The Serre Spectral Sequence and Applications — Examples

By [F1], the finite-stage quotients $S^{2N+1}\mathrm{/}{S}^1={\mathbb{C}\mathbb{P}}^N$ assemble to the Milnor quotient $B{S}^1={\mathbb{C}\mathbb{P}}^{\mathrm{\infty }}$. The weak equivalence in [F2], together with ${\pi }_1(S^1)=\mathbb{Z}$ and vanishing higher homotopy, gives ${\pi }_i({\mathbb{C}\mathbb{P}}^{\mathrm{\infty }})=\{\begin{array}{ll}\mathbb{Z},& i=2,\\ 0,& i\mathrm{\ne }2,\end{array}$ for $i\ge 1$. Thus the displayed CW colimit is a marked $K(\mathbb{Z},2)$, so [F3] applies and makes the displayed strict path-loop fibration legitimate, with contractible total space and fiber ring $\mathrm{\Lambda }(u)$.

Put $A=H^{\ast }({\mathbb{C}\mathbb{P}}^{\mathrm{\infty }};\mathbb{Z})$. The base is simply connected, hence the fiber cohomology system is constant. Because its two nonzero stalks are the free rank-one group $\mathbb{Z}$, the constant-system comparison is literal, and [F5] gives $E_2^{p,q}=A^p\otimes H^q(S^1;\mathbb{Z}),$ with nonzero rows only at $q=0,1$. Consequently the only possibly nonzero differential is $d_2:E_2^{p,1}=A^{p}u⟶E_2^{p+2,0}=A^{p+2}\mathrm{.}$ By [F6], every positive-total-degree stable term is zero.

Define $c=d_2(u)\in A^2$. It is the transgression of $u$ by [F5]. The upper row has no incoming differential, the bottom row has no outgoing differential, and there are no differentials after $d_2$. Therefore the vanishing stable page says simultaneously that $A^p⟶A^{p+2},a⟼ac,$ is injective (to kill every $au$) and surjective (to hit every positive-degree bottom class), for every $p\ge 0$. Here the Serre Leibniz sign is $(-1{)}^{\mathrm{\mid }a\mathrm{\mid }}$; the induction below makes $A^{\mathrm{o}\mathrm{d}\mathrm{d}}=0$, so this is the displayed multiplication map.

Since $A^0=\mathbb{Z}$, step 2.1 gives $A^2=\mathbb{Z}c$ and shows that $c$ is primitive, not a nonunit multiple of a generator. Since $A^1$ has no possible incoming differential and must vanish at infinity, $A^1=0$. Inductively, multiplication by $c$ identifies $A^{2k}$ with $\mathbb{Z}{c}^{k+1}$ and identifies $A^{2k+1}=0$ with $A^{2k+3}=0$. Hence the homomorphism $\mathbb{Z}[c]\to A$ sending the formal generator to the transgression is an isomorphism in every degree, both injectively and surjectively. There is no additive or multiplicative extension ambiguity: each associated-graded total degree has at most the single bottom-row term.

The base, fiber, and path space are nonempty. The zero and degree-one groups, the unit, $c^0$, the first differential, both rows, both endpoints of every possible differential, and both directions of the ring isomorphism were checked above. Changing $u$ to $-u$ changes $c=d_2(u)$ to $-c$. AC is used through [F2]–[F6]; the finite-stage identification and the integer induction add no choice. This is a computation, not a converse characterization.

Let $U_j=\{[z]:z_j\mathrm{\ne }0\}$. Every line in $U_j$ has the unique unit representative $s_j([z])$ whose $j$th coordinate is positive real. Explicitly, if $z$ is already unit, then $s_j([z])=z\mathrm{\mid }{z}_j\mathrm{\mid }\mathrm{/}{z}_j$. Hence $p^{-1}(U_j)⟶U_j\times S^1,z⟼([z],z_j\mathrm{/}\mathrm{\mid }{z}_j\mathrm{\mid })$ is a bundle chart, with inverse $([z],\lambda )\mapsto \lambda s_j([z])$. For $a=1\mathrm{/}(2(n+1))$, set $r_j([z])=\max \{\mathrm{\mid }{z}_j{\mathrm{\mid }}^2-a,0\},{\rho }_j=r_j\mathrm{/}{\sum }_i{r}_i\mathrm{.}$ Some $\mathrm{\mid }{z}_j{\mathrm{\mid }}^2\ge 1\mathrm{/}(n+1)>a$, so the denominator is positive; moreover $\mathrm{supp}({\rho }_j)\subseteq \{\mathrm{\mid }{z}_j{\mathrm{\mid }}^2\ge a\}\subset U_j$. Thus these finitely many charts are support-subordinate numerating data, and [F1] makes $p$ a Serre fibration.

