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Given a semisimple Lie algebra $\frak{g}$ of type $\Phi$ with a Lie algebra representation $\rho:\frak{g}\to \frak{gl}(v)$ and an arbitrary commutative ring one can associate the following gadgets:

  • A simply connected split reductive group scheme $G(\Phi)$ over $R$.

  • An abstract group $G(\Phi,R)$ which is the $R$-points of the reductive group scheme $G(\Phi)$ which comes with an action on $V\otimes_\mathbb{Z}R$ (where $V$ is an appropriate $\mathbb{Z}$ lattice inside of $\frak{v}$ corresponding to some Chevalley basis).

  • An abstract subgroup $E(\Phi,R)\leq G(\Phi,R)$ which is the subgroup generated by the root groups, that is, generated by elements of the form $U_\alpha(R)$ where $\alpha\in \Phi$ and $U_\alpha$ the root group associated to $\alpha$.

Whenever $R=k$ is a field one can also think of $E(\Phi,k)$ as an abstract group presented by symbols $x_\alpha(r)$ with $\alpha\in \Phi$ and $r\in R$ subject to the so called "Steinberg Relations". This is Corollary 3 on page 21 of this pdf.

For arbitrary rings $R$ the group $E(\Phi,R)$ will also satisfy the Steinberg relations and my question is wether or not those relations are sufficient to present $E(\Phi,R)$.

One approach to address this question is of course to follow this reference and verify that the hypothesis of $R$ being a field was not used on the process of proving Corollary 3. Skimming through the argument it seems quite possible to me, but since I want to use this result it would be very useful to me if there is already a reference that deals with this explicitly for arbitrary ring $R$, so that I do not have to repeat the argument as an appendix.

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    $\begingroup$ It seems you are asking if $K_2(\Phi,R)$ is always generated by Steinberg symbols. Definitely not. $\endgroup$
    – Wilberd van der Kallen
    Commented Oct 10, 2019 at 9:19
  • $\begingroup$ The answer is no, for example for $\Phi = A_1$ and polynomial ring $R = \mathbb{Z}[T]$. (I think this is due to P. M. Cohn.) $\endgroup$
    – spin
    Commented Oct 10, 2019 at 13:43
  • $\begingroup$ It might help to point out that Steinberg's mimeographed lecture notes were distributed for many yeers by the Yae mathematics department, but a typeset and slightly revised set of notes was published after Steinberg's death by the AMS as part of their University Lecture Series $\endgroup$
    – Jim Humphreys
    Commented Oct 10, 2019 at 19:48
  • $\begingroup$ I'm not sure I understand the claim about $V$. Do we start working over $\mathbb C$, or something less (any field? Algebraically closed? Characteristic 0?)? Is the claim that any $\mathfrak v$ carries a canonical such lattice $V$ (corresponding somehow to a choice of Chevalley basis for $G(\Phi)$?), or are you only considering certain 'natural' representations, say the ones on classical groups that arise by considering them as fixed points of involutions? $\endgroup$
    – LSpice
    Commented Oct 10, 2019 at 20:08
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    $\begingroup$ Corollary 3 does not, indeed, use the assumption that the base ring is a field, but the preceding Proposition definitely does (in statement (b), which fails dramatically in general setting). $\endgroup$
    – Andrei Smolensky
    Commented Oct 10, 2019 at 21:30

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