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Can anyone help me to solve the following problem.

We know if $B$ is a unital sub-algebra of a unital Banach algebra $A$ and if $a\in B$, then $\sigma _{A}[a] \subset \sigma _{B}[a] $(Here $\sigma _{T}$ denotes the spectrum with respect to the algebra $T$). What happens if we choose $B$ as an maximal sub-algebra. Can we say that corresponding spectrum will be equal?

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    Sorry for my ignorance, but is it obvious that maximal subalgebras exist?2012-10-07
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    I don't know whether maximal sub-algebra exist obviously or not. But in the question, I am supposing there exist a maximal sub-algebra. At this moment I guess (I am not sure),one can use zorn's lemma to argue for existence of a maximal algebra. If I am wrong please let me know.2012-10-07
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    You can use Zorn, but you cannot guarantee that your maximal subalgebra is not the whole algebra. If you add "proper" and try to do Zorn, you get stuck because there is no reason that a union of proper subalgebras is proper.2012-10-07

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I cannot say much about the case requested, as I know very few examples of maximal subalgebras that are not C$^*$-subalgebras (and these preserve the spectrum). The case that is easy is the case where $B$ is maximal abelian. In that case, if $\lambda\in\rho_A[a]$, then there exists $c\in A$ with $c(a-\lambda 1)=1$. For any $x\in B$, we have $$ xc=c(a-\lambda 1)xc=cx(a-\lambda 1)c=cx, $$ so $c$ commutes with every $x\in B$ and so $c\in B$. This shows that $\lambda\in\rho_B[a]$, i.e. $\rho_A[a]\subset\rho_B[a]$, or $\sigma_B[a]\subset\sigma_A[a]$.