group cohomology, nonabelian group cohomology, Lie group cohomology
cohomology with constant coefficients / with a local system of coefficients
differential cohomology
algebraic topology – application of higher algebra and higher category theory to the study of (stable) homotopy theory
(also nonabelian homological algebra)
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.
Let $R$ be a commutative ring and let
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
(e.g. Switzer 75, section 15.30, Kochman 96, corollary 2.2.6)
Under the given assumptions there is the corresponding Serre spectral sequence
Since the ordinary cohomology of the n-sphere fiber is concentrated in just two degees
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 $(\bullet,n)$:
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
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
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
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 $c$:
This shows that $d_{n+1}$ is identified with the cup product operation in question:
In summary, the non-vanishing entries of the $E_\infty$-page of the spectral sequence sit in exact sequences like so
Finally observe (lemma ) that due to the sparseness of the $E_\infty$-page, there are also short exact sequences of the form
Concatenating these with the above exact sequences yields the desired long exact sequence.
Consider a cohomology spectral sequence converging to some filtered graded abelian group $F^\bullet C^\bullet$ such that
$F^0 C^\bullet = C^\bullet$;
$F^{s} C^{\lt s} = 0$;
$E_\infty^{s,t} = 0$ unless $t = 0$ or $t = n$,
for some $n \in \mathbb{N}$, $n \geq 1$. Then there are short exact sequences of the form
(e.g. Switzer 75, p. 356)
By definition of convergence of a spectral sequence, the $E_{\infty}^{s,t}$ sit in short exact sequences of the form
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
or to shift away the offset of the filtering to the total degree:
Moreover, by the assumption that if $t \lt 0$ then $F^{s}C^{s+t} = 0$, we also get
In summary this yields the vertical isomorphisms
and hence with the top sequence here being exact, so is the bottom sequence.
Robert Switzer, section 15.30 of Algebraic Topology - Homotopy and Homology, Die Grundlehren der Mathematischen Wissenschaften in Einzeldarstellungen, Vol. 212, Springer-Verlag, New York, N. Y., 1975 (doi:10.1007/978-3-642-61923-6)
Stanley Kochmann, section 2.2. of Bordism, Stable Homotopy and Adams Spectral Sequences, AMS 1996
Wikipedia, Gysin sequence
James Milne, section 16 of Lectures on Étale Cohomology
Formalization in homotopy type theory is discussed in
Applications:
Last revised on June 12, 2019 at 16:48:17. See the history of this page for a list of all contributions to it.