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I'm reading section 2.1 of Lawson's book, Spin Geometry. The book states the following fact. Let $X$ be a manifold and $E$ a vector bundle over it. Equip $E$ with a Riemannian structure. Let $P_O$ be the bundle of orthonormal frames in $E$ which is a principal $O_n$ bundle. The fibration $O_n \rightarrow P_O(E) \rightarrow X$ gives an exact sequence $0 \rightarrow H^{0}(X;\mathbb{Z}_2) \rightarrow H^{0}(P_O(E);\mathbb{Z}_2) \rightarrow H^{0}(O_n;\mathbb{Z}_2) \rightarrow H^{1}(X;\mathbb{Z}_2) $ and the fibration $SO_n \rightarrow P_{SO}(E) \rightarrow X$ gives another exact sequence $0 \rightarrow H^{1}(X;\mathbb{Z}_2) \rightarrow H^{1}(P_{SO}(E);\mathbb{Z}_2) \rightarrow H^{1}(SO_n;\mathbb{Z}_2) \rightarrow H^{2}(X;\mathbb{Z}_2) $. Lawson only says that we can deduce them from Serre spectral sequence but I don't know how. Could someone give an explicit recipe? (By the way, we are around page 79 to page 81.)

Thank you.

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  • $\begingroup$ The space $P_{SO}(E)$ in the second statement becomes defined only when when the vector is oriented, so the second statement makes sense only in that case. It is true if in addition $X$ is connected (which I imagine is being assumed). $\endgroup$
    – Tom Goodwillie
    Oct 13, 2013 at 16:04
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    $\begingroup$ The first statement is false when $X$ is a point. $\endgroup$
    – Tom Goodwillie
    Oct 13, 2013 at 16:05
  • $\begingroup$ ("when the vector bundle is oriented", I meant in my first comment) $\endgroup$
    – Tom Goodwillie
    Oct 13, 2013 at 17:21
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    $\begingroup$ The second exact sequence is (part of) what's sometimes called the Serre spectral sequence. It's deduced from the Serre spectral sequence as in Example 1.A of McCleary's "User's guide to spectral sequences". $\endgroup$
    – Mark Grant
    Oct 13, 2013 at 20:15
  • $\begingroup$ Can you elaborate how this is related to example 1.A? $\endgroup$
    – Matthias Ludewig
    Oct 14, 2013 at 5:43

1 Answer 1

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What you are asking about is a consequence of a more general statement about fibrations. Let $p: E \to B$ be a fibration with $B$ path connected and based. Set $F = p^{-1}(*)$. Assume $B$ is $r$-connected and $p$ is $s$-connected. Then there's a exact sequence $$ 0 \to H^0(B) \to H^0(E) \to H^0(F) \to H^1(B) \to \cdots \to H^{r+s}(F) \to H^{r+s+1}(B) $$ One way to prove this is to show that the evident map $E \cup CF \to B$, whose domain is the mapping cone of $F\to E$, is $(r+s+2)$-connected. There are a variety of ways to show this, one of which is called the "dual Blakers-Massey Theorem." Then the long exact cohomology sequence of the cofiber sequence $F \to E \to E\cup CF$ combined with the connectivity statement gives what you want.

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  • $\begingroup$ In the case in the OP, neither $F$ nor $B$ are 1-connected, so how does one get to $H^2(B)$? $\endgroup$
    – Ryan Unger
    Jul 2, 2018 at 16:01
  • $\begingroup$ If $F,B$ are $0$-connected, then the map $H^2(B) \to H^2(E\cup CF)$ is injective. The exactness on the right is $\cdots \to H^1(F) \to H^2(E\cup CF)$. There is no a priori reason in general for the image of this last map to factor through $H^2(B)$. It does factor through $H^1(\Omega B)$. So there is a question about a transgression in this case. $\endgroup$
    – John Klein
    Jul 2, 2018 at 23:37
  • $\begingroup$ Alternatively, there is a map $H^1(F) \to H^2(\tilde B)$ which fits the exact sequence above, where $\tilde B$ is the universal cover of $B$. All this makes me slightly skeptical that Lawson's second sequence is correct. But he's working mod 2. Maybe that makes all the difference. $\endgroup$
    – John Klein
    Jul 2, 2018 at 23:43
  • $\begingroup$ It's unclear to me what are the maps. In particular, I have a problem with the $0$-th dimension situation. Suppose $r,s\geq 0$, then $F,E,B$ are all connected and thus $H^0$ is free of rank 1, and the beginning of this sequence can't happen with the obvious maps. In n-lab the homological sequence stops in degree 1, is this also the case here? $\endgroup$
    – Noam Zimhoni
    Jul 8, 2023 at 9:56
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    $\begingroup$ @NoamZimhoni In the sequence I exhibited $H^\ast(-)$ means reduced cohomology. In particular, $H^0(X)$ is trivial if $X$ is connected. $\endgroup$
    – John Klein
    Jul 9, 2023 at 13:29

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