# Comptes Rendus
Mathématique

Akaki Tikaradze

Generic simplicity of quantum Hamiltonian reductions

Volume 359, issue 6 (2021), p. 739-742

Published online: 2 September 2021

https://doi.org/10.5802/crmath.214

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# Generic simplicity of quantum Hamiltonian reductions

Akaki Tikaradze<sup>a</sup>

**Abstract.** Let a reductive group $G$ act on a smooth affine complex algebraic variety $X$. Let $g$ be the Lie algebra of $G$ and $\mu: T^*(X) \to g^*$ be the moment map. If the moment map is flat, and for a generic character $\chi: g \to \mathbb{C}$, the action of $G$ on $\mu^{-1}(\chi)$ is free, then we show that for very generic characters $\chi$ the corresponding quantum Hamiltonian reduction of the ring of differential operators $D(X)$ is simple.

Manuscript received 16th September 2020, revised 1st April 2021 and 12th April 2021, accepted 13th April 2021.

Let a reductive algebraic group $G$ act on a smooth affine algebraic variety $X$ over $\mathbb{C}$. Let $g$ be the Lie algebra of $G$. Let $\mu: T^*(X) \to g^*$ be the corresponding moment map. We will assume that this map is flat, and for generic $G$-invariant character $\chi \in g^*$ the action of $G$ on $\mu^{-1}(\chi)$ is free.

Given a $G$-invariant character $\chi \in g^*$, denote by $U_\chi(G, X)$ the quantum Hamiltonian reduction of $D(X)$ with respect to $\chi$. So,

$$U_{\chi}(G, X) = (D(X)/D(X)g^{\chi})^{G},$$

where $g^{\chi} = \{g - \chi(g) \in D(X), g \in g\}$. The usual filtration on $D(X)$ by the order of differential operators induces the corresponding filtration on $U_{\chi}(G, X)$. Then it follows from the flatness of the moment map that

$$\mathrm{gr} U_{\chi}(G, X) = \mathcal{O}(\mu^{-1}(0) // G).$$

In what follows by a very generic subset we mean a complement of a union of countably many proper closed Zariski subsets. Under these assumptions we have the following result.

**Theorem 1.** For very generic values of a $G$-invariant character $\chi \in g^*$, the corresponding quantum Hamiltonian reduction $U_\chi(G, X)$ is simple. Moreover, if $f \in g_Z / [g_Z, g_Z]$ is so that $G$ acts freely on $\mu^{-1}(\chi)$ whenever $\chi(f) \neq 0$, then $U_\chi(G, X)$ is simple for all $\chi$ such that $\chi(f) \notin \mathbb{Q}$.

The proof is based on the reduction modulo $p^n$ technique for a large prime $p$.

At first, we recall that given a ring $R$ such that $p$ is not a zero divisor, then the center of its reduction modulo $p, R_p = R/pR$ acquires a natural Poisson bracket, to be referred to as the reduction modulo $p$ Poisson bracket, defined as follows. Given central elements $x, y \in Z(R_p)$, let $x', y' \in R$ be their lifts. Then

$$\{x, y\} = \left( \frac{1}{p} [x', y'] \right) \mod p \in Z(R_p).$$

<footnote><sup>a</sup> University of Toledo, Department of Mathematics & Statistics, Toledo, OH 43606, USA.</footnote>

<footnote>E-mail: tikar06@gmail.com</footnote>
We use the following result [5, Corollary 8].

**Lemma 2.** Let $\mathbf{k}$ be a perfect field of characteristic $p$. Let $A$ be a $p$-adically complete topologically free $W(\mathbf{k})$-algebra, such that $A_1 = A/pA$ is an Azumaya algebra over its center $Z_1$. Assume that $\text{Spec}(Z_1)$ is a smooth symplectic $\mathbf{k}$-variety under the reduction modulo $p$ Poisson bracket. Then $A[p^{-1}]$ is topologically simple.

