# nLab Thom-Gysin sequence

Contents

cohomology

### Theorems

#### Homological algebra

homological algebra

Introduction

diagram chasing

# Contents

## Idea

The Thom-Gysin sequence is a type of long exact sequence in cohomology induced by a spherical fibration and expressing the cohomology groups of the total space in terms of those of the base plus correction. The sequence may be obtained as a corollary of the Serre spectral sequence for the given fibration. It induces, and is induced by, the Thom isomorphism.

## Statement

###### Proposition

Let $R$ be a commutative ring and let

$\array{ S^n &\longrightarrow& E \\ && \downarrow^{\mathrlap{\pi}} \\ && B }$

be a Serre fibration over a simply connected CW-complex with typical fiber (exmpl.) the n-sphere.

Then there exists an element $c \in H^{n+1}(E; R)$ (in the ordinary cohomology of the total space with coefficients in $R$, called the Euler class of $\pi$) such that the cup product operation $c \cup (-)$ sits in a long exact sequence of cohomology groups of the form

$\cdots \to H^k(B; R) \stackrel{\pi^\ast}{\longrightarrow} H^k(E; R) \stackrel{}{\longrightarrow} H^{k-n}(B;R) \stackrel{c \cup (-)}{\longrightarrow} H^{k+1}(B; R) \to \cdots \,.$
###### Proof

Under the given assumptions there is the corresponding Serre spectral sequence

$E_2^{s,t} \;=\; H^s(B; H^t(S^n;R)) \;\Rightarrow\; H^{s+t}(E; R) \,.$

Since the ordinary cohomology of the n-sphere fiber is concentrated in just two degees

$H^t(S^n; R) = \left\{ \array{ R & for \; t= 0 \; and \; t = n \\ 0 & otherwise } \right.$

the only possibly non-vanishing terms on the $E_2$ page of this spectral sequence, and hence on all the further pages, are in bidegrees $(\bullet,0)$ and $(\bullet,n)$:

$E^{\bullet,0}_2 \simeq H^\bullet(B; R) \,, \;\;\;\; and \;\;\; E^{\bullet,n}_2 \simeq H^\bullet(B; R) \,.$

As a consequence, since the differentials $d_r$ on the $r$th page of the Serre spectral sequence have bidegree $(r+1,-r)$, the only possibly non-vanishing differentials are those on the $(n+1)$-page of the form

$\array{ E_{n+1}^{\bullet,n} & \simeq & H^\bullet(B;R) \\ {}^{\mathllap{d_{n+1}}}\downarrow \\ E_{n+1}^{\bullet+n+1,0} & \simeq & H^{\bullet+n+1}(B;R) } \,.$

Now since the coefficients $R$ is a ring, the Serre spectral sequence is multiplicative under cup product and the differential is a derivation (of total degree 1) with respect to this product. (See at multiplicative spectral sequence – Examples – AHSS for multiplicative cohomology.)

To make use of this, write

$\iota \coloneqq 1 \in H^0(B;R) \stackrel{\simeq}{\longrightarrow} E_{n+1}^{0,n}$

for the unit in the cohomology ring $H^\bullet(B;R)$, but regarded as an element in bidegree $(0,n)$ on the $(n+1)$-page of the spectral sequence. (In particular $\iota$ does not denote the unit in bidegree $(0,0)$, and hence $d_{n+1}(\iota)$ need not vanish; while by the derivation property, it does vanish on the actual unit $1 \in H^0(B;R) \simeq E_{n+1}^{0,0}$.)

Write

$c \coloneqq d_{n+1}(\iota) \;\; \in E_{n+1}^{n+1,0} \stackrel{\simeq}{\longrightarrow} H^{n+1}(B; R)$

for the image of this element under the differential. We will show that this is the Euler class in question.

To that end, notice that every element in $E_{n+1}^{\bullet,n}$ is of the form $\iota \cdot b$ for $b\in E_{n+1}^{\bullet,0} \simeq H^\bullet(B;R)$.

(Because the multiplicative structure gives a group homomorphism $\iota \cdot(-) \colon H^\bullet(B;R) \simeq E_{n+1}^{0,0} \to E^{0,n}_{n+1} \simeq H^\bullet(B;R)$, which is an isomorphism because the product in the spectral sequence does come from the cup product in the cohomology ring, see for instance Kochman 96, first equation in the proof of prop. 4.2.9, and since hence $\iota$ does act like the unit that it is in $H^\bullet(B;R)$).

