Affine Steinberg groups vs Steinberg groups over Laurent polynomials

Let R be a commutative ring and Φ be a finite (also called spherical) reduced irreducible root system of rank 2. I will denote by St(Φ,R) the Steinberg group of type Φ over R, i. e. the quotient of the free product αΦXα of root subgroups Xα=xα(a)aR modulo Chevalley commutator formulae:

  • [xα(a);xβ(b)]=iα+jβΦxiα+jβ(Ni,jα,βaibj), i,jN;
    (here Ni,jα,β are integers equal to ±1,2,3).

There is a more general definition of Steinberg groups due to J. Tits which works for infinite root systems. It is very similar to the above one with the only difference that this time we only consider root subgroups corresponding to real roots of Ψ and Chevalley commutator formulae are now imposed only for prenilpotent pairs of real roots α,β (so that (Nα+Nβ)Ψre is finite).

If R=k is a field we can obtain the corresponding Kac-Moody group from this generalized Steinberg group by modding out Steinberg symbols {u,v}=hα(u)hα(v)hα(uv)1, u,vk×.

Question: Let ˜Φ be the affine root system corresponding to Φ.
There is an arrow St(˜Φ,R)St(Φ,R[t,t1]) (sending x(α,m)(a)xα(atm), see below).
Is this arrow an isomorphism?

I suspect that the answer to my question is affirmative. Below I’ll try to sketch what I’ve learned from section 4 of this paper.

The real roots of the affine root system ˜Φ can be thought of as pairs ˜Φre=Φ×Z. A direct check shows that the roots (α,m), (β,n) form a prenilpotent pair iff αβ.

Thus, by the definition, the group St(˜Φ,R) is the quotient of (α,m)Φ×ZX(α,m), modulo the following two families of relations:

  1. commutator formulae for (α,m), (β,n) in the case when α and β form a classically prenilpotent pair, i. e. α±β;
  2. commutator formulae of the form [x(α,m)(a);x(α,n)(b)]=1, for m,nZ, a,bR (the case of a prenilpotent but not classically prenilpotent pair).

If we identify x(α,m)(a) with xα(atm) the latter presentation looks just like a variant of presentation for
St(Φ,R[t,t1]) formulated in terms of monomials rather than polynomials.

Answer

Attribution
Source : Link , Question Author : Sergey Sinchuk , Answer Author : Community

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