The discussion centers on proving the existence of a homomorphism in commutative unital subalgebras of bounded operators, specifically within the context of the spectral theorem. Participants confirm that if A is a self-adjoint commutative unital C*-algebra, then for any operator T in A, the spectrum of T corresponds to the range of the Gelfand transform. The Gelfand–Naimark theorem is pivotal, establishing an isometric isomorphism between A and the algebra of continuous functions on a compact Hausdorff space. This leads to the conclusion that homomorphisms from A to the scalars are represented as point evaluation functions on the spectrum.
PREREQUISITESMathematicians, particularly those specializing in functional analysis, operator theory, and algebra, will benefit from this discussion. It is also relevant for graduate students studying advanced topics in mathematics related to bounded operators and C*-algebras.
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 $p(g(T,T^*))$.