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Given an integral domain $R$, the Steinberg group $St_n(R)$ is the group given by generators
$e_{pq}(\lambda) := \mathbf{1} + a_{pq}(\lambda)$, $p\neq q$, $1\leq p,q \leq n$

Subject to the relations
$$\begin{align} e_{ij}(\lambda) e_{ij}(\mu) &= e_{ij}(\lambda+\mu) \\ \left[ e_{ij}(\lambda),e_{jk}(\mu) \right] &= e_{ik}(\lambda \mu) && \mbox{for } i \neq k\\ \left[ e_{ij}(\lambda),e_{kl}(\mu) \right] &= \mathbf{1} && \mbox{for } i \neq l, j \neq k\\ \end{align}$$

The Steinberg group is the universal central extension of the special linear group over $R$; $Sl_n(R)$.

Is there a description of the Steinberg group $St_n(Z)$, the special linear group over the integers as a lattice in some lie group, and some covering map realizing the universal central extension of $Sln(R)$ ( real coefficients), which restricts to the integral universal central extension of $Sln(Z)$ given by the Steinberg group ?

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2 Answers

For $n\ge 3$ the Steinberg group $\text{St}_n(\mathbf{Z})$ is a lattice in the universal covering of $\text{SL}_n(\mathbf{R})$ (which has fundamental group the cyclic group of order 2).

To see that the inverse image of $\text{St}_n(\mathbf{Z})$ in the universal covering of $\text{SL}_n(\mathbf{R})$ is indeed a non trivial central extension, essentially amounts to check that the image of $K_2(\mathbf{Z})$ in the topological $K_2$ or the reals is injective (but I guess that a suitable direct argument, for instance using an element of order 2, can work as well).

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I mean "of the reals", sorry. –  Yves Cornulier Aug 21 '12 at 22:09
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EDIT: This started to be an answer, but my recollections were inaccurate. It's essential here to distinguish carefully between what happens when $n=2$ (where life is much more complicated) and when $n \geq 3$ (where the situation stabilizes). The definition of the Steinberg group differs in these situations.

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@Jim: I respectfully disagree: $SL_n(\mathbb{R})$ is NOT simply connected for $n\geq 3$ (it is homotopy equivalent to $SO(n)$), so its universal cover is precisely of degree 2, and therefore the answer for the Steinberg group seems to be affirmative. –  Alain Valette Aug 21 '12 at 21:54
    
@Alain: On further thought, I see that I was too hasty in my "answer", so I've edited most of it out. Sorry for the confusion. –  Jim Humphreys Aug 22 '12 at 0:00
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