I am trying to derive the following result of Serre's:

Let $F \hookrightarrow E \stackrel{p}{\to} B$ be a fibration with $B$ simply connected. Suppose $H_i(B)=0$ for $0 < i < p $ and that $H_j(F)=0$ for $0 < j < q$. There is an exact sequence $$H_{p+q-1}(F) \to H_{p+q-1}(E) \to H_{p+q-1}(B) \to H_{p+q-2}(F) \to \cdots \to H_1(E) \to 0$$

The proof is via the Serre spectral sequence.

The $E_2$-term is zero whenever $0 < i < p$ or $0 < j < q$, so we essentially end up with three blocks of non-zero entries (and I guess a non-zero in the (0,0) spot).

Now if we consider the differential $d_n:E^n_{n,0} \to E^n_{0,n-1}$ we see that there are no more (non-zero) differentials entering or leaving these positions and so $E^\infty_{0,n-1} \simeq E^n_{0,n-1}/\text{im }d_n$ and $E^\infty_{n,0} \simeq \ker d_n$. In fact if we restrict to $n<p+q$ these are the only non-zero differentials, and so (using the fact that that $H_0(F)\simeq H_0(B) \simeq \mathbb{Z}$) and so we get the exact sequence $$0 \to E^\infty_{n,0} \to H_n(B) \to H_{n-1}(F) \to E^\infty_{0,n-1}\to 0$$

This is where i got stuck. Looking up the solution in Mosher and Tangora they state:

The normal series for $H_n(E)$, which consists of the $E^\infty_{i,j}$ terms for which $i+j=n$, collapses to the exact sequence $$O \to E^\infty_{0,n} \to H_n(E) \to E^\infty_{n,0}$$

The result then follows from splicing together the exact sequences. So my basic question is: where does this second SES arise? Does this hold for all $n$ or only $n< p + q$? (I do not know what M&T mean by 'the normal series' either)


1 Answer 1


The short exact sequence $$O \to E^\infty_{0,n} \to H_n(E) \to E^\infty_{n,0}\to 0$$ comes from the fact that the spectal sequence converges. Remember that this means that the filtration on the complex $E$ induces a filtration on its homology such that the corresponding subquotients of $H_n(N)$ are isomorphic to the $E^\infty_{p,q}$ with $p+q=n$. In your case, only two of these are non-zero, so the filtration on $H_n(E)$ is a two-step filtration, which is the same thing as a submodule of $H_n(E)$ —in this case, the submodule is $E^\infty_{0,n}$ and the corresponding subquotient is $E^\infty_{n,0}$.

This is explained in detail, if I recall correctly, in the first chapter of McCleary's book on spectral sequences.

  • $\begingroup$ Thanks Mariano. Why are only two non-zero? What about the upper right hand block - i.e. when $n> p+q$. I've checked McClearly and there is a similar argument there for the Gysin sequence, when the spectral sequence is only non-zero in two columns $\endgroup$
    – Juan S
    Sep 2, 2011 at 1:01
  • $\begingroup$ In the range involved in the long exact sequence you mention in the question, there are two non-zero groups in each anti-diagonal. Notice that the short exact sequence stops on the left at $H_{p+q-1}(F)$. $\endgroup$ Sep 2, 2011 at 1:04
  • $\begingroup$ Ok. So the claim is only for $n<p+q$? I am much happier with that statement! (I know the sequence is only over that range, but the text seemed to imply it was for any $n$, hence my confusion) $\endgroup$
    – Juan S
    Sep 2, 2011 at 1:12

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