2
$\begingroup$

Let $A$ be a unital $C^\ast$ algebra and let $B$ be a $\ast$ subalgebra such that $B \oplus \mathbb C = A$ and such that the unit in $B$, $1_B$, is not equal to the unit in $A$. I am trying to show:

If $\lambda \in \mathbb C$ is non-zero then $b-\lambda\cdot 1_B$ is invertible in $B$ if and only if $b-\lambda \cdot 1_A$ is invertible in $A$.

(see Murphy's book at the top of page 45).

I started the proof like this: Let $b-\lambda\cdot 1_B$ be invertible in $B$. Then it is also invertible in $A$. Let $a \in A$ denote its inverse. Then

$$ ba - \lambda 1_B a = 1_A$$

Now the goal is to find $c \in A$ such that $(b -\lambda 1_A)c = 1_A$. Somehow I have to show that $1_B a = 1_A c$ but I can't seem to do it. So this leads nowhere.

Can someone help me prove this please?

$\endgroup$
  • $\begingroup$ Assuming you mean $b = b 1_B \in B$, then \begin{align} (b-\lambda 1_B)c = 1_B & \Leftrightarrow (c \oplus (- \lambda)^{-1})\cdot (b-\lambda 1_B) \oplus (- \lambda) = 1_A. \end{align} $\endgroup$ – Michael Nov 8 '14 at 15:26
2
$\begingroup$

Assume $(b-\lambda 1_B)c=1_B$.

When we use $1_A=(0,1)$, $$ [(b, 0) - \lambda (0,1)](x,t)=(b,-\lambda)(x,t)=\left(bx-\lambda x+tb, -\lambda t\right). $$ Looking at the second coordinate, $t=-1/\lambda$. Then the equality in the first coordinate becomes $$ 0=bx-\lambda x -\frac1\lambda b=(b-\lambda 1_B)x-\frac1\lambda b, $$ So $x=\frac1\lambda cb=\frac1\lambda 1_B+c $. That is, in $A$ we have $(b-\lambda 1_A)^{-1}=\frac1\lambda cb=\frac1\lambda 1_B+c-\frac1\lambda 1_A$.

The computations above also allow us to do the converse: if $b-\lambda 1_A$ is invertible in $A$, then $x$ above exists. Now take $c=x-\frac1\lambda 1_B$, and a straightforward calculation shows that $(b-\lambda 1_B)c=1_B$.

$\endgroup$
  • $\begingroup$ Why can you use $(0,1) = 1_A$? Is the argument that every unital $C^\ast$-algebra is isomorphic to its unitisation? $\endgroup$ – user167889 Nov 10 '14 at 0:46
  • $\begingroup$ It's not about the unitization. In a unital C$^*$-algebra, the unit is identified with the complex number $1$. So if $A=B+\mathbb C$ with $B$ non-unital, $1_A$ is necessarily $0+1$. $\endgroup$ – Martin Argerami Nov 10 '14 at 1:20
  • $\begingroup$ How do I see this? Is it because a unital $C^\ast$-algebra contains $\mathbb C$ as a subalgebra? (but wouldn't then the unit in any commutative unital Banach algebra be identified with $1 \in \mathbb C$?) $\endgroup$ – user167889 Nov 10 '14 at 1:34
  • 1
    $\begingroup$ Of course. And you don't need commutative. $\endgroup$ – Martin Argerami Nov 10 '14 at 1:41
  • 1
    $\begingroup$ When I said "non-unital" I meant "without the unit of $A$" $\endgroup$ – Martin Argerami Nov 21 '14 at 3:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy