13
$\begingroup$

$\DeclareMathOperator\Top{Top}$I would like to compute $\pi_3\Top(4)$ and $\pi_4\Top(4)$. It is known that $\Top(4)/O(4) \rightarrow \Top/O$ is 5-connected and $$ \pi_k(\Top/O) = \begin{cases} 0, & k = 0,1,2,4; \\ \mathbb{Z}/2, & k = 3 . \end{cases} $$ The long exact sequence of homotopy groups for fibre sequence $\Top(4)/O(4) \rightarrow BO(4) \rightarrow B\Top(4)$ is: $$\pi_5\Top(4)/O(4) \rightarrow \pi_4 O(4) \rightarrow \pi_4 \Top(4)\rightarrow \pi_4\Top(4)/O(4) \rightarrow \pi_3 O(4) \rightarrow \pi_3 \Top(4) \rightarrow\pi_3\Top(4)/O(4) \rightarrow \pi_2 O(4)$$ By plugging in the known groups ($SO(4) \cong S^3 \times \mathbb{R}P^3$) we get two exact sequences: $$\pi_5\Top(4)/O(4) \rightarrow \mathbb{Z}/2 \times \mathbb{Z}/2 \rightarrow \pi_4 \Top(4) \rightarrow 0$$ $$0 \rightarrow \mathbb{Z} \times \mathbb{Z} \rightarrow \pi_3 \Top(4) \rightarrow \mathbb{Z}/2 \rightarrow 0$$ I can also show that $\pi_4\Top(4)$ is at least $\mathbb{Z}/2$ by comparing to the same long exact sequence for $O(5), \Top(5)$, but it still may be $\mathbb{Z}/2 \times \mathbb{Z}/2$. Is there a way to finish the computation?

Improve this question
$\endgroup$
1
  • 2
    $\begingroup$ You might like to look at Randall (On 4-dimensional bundle theories. Differential topology, foliations, and group actions (Rio de Janeiro, 1992), 217–233. Contemp. Math., 161, 1994) says that $\pi_3(STOP(4))$ is $\mathbb{Z} \oplus \mathbb{Z} \oplus \mathbb{Z}_2$. In other words, it seems that your exact sequence at the bottom splits. There's some more information in his related publication with Schweitzer (On Foliations, Concordance Spaces, and the Smale Conjectures) in the same volume. $\endgroup$
    – Danny Ruberman
    Commented Jul 8 at 1:36

1 Answer 1

Reset to default
14
$\begingroup$

It is $\mathbb{Z}/2 \oplus \mathbb{Z}/2$, see this note.

Improve this answer
$\endgroup$
3
  • 1
    $\begingroup$ Did you write this specifically to answer the OP's question?? $\endgroup$
    – David Roberts ♦
    Commented Jul 8 at 12:59
  • 1
    $\begingroup$ No, it's an unpublished note that has been sitting on my laptop for years. $\endgroup$
    – skupers
    Commented Jul 8 at 13:54
  • $\begingroup$ Thank you for sharing the note! It answers my question together with the reference from Danny Ruberman's comment. $\endgroup$
    – Oleksandr Kharchenko
    Commented Jul 9 at 10:23

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .