Let $R$ be a discrete valuation ring containing an algebraically closed field $K$ of characteristic zero and let $L$ be a Lie algebra over $R$ whose underlying $R$-module is finitely generated and free.

Let $\mathfrak{m}$ denote the maximal ideal of $R$ and $\hat{R}$ the $\mathfrak{m}$-adic completion of $R$. While thinking about something else I came up with the following question, which I cannot answer:

Is it true that if $L \otimes_R \hat{R}$ is isomorphic (as Lie algebras) to $\mathfrak{g} \otimes_K \hat{R}$ for some simple Lie algebra $\mathfrak{g}$ over $K$, then $L$ is already isomorphic to $\mathfrak{g} \otimes_K R$ ?

I have the feeling that this is either well-known or false (or both), but since I am no expert in Lie-theory, I thought I might just ask it here. Thanks in advance.


2 Answers 2


One can also give explicit high-rank counterexamples using special orthogonal groups of quadratic lattices over $R$ with non-degenerate reduction, and we can also arrange that $R$ has $K$ as its residue field (likewise for Jason Starr's examples).

To give wider context, let $K$ be any field of characteristic 0, $R$ any $K$-algebra, $\mathfrak{g}$ any semisimple Lie algebra over $K$ (we'll use $\mathfrak{so}_n$ in the end for any odd $n \ge 3$), and consider in place of $\widehat{R}$ any faithfully flat $R$-algebra $R'$.

Now consider a projective $R$-module $\Lambda$ of constant odd rank $n \ge 3$ and a quadratic form $q:\Lambda \rightarrow R$ over $R$ that is non-degenerate at every point of ${\rm{Spec}}(R)$. Let $H = {\rm{SO}}(q)$ be the associated special orthogonal group scheme; this is a smooth affine $R$-group whose geometric fibers over Spec($R$) are the classical special orthogonal groups of the corresponding fibers of $q$. In particular, since $n$ is odd it follows that such fibers are of adjoint type (and $\mu_2 \times H = {\rm{O}}(q)$ is the orthogonal group scheme of $q$).

Let $L = \mathfrak{so}(q)$ be the Lie algebra of $H$, and let $\mathfrak{g}$ be the Lie algebra for the special orthogonal group ${\rm{SO}}_n := {\rm{SO}}(q_n)$ of the corresponding "split" rank-$n$ quadratic form $$q_n := x_0^2 + x_1 x_2 + \dots + x_{2m-1}x_{2m}$$ where $2m+1=n$. Let $R'$ be a faithfully flat $R$-algebra such that $q$ becomes homothetic to $q_n$ over $R'$; such an $R'$ always exists (can even be chosen to be etale over $R$ since $R$ is a $\mathbf{Z}[1/2]$-algebra). Thus, $L$ and $\mathfrak{so}_n$ become isomorphic over $R'$. But I claim that if $L$ and $\mathfrak{so}_n$ are $R$-isomorphic then necessarily $q$ and $q_n$ are homothetic over $R$ (i.e., $q$ is isometric to $u q_n$ for some $u \in R^{\times}$).

It is a general fact that for odd $n \ge 3$, ${\rm{SO}}_n$ represents the automorphism functor of its Lie algebra on the category of all $K$-algebras, as may be checked on $\overline{K}$-points since ${\rm{char}}(K)=0$. (This is a special case of a general fact about connected semisimple groups of adjoint type whose Dynkin diagram has no nontrivial automorphisms.) Thus, $L$ as an fpqc $R$-form of $\mathfrak{so}_n$ corresponds to an element in the cohomology set ${\rm{H}}^1(R'/R, {\rm{SO}}_n)$.

But since $n$ is odd, ${\rm{SO}}_n$ represents the functor of conformal isometries of $q_n$ (i.e., pairs $(T,u)$ consisting of a unit $u$ and $T \in {\rm{GL}}_n$ such that $q_n \circ T = u q_n$). Unraveling everything, this says that $L := {\mathfrak{so}}(q)$ is $R$-isomorphic to $\mathfrak{so}_n$ if and only if $q$ is homothetic to $q_n$.

Now we can give explicit counterexamples over discrete valuation rings $R$ containing an algebraically closed field $K$ of characteristic 0 (and even with $K$ as its residue field). Letting $\mathfrak{m}$ be the maximal ideal of such an $R$, consider a quadratic lattice $(R^n, q)$ of odd rank $n \ge 3$ with $\overline{q} := q \bmod \mathfrak{m}$ non-degenerate over $R/\mathfrak{m}$ such that $\overline{q}$ is isometric to $q_n$. There is no infinitesimal obstruction to lifting an isomorphism between non-degenerate quadratic spaces $(M,Q)$ and $(M',Q')$ over a $\mathbf{Z}[1/2]$-algebra $A$ (indeed, the Isom-scheme ${\rm{Isom}}(Q',Q)$ is an ${\rm{O}}(Q)$-torsor and hence is smooth since $A$ is a $\mathbf{Z}[1/2]$-algebra), so $q$ becomes isometric to $q_n$ over the completion $R' := \widehat{R}$. Thus, $L := \mathfrak{so}(q)$ is a counterexample (using $\mathfrak{g} = \mathfrak{so}_n$) provided that $q$ is not homothetic to $q_n$ over $R$.