Next we need to recall some results and notations associated with quantum Hamiltonian reduction of the ring of crystalline differential operators in characteristic $p$ from [1].

Let $X$ be a smooth affine variety over an algebraically closed field $\mathbf{k}$ of characteristic $p$, and $G$ be a reductive algebraic group over $\mathbf{k}$ with the Lie algebra $\mathfrak{g}$. Denote by $D(X)$ the ring of crystalline differential operators on $X$. As before, we have the moment map $\mu: T^*(X) \to \mathfrak{g}^*$ and the algebra homomorphism $U(\mathfrak{g}) \to D(X)$. Now recall that the $p$-center of $U(\mathfrak{g})$, denoted by $Z_p(\mathfrak{g})$, is generated by $g^p - g^{[p]}, g \in \mathfrak{g}$. We get an isomorphism

$$i : \text{Sym}(\mathfrak{g})^{(1)} \to Z_p(\mathfrak{g}).$$

On the other hand, the center of $D(X)$ is generated by $\mathcal{O}(X)^p$ and $\xi^p - \xi^{[p]}, \xi \in T_X$ and this leads to an isomorphism

$$\mathcal{O}(T^*(X))^{(1)} \to Z(D(X)).$$

We have $\eta' : Z_p(\mathfrak{g}) \to Z(D(X))$ and the corresponding homomorphism

$$\eta : \text{Sym}(\mathfrak{g})^{(1)} \to \mathcal{O}(T^*(X))^{(1)}.$$

Given $\chi \in \mathfrak{g}^*$, then $\chi^{[1]} \in \mathfrak{g}^*$ is defined as follows:

$$\chi^{[1]}(g) = \chi(g)^p - \chi(g^{[p]}), \quad g \in \mathfrak{g}.$$

Using the above homomorphisms it follows that the center of $U_\chi(G, X)$ contains $\mathcal{O}(\mu^{-1}(\chi^{[1]})//G)$. In this setting the following holds.

**Lemma 3 ([1]).** Let $\chi \in (\mathfrak{g}^*)^G$ be a character. Then $U_\chi(G, X)$ is a finite algebra over $\mu^{-1}(\chi^{[1]})//G$. If $G$ acts freely of $\mu^{-1}(\chi^{[1]})$, then $U_\chi(G, X)$ is an Azumaya algebra over $\mu^{-1}(\chi^{[1]})//G$.

We need the following criterion of simplicity of certain filtered quantizations.

**Lemma 4.** Let $S \subset \mathbb{C}$ be a finitely generated ring, and let $R$ be a filtered $S$-algebra, such that $\text{gr}(R)$ is a finitely generated commutative ring over $S$. Assume that for all large enough primes $p$ the algebra $R_p = R/pR$ is an Azumaya algebra over its center $Z_p$, moreover $\text{Spec}(Z_p)$ is a smooth symplectic variety over $S_p$ under the reduction modulo $p$ Poisson bracket. Let $F$ be the field of fractions of $S$. Then $R_F = R \otimes_S F$ is a simple ring.

**Proof.** Let $I$ be a nonzero two sided ideal of $R$ such that $(R/I)_F \neq 0$. After localizing $S$ further, we may assume using the generic flatness theorem that $\text{gr}(R/I)$ and $R/I, R$ are free $S$-modules. Hence for $p \gg 0$, $\bar{I}_p$ (the $p$-adic completion of $I$) is a topologically free nontrivial two-sided ideal of $\bar{R}_p$ (the $p$-adic completion of $R$). Now Lemma 2 yields a contradiction. $\square$

Next we state a result implying that taking quantum Hamiltonian reduction and reducing modulo a large prime commute. The statement and its proof were kindly provided by W. van der Kallen (via mathoverflow.org.) Possible mistakes in the proof below are solely due to the author.

