Serre spectral sequence

Recall that the Hopf fibration is given by ${\displaystyle S^{1}\hookrightarrow S^{3}\to S^{2}}$. The ${\displaystyle E_{2}}$-page of the Leray–Serre Spectral sequence reads

${\displaystyle {\begin{array}{c|ccc}1&H^{0}(S^{2};\mathbb {Z} )&0&H^{2}(S^{2};\mathbb {Z} )\\0&H^{0}(S^{2};\mathbb {Z} )&0&H^{2}(S^{2};\mathbb {Z} )\\\hline &0&1&2\end{array}}}$

The differential ${\displaystyle d_{2+i}}$ goes ${\displaystyle 1+i}$ down and ${\displaystyle 2+i}$ right. Thus the only differential which is not necessarily 0 is d^0,1₂, because the rest have domain or codomain 0 (since they are 0 on the E₂-page). In particular, this sequence degenerates at E₂ = E_∞. The E₃-page reads

${\displaystyle {\begin{array}{c|ccc}1&\ker d_{2}^{0,1}&0&H^{2}(S^{2};\mathbb {Z} )\\0&H^{0}(S^{2};\mathbb {Z} )&0&\operatorname {coker} d_{2}^{0,1}\\\hline &0&1&2\end{array}}}$

The spectral sequence abuts to ${\displaystyle H^{p+q}(S^{3}),}$ i.e. ${\displaystyle E_{3}^{p,q}=Gr^{p}H^{p+q}(S^{3}).}$ Evaluating at the interesting parts, we have ${\displaystyle \ker d_{2}^{0,1}=Gr^{1}H^{1}(S^{3})}$ and ${\displaystyle \operatorname {coker} d_{2}^{0,1}=Gr^{0}H^{2}(S^{3}).}$ Knowing the cohomology of ${\displaystyle S^{3},}$ both are zero, so the differential ${\displaystyle d_{2}^{0,1}}$ is an isomorphism.

Given a complex n-dimensional projective variety X there is a canonical family of line bundles ${\displaystyle {\mathcal {O}}_{X}(k)}$ for ${\displaystyle k\in \mathbb {Z} }$ coming from the embedding ${\displaystyle X\to \mathbb {CP} ^{m}}$. This is given by the global sections ${\displaystyle s_{0},\ldots ,s_{m}\in \Gamma (X,{\mathcal {O}}_{X}(1))}$ which send

${\displaystyle x\mapsto [s_{0}(x):\ldots :s_{m}(x)]}$

If we construct a rank r vector bundle ${\displaystyle {\mathcal {E}}}$ which is a finite whitney sum of vector bundles we can construct a sphere bundle ${\displaystyle S\to X}$ whose fibers are the spheres ${\displaystyle S^{2r-1}\subset \mathbb {C} ^{r}}$. Then, we can use the Serre spectral sequence along with the Euler class to compute the integral cohomology of S. The ${\displaystyle E_{2}}$-page is given by ${\displaystyle E_{2}^{p,q}=H^{p}(X;H^{q}(S^{2r-1}))}$. We see that the only non-trivial differentials are given on the ${\displaystyle E_{2r}}$-page and are defined by cupping with the Euler class ${\displaystyle e({\mathcal {E}})}$. In this case it is given by the top chern class of ${\displaystyle {\mathcal {E}}}$. For example, consider the vector bundle ${\displaystyle {\mathcal {O}}_{X}(a)\oplus {\mathcal {O}}_{X}(b)}$ for X a K3 surface. Then, the spectral sequence reads as

${\displaystyle E_{2}=E_{3}=E_{4}={\begin{array}{c|ccccc}3&H^{0}(X;\mathbb {Z} )&H^{1}(H;\mathbb {Z} )&H^{2}(H;\mathbb {Z} )&H^{3}(H;\mathbb {Z} )&H^{4}(H;\mathbb {Z} )\\2&0&0&0&0&0\\1&0&0&0&0&0\\0&H^{0}(X;\mathbb {Z} )&H^{1}(H;\mathbb {Z} )&H^{2}(H;\mathbb {Z} )&H^{3}(H;\mathbb {Z} )&H^{4}(H;\mathbb {Z} )\\\hline &0&1&2&3&4\end{array}}}$

The differential ${\displaystyle d_{4}=\cup a\cdot bH^{2}}$ for ${\displaystyle H^{2}}$ is the square of the Lefschetz class. In this case, the only non-trivial differential is then

${\displaystyle d_{4}^{0,3}:H^{0}(X;\mathbb {Z} )\to H^{4}(X;\mathbb {Z} )}$

We can finish this computation by noting the only nontrivial cohomology groups are

${\displaystyle H^{k}(X:\mathbb {Z} )={\begin{cases}\mathbb {Z} &k\in \{0,4\}\\\mathbb {Z} ^{\oplus 22}&k=2\end{cases}}}$

We begin first with a basic example; consider the path space fibration

${\displaystyle \Omega S^{n+1}\to PS^{n+1}\to S^{n+1}.}$

We know the homology of the base and total space, so our intuition tells us that the Serre spectral sequence should be able to tell us the homology of the loop space. This is an example of a case where we can study the homology of a fibration by using the E^∞ page (the homology of the total space) to control what can happen on the E² page. So recall that

