functions (polynomials) but composition of operators. Case closed.Opalg said:The product of the polynomials $p(T,T^*)$ and $g(T,T^*)$ is $p(T,T^*)g(T,T^*)$, not $p(g(T,T^*))$.