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Assume $\mathcal{A}$ is a $C^*$-algebra with unit $1$ and $\mathcal{B}\subset\mathcal{A}$ is a $C^*$-subalgebra (i.e. a closed $*$-subalgebra) such that $1\in\mathcal{B}$. It is said that under these assumptions, for any $a\in\mathcal{B}$ the spectrum $\sigma_\mathcal{B}(a)$ of $a$ in $\mathcal{B}$ coincides with $\sigma_\mathcal{A}(a)$, that is: If $a-\lambda 1$ has an inverse $b\in\mathcal{A}$, then $b\in\mathcal{B}$.

Now, my questions are the following:

  1. Do you know a nice proof of the above statement? I have found a proof that goes like this: If $b=(a-\lambda)^{-1}$ exists in $\mathcal{A}$, it can be expressed as a convergent power series, i.e. it is the norm limit of partial sums each belonging to $\mathcal{B}$, hene also $b\in\mathcal{B}$. Although this argument looks really nice and I'm aware of Neumann series, I do not see why $b$ can be expressed as a power series. Do you?
  2. Under the above assumptions, is the more general statement $\mathcal{B}^\times=\mathcal{A}^\times\cap B$ true? (here, we denote by $\mathcal{A}^\times$ and $\mathcal{B}^\times$ the set of invertible elements in $\mathcal{A}$ and $\mathcal{B}$, respectively)
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Do you assume $\lVert 1\rVert=1$? – Davide Giraudo Dec 6 '12 at 11:03
You can assume that for now. Interesting that this could make a difference... – Robert Rauch Dec 6 '12 at 11:07
It's actually true in the context of $C^*$-algebras, but it seems we don't use this structure. So it's also interesting to see whether it works in a Banach algebra. – Davide Giraudo Dec 6 '12 at 11:13
up vote 1 down vote accepted

We essentially want to prove that $a \in B$ is invertible in $A$ iff it is invertible in $B$. Let us assume that $a$ is selfadjoint; after all, if $a^*a$ has an inverse in $B$, then so does $a$.

I think one has to use the Stone-Weierstrass theorem in some form. The argument you indicate basically does this in the algebra generated by $a$, which is commutative and actually of the form $C(Spec(a))$. I am slightly worried about the non-circularity of such a proof, however: That the spectrum of an element does not change when passing to a subalgebra is quite deeply ingrained into the continuous function calculus.

I would prove it as follows: Let $X$ be the $C^*$-algebra generated by $1,a,a^{-1}$ inside $A$ and $Y$ the algebra generated by $1,a$ inside $A$ and hence inside $B$ since $1, a \in B$. Since $a$ is self-adjoint, both $X$ and $Y$ are commutative and we have $Y \subset X$. We seek to prove $X = Y$.

Since $X$ is commutative, the Gelfand transform yields an isomorphism $X \rightarrow C(Spec(X))$. Let $Y' \subset C(Spec(X))$ be the image of $Y$ under this map. $Y'$ is a closed $*$-subalgebra of $C(Spec(X))$. Furthermore, if $l,k \in Spec(X)$ are different, we must have $l(a) \neq k(a)$ since otherwise $l, k$ agree on $1, a , a^{-1}$ and then everywhere. But since $a \in Y$, the evaluation at $a$-map $Spec(X) \rightarrow \mathbb{C}$ lies in $Y'$. In other words, $Y'$ separates the points of $Spec(X)$ and is hence dense in $C(Spec(X))$ by Stone-Weierstrass. Since $Y'$ is also closed, it follows $Y' = C(Spec(X))$ and hence $Y = X$ as desired.

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Three little questions/remarks: 1) Why does $a^*a\in B^\times$ imply $a\in B^\times$ and 2) don't we also need that $Y'$ does not vanish in any point (i.e. $\forall l\in Spec X\exists b\in Y:l(b)\ne 0$) before we can invoke the Stone-Weierstraß theorem? 3) I think $Y'$ is not closed, however your proof shows that it is at least dense in $X$, which is sufficient – Robert Rauch Dec 6 '12 at 12:45
1) If a is invertible in A, then so is aa. Then aa is invertible in B. But $a^{-1} = (a*a)^{-1}a^*$ is then also in $B$. 2)& 3) Y is a closed subalgebra of $X$. Then we transport everything via an isometry to C(Spec(X)) and call the image of Y Y'. Since $Y$ was closed in $X$, $Y'$ is closed in $C(Spec(X))$. This also says that no element of $Y$ acts as the $0$-function on Spec(X). – Fabian Lenhardt Dec 6 '12 at 12:58
Ad 2) I think my confusion was because you defined $Y$ to be the algebra generated by $1,a$, but if you actually meant the $C^*$-algebra generated by $1,a$, then everything is fine. Ad 3) I am not convinced from your answer. You say that no element of $Y$ acts as the $0$-function on $Spec(X)$. First, this is not true (take $0\in Y$) and second it does not imply that $Y'$ does not vanish in any point: For any $l\in Spec X$ we need to specify an $b\in Y$, such that $l(b)\ne 0$! – Robert Rauch Dec 6 '12 at 13:49
Ok, $Y'$ clearly contains the constant functions, thus it does not vanish in any point -- stupid me. Your proof then works perfectly, thank you very much! – Robert Rauch Dec 6 '12 at 17:19

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