Zeeman's Comparison Theorem

In homological algebra, Zeeman's comparison theorem, introduced by Christopher Zeeman, gives conditions for a morphism of spectral sequences to be an isomorphism.

Statement

Illustrative example

As an illustration, we sketch the proof of Borel's theorem, which says the cohomology ring of a classifying space is a polynomial ring.

First of all, with ''G'' as a Lie group and with $\mathbb$ as coefficient ring, we have the Serre spectral sequence $E_2^$ for the fibration $G \to EG \to BG$. We have: $E_ \simeq \mathbb$ since ''EG'' is contractible. We also have a theorem of Hopf stating that $H^*(G; \mathbb) \simeq \Lambda(u_1, \dots, u_n)$, an exterior algebra generated by finitely many homogeneous elements.

Next, we let $E(i)$ be the spectral sequence whose second page is E(i)_2 = \Lambda(x_i) \otimes \mathbb _i/math> and whose nontrivial differentials on the ''r''-th page are given by $d(x_i) = y_i$ and the graded Leibniz rule. Let $^ E_ = \otimes_i E_(i)$. Since the cohomology commutes with tensor products as we are working over a field, $^ E_$ is again a spectral sequence such that $^ E_ \simeq \mathbb \otimes \dots \otimes \mathbb \simeq \mathbb$. Then we let
:$f: ^ E_r \to E_r, \, x_i \mapsto u_i.$
Note, by definition, ''f'' gives the isomorphism $^ E_r^ \simeq E_r^ = H^q(G; \mathbb).$ A crucial point is that ''f'' is a "ring homomorphism"; this rests on the technical conditions that $u_i$ are "transgressive" (cf. Hatcher for detailed discussion on this matter.) After this technical point is taken care, we conclude: $E_2^ \simeq ^ E_2^$ as ring by the comparison theorem; that is, E_2^ = H^p(BG; \mathbb) \simeq \mathbb _1, \dots, y_n

References

Bibliography

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Spectral sequences
Theorems in algebraic topology

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