Now writing $n=2m+1$, observe that the ring $$S := K[t_0, \dots, t_{n-1},v_1, \dots, v_{n-1}]/(v_i^2 + t_0 v_i - t_i)$$ satisfies $S/t_0 S = K[v_1, \dots, v_{n-1}]$ (in which $t_i = v_i^2$), so $t_0 S$ is prime in $S$ and $S$ is normal (e.g., by Serre's criterion). Thus, $R = S_{(t_0)}$ is a dvr and $$q = x_0^2 + t_1 x_1^2 + \dots + t_{n-1} x_{n-1}^2$$ over $R$ has reduction isometric to $q_n$. To show that $q$ is not homothetic to $q_n$ over $R$ it suffices to prove that $q$ is even anisotropic over $F = {\rm{Frac}}(R) = {\rm{Frac}}(S) = K(t_0, v_1, \dots, v_{n-1})$. This anisotropicity is "obvious", and is rigorously proved by induction on $n$ (allowed to have any parity) or perhaps in other ways too.

That $R$ does not have residue field $K$. But consider the field extension $K(\!(t_0)\!)$ of $K(t_0)$. This has countably infinite transcendence degree, so we can find elements $t_1, \dots, t_{n-1} \in K(\!(t_0)\!)$ algebraically independent over $K(t_0)$ that are all 1-units. Thus, $X^2 + t_0 X - t_i$ has a root $v_i \in K[\![t_0]\!]$. Thus, the ring $S$ above can be found inside $K[\![t_0]\!]$, so the restriction to ${\rm{Frac}}(S)$ of the $t_0$-adic valuation on $K(\!(t_0)\!)$ is a discrete valuation whose valuation ring $R$ contains $S$ (it is not $S_{(t_0)}$!) and has the same fraction field as $S$, and has residue field $K$. Clearly $q$ makes sense as a quadratic form over $R$, and as such it does the job (by the same anisotropicity argument over ${\rm{Frac}}(R) = {\rm{Frac}}(S)$!).

Remark. Theorem 3.5 of http://arxiv.org/pdf/math/0406508v5.pdf provides a much more far-reaching result: over any ring whatsoever (no field of char. 0 in the picture), the groupoid of semisimple group schemes of adjoint type is equivalent (via the Lie functor) to the groupoid of vector-bundle Lie algebras whose Killing form is fiberwise non-degenerate!

  • $\begingroup$ It is very detailed and easy to understand - great! $\endgroup$
    – user8249
    Commented Jul 21, 2015 at 7:03

That is false. Let $A_R$ be an Azumaya algebra with center $R$ and that is a free $R$-module of rank $n^2$. Consider $\mathfrak{g}_R = A_R$ with bracket the usual commutator. Assume that $A_R/\mathfrak{m}_RA_R$ is isomorphic to $\text{Mat}_{n\times n}(k)$ as an Azumaya algebra over $k$. Then $\mathfrak{g}_{\widehat{R}}$ is isomorphic to $\text{gl}_{n,\widehat{R}}$ by Prop. 2.1 and the discussion immediately following, p. 93 of Brauer III.

MR0244271 (39 #5586c) Reviewed
Grothendieck, Alexander
Le groupe de Brauer. III. Exemples et complèments. (French) 1968
Dix Exposés sur la Cohomologie des Schémas pp. 88–188
North-Holland, Amsterdam; Masson, Paris

On the other hand, if $A\otimes_R \text{Frac}(R)$ is a division algebra, then there is no nilpotent element in $\mathfrak{g}_R\otimes_R \text{Frac}(R)$. For instance, let $k$ be $\mathbb{C}(s)$, let $R$ be $k[t]_{\langle t \rangle}$, and let $A_R$ be the quaternion algebra $$A_R = (R\cdot 1) \oplus (R\cdot x) \oplus (R\cdot y) \oplus (R\cdot xy),$$ with multiplication rule $x^2=s\cdot 1$, $y^2 = 1+t$, $yx=-xy$. By Tsen's theorem (or direct computation), $A_R\otimes_R R/\mathfrak{m}_R$ is a matrix algebra. Yet $A_R\otimes_R \text{Frac}(R)$ is a division algebra.


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