Complex Topological K Theory and Bott Periodicity — Examples

A complex bundle over a point is a finite-dimensional complex vector space, classified by its dimension. Whitney sum adds dimensions, so [F1] completes $\mathbb{N}$ to $\mathbb{Z}$, with the trivial line representing $1$.

Over $\mathrm{\varnothing }$, every bundle has empty total space and all are isomorphic, so the bundle monoid has one element and [F1] gives the zero group. For the point, the restriction map in [F2] is the identity of $K^0(\ast )$, so its kernel is zero. These calculations make no choices.

By [F3], the degree-zero group in step 1.1 repeats in every even degree. By [F4], $K^{-1}(\ast )={\tilde{K}}^0(S^1)=0$, and [F3] repeats this zero group in every odd degree. Thus the stated graded groups follow under AC.

By [F1], $K^0(S^2)$ is generated by $1$ and $\beta$, the only polynomial relation is ${\beta }^2=0$, and both generators are additively independent. This proves both displayed descriptions, including the zero reduced summand only when its integer coefficient is zero.

The class of $\gamma$ is $1+\beta$. Since ${\gamma }^{\ast }$ is the tensor inverse of $\gamma$, its class is the multiplicative inverse of $1+\beta$. The relation ${\beta }^2=0$ gives $(1+\beta )(1-\beta )=1-{\beta }^2=1$, so $[{\gamma }^{\ast }]=1-\beta$ and $[{\gamma }^{\ast }]-1=-\beta$.

By [F2], reversing the hemispheres changes the loop $z$ to $z^{-1}$, which clutches the dual line ${\gamma }^{\ast }$. Step 2.1 therefore proves that this convention reverses the Bott generator and leaves the square-zero presentation unchanged.

By [F1], ${\tilde{K}}^0(S^{2n})\cong \mathbb{Z}$ and ${\tilde{K}}^0(S^{2n+1})=0$ for every $n\ge 1$. Since each such sphere is nonempty, connected, and based, restriction to the basepoint is split by pullback along the collapse $S^m\to \ast$. Hence $K^0(S^m)\cong K^0(\ast )\oplus {\tilde{K}}^0(S^m)$. Substitution of [F2] gives the two displayed unreduced groups.

Suspending the coefficient calculation gives ${\tilde{K}}^q(S^n)\cong K^{q-n}(\ast )$. By [F2], this group is $\mathbb{Z}$ precisely when $q-n$ is even and is zero precisely when $q-n$ is odd, proving both exhaustive parity cases.

At $n=0$, the based sphere $S^0$ is the disjoint union of the basepoint and one further point. Its reduced group is the difference between the two coefficient copies and hence is one copy of $K^q(\ast )$, agreeing with step 1.2. This is why the unreduced formulas were stated only for $n\ge 1$.

For $r=0$, ${\mathbb{C}\mathbb{P}}^0=\ast$, its tautological line is trivial, and [F5] gives $K^0(\ast )=\mathbb{Z}$ and $K^1(\ast )=0$. Thus $x=0$ and the asserted presentation holds in the base case.

Fix $r\ge 1$ and assume that $K^1({\mathbb{C}\mathbb{P}}^{r-1})=0$ and that $1,x,\dots ,x^{r-1}$ is a basis there. By [F6], ${\mathbb{C}\mathbb{P}}^{r-1}\hookrightarrow {\mathbb{C}\mathbb{P}}^r$ has quotient $S^{2r}$. The long exact sequence [F1] and the even-sphere groups [F5] then give $K^1({\mathbb{C}\mathbb{P}}^r)=0$ and a short exact sequence $0\to {\tilde{K}}^0(S^{2r})\to K^0({\mathbb{C}\mathbb{P}}^r)\to K^0({\mathbb{C}\mathbb{P}}^{r-1})\to 0$. In particular, the restriction kernel is infinite cyclic.

