- #1

Bacle

- 662

- 1

I am a bit confused about this result:

Mg/Mg^(2) ~ Sp(2g,Z) (group iso.)

Where:

i) Mg is the mapping class group of the genus-g surface, i.e., the collection of diffeomorphisms: f:Sg-->Sg , up to isotopy.

ii)Mg^(2) is the subgroup

of Mg of maps that induce the identity in H_1(Sg,Z/2), and

iii)Sp(2g,Z) is the symplectic

group associated with the intersection form over Z, i.e., we consider the pair (Z-module,

Symplectic form) given by: (H_1(Sg,Z), (a,b)_2), where (a,b)_2 is the intersection form

in in H_1(Sg,Z/2), so that Sp(2g,Z) is the subgroup of Gl( H_1(Sg,Z)) that preserves

(, )_2.

For one thing, the mapping class group is in a different "category" (used in the informal

sense) than Sp(2g,Z) ; Mg and Mg^2 are maps f,g :Sg-->Sg , and h in Sp(2g,Z) is a linear

map m: H_1(Sg,Z)-->H_1(Sg,Z) (with (x,y)_2 =(m(x),m(y))_2.

I understand being in different categories does not make an iso. impossible, but I

don't see what the isomorphism would be.

I don't know if these are Lie group isos. or just standard group isos. Any ideas?

Thanks.