Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 8 / 7 / b / Solution

Let $u:X\to U_{\mathcal{K}}$ be continuous. The family $\{u(x)-I:x\in X\}$ is norm compact in the compact-operator ideal. For any $\epsilon >0$ there is one finite-rank orthogonal projection $P$ withTo prove it, take a finite $\epsilon$-net of compact operators, approximate each by a finite-rank operator, and include both their ranges and adjoint ranges in $P\mathcal{H}$. Contractivity of compression transfers the estimate to the whole family.

Take $\epsilon <1$ and set $g(x)=I+P(u(x)-I)P$. It is invertible because it is within distance one of the unitary $u(x)$. It has the form $g_V(x)\oplus I$ on $V\oplus V^{\perp }$, where $V=P\mathcal{H}$ is finite-dimensional and $g_V:X\to GL(V)$ is continuous. The straight segment from $u$ to $g$ stays invertible; taking its unitary polar factors supplies a homotopy in $U_{\mathcal{K}}$, since inverse square root depends continuously on the positive invertible factors and preserves identity modulo compacts.

Conversely a map $g:X\to GL(V)$ defines such a class by the continuous deformation $g(g^{\ast }g{)}^{-t/2}$ to its unitary polar factor, then extension by the identity on $V^{\perp }$. Apply the same uniform finite-rank approximation to a homotopy on the compact space $X\times [0,1]$. Choosing $P$ to include the original endpoint supports keeps those endpoints unchanged under compression. This proves that two finite-dimensional maps give the same class exactly when, after adding identity blocks, they are homotopic through a common finite-dimensional general linear group. Hencewith stabilization $g\mapsto \mathrm{diag}(g,1)$.