${\displaystyle E_{p,q}^{2}=H_{p}(S^{n+1};H_{q}(\Omega S^{n+1})).}$

Thus we know when q = 0, we are just looking at the regular integer valued homology groups H_p(Sⁿ⁺¹) which has value ${\displaystyle \mathbb {Z} }$ in degrees 0 and n+1 and value 0 everywhere else. However, since the path space is contractible, we know that by the time the sequence gets to E^∞, everything becomes 0 except for the group at p = q = 0. The only way this can happen is if there is an isomorphism from ${\displaystyle H_{n+1}(S^{n+1};H_{0}(F))=\mathbb {Z} }$ to another group. However, the only places a group can be nonzero are in the columns p = 0 or p = n+1 so this isomorphism must occur on the page Eⁿ⁺¹ with codomain ${\displaystyle H_{0}(S^{n+1};H_{n}(F))=\mathbb {Z} .}$ However, putting a ${\displaystyle \mathbb {Z} }$ in this group means there must be a ${\displaystyle \mathbb {Z} }$ at H_n+1(Sⁿ⁺¹; H_n(F)). Inductively repeating this process shows that H_i(ΩSⁿ⁺¹) has value ${\displaystyle \mathbb {Z} }$ at integer multiples of n and 0 everywhere else.

We compute the cohomology of ${\displaystyle \mathbb {CP} ^{n}}$ using the fibration:

${\displaystyle S^{1}\hookrightarrow S^{2n+1}\to \mathbb {CP} ^{n}}$

Now, on the E₂ page, in the 0,0 coordinate we have the identity of the ring. In the 0,1 coordinate, we have an element i that generates ${\displaystyle \mathbb {Z} .}$ However, we know that by the limit page, there can only be nontrivial generators in degree 2n+1 telling us that the generator i must transgress to some element x in the 2,0 coordinate. Now, this tells us that there must be an element ix in the 2,1 coordinate. We then see that d(ix) = x² by the Leibniz rule telling us that the 4,0 coordinate must be x² since there can be no nontrivial homology until degree 2n+1. Repeating this argument inductively until 2n + 1 gives ixⁿ in coordinate 2n,1 which must then be the only generator of ${\displaystyle \mathbb {Z} }$ in that degree thus telling us that the 2n + 1,0 coordinate must be 0. Reading off the horizontal bottom row of the spectral sequence gives us the cohomology ring of ${\displaystyle \mathbb {CP} ^{n}}$ and it tells us that the answer is ${\displaystyle \mathbb {Z} [x]/x^{n+1}.}$

In the case of infinite complex projective space, taking limits gives the answer ${\displaystyle \mathbb {Z} [x].}$

A more sophisticated application of the Serre spectral sequence is the computation ${\displaystyle \pi _{4}(S^{3})=\mathbb {Z} /2\mathbb {Z} .}$ This particular example illustrates a systematic technique which one can use in order to deduce information about the higher homotopy groups of spheres. Consider the following fibration which is an isomorphism on ${\displaystyle \pi _{3}}$

${\displaystyle X\to S^{3}\to K(\mathbb {Z} ,3),}$

where ${\displaystyle K(\pi ,n)}$ is an Eilenberg–MacLane space. We then further convert the map ${\displaystyle X\to S^{3}}$ to a fibration; it is general knowledge that the iterated fiber is the loop space of the base space so in our example we get that the fiber is ${\displaystyle \Omega K(\mathbb {Z} ,3)=K(\mathbb {Z} ,2).}$ But we know that ${\displaystyle K(\mathbb {Z} ,2)=\mathbb {CP} ^{\infty }.}$ Now we look at the cohomological Serre spectral sequence: we suppose we have a generator for the degree 3 cohomology of ${\displaystyle S^{3}}$, called ${\displaystyle \iota }$. Since there is nothing in degree 3 in the total cohomology, we know this must be killed by an isomorphism. But the only element that can map to it is the generator a of the cohomology ring of ${\displaystyle \mathbb {CP} ^{\infty }}$, so we have ${\displaystyle d(a)=\iota }$. Therefore by the cup product structure, the generator in degree 4, ${\displaystyle a^{2}}$, maps to the generator ${\displaystyle \iota a}$ by multiplication by 2 and that the generator of cohomology in degree 6 maps to ${\displaystyle \iota a^{2}}$ by multiplication by 3, etc. In particular we find that ${\displaystyle H_{4}(X)=\mathbb {Z} /2\mathbb {Z} .}$ But now since we killed off the lower homotopy groups of X (i.e., the groups in degrees less than 4) by using the iterated fibration, we know that ${\displaystyle H_{4}(X)=\pi _{4}(X)}$ by the Hurewicz theorem, telling us that ${\displaystyle \pi _{4}(S^{3})=\mathbb {Z} /2\mathbb {Z} .}$

Corollary: ${\displaystyle \pi _{4}(S^{2})=\mathbb {Z} /2\mathbb {Z} .}$

Proof: Take the long exact sequence of homotopy groups for the Hopf fibration ${\displaystyle S^{1}\to S^{3}\to S^{2}}$.