Generalized Cohomology and the Atiyah Hirzebruch Spectral Sequence — Examples

The transition functions of a real line bundle take values in $\{\pm 1\}$, so the transition functions of $\lambda {\otimes }_{\mathbb{R}}\lambda$ are squares of $\pm 1$, hence equal to $1$, and $\lambda {\otimes }_{\mathbb{R}}\lambda$ is trivial.

The $K$-AHSS of ${\mathbb{R}\mathbb{P}}^r$ has nonzero entries only in even coefficient rows. A differential $d_s$ with $s$ even has odd target row and vanishes. Let $s$ be odd and let its source column satisfy $p\ge 1$. If $p=r$, as can occur in the top integral column when $r$ is odd, then the target column $r+s$ is outside the complex. If $0<p<r$, a nonzero source has $p$ even by [A3], so its target column $p+s$ is odd and the target vanishes unless $p+s=r$; in that remaining case the source is a torsion group while $H^r({\mathbb{R}\mathbb{P}}^r;\mathbb{Z})=\mathbb{Z}$ is torsion-free, so the differential vanishes as well. Finally, a differential with source in the $0$-column cannot be hit, and the $0$-column edge identifies $E_{\mathrm{\infty }}^{0,q}\cong h^q(\mathrm{p}\mathrm{t})=\mathbb{Z}=E_2^{0,q}$, so those differentials vanish too. Hence all differentials vanish and $E_2=E_{\mathrm{\infty }}$, so the associated graded of $K^0({\mathbb{R}\mathbb{P}}^r)$ consists of the integral cohomology of the projective space in even degrees, with the degree-two class in filtration two.

Complexifying the triviality of the transition functions gives $\xi {\otimes }_{\mathbb{C}}\xi \cong {\epsilon }_{\mathbb{C}}^1$; writing $\alpha =[\xi ]-1$ in the ring of [A2] therefore gives $(1+\alpha {)}^2=1$, that is ${\alpha }^2=-2\alpha$, and multiplying repeatedly gives ${\alpha }^k=(-2{)}^{k-1}\alpha$ for every $k\ge 1$.

If $m=0$, both ${\mathbb{R}\mathbb{P}}^0$ and ${\mathbb{R}\mathbb{P}}^1\cong S^1$ have trivial reduced $K^0$ by [A5], so $\alpha =0$ has the asserted order $1$. Now suppose $m\ge 1$. On ${\mathbb{R}\mathbb{P}}^{2m}$, [A5] gives ${\alpha }^m\mathrm{\ne }0$ and ${\alpha }^{m+1}=0$. On ${\mathbb{R}\mathbb{P}}^{2m+1}$, the restriction isomorphism of [A5] carries the tautological $\alpha$ to the even-dimensional one, so the same two power statements hold there as well. With ${\alpha }^2=-2\alpha$ from step 2.1, consequently $2^m\alpha =(-1{)}^m{\alpha }^{m+1}=0$ while $2^{m-1}\alpha =(-1{)}^{m-1}{\alpha }^m\mathrm{\ne }0$. Thus $\alpha$ has exact order $2^m$ and generates ${\tilde{K}}^0({\mathbb{R}\mathbb{P}}^r)$; the two $K^1$ calculations are the final clauses of [A5].

Steps 1.2, 2.1 and 3.1 give the asserted relations, the exact order of $\alpha$, the cyclicity of the reduced group and the two odd-group computations.

Lift $c$ coefficientwise to the integral cochain $\tilde{c}$ given by the least nonnegative residues. Then $\delta \tilde{c}$ is coefficientwise divisible by $2$ because $\delta c=0$ modulo two, so there is a unique integral cochain $a$ with $\delta \tilde{c}=2a$. Moreover $\delta a=0$, since $2\delta a={\delta }^2\tilde{c}=0$ in the torsion-free group of integral cochains.