Put $P^r={\mathbb{C}\mathbb{P}}^r$. The characteristic map $D^{2r}⟶P^r,w⟼[w_0:\cdots :w_{r-1}:\sqrt{1-\mathrm{\parallel }w{\mathrm{\parallel }}^2}]$ sends the boundary into $P^{r-1}$ and its interior homeomorphically onto $P^r\setminus P^{r-1}$: there the last homogeneous coordinate has a unique positive-real unit representative. Compact-to-Hausdorff quotient descent gives the attachment homeomorphism. Induction produces one cell in dimensions $0,2,\dots ,2n$ and none elsewhere. Adjacent cellular chain groups never both occur, so every cellular differential is zero. By [F2, F3], $H^k(P^n;\mathbb{Z})=\{\begin{array}{ll}\mathbb{Z},& k=0,2,\dots ,2n,\\ 0,& \text{otherwise}\mathrm{.}\end{array}$

The base is simply connected: its CW structure has one vertex and no one-cells. Thus the fiber system is constant. Since all base and fiber groups are free, [F4] has $E_2^{p,q}=H^p(P^n;\mathbb{Z})\otimes H^q(S^1;\mathbb{Z}),$ with only rows $q=0,1$ and columns $p=0,2,\dots ,2n$. The only possible differential is $d_2:E_2^{p,1}\to E_2^{p+2,0}$. The total sphere has cohomology only in total degrees $0$ and $2n+1$ by [F3]. Therefore every displayed $d_2$ for $0\le p<2n$ is an isomorphism between infinite cyclic groups, while the class in $(2n,1)$ survives as the total sphere's top class.

Choose $u$ as in the statement and put $x=d_2(u)$. The $p=0$ isomorphism in step 2.1 makes $x$ a primitive generator of $H^2(P^n;\mathbb{Z})$. Since $d_2$ vanishes on the bottom row, the derivation rule gives $d_2(u{x}^k)=x^{k+1}(0\le k<n)\mathrm{.}$ Inductively the isomorphisms in step 2.1 make $x^k$ a generator of $H^{2k}(P^n;\mathbb{Z})$ for every $0\le k\le n$. The CW dimension gives $x^{n+1}=0$. Hence evaluation induces a surjection $\mathbb{Z}[X]\mathrm{/}(X^{n+1})\to H^{\ast }(P^n;\mathbb{Z})$; it is injective because in each allowed degree the image $x^k$ has infinite order. This proves both directions of the ring identification.

There is no extension ambiguity: every total degree on the base ring has one group, and the actual cup powers were identified before passage to the stable page. The top element $u{x}^n$ survives because its target column is $2n+2$, outside the base, exactly accounting for $H^{2n+1}(S^{2n+1})$. For $n=1$ the sole differential is $d_2(u)=x$ and the survivor is $ux$; the zero groups, unit, first and last columns, both rows, and both differential endpoints are therefore included. Reversing $u$ reverses $x$. AC is used only through [F1], [F3], and [F4]; all charts, the finite partition, and the cellular calculation are choice-free. No converse or splitting is asserted.

For $U_j=\{[z]:z_j\mathrm{\ne }0\}$ and unit $z$, right multiplication by $z_j^{-1}\mathrm{\mid }{z}_j\mathrm{\mid }$ gives the unique unit representative $s_j([z])$ whose $j$th coordinate is positive real. The formulas $z⟼([z],z_j\mathrm{/}\mathrm{\mid }{z}_j\mathrm{\mid }),([z],\lambda )⟼s_j([z])\lambda$ are inverse bundle charts $p^{-1}(U_j)\cong U_j\times S^3$. Their order is essential: all scalar multiplication is on the right. With $a=1\mathrm{/}(2(n+1))$, the normalized functions ${\rho }_j([z])=\frac{\max \{\mathrm{\mid }{z}_j{\mathrm{\mid }}^2-a,0\}}{{\sum }_i\max \{\mathrm{\mid }{z}_i{\mathrm{\mid }}^2-a,0\}}$ are defined because some norm square is at least $1\mathrm{/}(n+1)$, and their supports lie in $U_j$. Thus the finite bundle is numerable and [F1] makes it a Serre fibration.

