Topological K-theory ($K^{\ast }(X)$)

Complex topological K-theory is the generalized cohomology theory built from stable equivalence classes of complex vector bundles. Its degree-zero group is the Grothendieck group $K^0(X)$, and its reduced theory is denoted ${\tilde{K}}^{\ast }(X)$.

For compact Hausdorff topological spaces $X,Y$ and complex vector bundles $E\to X$, $F\to Y$, the external product in Topological K-theory iswhere $p_X,p_Y$ are the two projections. Pullback places both bundles over the same product space, so their fibrewise tensor product is defined. The construction respects direct sums and therefore extends bilinearly to differences of bundle classes in the Grothendieck group.

For example, external multiplication by the trivial line bundle on a point just pulls the other class back along $X\times \{\ast \}\to X$. When $X=Y$, pulling the external product back along the diagonal gives the internal ring product on $K^0(X)$. The external product combines information from separate spaces; its existence alone does not assert a Künneth isomorphism for every pair of spaces.

For compact metric $X$, odd topological K-theory is the homotopy set of norm-continuous maps into unitaries equal to the identity modulo compact operators on a fixed separable infinite-dimensional Hilbert space. Uniform finite-rank approximation over $X$ and over homotopies gives the equivalent stable model ${\mathrm{colim}}_N[X,G{L}_N(\mathbb{C})]$. Block sum gives an abelian group: a finite-dimensional unitary path swaps blocks, and a two-block rotation contracts $\mathrm{diag}(u,u^{-1})$ to the identity. On the circle the determinant winding identifies $K^1(S^1)$ with $\mathbb{Z}$; Bott periodicity gives $K^1(S^n)=\mathbb{Z}$ in odd dimensions and zero in even dimensions.

If $L$ is the complexification of a real vector bundle of the tautological bundle and $\mu =[L]-1$, thenRealizing Real projective space as the circle bundle of the square of the tautological bundle on Complex projective space gives this through the K-theory Gysin sequence of a sphere bundle.

For a finite covering map $p:Y\to X$, define the fiber of the transferred vector bundle by $(p_{!}V{)}_x={⨁}_{y\in p^{-1}(x)}{V}_y$. Local trivializations of the covering map make this a vector bundle, and compatibility with direct sums extends it to Topological K-theory.

For a finite covering map $p:Y\to X$, the K-theory transfer of a finite covering obeysOn each fibre, this is the natural linear isomorphism distributing a tensor product over the direct sum of the fibers of $b$. It gives the formula for vector bundles and hence for their virtual differences.

Now suppose $X$ is a finite CW complex and the covering map has constant degree $n\ge 1$. The class $p_{!}(1)$ is the permutation vector bundle of rank $n$, so $\eta =p_{!}(1)-n$ has rank zero. By nilpotence of rank-zero K-theory classes, some power of $\eta$ is zero. After localization of a ring at $n$, the class $p_{!}(1)=n+\eta$ is a unit, with inverse given by a finite geometric sum in $-\eta /n$. If $p^{\ast }a=0$, the projection formula gives $a{p}_{!}(1)=0$, hence $a=0$ after this localization. Thus pullback is injective after inverting $n$. The finite-CW hypothesis supplies the nilpotence step.

The rank of a vector bundle is locally constant on the base. Taking the difference of the ranks extends to the Grothendieck group of vector bundles, giving the rank map in Topological K-theory. For a finite CW complex, its kernel consists of classes of rank zero on every connected component.

For a finite CW complex, every element in the kernel of the rank map in topological K-theory is a nilpotent element. One proof adjoins one positive-dimensional cell at a time: the restriction kernel is square-zero by the relative product in topological K-theory. For connected spaces the rank-zero kernel is Reduced topological K-theory; for disconnected spaces the kernel of restriction to just one basepoint can contain nonzero idempotents.

For a nonempty subcomplex $A\subset X$, relative Topological K-theory is $K^i(X,A)={\tilde{K}}^i(X/A)$, where the collapsed subcomplex is the basepoint. It fits into the K-theory six-term exact sequence.

