# Schur multiplier of a Chevalley group of type $D_5$

$$\DeclareMathOperator\EO{EO}\DeclareMathOperator\SO{SO}\DeclareMathOperator\St{St}\DeclareMathOperator\Sp{Sp}$$This is sort of a follow up question to my post here regarding the commutator subgroup of $$\SO(5,5,\mathbb{Z})$$. As discussed there, this group can be identified with the group of $$\mathbb{Z}$$ points of a Chevalley–Demazure group scheme of type $$D_5$$. Now, upon reading a little further into "Generators, relations, and coverings of Chevalley groups over commutative rings" by Stein, it seems that we may be able to determine the second integral homology of $$\SO(5,5,\mathbb{Z})$$ as well. In particular, from Corollary 5.4, we know that if $$K_2(D_5,\mathbb{Z}) = \ker(\St(D_5,\mathbb{Z})\rightarrow \EO(5,5,\mathbb{Z}))$$ is central, then the second homology of $$EO(5,5,\mathbb{Z})$$ is isomorphic to $$K_2(D_5,\mathbb{Z})$$. Here we have let $$\St(D_5,\mathbb{Z})$$ denote the Steinberg group and $$E(D_5,\mathbb{Z})$$ the elementary subgroup of our Chevalley group $$\SO(5,5,\mathbb{Z})$$. As was shown in the linked post, $$\EO(5,5,\mathbb{Z}) = \SO(5,5,\mathbb{Z})$$.

If our Chevalley group were to be simply connected, then $$K_2(D_5,\mathbb{Z})$$ is central and has been computed. This is discussed in the introduction to "The Schur Multipliers of $$\Sp_6(\mathbb{Z})$$, $$\text{Spin}_8(\mathbb{Z})$$, $$\text{Spin}_7(\mathbb{Z})$$, and $$F_4(\mathbb{Z})$$" by Stein. From Corollary 3.4 and the remarks at the end of section 4 in "Stability theorems for $$K_1$$, $$K_2$$ and related functors modeled on Chevalley groups" by Stein, it seems like $$K_2(D_5,\mathbb{Z})$$ is central. The problem then becomes computing $$K_2(D_5,\mathbb{Z})$$ in our case.

I'm not sure how to do this. Corollary 4.3.5 of "Clifford Algebras and Spinor Norms Over a Commutative Ring" by Bass seems relevant here, but I don't think it applies exactly here.

According to Theorem 5.3 of the first paper of Mike Stein that you mention, the universal central extension of the Chevalley group of type $$D_5$$ over the integers is given by the Steinberg generators and relations. The centre of this group is $$\mathbb{Z}/4$$. This implies that if $$G$$ is the group of adjoint type then $$H_2(G)$$ is $$\mathbb{Z}/4$$. The special orthogonal version is already a $$2$$-fold cover of the adjoint group, with minus the identity as the non-trivial central element, so $$H_2$$ of that group is just $$\mathbb{Z}/2$$.

If you're interested in other groups of type $$D_n$$ for $$n\geqslant 5$$, the behaviour depends on the parity of $$n$$. For $$n$$ even, the centre of the Steinberg group is $$\mathbb{Z}/2\times\mathbb{Z}/2$$, while for $$n$$ odd it is $$\mathbb{Z}/4$$.

The case $$n=4$$ requires special treatment, which can be found in Theorem (2.10) of the paper of Stein and van der Kallen, "On the Schur multipliers of Steinberg and Chevalley groups over commutative rings". In this case the Steinberg group is not the universal central extension.

• Thank you for the response! That's interesting about the dependence on parity. Does that matter for the first homology? That is, do we have $EO(n,n,\mathbb{Z}) = SO(n,n,\mathbb{Z})$ for all $n\geq 5$? Or do we have to worry about parity? Commented May 23, 2023 at 18:30
• Also, could you please provide a little bit more details about your last two deductions in the first paragraph regarding $H_2(G)$ when $G$ is of adjoint type and $H_2$ of the special orthogonal group? Commented May 23, 2023 at 21:11
• Can I refer you to the appendix C in Conrad, "Reductive group schemes" in "Autour des schémas en groupes, I" where this is discussed in detail. In particular, a great deal of care is needed in the very definition of the special orthogonal group if $n$ is even. With the appropriate definition, it is perfect as a group scheme over Spec($\mathbb{Z})$. For the centres, see Example 1.5.2 in the same article. You might also enjoy looking at the follow-up article of Conrad, "Non-split reductive groups over $\mathbb{Z}$" in volume II of the same set. Commented May 24, 2023 at 8:50