Are S(g) and U(g) isomorphic as g-modules for g Lie algebra over F_p ? Are S(g)^g and U(g)^g isomorphic as com.algs ?

Question

If $g$ is Lie algebra over field char(k)=0, then the following facts are well-known:

1) S(g) and U(g) are isomorphic as $g$-modules. (Symmetrization map S(g)->U(g) gives isomorphism).

2) S(g)^g and ZU(g)=U(g)^g are isomorphic as commutative algebras. (The Duflo map defines is isomorphism which is combination of symmetrization map with some intricate corrections by terms of smaller degree).

Both facts are based on the symmetrization. If we consider $g$ over field char(k)$\ne$0, there is NO symmetrization map. So I wonder the following:

Question are the facts above true for Lie algebras over char(k)$\ne$0 ?

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Notations

U(g) - universal enveloping algebra (non-commutative associative algebra defined by relations $[x_i, x_j] = \sum_k c_{ij}^k x_k$, for any linear bases $x_k$ of $g$.

S(g) - symmetric algebra of $g$ (defined as $k[x_1...x_n]$ for any bases $x_k$ of $g$.

S(g)^g, U(g)^g means subspaces of g-invariants (g act by zero).

Symmetrization map is defined as $S(x_1..x_k) = 1/k! \sum_{\sigma} \prod_l {x_{\sigma(l)} $.

Duflo map is not so easy to write so let me just mention some MO questions:

Is the Duflo map for Lie algs. unique ?

Capelli determinant = Duflo ( determinant) - was it known ?

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Remark

For the case of gl_n there are so-called Capelli generators of the center of U(g), which can be defined over any char. So I think that the center of U(gl) is the same as for char(k)=0 the result is true. (It is NOT the same as in char=0, as I mistakenly wrote first, since there should be generators corresponding to a^p (which are S(g)^g for any a)).

I think the same is true for other classical semi-simples - there are analogous Capelli like formulas.

Answer 1

Question 2 has a negative answer. Indeed, let $L =sl(2,k)$ where $k$ is an algebraically closed field of characteristic $p>3$ and let {$e,h,f$} be the standard basis of $L$. Then it is well-known (and was first proved by Rudakov-Shafarevich in the late 60s) that the centre $Z(L)$ of $U(L)$ is generated by $E=e^p$, $F=f^p$, $H=h^p-h$ and a Casimir element $C$ subject to the relation $$4EF+H^2=\prod_{i\in{\mathbb F}_p}(C-i^2)=C^p-2C^{(p+1)/2}+C.$$ Furthermore, the maximal spectrum of $Z(L)$ is a hypersurface in ${\mathbb A}^4$ given by the above equation. On the other hand, it follows from a result of Brown-Goodearl that the Azumaya locus of $U(L)$ coincides with the smooth locus of ${\rm Specm\ }Z(L)$. So the modular representation theory of $sl(2,k)$ yields that ${\rm Specm\ }Z(L)$ has exactly $(p-1)/2$ singular points (one can also see this directly by differentiating the above equation).

On the other hand, it is also known (and not difficult to see directly) that $S(L)^L$ is generated by $X=e^p$, $Y=f^p$, $Z=h^p$ and $\Omega=4ef+h^2$ subject to the relation $4XY+Z^2=\Omega^p$. Moreover, ${\rm Specm\ }S(L)^L$ is a hypersurface in ${\mathbb A}^4$ given by the above equation. This hypersurface has a unique singular point at the origin. Since $(p-1)/2>1$ under our assumptions on $p$, we see that the $k$-algebras $Z(L)$ and $S(L)^L$ cannot be isomorphic.

As far as I know the answer to Question 1 is unknown, in general, although there do exist finite dimensional Lie algebras $L$ for which $L\subset U_1(L)$ is NOT a direct summand of $U(L)$ (if $U(L)$ is isomorphic to $S(L)$ as $ad(L)$-modules, then $U(L)$ must contain an isomorphic copy of $L$ as a direct summand, but that copy doesn't have to lie in the first component $U_1(L)$ of the canonical filtration of $U(L)$).

If we require the stronger condition that an isomrphism $S(L)\rightarrow U(L)$ sends $S^1(L)$ onto $L\subset U_1(L)$ (which is obviously true for the symmetrisation map), then the answer to Question 1 is NO. For that condition to hold the Lie algebra $L$ must admit at least one $[p]$-th power map, i.e. must be ${\it restrictable}$ (this observation is due to Michel Duflo). The majority of finite dimensional simple Lie algebras are non-restrictable, and the smallest example is the $3$-dimensional simple Lie algebra over an algebraically closed field of characteristic $2$.

Answer 2

Your question may be related to this question about hyperalgebras. In my answer there, I gave a reference to a paper of Friedlander and Parshall (Rational actions associated with the adjoint representation, Ann. scient. Ec. Norm. Sup., 4e serie, t. 20, 1987, p. 215 a 226), which shows that the answer to your question (1) is yes if $g$ is the Lie algebra of a simple algebraic group.

For a simple algebraic group $G$ and Lie algebra $g = Lie(G)$, I think that looking at the spaces of $g$-invariants in $S(g)$ and $U(g)$ are probably the same as the spaces of $G_1$-invariants, where $G_1$ is the first Frobenius kernel of $G$. Then the results in Section 4 of the Friedlander-Parshall paper may be relevant to your question (2).

Answer 3

It seems that in some particular cases it is known, see e.g. this recent result.

I was looking at that a while ago, and at least it became clear to me that one has to be very careful with lifting the $p$th powers from the Poisson centre of $S(\mathfrak{g})$ (which is where symmetrisation fails most dramatically), e.g. for $\mathfrak{g}=\mathbb{F}_3\{x,y\}$ with $[x,y]=y$ the Poisson central elements $x^3$ and $y^3$ lift to central elements $x^3-x$ and $y^3$ respectively.

Answer 4

A sort of involved and (seemingly very partial) variant of Duflo isomorphism in characretristic $p$ is claimed in the following paper: N.A. Koreshkov, Central elements and invariants in modular Lie algebras, Russ. Math. (Izv. VUZ) 46 (2002), N7, 20-24 (Russian original is available, for example, at http://www.ksu.ru/journals/izv_vuz/arch/2002/07/05-7.PDF ). Roughly, it is proved there that for a finite-dimensional Lie algebra over a finite field $\mathbb Z_p$, a certain modification of $S(\overline{L})^{\overline L}$ is isomorphic to a certain subring of $U(\overline L)^{\overline L}$, where $\overline L$ is obtained from $L$ by some (infinite) field extension.