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The classifying space $BDiff(S^1 \times S^1)$ of the diffeomorphism group of the torus is a 2-type with $\pi_1 = GL(2, \mathbb{Z})$ and $\pi_2 = \mathbb{Z} \times \mathbb{Z}$ and all higher $\pi_i$ = 0.

Presumably the action of $\pi_1$ on $\pi_2$ is the standard action of $GL(2, \mathbb{Z})$ on $\mathbb{Z} \times \mathbb{Z}$.

What is its first Postnikov invariant $k \in H^3_{grp}(GL(2, \mathbb{Z}), \mathbb{Z} \times \mathbb{Z})$?

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It is zero, because $BDiff(S^1 \times S^1) \to BGL_2(\mathbb{Z})$ is split by the standard action of $GL_2(\mathbb{Z})$ on $S^1 \times S^1 = \mathbb{R}^2/\mathbb{Z}^2$. In other words $$BDiff(S^1 \times S^1) \simeq B(S^1 \times S^1 \rtimes GL_2(\mathbb{Z})).$$

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  • $\begingroup$ Ok, thanks. So - what are some examples of closed compact manifolds $M$ where $BDiff(M)$ has a nontrivial first Postnikov invariant? $\endgroup$
    – Bruce Bartlett
    Commented Nov 12, 2016 at 7:18
  • $\begingroup$ $M=S^2$ (by Smale's Theorem $Diff(M) \cong O(3)$) or $M=S^3$ (Hatcher's Theorem says $Diff(M) \cong O(4)$). $\endgroup$
    – Gustavo Granja
    Commented Nov 12, 2016 at 17:21
  • $\begingroup$ @Gustavo: The classifying spaces of both those groups have trivial first k-invariants. $\endgroup$
    – Oscar Randal-Williams
    Commented Nov 12, 2016 at 18:38
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    $\begingroup$ Maybe an easier question: Are there examples of closed compact manifolds $M$ where the mapping class group exact sequence $1 \rightarrow Diff_0(M) \rightarrow Diff(M) \rightarrow \Gamma(M) \rightarrow 1$ does not split (i.e. the projection map $Diff(M) \rightarrow \Gamma(M)$ does not have a section), but nevertheless the first Postnikov invariant is still zero? Is that even possible? $\endgroup$
    – Bruce Bartlett
    Commented Nov 12, 2016 at 20:44
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    $\begingroup$ I guess the first Postnikov invariant vanishes automatically whenever $\pi_1 Diff(M)$ is trivial, so in particular for surfaces with genus $g \geq 18$ the sequence above doesn't split (Morita I think?) but its Postnikov invariant vanishes. $\endgroup$
    – Bruce Bartlett
    Commented Nov 12, 2016 at 22:22

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