Put $P^r={\mathbb{H}\mathbb{P}}^r$. The map $D^{4r}⟶P^r,w⟼[w_0:\cdots :w_{r-1}:\sqrt{1-\mathrm{\parallel }w{\mathrm{\parallel }}^2}]$ attaches one $4r$-cell to $P^{r-1}$: the boundary lands in $P^{r-1}$, and right normalization of the last nonzero coordinate gives a unique positive-real representative on the complement. Compact-to-Hausdorff quotient descent proves the attachment map is a homeomorphism. Starting at a point gives one cell in dimensions $0,4,\dots ,4n$. Every cellular differential is zero, since adjacent cellular dimensions are never both occupied. Hence [F2, F3] give $H^k(P^n;\mathbb{Z})=\{\begin{array}{ll}\mathbb{Z},& k=0,4,\dots ,4n,\\ 0,& \text{otherwise}\mathrm{.}\end{array}$

The CW structure has one vertex and no one-cells, so the base is simply connected and the fiber system is constant. Its groups and the base groups are free, so the multiplicative sequence has $E_2^{p,q}=H^p(P^n;\mathbb{Z})\otimes H^q(S^3;\mathbb{Z}),$ with rows only at $q=0,3$ and columns only at $p=0,4,\dots ,4n$. Thus the only possible differential is $d_4:E_4^{p,3}\to E_4^{p+4,0}$. The total sphere has cohomology only in degrees $0$ and $4n+3$ by [F3], so every $d_4$ with $0\le p<4n$ is an isomorphism of infinite cyclic groups; the upper-right class at $(4n,3)$ survives.

Put $x=d_4(u)$. The first isomorphism makes $x$ a primitive generator of $H^4(P^n;\mathbb{Z})$. Since $d_4$ is zero on the bottom row and $\mathrm{\mid }x\mathrm{\mid }$ is even, the derivation rule gives $d_4(u{x}^k)=x^{k+1}(0\le k<n)\mathrm{.}$ It follows inductively that $x^k$ generates $H^{4k}(P^n;\mathbb{Z})$ for $0\le k\le n$. Dimension gives $x^{n+1}=0$, so evaluation defines a surjection $\mathbb{Z}[X]\mathrm{/}(X^{n+1})\to H^{\ast }(P^n;\mathbb{Z})$. It is injective degree by degree because every $x^k$ in the allowed range has infinite order.

The survivor $u{x}^n$ has no target because column $4n+4$ is absent and is the unique class accounting for $H^{4n+3}(S^{4n+3})$. This also eliminates any hidden extension: the actual base cup powers were computed and each relevant degree has one cyclic group. For $n=1$, $d_4(u)=x$ is the sole differential and $ux$ is the top survivor. The unit, zero groups, first and last columns, both rows, both differential endpoints, both ring-map directions, and reversal of the orientation generator are explicit. AC occurs only through [F1], [F3], and [F4]; the finite quaternionic charts and cell argument make no choices. There is no converse or splitting claim.

The path-space formula is $P{S}^N=\{\gamma :I\to S^N:\gamma (0)=\ast \},p(\gamma )=\gamma (1)\mathrm{.}$ The homotopy $H_s(\gamma )(t)=\gamma (st)$ contracts it to the constant path; the compact-open exponential law makes this formula continuous. By [F2], any based loop has a based nullhomotopy, whose slices give a path in $\mathrm{\Omega }{S}^N$, so $H_0(\mathrm{\Omega }{S}^N;\mathbb{Z})=\mathbb{Z}$.

Write $G_q=H_q(\mathrm{\Omega }{S}^N;\mathbb{Z})$. Since the base is simply connected, [F3, F4] give $E_{p,q}^2=\{\begin{array}{ll}{G}_q,& p=0,N,\\ 0,& \text{otherwise}\mathrm{.}\end{array}$ The homological bidegree $(-r,r-1)$ shows that every differential before page $N$ is zero and that the only possibly nonzero family is $d_N:E_{N,q}^N=G_q⟶E_{0,q+N-1}^N=G_{q+N-1}\mathrm{.}$ There are no nonzero differentials after page $N$. The contractibility in step 1.1 and [F4] say that the stable page is $\mathbb{Z}$ at $(0,0)$ and zero in every positive total degree.

For every $q\ge 0$, the source $(N,q)$ has zero stable term, so $d_N$ has zero kernel. Its target has positive total degree $q+N-1$, so that stable term is also zero and $d_N$ has zero cokernel. Thus $G_q\stackrel{\cong }{\to }{G}_{q+N-1}=G_{q+2m}(q\ge 0)\mathrm{.}$ For $0<j<N-1$, the group $G_j$ in position $(0,j)$ has no possible incoming differential and must vanish. Together with $G_0=\mathbb{Z}$, induction on the quotient and remainder of $k$ by $N-1=2m$ gives exactly the displayed groups.

When $m=1$, the recurrence has period two and gives $\mathbb{Z}$ in every even degree and zero in every odd degree. Degree zero is the surviving $G_0=\mathbb{Z}$, not a target required to vanish. The proof also covers the zero groups, the first gap $1\le j<2m$, the first isomorphism, both columns, both endpoints of every $d_N$, and every remainder class. Each construction uses one supplied basepoint or one finite chain representative; neither an infinite family of choices nor AC is used. There is no ring claim, converse, or extension splitting.

Away from the identified endpoint, the quotient projection has product charts. Across the endpoint, use one interval on the $t=1$ side and one on the $t=0$ side, changing the fiber coordinate by $f$ in one direction and by $f^{-1}$ in the other. Thus the displayed map is the standard mapping-torus fiber bundle. Lifting the positive base loop from $t=0$ to $t=1$ starts at $x$ and ends at the class of $(x,1)$, which equals the class of $(f(x),0)$. Hence [F1] gives $T_q=f_{\ast }:H_q(F)\to H_q(F)$.

Substitute $T_q=f_{\ast }$ into the five-term part of [F2]: $H_q(F)\stackrel{1-f_{\ast }}{\to }{H}_q(F)\stackrel{i_q}{\to }{H}_q(T_f)\stackrel{{\mathrm{\partial }}_q}{\to }{H}_{q-1}(F)\stackrel{1-f_{\ast }}{\to }{H}_{q-1}(F)\mathrm{.}$ Exactness gives $\ker i_q=\mathrm{im}(1-f_{\ast })$, so $i_q$ descends to an injection from the displayed cokernel, whose image is $\ker {\mathrm{\partial }}_q$. It also gives $\mathrm{im}{\mathrm{\partial }}_q=\ker (1-f_{\ast }{\mathrm{\mid }}_{H_{q-1}(F)})$, so ${\mathrm{\partial }}_q$ corestricts to a surjection onto the displayed kernel. These two conclusions are exactly the claimed short exact sequence.

For $q=0$, the right group is zero by the stated $H_{-1}=0$ convention, so the conclusion is the literal isomorphism $H_0(T_f)\cong \mathrm{coker}(1-f_{\ast }{\mathrm{\mid }}_{H_{0}F})$. If $F$ is empty, all three nonzero-index groups are zero and the same sequence is exact. If $f=\mathrm{id}$, both Wang maps vanish and the sequence becomes $0\to H_q(F)\to H_q(F\times S^1)\to H_{q-1}(F)\to 0$, still without selecting a splitting. Orientation reversal replaces $1-f_{\ast }$ by $f_{\ast }-1$; multiplication by $-1$ leaves its kernel, image, and cokernel canonically isomorphic. Thus zero maps, identity monodromy, both exactness endpoints, and both orientation signs are covered. The argument uses no choice principle and proves no converse.

Since $j_q$ is a homotopy equivalence, [F1, F2] identify ${\pi }_1(E_q)$ with $\mathbb{Z}$, all its higher homotopy groups with zero, and $(p_q{)}_{\ast }$ with the quotient $\mathbb{Z}\twoheadrightarrow C_n$. The component end of [F3] is $\mathbb{Z}\twoheadrightarrow C_n⟶{\pi }_0(F_q)⟶\ast \mathrm{.}$ The first map is onto, so exactness gives one fiber component. The group part then gives an injection ${\pi }_1(F_q)\hookrightarrow \mathbb{Z}$ with image the kernel $n\mathbb{Z}$. For $k>1$, the adjacent homotopy groups of both $E_q$ and $B{C}_n$ vanish, so [F3] gives ${\pi }_k(F_q)=0$.