Reducing $\tilde{c}$ modulo four gives a $\mathbb{Z}\mathrm{/}4$-cochain whose image modulo two is $c$ and whose coboundary is the reduction of $2a$, namely $2a$ modulo four; hence the mod-two Bockstein of the sequence $0\to {\mathbb{F}}_2\to \mathbb{Z}\mathrm{/}4\to {\mathbb{F}}_2\to 0$ assigns to $x$ the class of $\delta (\tilde{c})\mathrm{/}2=a$ modulo two.

The residue construction descends without any choice. If $c^{\mathrm{\prime }}=c+\delta u$ modulo two, let ${\tilde{c}}^{\mathrm{\prime }},\tilde{c},\tilde{u}$ be their integral residue lifts. The integral cochain ${\tilde{c}}^{\mathrm{\prime }}-\tilde{c}-\delta \tilde{u}$ reduces to zero modulo two, so it equals $2h$ for a unique integral cochain $h$. If $\delta {\tilde{c}}^{\mathrm{\prime }}=2a^{\mathrm{\prime }}$, applying $\delta$ gives $2a^{\mathrm{\prime }}=2a+2\delta h$, hence $a^{\mathrm{\prime }}=a+\delta h$. Thus $[a]$ depends only on $x$, using no simultaneous selection from fibres. Reducing $a$ modulo two gives the mod-two Bockstein by step 1.2, so ${\rho }_2{\beta }_{\mathbb{Z}}(x)=\beta (x)=S{q}^1(x)$ by [F2].

Steps 1.1, 1.2 and 2.1 prove the stated identity for every space and every degree; both the residue lift and the comparison cochain $h$ are uniquely specified coefficientwise, so no choice principle is used.

Reduce the cellular incidence numbers in [F3] modulo two. The cellular chain complex of ${\mathbb{R}\mathbb{P}}^m$ is then ${\mathbb{F}}_2$ in degrees $0,\dots ,m$, zero elsewhere, with zero differential. Thus its mod-two singular homology is one-dimensional in that range and zero above $m$. Field duality gives the same dimensions and vanishing for cohomology. By [F2], restriction sends $a^j$ to the nonzero power $a_m^j$ for $0\le j\le m$; restriction preserves products. Higher powers vanish by the just-proved cohomological vanishing. Therefore the finite ring is exactly ${\mathbb{F}}_2[a_m]\mathrm{/}(a_m^{m+1})$.

For a degree-one class $t$, [F5] gives $S{q}^{0}t=t$, $S{q}^{1}t=t^2$, and $S{q}^{i}t=0$ for $i>1$. Repeated Cartan says that in $S{q}^i(t^j)$ only choices of $i$ among the $j$ factors to receive $S{q}^1$ contribute; each contributes $t^{j+i}$. Thus $S{q}^i(t^j)=\left(\genfrac{}{}{0px}{}{j}{i}\right)t^{j+i}$, with the binomial coefficient reduced modulo two. This is a finite product computation, valid directly for the classes $u,v$ on $Y$. In particular $S{q}^1(t^2)=0$, $S{q}^2(t^2)=t^4$, and $S{q}^1(t^4)=0$. [F5, algebra] 2.1 The homology groups of both finite projective factors are finite free over ${\mathbb{F}}_2$ by step 1.1, so the full hypothesis of [F4] holds. Its cross-product ring isomorphism gives $H^{\ast }(Y;{\mathbb{F}}_2)={\mathbb{F}}_2[u,v]\mathrm{/}(u^3,v^5)$, with basis $u^i{v}^j$ for $0\le i\le 2$, $0\le j\le 4$. In particular $uv$, $u{v}^4$ and $u^2{v}^4$ are nonzero. The two summands $u^{2}v$ and $u{v}^2$ are distinct basis elements.

Cartan gives $S{q}^1(uv)=u^{2}v+u{v}^2$. Also $S{q}^2(u^{2}v)=u^{4}v+0+0=0$ since $u^3=0$, whereas $S{q}^2(u{v}^2)=0+0+u{v}^4=u{v}^4$. Additivity therefore gives $S{q}^2(u^{2}v+u{v}^2)=u{v}^4$.

