Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 106 / 5

A character of an algebra is a nonzero algebra homomorphism $\phi :A\to \mathbb{C}$. The character space isFor a unital algebra, $\phi (1)=1$. Moreover $\phi (a)\in {\sigma }_A(a)$, since applying $\phi$ shows that $a-\phi (a)1$ cannot be invertible. In a Banach algebra,Thus $\Vert \phi \Vert \le 1$, while $\phi (1)=1$ gives equality.

If $A$ is commutative and $\lambda \in {\sigma }_A(a)$, the proper ideal generated by $a-\lambda 1$ lies in a maximal ideal. The quotient by that maximal ideal is $\mathbb{C}$, and its quotient map is a character taking $a$ to $\lambda$. ThereforeSince every Banach-algebra element has nonempty spectrum, ${\Phi }_A$ is nonempty.

For $X=Y\oplus Z$, let $P,Q$ be the complementary coordinate projections. A character must send each idempotent to zero or one, and $P+Q=I$ forces their values to be different. Choose mutually inverse isomorphisms between $Y$ and $Z$ and regard them as off-diagonal operators $S,T$ on $X$. ThenMultiplicativity would give $\phi (P)=\phi (S)\phi (T)=\phi (T)\phi (S)=\phi (Q)$, a contradiction. Hence ${\Phi }_{\mathcal{B}(X)}$ is empty. This is the absence of characters on an operator algebra with isomorphic complementary summands.

If $a=a^{\ast }$ in a unital C*-algebra, then $e^{ita}$ is unitary for real $t$. If $\lambda \in \sigma (a)$, the spectral mapping theorem gives $e^{it\lambda }\in \sigma (e^{ita})$, whose modulus is one. Varying positive and negative $t$ forces $\mathrm{Im}\lambda =0$, so $\sigma (a)\subset \mathbb{R}$.

If $B\subseteq A$ is a unital C*-subalgebra and $x\in B$ is normal, spectral permanence holds. Indeed, when $x-\lambda$ is invertible in $A$, the positive normal elementhas spectrum bounded away from zero. Continuous functional calculus uniformly approximates its reciprocal by polynomials, placing the reciprocal in $B$. It follows that $(x-\lambda {)}^{-1}\in B$. Thus ${\sigma }_B(x)={\sigma }_A(x)$.

The Commutative Gelfand--Naimark theorem says that the Gelfand transform is an isometric unital star-isomorphismfor every commutative unital C*-algebra. If $a$ is positive, continuous functional calculus for the function $t\mapsto \sqrt{t}$ on $\sigma (a)\subset [0,\infty )$ gives a positive $b$ with $b^2=a$. Pointwise uniqueness in $C^{\ast }(1,a)\cong C(\sigma (a))$ gives the unique positive square root. This is the positive square root in a C*-algebra.

SetBoth are hermitian and $x=h+ik$. If also $x=h^{\prime }+i{k}^{\prime }$ with hermitian $h^{\prime },k^{\prime }$, taking adjoints and then adding or subtracting the two equations gives $h=h^{\prime }$ and $k=k^{\prime }$.

For hermitian $x$ with $\Vert x\Vert \le 1$, its spectrum lies in $[-1,1]$. Continuous functional calculus therefore shows that $1-x^2$ is positive. LetIt commutes with $x$, andsatisfies $u^{\ast }u=u{u}^{\ast }=x^2+v^2=1$. Thus $u$ is unitary and

The hypothesis is possible only for $t\ge 0$. For every $\lambda \in \sigma (x)\subset \mathbb{R}$,so $0\le \lambda \le 2t$. The spectral characterization of a positive element in a C*-algebra gives $x\ge 0$.

Positivity and $\Vert x\Vert \le t$ give $\sigma (x)\subset [0,t]$. HenceThe element $t1-x$ is hermitian, so its norm equals its spectral radius and is at most $t$.

Put $s=\Vert x\Vert +\Vert y\Vert$. Part 4 givesThereforePart 3 now proves that $x+y$ is positive.