All three spaces used here are connected. Their fundamental groups $n\mathbb{Z}$, $\mathbb{Z}$, and $C_n$ are abelian, so [F4] identifies their first homology groups with those same groups. Naturality of Hurewicz identifies the fiber-inclusion map on first homology with $n\mathbb{Z}\hookrightarrow \mathbb{Z}\mathrm{.}$ By [F5], its image is precisely the first Serre filtration term $F_0{H}_1(E_q)=n\mathbb{Z}$. Strong convergence in total degree one therefore gives $E_{0,1}^{\mathrm{\infty }}=F_0{H}_1(E_q)=n\mathbb{Z},E_{1,0}^{\mathrm{\infty }}=H_1(E_q)\mathrm{/}{F}_0{H}_1(E_q)=\mathbb{Z}\mathrm{/}n\mathrm{.}$ This uses the actual edge image, so it does not assume that either stable term already splits off.

If $0\subset n\mathbb{Z}\subset \mathbb{Z}$ split, a section of $\mathbb{Z}\twoheadrightarrow \mathbb{Z}\mathrm{/}n$ would embed a nonzero element of order $n$ in the torsion-free group $\mathbb{Z}$, which is impossible for $n\ge 2$. Thus the abutment is not the direct sum $n\mathbb{Z}\oplus \mathbb{Z}\mathrm{/}n$, even though these are its two stable pieces. The witness is explicit, not merely an appeal to the existence of extension problems.

The fiber and total space are nonempty and connected. The zero higher homotopy groups, zero and first filtration terms, both total-degree-one stable positions, both maps in the short exact filtration, and the first permitted case $n=2$ were checked. The excluded value $n=1$ would give a trivial quotient and no nonsplit extension. AC is used only by [F1]; mapping paths, the homotopy exact sequence, Hurewicz, Serre filtration, and the finite torsion argument add no choice. No converse is asserted.

Reflection sends the positive loop $t\mapsto [t]$ to the negative loop $t\mapsto [-t]$. Hence [F1] gives $r_{\ast }=-1:H_1(S^1;\mathbb{Z})⟶H_1(S^1;\mathbb{Z})\mathrm{.}$ By [F2], this is exactly the degree-one monodromy of the mapping-torus fibration. On $H_0(S^1;\mathbb{Z})=\mathbb{Z}$ the monodromy is the identity, because the fiber is connected.

In fiber degree one, the cellular local chain complex of the one-vertex, one-edge base circle is $0⟶\mathbb{Z}\stackrel{1-(-1)=2}{\to }\mathbb{Z}⟶0.$ Changing the edge convention multiplies this boundary by $-1$ and changes neither group. Thus [F3] gives $E_{0,1}^2=\mathrm{coker}(2)=\mathbb{Z}\mathrm{/}2,E_{1,1}^2=\ker (2)=0.$ If one falsely makes the coefficient system constant, the boundary becomes $1-1=0$, so the same two entries become $\mathbb{Z}$ and $\mathbb{Z}$.

Apply [F2] in degree one. On fiber $H_1$, $1-r_{\ast }=2$; on fiber $H_0$, $1-r_{\ast }=0$. Hence $0⟶\mathbb{Z}\mathrm{/}2⟶H_1(K;\mathbb{Z})⟶\mathbb{Z}⟶0.$ This sequence splits explicitly: the point $[0]\in S^1$ is fixed by $r$, so $[t]\mapsto [([0],t)]$ is a section of $K\to S^1$ and its first-homology map splits the displayed projection. Therefore $H_1(K;\mathbb{Z})\cong \mathbb{Z}\oplus \mathbb{Z}\mathrm{/}2$.

The base has cellular dimension one, so no Serre differential $d_s$ with $s\ge 2$ can leave or enter its columns. The correct $E^2$ table therefore already displays the torsion piece in total degree one. The false constant table has $\mathbb{Z}$ instead at $(0,1)$ and would have two infinite cyclic total-degree-one pieces, so it cannot yield the actual $H_1(K)$. Degree zero, the zero kernel of multiplication by two, identity action on $H_0$, both base cells, both orientations, both row degrees, and the fixed-point section are all explicit above. No AC is used, and there is no converse assertion.