By [F1], ${\rho }_{2}z=S{q}^1(uv)=u^{2}v+u{v}^2$, which is nonzero by step 2.1, so $z$ is also nonzero. Step 3.1 gives $S{q}^2{\rho }_{2}z=u{v}^4$, and the specified definition implies $S{q}_{\mathbb{Z}}^3(z)={\beta }_{\mathbb{Z}}(u{v}^4)$. Finally ${\rho }_2{\beta }_{\mathbb{Z}}(u{v}^4)=S{q}^1(u{v}^4)=u^2{v}^4+uS{q}^1(v^4)=u^2{v}^4\mathrm{\ne }0$ by steps 2.1 and 1.2 and Cartan. A zero integral class would have zero reduction, so this proves the claimed integral nonvanishing without asserting injectivity of reduction.

By [A3] the $E_2$ page has $E_2^{0,q}=K^q(\ast )$ and $E_2^{n,q}=K^q(\ast )$ for every even $q$, and vanishes in all other positions.

For $n$ even, consider a differential $d_r:E_r^{p,q}\to E_r^{p+r,q-r+1}$ with source in an even coefficient row $q$ at a nonzero column $p\in \{0,n\}$. If $r$ is even then the target row $q-r+1$ is odd, so the target lies in a vanishing coefficient row. If $r$ is odd then the target column $p+r$ is odd, hence is neither $0$ nor $n$ when $p=0$ (the number $n$ being even) and exceeds $n$ when $p=n$; in both cases the target column lies outside the support $\{0,n\}$ of $H^{\ast }(S^n;\mathbb{Z})$, and the parity of the target row is irrelevant. In either case the target vanishes, so every differential is zero.

Suppose $n$ is odd. If $n=1$, every differential has $r\ge 2$ by [A1], so a source in column $0$ or $1$ has target column $p+r>1$; hence every target is zero. Now let $n\ge 3$. The only possibly nonzero differentials are $d_n:E_n^{0,q}\to E_n^{n,q-n+1}$ on even rows $q$. There is no incoming differential at column $0$, so the surviving subgroup there is $\ker d_n$. On the other hand, the edge quotient $E_{\mathrm{\infty }}^{0,q}=F^0{K}^q(S^n)\mathrm{/}{F}^1{K}^q(S^n)$ identifies with the image of restriction to the basepoint. For even $q$, [A4] and Bott periodicity give $F^1={\tilde{K}}^q(S^n)=0$, while pullback along $S^n\to \ast$ splits restriction, so that image is all of $K^q(\ast )=\mathbb{Z}$. Thus $\ker d_n=\mathbb{Z}$ inside its source $\mathbb{Z}$, forcing $d_n=0$.

Steps 2.1 and 2.2 show that all differentials vanish and $E_2=E_{\mathrm{\infty }}$. If $n$ is odd, each total-degree diagonal has only one nonzero term, so there is no extension problem. If $n$ is even, an even total degree $t$ has two graded pieces, at $p=0$ and $p=n$, and the filtration gives $0\to F^n{K}^t(S^n)\cong \mathbb{Z}\to K^t(S^n)\to K^t(\ast )\cong \mathbb{Z}\to 0$. The quotient map is restriction to the basepoint and is split by pullback along $S^n\to \ast$, since the composite $\ast \to S^n\to \ast$ is the identity. Thus $K^t(S^n)\cong \mathbb{Z}\oplus \mathbb{Z}$ in even degree, with the reduced summand $F^n\cong {\tilde{K}}^t(S^n)\cong \mathbb{Z}$ from [A4]; odd total degrees vanish. Hence the only two-piece extension is split, while the odd-dimensional cases have a single graded piece.

Steps 2.1, 2.2 and 3.1 verify the displayed reduced groups and show that the sphere $K$-AHSS has no nonzero differential and no nontrivial extension.

By [F2], the integral cellular cochains are $\mathbb{Z}$ in degrees $0,2,\dots ,2n$ and zero elsewhere; every cellular coboundary is zero. By [F3], $H^p(X;\mathbb{Z})$ has exactly these groups. This calculation needs no cohomology ring presentation.