We first construct the relative product used here. For a compact cofibration pair $(X,A)$, write $K^0(X,A)={\tilde{K}}^0(X\mathrm{/}A)$; its quotient is based and well-pointed. For two such pairs $(X,A)$ and $(Y,B)$, the natural homeomorphism $\frac{X\times Y}{X\times B\cup A\times Y}\cong (X\mathrm{/}A)\wedge (Y\mathrm{/}B)$ and the reduced external product [F3] define $K^0(X,A)\otimes K^0(Y,B)⟶K^0(X\times Y,X\times B\cup A\times Y)\mathrm{.}$ For two closed subspaces $A,B\subseteq X$, the relative diagonal ${\delta }_{A,B}:X\mathrm{/}(A\cup B)⟶(X\mathrm{/}A)\wedge (X\mathrm{/}B),[u]⟼[u]\wedge [u],$ is well-defined since a point of $A\cup B$ maps to the smash basepoint. Pullback along ${\delta }_{A,B}$ therefore gives $K^0(X,A)\otimes K^0(X,B)\to K^0(X,A\cup B)$. The square formed by ${\delta }_{A,B}$, the ordinary diagonal of $X$, and the quotient maps $X_{+}\to X\mathrm{/}A$, $X_{+}\to X\mathrm{/}B$, and $X_{+}\to X\mathrm{/}(A\cup B)$ commutes. Thus forgetting relative support sends this product to the ordinary product in $K^0(X)$; the same quotient-square argument, functoriality of pullback, and uniqueness in [F3] show that maps of pairs preserve these relative products. All pairs used below are finite ball or CW cofibration pairs. Now realize ${\mathbb{C}\mathbb{P}}^r$ as the scalar-orbit space of the boundary of $D_0^2\times \cdots \times D_r^2$, and let $C_i$ be the image of the face with its $i$th coordinate on $\mathrm{\partial }{D}_i^2$. Normalizing that coordinate to $1$ identifies $C_i$ with the product of the other $r$ disks, so $C_i$ is a closed $2r$-ball, ${\mathbb{C}\mathbb{P}}^r={\bigcup }_{i=0}^r{C}_i$, and $C_i\cap C_j=\mathrm{\partial }{C}_i\cap \mathrm{\partial }{C}_j$. The tautological line is trivial on $C_i$, hence exactness [F1] supplies a lift $x_i\in K^0({\mathbb{C}\mathbb{P}}^r,C_i)$ of $x$. For $C_0=D_1^2\times \cdots \times D_r^2$, restriction along the map of pairs $(C_0,{\mathrm{\partial }}_i{C}_0)\to ({\mathbb{C}\mathbb{P}}^r,C_i)$ sends $x_i$, up to the fixed disk-orientation sign, to the $i$th disk class clutched by $z$, hence to a generator by [F4]. Relative-product naturality now puts $x_1\cdots x_r$ in $K^0({\mathbb{C}\mathbb{P}}^r,C_1\cup \cdots \cup C_r)$, and the homeomorphism $C_0\mathrm{/}\mathrm{\partial }{C}_0\cong {\mathbb{C}\mathbb{P}}^r\mathrm{/}(C_1\cup \cdots \cup C_r)\cong (D^2\mathrm{/}\mathrm{\partial }{D}^2{)}^{\wedge r}$ identifies its restriction with the $r$-fold reduced external product of the disk generators. This is a generator by [F2]. Finally let $P={\mathbb{C}\mathbb{P}}^{r-1}$ be the standard subspace in the last $r$ coordinates. In Hatcher's ball model, $P\subseteq C_1\cup \cdots \cup C_r$ is disjoint from the interior of $C_0$, and the induced quotient map ${\mathbb{C}\mathbb{P}}^r\mathrm{/}P⟶{\mathbb{C}\mathbb{P}}^r\mathrm{/}(C_1\cup \cdots \cup C_r)$ is a homotopy equivalence. Its pullback therefore identifies the generator $x_1\cdots x_r$ with a generator of $K^0({\mathbb{C}\mathbb{P}}^r,P)$. The commuting forget-support maps send this class to the ordinary product $x^r$. Hence the image of $K^0({\mathbb{C}\mathbb{P}}^r,P)\to K^0({\mathbb{C}\mathbb{P}}^r)$, which is the restriction kernel from Step 2.1, is generated by the nonzero class $x^r$.

By the induction hypothesis in step 2.1, the short exact sequence there and the kernel generator in step 3.1 show that $1,x,\dots ,x^r$ is a basis on ${\mathbb{C}\mathbb{P}}^r$. Apply the independently proved step 3.1 with $r+1$: the class $x^{r+1}$ on ${\mathbb{C}\mathbb{P}}^{r+1}$ belongs to the kernel of restriction to ${\mathbb{C}\mathbb{P}}^r$, so its restriction, namely $x^{r+1}$ on ${\mathbb{C}\mathbb{P}}^r$, is zero. Evaluation therefore induces $\mathbb{Z}[x]\mathrm{/}(x^{r+1})\to K^0({\mathbb{C}\mathbb{P}}^r)$, and the two displayed bases make it an isomorphism. Together with $K^1({\mathbb{C}\mathbb{P}}^r)=0$ from step 2.1, this discharges the induction and proves the assertion for every $r=n$, including both the relation and the absence of further additive relations.

Form $E=(X_1\times \mathbb{C})⨿(X_2\times {\mathbb{C}}^2)$ with projection to $X$. Since $X_1$ and $X_2$ are disjoint open subsets, these product charts make $E$ a complex vector bundle whose fiber dimension is $1$ on $X_1$ and $2$ on $X_2$.

Degree-zero cohomology of a disjoint union is the product of the degree-zero groups, and the fiber-dimension function in step 1.1 corresponds to $(1_{X_1},2_{X_2})$. Therefore [F1] gives the displayed rank class.

If this class came from one integer $m$, its restrictions to both nonempty pieces would be the same constant $m$. Step 2.1 would force simultaneously $m=1$ and $m=2$, a contradiction. Hence a single global integer cannot encode rank in general.

Write $T{S}^2=\{(x,v)\in S^2\times {\mathbb{R}}^3:⟨x,v⟩=0\}$. Its normal line is trivialized by $x\mapsto (x,x)$. Using [F2], define $\mathrm{\Phi }:T{S}^2\oplus {\epsilon }_{\mathbb{R}}^1\to S^2\times {\mathbb{R}}^3$ by $\mathrm{\Phi }((x,v),(x,t))=(x,v+tx)$. The continuous inverse sends $(x,w)$ to $((x,w-⟨w,x⟩x),(x,⟨w,x⟩))$. Thus [F1] verifies the displayed stable bundle isomorphism fiber by fiber, including at $t=0$.

Suppose for contradiction that an actual bundle isomorphism $\mathrm{\Psi }:{\epsilon }_{\mathbb{R}}^2\to T{S}^2$ exists.

The constant section $x\mapsto (x,(1,0))$ of ${\epsilon }_{\mathbb{R}}^2$ is nowhere zero. Composing it with $\mathrm{\Psi }$ gives a continuous nowhere-zero section of $T{S}^2$, hence a nowhere-zero tangent vector field in the sense of [F1].

This contradicts [F3]. Therefore $T{S}^2$ is not actually trivial, while step 1.1 proves that it becomes trivial after adding one trivial real line. The two bundles in the false claim are $T{S}^2$ and ${\epsilon }_{\mathbb{R}}^2$, with the same summand ${\epsilon }_{\mathbb{R}}^1$ added to both.