Talk:5-manifolds: 1-connected

1 Conjecture about mapping class groups of 1-connected 5-manifolds

Let $M$$== Conjecture about mapping class groups of 1-connected 5-manifolds == <wikitex>; Let $M$ be a closed, smooth, 1-connected 5-manifold, [[User:Matthias Kreck|Matthias Kreck]] and [[User:Diarmuid Crowley|Diarmuid Crowley]] conjecture that there is an isomorphism of abelian groups $$ \pi_0(\SDiff(M)) \cong \Omega_6(B^2(M)) $$ where $B^2(M)$ is the normal $-type of $M$ as defined in {{cite|Kreck1999}}. For example, if $M$ is Spinable with $H_2(M) \cong H$ then $$ \Omega_6(B^2(M)) \cong \Omega_6^{\Spin}(K(H, 2)).$$ At present we are checking the details of the proof of this conjecture using the methods of {{cite|Kreck1999}}. </wikitex> [[User:Diarmuid Crowley|Diarmuid Crowley]] 10:02, 29 September 2009 (UTC) == Earlier work of Fang == <wikitex>; The group $\pi_0\SDiff(M)$ was computed in \cite{Fang1993} provided that $H_2(M)$ has no $-torsion and no $-torsion. <wikitex> == Up-date of conjecture: module structure == <wikitex>; If the conjecture above holds, then from the short exact sequence $$ 0 \rightarrow \pi_0\SDiff(M) \rightarrow \pi_0\text{Diff}_+(M) \rightarrow \Aut(H_2(M)) \rightarrow 0 \quad (\ast)$$ we obtain an action of $\Aut(H_2(M))$ on the abelian group $\pi_0\SDiff(M) \cong \Omega_6(B^2(M))$. [[User:Diarmuid Crowley|Diarmuid Crowley]] and [[User:Matthias Kreck|Matthias Kreck]] also conjecture that the action of $\Aut(H_2(M))$ is via the induced action on $B^2(M)$. In particular, if $M$ is spinable with $H = H_2(M)$, then $\Aut(H_2(M))$ acts on $K(H, 2)$ in the obvious way and so on $\Omega_6^{\Spin}(K(H, 2))$. </wikitex>M$ be a closed, smooth, 1-connected 5-manifold, Matthias Kreck and Diarmuid Crowley conjecture that there is an isomorphism of abelian groups

$\displaystyle \pi_0(\SDiff(M)) \cong \Omega_6(B^2(M))$ $$\displaystyle \pi_0(\SDiff(M)) \cong \Omega_6(B^2(M))$$

where $B^2(M)$$B^2(M)$ is the normal $2$$2$-type of $M$$M$ as defined in [Kreck1999]. For example, if $M$$M$ is Spinable with $H_2(M) \cong H$$H_2(M) \cong H$ then

$\displaystyle \Omega_6(B^2(M)) \cong \Omega_6^{\Spin}(K(H, 2)).$ $$\displaystyle \Omega_6(B^2(M)) \cong \Omega_6^{\Spin}(K(H, 2)).$$

At present we are checking the details of the proof of this conjecture using the methods of [Kreck1999].

Diarmuid Crowley 10:02, 29 September 2009 (UTC)

2 Earlier work of Fang

The group $\pi_0\SDiff(M)$$\pi_0\SDiff(M)$ was computed in [Fang1993] provided that

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$H_2(M)$ has no $2$$2$-torsion and no $3$$3$-torsion.

Up-date of conjecture: module structure

If the conjecture above holds, then from the short exact sequence

$\displaystyle 0 \rightarrow \pi_0\SDiff(M) \rightarrow \pi_0\text{Diff}_+(M) \rightarrow \Aut(H_2(M)) \rightarrow 0 \quad (\ast)$ $$\displaystyle 0 \rightarrow \pi_0\SDiff(M) \rightarrow \pi_0\text{Diff}_+(M) \rightarrow \Aut(H_2(M)) \rightarrow 0 \quad (\ast)$$

we obtain an action of $\Aut(H_2(M))$$\Aut(H_2(M))$ on the abelian group $\pi_0\SDiff(M) \cong \Omega_6(B^2(M))$$\pi_0\SDiff(M) \cong \Omega_6(B^2(M))$. Diarmuid Crowley and Matthias Kreck also conjecture that the action of $\Aut(H_2(M))$$\Aut(H_2(M))$ is via the induced action on $B^2(M)$$B^2(M)$. In particular, if $M$$M$ is spinable with $H = H_2(M)$$H = H_2(M)$, then $\Aut(H_2(M))$$\Aut(H_2(M))$ acts on $K(H, 2)$$K(H, 2)$ in the obvious way and so on $\Omega_6^{\Spin}(K(H, 2))$$\Omega_6^{\Spin}(K(H, 2))$.