Construct the additive skeletal sequence directly using the actual $K$-pair sequences of [F1]. In homological indexing put $D_{a,b}=K^{-a-b-1}(X^{-a-1})$, $E_{a,b}=K^{-a-b}(X^{-a},X^{-a-1})$; let $i$ be restriction, $j$ the pair connecting map, and $k$ the forget-relative map. The pair long exact sequences give $\mathrm{im}i=\ker j$, $\mathrm{im}j=\ker k$, $\mathrm{im}k=\ker i$ with exactly the shifts in [F4]. Thus [F4] gives a spectral sequence. Reindex $(p,q)=(-a,-b)$ to obtain $E_1^{p,q}=K^{p+q}(X^p,X^{p-1})$ and $d_r:(p,q)\mapsto (p+r,q-r+1)$.

We compute the ring independently of the spectral sequence. Induct on $r$. For $r=0$, the tautological line on the point is trivial, so $\alpha =0$ and $K^0({\mathbb{C}\mathbb{P}}^0)=\mathbb{Z}$. Induct simultaneously that $K^1({\mathbb{C}\mathbb{P}}^{r-1})=0$ and that $1,\alpha ,\dots ,{\alpha }^{r-1}$ is a basis there. The odd part of the pair sequence and the odd sphere coefficient give $K^1({\mathbb{C}\mathbb{P}}^r)=0$. The cofibration ${\mathbb{C}\mathbb{P}}^{r-1}\subset {\mathbb{C}\mathbb{P}}^r$ has quotient $S^{2r}$ by [F2] and [F7]; [F1] and the even-sphere coefficients give a short exact sequence $0⟶{\tilde{K}}^0(S^{2r})⟶K^0({\mathbb{C}\mathbb{P}}^r)⟶K^0({\mathbb{C}\mathbb{P}}^{r-1})⟶0.$ It remains to identify the kernel generator. 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 the $i$th coordinate on $\mathrm{\partial }{D}_i^2$. Normalizing that coordinate to $1$ identifies $C_i$ with the product of the other disks, so $C_i$ is a closed $2r$-ball, ${\mathbb{C}\mathbb{P}}^r={\bigcup }_i{C}_i$, and $C_i\cap C_j=\mathrm{\partial }{C}_i\cap \mathrm{\partial }{C}_j$. The radial collars of the polydisk faces descend through scalar multiplication and give neighborhood deformation retractions for every $C_i$ and every finite union used below. Thus their inclusions are closed cofibrations of compact Hausdorff CGWH spaces, and the corresponding quotient basepoints are well-pointed. The compact-cofibration exact sequence in [F1], rather than the finite-CW-pair clause, therefore applies. The tautological line has the section obtained by setting its $i$th coordinate equal to $1$ on $C_i$, so exactness gives a relative lift ${\alpha }_i\in K^0({\mathbb{C}\mathbb{P}}^r,C_i)$ of $\alpha$. For a compact Hausdorff $Y$ and closed cofibration subspaces $A,B$ whose union is also collared as above, the quotient spaces are based well-pointed and [F5] supplies the reduced product. Pulling it back along the based diagonal $Y\mathrm{/}(A\cup B)\to (Y\mathrm{/}A)\wedge (Y\mathrm{/}B)$ gives $K^0(Y,A)\otimes K^0(Y,B)⟶K^0(Y,A\cup B),$ whose forget-support image is the ordinary product by naturality of the external product. On $C_0=D_1^2\times \cdots \times D_r^2$, put ${\mathrm{\partial }}_i{C}_0=\{\mathrm{\mid }{z}_i\mathrm{\mid }=1\}$. Contracting the other disk coordinates gives a pair equivalence $(C_0,{\mathrm{\partial }}_i{C}_0)\simeq (D_i^2,\mathrm{\partial }{D}_i^2)$. The two line sections normalized in coordinates $0$ and $i$ differ on this boundary by $z_i$ or its inverse according to clutching direction. Its winding is $\pm 1$, so the relative restriction of ${\alpha }_i$ is the Hopf generator up to sign by [F5]. The lift is unambiguous because $K^{-1}(C_i)=K^{-1}(\ast )=0$. Hence the relative product ${\alpha }_1\cdots {\alpha }_r$ restricts under $C_0\mathrm{/}\mathrm{\partial }{C}_0\cong (D^2\mathrm{/}\mathrm{\partial }{D}^2{)}^{\wedge r}\cong S^{2r}$ to the $r$-fold Bott generator and is therefore a generator by [F5]. Put $U=C_1\cup \cdots \cup C_r$. In the scalar-orbit coordinates a point of $U$ has ${\max }_{j\ge 1}\mathrm{\mid }{z}_j\mathrm{\mid }=1$ and $\mathrm{\mid }{z}_0\mathrm{\mid }\le 1$. The equivariant homotopy $(z_0,z_1,\dots ,z_r)\mapsto ((1-t)z_0,z_1,\dots ,z_r)$ retracts $U$ onto the standard ${\mathbb{C}\mathbb{P}}^{r-1}$ given by $z_0=0$. It fixes that subspace. The induced map of relative pair sequences therefore identifies $K^0({\mathbb{C}\mathbb{P}}^r,U)$ with $K^0({\mathbb{C}\mathbb{P}}^r,{\mathbb{C}\mathbb{P}}^{r-1})$: on absolute groups it is identity and on subspace groups it is the retraction isomorphism, so exactness gives the relative comparison. Thus this relative generator maps to the kernel generator for restriction to ${\mathbb{C}\mathbb{P}}^{r-1}$, while forgetting support maps it to ${\alpha }^r$. Therefore $1,\alpha ,\dots ,{\alpha }^r$ is a basis. Take the relative product of all $r+1$ lifts ${\alpha }_0,\dots ,{\alpha }_r$. It lies in $K^0({\mathbb{C}\mathbb{P}}^r,{\bigcup }_i{C}_i)=K^0({\mathbb{C}\mathbb{P}}^r,{\mathbb{C}\mathbb{P}}^r)=0$, and its absolute image is ${\alpha }^{r+1}$, proving nilpotence without a forward induction. This completes the induction and proves $K^0({\mathbb{C}\mathbb{P}}^n)\cong \mathbb{Z}[\alpha ]\mathrm{/}({\alpha }^{n+1})\mathrm{.}$ No multiplicative spectral-sequence theorem is used.

For even $p$ in $0\le p\le 2n$, the relative quotient is $S^p$, with the $p=0$ term interpreted as the absolute group of the single vertex. For odd $p$ the successive skeleta agree, and outside this range the relative groups vanish. The quotient identifications [F7], suspension and coefficients [F1] therefore give $E_1^{p,q}=\mathbb{Z}$ precisely when $p$ is in that even range and $q$ is even, and zero otherwise. Every differential raises total degree by one, so every possible source of a nonzero differential has a zero target. Induction on the page gives $d_r=0$ for every $r\ge 1$ and the same support on all pages. By step 1.1, the resulting second page has $E_2^{p,q}\cong H^p(X;\mathbb{Z})$ for even $q$ and zero for odd $q$. In particular the K-AHSS collapses at $E_2$.

For completeness verify the finite abutment from the actual couple. Fix $p,q$, $t=p+q$, and use $(a,b)=(-p,-q)$. The formulas of [F4] give the stable numerator $N^{\mathrm{\infty }}=k^{-1}\mathrm{im}(K^t(X)\to K^t(X^p))$ once the upper skeleton is $X$, and the stable denominator $B^{\mathrm{\infty }}=j{K}^{t-1}(X^{p-1})=\ker k$ once the lower skeleton is empty. Thus $k$ identifies $E_{\mathrm{\infty }}^{p,q}$ with $\mathrm{im}(K^t(X)\to K^t(X^p))\cap \ker (K^t(X^p)\to K^t(X^{p-1}))$. Restriction from $F^p{K}^t(X):=\ker (K^t(X)\to K^t(X^{p-1}))$ surjects onto this intersection and has kernel $F^{p+1}{K}^t(X)$. Hence $E_{\mathrm{\infty }}^{p,q}\cong F^p{K}^t(X)\mathrm{/}{F}^{p+1}{K}^t(X)$. The filtration is nested by functoriality, equals the whole group for $p\le 0$, and is zero for $p>2n$.

