3
$\begingroup$

I am trying to follow the proof in Kochman's Introduction to Stable Homotopy Theory, page 59. This will be first part of a series of post.

(Serre Spectral Sequence) Let $R$ be a commutative ring and let $$F \rightarrow E \xrightarrow{\pi} B$$
be a fibration. Assume that $B$ is a simplicial complex which is either simply connected or connected with char $R=2$. Then there is a multiplicativr spectral sequence $$E^{s,t}_2 = H^s(B;H^t(F;R)) \Rightarrow H^*(E;R) $$

Q1. From statement, the first bulletted scenario, it seems that $\pi$ is a fiber bundle and not just a Serre Fibration? Also are $E,F$ also CW complexes?

Let $C(n)$ denote the set of $n$ simplices $\Delta$ of $B$. Subdivide $B$ if necessary so that then we have $(\pi^{-1}(\Delta), \pi^{-1}(\partial \Delta)) \simeq (D^n, S^{n-1}) \times F$ for each $\Delta \in C(n)$.

It then claims to have an isomorphism

$$\tilde{H}^{n+t} (E^{(n)}/E^{(n-1)};k) \cong \tilde{H}^{n+t}( \bigvee _{\Delta \in C(n)} S^n \ltimes F ;k ) \cong \bigoplus_{\Delta \in C(n)} H^t(F;k)$$


Q2:What exactly is $S^n\ltimes F$?

I am completely lost by the second isomoprhism. It doesn't seem to be the smash product.

$\endgroup$
4
  • $\begingroup$ $S^n\ltimes F$ is called the half-smash and is defined as $(S^n\times F)/(\ast\times F)$. You can check that this quotient is homeomorphic to the smash $S^n\wedge F_+$ where $F_+$ is $F$ with a disjoint basepoint. The extra basepoint is why the reduced homology group in the middle becomes the unreduced group under the final isomorphism. $\endgroup$
    – Tyrone
    Apr 19 '19 at 14:21
  • $\begingroup$ So is $\tilde{H}^*(F_+) \cong H^*(F)$ for all $*$? $\endgroup$
    – Bryan Shih
    Apr 19 '19 at 15:36
  • $\begingroup$ Just check the definitions. $\endgroup$
    – Tyrone
    Apr 19 '19 at 16:02
  • $\begingroup$ Hi Tyrone, I believe it is so! (I used LES). Then we use the suspension isomorphism, I believe. I wonder if you mind commenting to on my other post, regarding a topological construction. $\endgroup$
    – Bryan Shih
    Apr 19 '19 at 17:13
1
$\begingroup$

For now I could reply Q2. I am not totally sure my argument is correct.

I do not know what $S^n \ltimes F$ is. But let us break down the maps. Let me suppress $R$ notation.

  1. There are pointed isomorphisms. $$ E^{(n)}/E^{(n-1)} \cong \bigvee (D^n \times F/S^{n-1} \times F) $$ So your guess should be correct.
  2. Indeed, by the wedge axiom it suffices to compute $$\tilde{H}^{n+t}(D^n \times F/S^{n-1} \times F)\cong H^{n+t}(D^n\times F, S^{n-1} \times F )$$
  3. Now we could utilize Theorem 3.18 in Page 219 of Hatcher's, observing that $H^n(S^n)=H^n(S^0)\cong R$.
    $$H^{n+s}(S^{n-1} \times F) \cong H^{n+s}(F) \oplus H^{s-1}(F) $$ $$ H^{n+s}(D^n \times F) \cong H^{n+s}(F) $$ Hence fit into LES, we have $$H^{n+t+1}(F) \rightarrow H^{n+t+1}(F) \oplus H^{t}(F) \rightarrow H^{n+t}(D^n \times F, S^{n-1} \times F) \rightarrow H^{n+t}(F) \rightarrow H^{n+t}(F) \oplus H^{t-1}(F)$$ The first and last map are simply isomorphisms into the corresponding component. To verify this, we check the maps in the definition of cup product. Hence, $$H^t(F) \cong H^{n+t}(D^n \times F, S^{n-1} \times F) $$
$\endgroup$
1
  • $\begingroup$ That funny symbol I think is the "half-smash" product. You take A x B and you only "smash" one of them. I.e. you have (A x B) / (A x *) I think... See page 61 of John Harper's book Cohomology Operations $\endgroup$
    – user281395
    Feb 25 '20 at 17:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.