MHB Proving Homomorphism in Commutative Subalgebras of Bounded Operators

[Boromir]

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]

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.

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.

 

[Boromir]

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.

Thanks for the welcome Opalg

I'm afraid I'm in the process of trying to prove the spectral theorem. I will give you more context. I'm trying to prove that for T in A, the spectrum of T is {h(T):h is a non-zero homomorphism A->C}. I have proved that h(T) is in the spectrum so now I need the reverse inclusion. That is, that every member of the spectrum is equal to h(T) for some (non-zero) homomorphism h. I don't want to go down the road of ideals as the proofs on the internet do as I'm specifically focusing on the space B(H).
Nearly forgot, the algebra is adjoint preserving

 

[Boromir]

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.

Do you think I am trying to 'square the circle' as in there needs to be an assumption of commutativity which is not present in B(H)?

 

[Opalg]

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.

 

[Boromir]

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.

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]

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

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]

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.

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?

 

[Opalg]

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?

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]

The product of the polynomials $p(T,T^*)$ and $g(T,T^*)$ is $p(T,T^*)g(T,T^*)$, not $p(g(T,T^*))$.

Yes of course the operation is not composition of functions (polynomials) but composition of operators. Case closed.