Now since $d_{n+1}$ is a graded derivation and vanishes on $E_{n+1}^{\bullet,0}$ (by the above degree reasoning), it follows that its action on any element is uniquely fixed to be given by the product with $c$:

\begin{aligned} d_{n+1}(\iota \cdot b) & = d_{n+1}(\iota) \cdot b + (-1)^{n}\, \iota \cdot \underset{= 0}{\underbrace{d_{n+1}(b)}} \\ & = c \cdot b \end{aligned} \,.

This shows that $d_{n+1}$ is identified with the cup product operation in question:

$\array{ E_{n+1}^{s,n} & \simeq & H^s(B;R) \\ {}^{\mathllap{d_{n+1}}}\downarrow && \downarrow^{\mathrlap{c \cup (-)}} \\ E_{n+1}^{s+n+1, 0} & \simeq & H^{s+n+1}(B;R) } \,.$

In summary, the non-vanishing entries of the $E_\infty$-page of the spectral sequence sit in exact sequences like so

$\array{ 0 \\ \downarrow \\ E_\infty^{s,n} \\ {}^{\mathllap{ker(d_{n+1})}}\downarrow \\ E_{n+1}^{s,n} & \simeq & H^s(B;R) \\ {}^{\mathllap{d_{n+1}}}\downarrow && \downarrow^{\mathrlap{c \cup (-)}} \\ E_{n+1}^{s+n+1, 0} & \simeq & H^{s+n+1}(B;R) \\ {}^{\mathllap{coker(d_{n+1})}}\downarrow \\ E_\infty^{s+n+1,0} \\ \downarrow \\ 0 } \,.$

Finally observe (lemma ) that due to the sparseness of the $E_\infty$-page, there are also short exact sequences of the form

$0 \to E_\infty^{s,0} \longrightarrow H^s(E; R) \longrightarrow E_\infty^{s-n,n} \to 0 \,.$

Concatenating these with the above exact sequences yields the desired long exact sequence.

###### Lemma

Consider a cohomology spectral sequence converging to some filtered graded abelian group $F^\bullet C^\bullet$ such that

1. $F^0 C^\bullet = C^\bullet$;

2. $F^{s} C^{\lt s} = 0$;

3. $E_\infty^{s,t} = 0$ unless $t = 0$ or $t = n$,

for some $n \in \mathbb{N}$, $n \geq 1$. Then there are short exact sequences of the form

$0 \to E_\infty^{s,0} \overset{}{\longrightarrow} C^s \longrightarrow E_\infty^{s-n,n} \to 0 \,.$

(e.g. Switzer 75, p. 356)

###### Proof

By definition of convergence of a spectral sequence, the $E_{\infty}^{s,t}$ sit in short exact sequences of the form

$0 \to F^{s+1}C^{s+t} \overset{i}{\longrightarrow} F^s C^{s+t} \longrightarrow E_\infty^{s,t} \to 0 \,.$

So when $E_\infty^{s,t} = 0$ then the morphism $i$ above is an isomorphism.

We may use this to either shift away the filtering degree

• if $t \geq n$ then $F^s C^{s+t} = F^{(s-1)+1}C^{(s-1)+(t+1)} \underoverset{\simeq}{i^{s-1}}{\longrightarrow} F^0 C^{(s-1)+(t+1)} = F^0 C^{s+t} \simeq C^{s+t}$;

or to shift away the offset of the filtering to the total degree:

• if $0 \leq t-1 \leq n-1$ then $F^{s+1}C^{s+t} = F^{s+1}C^{(s+1)+(t-1)} \underoverset{\simeq}{i^{-(t-1)}}{\longrightarrow} F^{s+t}C^{(s+1)+(t-1)} = F^{s+t}C^{s+t}$

Moreover, by the assumption that if $t \lt 0$ then $F^{s}C^{s+t} = 0$, we also get

$F^{s}C^{s} \simeq E_\infty^{s,0} \,.$

In summary this yields the vertical isomorphisms

$\array{ 0 &\to& F^{s+1}C^{s+n} &\longrightarrow& F^{s}C^{s+n} &\longrightarrow& E_\infty^{s,n} &\to& 0 \\ && {}^{\mathllap{i^{-(n-1)}}}\downarrow^{\mathrlap{\simeq}} && {}^{\mathllap{i^{s-1}}}\downarrow^{\mathrlap{\simeq}} && \downarrow^{\mathrlap{=}} \\ 0 &\to& F^{s+n}C^{s+n} \simeq E_\infty^{s+n,0} &\longrightarrow& C^{s+n} &\longrightarrow& E_\infty^{s,n} &\to& 0 }$

and hence with the top sequence here being exact, so is the bottom sequence.

## References

• Martin Saralegi, A Gysin Sequence for Semifree Actions of $S^3$, Proceedings of the American Mathematical Society Vol. 118, No. 4 (Aug., 1993), pp. 1335-1345 (jstor)