ExampleOfBanachAlgebraWhichIsNotACAlgebraForAnyInvolution 2007-07-28 10:55:39 2007-08-17 14:12:52 Example $C^*$-algebra finite dimensional $C^*$-algebras are semi-simple % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here Consider the Banach algebra $\mathcal{A}= \left\{ \begin{bmatrix} \lambda I_n & A \\ 0 & \lambda I_n \end{bmatrix} :\; \lambda \in \mathbb{C},\;\; A \in Mat_{n \times n}(\mathbb{C}) \right\}$ with the usual matrix operations and matrix norm, where $I_n$ denotes the identity matrix in $Mat_{n \times n}(\mathbb{C})$. {\bf Claim -} $\mathcal{A}$ is not a \PMlinkname{$C^*$-algebra}{CAlgebra} for any involution $*$. To prove the above claim we will give a \PMlinkescapetext{simple} proof of a more general fact about finite dimensional $C^*$-algebras, which clearly shows the \PMlinkescapetext{algebraic restrictions} for a Banach algebra to be a $C^*$-algebra for some involution. {\bf Theorem -} Every finite dimensional $C^*$-algebra is semi-simple, i.e. its Jacobson radical is $\{0\}$. {\bf Proof :} Let $\mathcal{B}$ be a finite dimensional $C^*$-algebra. Let $a$ be an element of $J(\mathcal{B})$, the Jacobson radical of $\mathcal{B}$. $J(\mathcal{B})$ is an ideal of $\mathcal{B}$, so $a^*a \in J(\mathcal{B})$. The Jacobson radical of a finite dimensional algebra is nilpotent, therefore there exists $n \in \mathbb{N}$ such that $(a^*a)^n=0$. Then, by the $C^*$ condition and the fact that $a^*a$ is selfadjoint, \begin{displaymath} 0 = \|(a^*a)^{2^n}\| = \|a^*a\|^{2^n} = \|a\|^{2^{n+1}} \end{displaymath} so $a = 0$ and $J(\mathcal{B})$ is trivial. $\square$ We now prove the above claim. {\bf Proof of the claim:} It is easy to see that $\left\{ \begin{bmatrix} 0 & A \\ 0 & 0 \end{bmatrix} : \; A \in Mat_{n \times n}(\mathbb{C}) \right\}$ is the only maximal ideal of $\mathcal{A}$. Therefore the Jacobson radical of $\mathcal{A}$ is not trivial. By the theorem we conclude that there is no involution $*$ that makes $\mathcal{A}$ into a $C^*$-algebra.$\square$ {\bf Remark -} It could happen that there were no involutions in $\mathcal{A}$ and so the above claim would be uninteresting. That's not the case here. For example, one can see that $[a_{i,j}] \longrightarrow [\overline{a}_{2n+1-j,2n+1-i}]$ defines an involution in $\mathcal{A}$ (this is just the \PMlinkescapetext{conjugate transpose} taken over the other diagonal of the matrix).