2
$\begingroup$

Consider the map $L_0(\mathbb{Z})\to L_0(\mathbb{Z}_2)$ of Wall L-groups induced by the surjection $\mathbb{Z}\to \mathbb{Z}_2$. Here $\mathbb{Z}_2$ is the cyclic group (field) of order 2. We know that $L_0(\mathbb{Z})\cong \mathbb{Z}$ (generated by the $E_8$ form) and $L_0(\mathbb{Z}_2)\cong\mathbb{Z}_2$. Is this map the zero map or is it a surjection?

Note: the notation in the literature can a little confusing. By $L_0(\mathbb{Z})$ I mean $L_0(\mathbb{Z}[e])$ and not $L_0(\mathbb{Z}[\mathbb{Z}])$, and by $L_0(\mathbb{Z}_2)$ I mean $L_0(\mathbb{F}_2)$ and not $L_0(\mathbb{Z}[\mathbb{Z}_2])$.

Improve this question
$\endgroup$
2
  • $\begingroup$ What is your reference for $L_0(Z_2)$? If there is a reasonable invariant, perhaps you can evaluate it on the mod 2 reduction of the $E_8$ form. $\endgroup$
    – Danny Ruberman
    Commented Aug 2, 2016 at 1:39
  • $\begingroup$ If a field $\mathbb{F}$ is square-root closed, then there is an isomorphism $L_0(\mathbb{F})=\mathbb{Z}_2$, so in fact $L_0(\mathbb{F}_{2^n})=\mathbb{Z}_2$ for all $n$. $\endgroup$
    – Stanley Chang
    Commented Aug 2, 2016 at 2:05

1 Answer 1

Reset to default
6
$\begingroup$

The map is the zero map. The isomorphism $L_0(\mathbb Z/2)=\mathbb Z/2$ is given by the Arf invariant, so one needs to compute the Arf invariant of the mod 2 reduction of $E_8$, and this is zero.

(With the usual basis for $E_8$, one can take $$e_1+e_3,\ e_2,\ e_1,\ e_4,\ e_1+e_5,\ e_6+e_8,\ e_7,\ e_8$$ as a symplectic basis mod 2.)

Improve this answer
$\endgroup$

You must log in to answer this question.

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