The relative topological K-theory productis induced by the reduced diagonal $X/A\to (X/A)\wedge (X/A)$. If $X/A$ is a positive-dimensional sphere, this diagonal is null-homotopic, so the kernel of $K^0(X)\to K^0(A)$ is square-zero.

For a pair $(X,A)$ of CW complexes, relative topological K-theory and Bott periodicity give the cyclic exact sequence

The Grothendieck group of a commutative monoid is its universal abelian-group completion. For vector bundles under direct sum, its elements are formal differences $[E]-[F]$ modulo stabilization.

For a based compact space, reduced K-theory is the kernel of restriction to the basepoint. It satisfies

Multiplication by the Bott element $\beta \in {\tilde{K}}^0(S^2)$ gives the Bott isomorphism

Let $H$ be the tautological bundle on ${\mathbb{CP}}^1\cong S^2$. One sign convention for the Bott element iswhere $[1]$ is the rank-one trivial vector bundle. It has rank zero, and its Chern character has degree-two term $c_1(H)$, a generator up to sign. Choosing the dual line bundle reverses that sign.

The $n$-fold reduced external product in topological K-theory gives a generator ofThis product uses the smash product of different sphere factors. It is distinct from taking exterior powers of a vector bundle over one fixed base. Multiplication by the Bott element realizes Bott periodicity.^[ref]

Complex Bott periodicity gives

If a finite CW-complex $Y$ has only even-dimensional cells, thenand $K^0(Y)$ is free abelian, with one generator for each cell. This follows by induction from the six-term exact sequence for adjoining a wedge of even-dimensional cells.

If $Y$ is a finite even-cell complex, the exterior product is an isomorphismCellular induction proves this because $K^0(Y)$ is free and $K^{-1}(Y)=0$.

If $t=1-[\overline{\gamma }]$ for the tautological complex line bundle, then

Let $E_0,E_1$ be complex vector bundles of rank $d\ge 1$ over a finite CW complex $X$ with $\dim X\le 2d$. Iffor some $r$, then $E_0\cong E_1$: in this dimension range, stable isomorphism already determines the actual vector bundle.

The stabilization $BU(d)\to BU$ of the classifying space for rank-$d$ complex vector bundles is an n-connected map with $n=2d+1$. One sees the range from the fibration with fiber $U(d+1)/U(d)\cong S^{2d+1}$ and then iterates stabilization. The vanishing of the fiber's homotopy groups through degree $2d$ allows a stable homotopy between classifying maps to lift over $X\times [0,1]$, relative to its ends. Thus the original classifying maps are homotopic and the bundles are isomorphic.

The rank bound is useful when invariants such as the Chern character first identify stable classes: it supplies the final cancellation step in the stated rank and dimension range.

Two rank-$d$ complex vector bundles over ${\mathbb{CP}}^d$ are isomorphic exactly when their Chern classes agree. Equality of Chern classes gives equality of their Chern characters; the Chern character is injective here because $K^0({\mathbb{CP}}^d)$ is torsion-free. The bundles are therefore stably isomorphic, and stable cancellation for complex vector bundles applies because ${\mathbb{CP}}^d$ has real dimension $2d$.

For a self-map $f:Z\to Z$, the mapping torus has an exact sequence containingIf $K^{-1}(Z)=0$, then $K^0(T_f)\cong \ker (1-f^{\ast }:K^0(Z)\to K^0(Z))$.

For the factor swap on ${\mathbb{CP}}^2\times {\mathbb{CP}}^2$, the invariant subgroup ofhas basis $1,xy,x^2{y}^2,x+y,x^2+y^2,x{y}^2+x^{2}y$. Hence the mapping torus has $K^0\cong {\mathbb{Z}}^6$.

