Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 10 / 2

Write $X=I+E$. Use a submultiplicative Sobolev algebra norm for which the Fourier projections are contractive, and its compatible matrix/column norms. The numeric hypothesis $\Vert E\Vert <1$ is understood in this algebra norm; the unscaled Fourier Sobolev norm need not have multiplication constant one.

Let $A_{>0}$ and $A_{<0}$ denote the strictly positive and strictly negative Fourier subspaces. The equation $y_{+}=-P_{>0}(E+E{y}_{+})$ is a contraction on $M_n(A_{>0})$, so it has a unique solution. Put $Y_{+}=I+y_{+}$ and $Z_{-}=X{Y}_{+}$. The equation says $Z_{-}\in M_n(A_{-})$, and $Y_{+}(0)=I$. Similarly the contraction $v_{-}=-P_{<0}(E+v_{-}E)$ gives $V_{-}=I+v_{-}$ with $Z_{+}=V_{-}X\in M_n(A_{+})$. No invertibility of these solutions has yet been assumed.

Now $V_{-}{Z}_{-}=Z_{+}{Y}_{+}$ belongs to both Fourier subalgebras, hence is a constant matrix $C$. Taking constant coefficients on either side shows $C=Z_{-}(0)=Z_{+}(0)$. We claim $C$ is invertible. If $Cv=0$, then the Hardy compression $T_X=P_{\ge 0}X$ annihilates $Y_{+}v$, because $P_{\ge 0}{Z}_{-}v=Cv=0$. But $T_X=I+P_{\ge 0}E$ on the Hardy Sobolev space has inverse given by a Neumann series, since the perturbation has norm less than one. Thus $Y_{+}v=0$, and its constant coefficient is $v$, forcing $v=0$. Finite-dimensionality proves the claim.

Consequently $C^{-1}{Z}_{+}$ is a left inverse of $Y_{+}$, and $C^{-1}{V}_{-}$ is a left inverse of $Z_{-}$. Matrix entries are functions in a commutative algebra, so pointwise finite-dimensional inversion makes these left inverses two-sided inverses in their respective Fourier subalgebras. SetThen $X=X_{-}{X}_{+}$, both factors are invertible in the stated algebras, and $X_{+}(0)=I$. If two normalized factorizations exist, ${\tilde{X}}_{-}^{-1}{X}_{-}={\tilde{X}}_{+}{X}_{+}^{-1}$ lies in $A_{-}\cap A_{+}$ and is constant; evaluating the analytic side at zero makes it $I$. This proves uniqueness. The argument is near-identity Sobolev loop factorization and, crucially, proves invertibility rather than treating the contraction solution itself as automatically invertible.

Give $G(H)$ its natural trace-class topology: differentiability means $F(t+h)=F(t)+h\dot{F}(t)+o_1(h)$, with remainder small in trace norm. The Fredholm determinant is the continuous extension of finite-rank determinants; it is multiplicative by finite-rank approximation and satisfiesnear zero. For example its exterior-power series bounds the remainder by $e^{\Vert K{\Vert }_1}-1-\Vert K{\Vert }_1$, proving that estimate directly.

Factor $F(t+h)=F(t)[I+hF(t{)}^{-1}\dot{F}(t)+o_1(h)]$. Multiplicativity and the preceding expansion giveThe determinant is nonzero for invertible $F$, so division yields the required logarithmic derivative. Differentiability only in the bounded-operator norm would not justify this trace formula; the specified topology is essential.

Let $C=[A,B]$ be trace class and set $F(t)=e^{tA}{e}^B{e}^{-tA}{e}^{-B}$. The integral identity $e^{B}A{e}^{-B}-A={\int }_0^1{e}^{sB}[B,A]e^{-sB}ds$ is trace-norm convergent. It shows that this difference is trace class. The analogous Duhamel identity then makes $F(t)-I$ trace class and differentiable in that norm. Hence $F(t)\in G(H)$.

Compute the right logarithmic derivative without taking separate traces of $A$:Trace invariance under bounded invertible similarity and the integral identity give $\mathrm{Tr}(\dot{F}{F}^{-1})=-{\int }_0^1\mathrm{Tr}[B,A]ds=\mathrm{Tr}[A,B]$. Cyclicity with a trace-class factor makes this equal to the trace in part (b). As $F(0)=I$, integrating the scalar differential equation givesThis Fredholm determinant of an exponential commutator can be nontrivial in infinite dimension: it is not legitimate to set the trace of a trace-class commutator to zero when $A,B$ themselves need not be trace class.

Split $f=f_{-}+a_0+f_{+}$ into strictly negative and strictly positive Fourier parts, and let $A=T(f_{-})$, $B=T(f_{+})$. The Toeplitz multiplication rule is $T(g)T(h)=T(gh)$ when $g$ is coanalytic or $h$ is analytic. It follows directly because the discarded negative Fourier part cannot return to the Hardy space in those cases. Thus exponentiation within either subalgebra is exact, andThe constant terms cancel. If $S=T(z)$ is the unilateral shift, $A={\sum }_{m>0}{a}_{-m}(S^{\ast }{)}^m$ and $B={\sum }_{n>0}{a}_n{S}^n$. The commutator $[(S^{\ast }{)}^m,S^n]$ is finite rank, with trace $n$ when $m=n$ and zero otherwise. For equal powers it is the projection onto the first $n$ basis vectors; for unequal powers its matrix has no diagonal entries. Its trace norm is bounded by $2\min (m,n)$. Smoothness of $f$ makes the weighted double coefficient sum finite, so the commutator series converges in trace norm andApplying part (c) proves the Toeplitz exponential determinant identityIn particular the product is an invertible identity-plus-trace-class operator, so the determinant in the formula is well-defined.