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

Fix a separable infinite-dimensional complex Hilbert space $\mathcal{H}$ and letwith its operator-norm topology. The odd topological K-theory group of a compact metric space is defined by continuous-map homotopy classesHomotopies use the same norm topology. Its finite-dimensional stable realization and group operation are established in the following parts, so this definition does not yet assume that homotopy classes form a group.

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 $g\mapsto \mathrm{diag}(g,1)$.

Use the stable finite-dimensional description. Define addition by block sum, $[u]+[v]=[\mathrm{diag}(u,v)]$. Stabilizations and homotopies preserve block sums, so this is well-defined. Associativity is immediate after identifying the three blocks; the identity class is zero.

Swapping two blocks is conjugation by a permutation unitary. Every finite-dimensional unitary is joined to the identity, since the matrix spectral theorem writes it as $e^{iA}$ with $A$ self-adjoint. Conjugation along that path gives a homotopy swapping the blocks, proving commutativity.

The inverse of $[u]$ is $[u^{-1}]$. To verify it rather than assuming a group structure, first stabilize to equal-sized blocks and putThe invertible homotopystarts at $\mathrm{diag}(u,u^{-1})$ and ends at the identity. Thus $\boxed{K^1(X)\text{ is an abelian group}.}$ The constructions are continuous in $x$ and remain valid for maps into $G{L}_N$; using polar factors also makes them homotopies in the unitary model.

Polar deformation reduces maps from the circle into $G{L}_N$ to maps into $U(N)$. Define their integer invariant by the winding number of the determinant. It is unchanged under homotopy and stabilization, and determinants multiply under block sum, so it gives a homomorphism $K^1(S^1)\to \mathbb{Z}$. Every integer occurs, via $z\mapsto \mathrm{diag}(z^m,1,\dots ,1)$.

To prove injectivity, first consider a map with determinant winding zero. Its determinant has a continuous logarithm $ih(z)$ on the circle. Multiplying the first row by $e^{-ith(z)}$ gives a homotopy into $SU(N)$ at $t=1$. If $N=1$ that endpoint is already the identity. For $N\ge 2$, $SU(N)$ is simply connected: $SU(2)\cong S^3$, and the last-column fibrationwith locally trivializations from orthonormal completion, together with its homotopy exact sequence, proves this inductively because the sphere has zero first and second homotopy groups. Thus the remaining circle map is null-homotopic. For a map of winding $m$, multiply by $\mathrm{diag}(z^{-m},1,\dots ,1)$ and apply the zero-winding argument. This provesFor higher spheres, stable unitary homotopy gives $K^1(S^n)={\pi }_n(U)$: a map can be normalized by multiplying by the inverse of its value at the basepoint, a free homotopy because the unitary group is path connected. A free homotopy between based maps can be made based by replacing $H(x,t)$ with $H(x_0,t{)}^{-1}H(x,t)$, without altering its endpoints. The Bott periodicity theorem, in its unitary form ${\pi }_{j+2}(U)\cong {\pi }_j(U)$ with ${\pi }_0(U)=0$ and ${\pi }_1(U)=\mathbb{Z}$, givesThe circle case was proved directly above; the higher-sphere values use the stated general periodicity theorem.