For a complex vector bundle $E\to X$, there is a map $p:F(E)\to X$ such that $p^{\ast }$ is injective on cohomology and $p^{\ast }E$ splits as a sum of line bundles. Symmetric identities in the Chern roots can therefore be proved after this pullback.

If a pulled-back complex vector bundle splits as $L_1\oplus \cdots \oplus L_r$, its formal Chern roots are $x_i=c_1(L_i)$. The Chern classes are the elementary symmetric polynomials in these roots.

If the formal Chern roots of $E$ are $x_1,\dots ,x_r$, thenIt extends to virtual bundles and is a ring homomorphism because direct sum joins root lists while tensor product replaces them by all sums $x_i+y_j$.

Under the natural identificationthe Chern character maps ${\tilde{K}}^0(S^{2n})$ into integral cohomology. A Bott generator is an exterior product of $n$ degree-two Bott elements, whose Chern characters multiply to an integral top-dimensional generator.

For a complex vector bundle $E\to S^{2n}$,is divisible by $(n-1)!$. Since the lower Chern classes vanish, the Newton identity givesand the Chern character on an even-dimensional sphere is integral.

For a complex vector bundle $E\to X$, the K-theory Thom classgenerates the Thom isomorphism $K^i(X)\cong {\tilde{K}}^i(\mathrm{Th}(E))$.

Pulling the K-theory Thom class back along the zero section gives

Let $\gamma$ be the tautological line bundle and put $t=1-[\gamma ]$. The Euler sequence on complex projective space givesHenceand evaluating at $s=-1$ in $K^0({\mathbb{CP}}^n)=\mathbb{Z}[t]/(t^{n+1})$ gives

The cofibration $S(E)\to D(E)\to \mathrm{Th}(E)$ and the Thom isomorphism give

Let $L$ be the complexification of a real vector bundle of the tautological bundle over Real projective space ${\mathbb{RP}}^{2n+1}$. For $k\ge 1$, the K-theory Euler class of $L^{\oplus k}$ is $-2^{k-1}([L]-1)$, andThe K-theory Gysin sequence of a sphere bundle gives an extension by $\mathbb{Z}$, which splits as an exact sequence of abelian groups.

The K-theory Gysin sequence of a sphere bundle and $e^K(T{\mathbb{CP}}^n)=(n+1)t^n$ giveand

For $E=\gamma \oplus \gamma$ over ${\mathbb{CP}}^n$ and $t=1-[\overline{\gamma }]$, the K-theory Euler class is $t^2$. Hencefor $n\ge 1$.

For $n=0$, the base is a point, the sphere bundle is $S^3$, and its odd K-theory is $\mathbb{Z}$.

The Adams operations are natural ring endomorphisms of complex K-theory characterized on line bundles by

If $u\in {\tilde{K}}^0(S^{2m})$ is a Bott element generator, then

Let ${\mathbb{CP}}_k^{k+n}={\mathbb{CP}}^{k+n}/{\mathbb{CP}}^{k-1}$. If the bottom-cell inclusion $S^{2k}={\mathbb{CP}}_k^k\to {\mathbb{CP}}_k^{k+2}$ admits a retraction, then $24$ divides $k$. Writing $x=[\overline{\gamma }]-1$, a retracted Bott generator has the form $x^k+a{x}^{k+1}+b{x}^{k+2}$. Comparing it with its image under ${\psi }^2(x)=2x+x^2$ gives $a=-k/2$ and $b=k(3k+5)/24$; integrality forces both $3$ and $8$ to divide $k$.

The cannibalistic class is defined byIt satisfies ${\rho }^k(E\oplus F)={\rho }^k(E){\rho }^k(F)$. For a line bundle,

For $p_{!}:K^{-1}(S(E))\to K^0(X)$ in the K-theory Gysin sequence,This follows by applying ${\psi }^k$ to $\delta x={\lambda }_E{p}_{!}(x)$.

Choose $a,b$ with $p_{!}(a)=t^{n-1}$ and $p_{!}(b)=t^n$, where $t=1-[\overline{\gamma }]$. Then