Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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)

share|cite|improve this question
up vote 4 down vote accepted

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.

share|cite|improve this answer
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 – Juan S Sep 2 '11 at 1:01
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 $H_{p+q-1}(F)$. – Mariano Suárez-Alvarez Sep 2 '11 at 1:04
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) – Juan S Sep 2 '11 at 1:12

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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