By step 2.1 every stable quotient in total degree one vanishes. The finite filtration of step 2.2 then has equal adjacent stages, so its whole group is its zero final stage: $K^1(X)=0$. In total degree zero its nonzero quotients are $\mathbb{Z}$ in columns $0,2,\dots ,2n$, while step 1.3 supplies the actual ring with the exact tautological-line convention $\alpha =[L]-1$.

The collapse and vanishing are proved in steps 2.1 and 3.1, and the ring presentation is step 1.3, not an inference from the additive page, consistently with [F6]. For $n=0$ the point has only column zero, $K^1=0$, and $L$ is trivial, so $\alpha =0$ and $K^0=\mathbb{Z}$. AC enters only through [A1]. This proves all assertions.

By [A2] the page $E_2$ has nonzero entries $H^0,H^1,H^2$ in each even coefficient row, all free abelian, and vanishes in odd coefficient rows.

Every differential $d_r$ with $r\ge 2$ has target in the odd coefficient row $q-r+1$ when $r$ is even, and in the column $p+r>2$ when $r$ is odd; since the odd coefficient rows and the columns above the surface dimension $2$ vanish, all differentials $d_r$ for $r\ge 2$ vanish. The first differential is the cellular coboundary, so $E_2$ is already the cohomology page by construction.

The stable page has graded pieces $\mathbb{Z}$ in total degree zero for the rows contributing $H^0$ and $H^2$, and ${\mathbb{Z}}^{2g}$ in total degree one from $H^1$, so the associated graded of $K^0$ is ${\mathbb{Z}}^2$ and that of $K^1$ is ${\mathbb{Z}}^{2g}$.

Since all graded pieces are free, the finite filtrations split by [A3], so $K^0(S_g)\cong {\mathbb{Z}}^2\cong \mathbb{Z}\oplus {\tilde{K}}^0(S_g)$ with ${\tilde{K}}^0(S_g)\cong \mathbb{Z}$ and $K^1(S_g)\cong {\mathbb{Z}}^{2g}$; the extension data are not inferred from the collapse but from the splitting of free extensions.

This verifies the displayed groups $K^0\cong {\mathbb{Z}}^2$ and $K^1\cong {\mathbb{Z}}^{2g}$.

By [A1] the stable page has one $\mathbb{Z}\mathrm{/}2$ in each even cohomological degree $2,4,\dots$ up to the dimension and, for odd $r$, an extra $\mathbb{Z}$ in the top degree; all differentials vanish, so these are the graded pieces of $K^{\ast }({\mathbb{R}\mathbb{P}}^r)$.

The graded pieces in total degree zero are $m$ copies of $\mathbb{Z}\mathrm{/}2$ together with the $\mathbb{Z}$ from degree zero; the associated graded of ${\tilde{K}}^0$ is therefore $(\mathbb{Z}\mathrm{/}2{)}^m$ of order $2^m$.

By [A2] the class $\alpha$ has exact order $2^m$ and generates ${\tilde{K}}^0({\mathbb{R}\mathbb{P}}^r)$, so the $m$ copies of $\mathbb{Z}\mathrm{/}2$ assemble into the single cyclic group $\mathbb{Z}\mathrm{/}{2}^m$. For $m\ge 2$ this is a nonsplit extension because $(\mathbb{Z}\mathrm{/}2{)}^m$ is not cyclic; for $m=0$ the reduced group is zero, and for $m=1$ it is the lone graded piece $\mathbb{Z}\mathrm{/}2$. The odd-degree statement is the corresponding clause of [A2].

Steps 1.1, 2.1 and 3.1 verify the displayed groups and identify exactly when a nonsplit extension occurs.