**Theorem 5 (van der Kallen).** Let $S$ be a commutative Noetherian ring of finite homological dimension, let $R$ be a commutative $S$-algebra flat over $S$. Let $G$ be a split reductive group over $S$ acting on $R$. Then for all $p \gg 0$ and a base change to a characteristic $p$ field $S \to \mathbf{k}$, the map $R^G \otimes_S \mathbf{k} \to R^G_{\mathbf{k}}$ is surjective.
**Proof.** At first, recall that there exists an integer $n \ge 1$ so that $H^i(G, S[\frac{1}{n}]) = 0$ for all $i$ [3, Theorem 33]. This implies $H^i(G, S[\frac{1}{n}] \otimes_S N) = 0$ for any $S$-module $N$ with the trivial $G$-action (since $S$ has a finite global dimension). Let $D, N$ be respectively the image and kernel of the map $S[\frac{1}{n}] \to \mathbf{k}$. As $H^1(G, S[\frac{1}{n}] \otimes_S N) = 0$, we get that $(R \otimes_S S[\frac{1}{n}])^G \to (R \otimes_S D)^G$ is surjective. Now flatness of $\mathbf{k}$ over $D$ yields that

$$ (R \otimes_S D)^G \otimes_D \mathbf{k} = R_{\mathbf{k}}^{G_{\mathbf{k}}}. $$

Therefore, we obtain the desired surjectivity $R^{G} \otimes_S \mathbf{k} \to R_{\mathbf{k}}^{G_{\mathbf{k}}}$ □

**Proof of Theorem 1.** Recall that some $0 \ne f \in \mathcal{O}((g^*)^G)$ has the property that for any $\chi \in (g/[g, g])^*$ such that $f(\chi) \ne 0$, the action of $G$ on $\mu^{-1}(\chi)$ is free. Let $S \subset \mathbb{C}$ be a large enough finitely generated subring over which $X, f$ and the action of $G$ on $X$ are defined. Let $U \subset (g/[g, g])^*_S$ denote the complement of the zero locus of $f$. Thus, $G_S$ acts freely on $\mu^{-1}(U)$. Localizing $S$ further and using the generic flatness theorem, we may assume that $\mathcal{O}(\mu^{-1}(0)//G)$ and $\mathcal{O}(\mu^{-1}(0)) / \mathcal{O}(\mu^{-1}(0)//G)$ is a flat $S$-module.

Let $e_1, \dots, e_l$ be a basis of $g_Z/[g_Z, g_Z]$ over $\mathbb{Z}$. Let $S \to \mathbf{k}$ be a base change to a characteristic $p$ field $\mathbf{k}$, let $\tilde{\chi}$ denote the image of $\chi$ in $g_{\mathbf{k}}^*$. Then

$$ \tilde{\chi}^{[1]}(\tilde{e}_i) = (\tilde{\chi}(\tilde{e}_i))^p - \tilde{\chi}(\tilde{e}_i). $$

Let $W \subset (g/[g, g])^*$ be the set of all $\chi$ so that $\chi(e_i)$ are algebraically independent over $S$. Clearly $W$ is a very generic subset. We will show that for any $\chi \in W$ algebra $U_\chi(G, X)$ is simple.

Put $R = U_\chi(G, X)$. We verify that $R$ satisfies assumptions in Lemma 4. Indeed, let $S \to \mathbf{k}$ be a base change to an algebraically closed field $\mathbf{k}$ of characteristic $\gg 0$, let $\tilde{\chi}$ denote the base change of $\chi$. Recall that $R$ is equipped with the filtration so that $\mathrm{gr}(R) = \mathcal{O}(\mu^{-1}(0)//G)$. In particular, $R$ is a free $S$-module. Similarly, $U_{\tilde{\chi}}(G_{\mathbf{k}}, X_{\mathbf{k}})$ is equipped with the filtration such that $\mathrm{gr}(U_{\tilde{\chi}}(G_{\mathbf{k}}, X_{\mathbf{k}}))$ is a subring of $\mathcal{O}(\mu_{\mathbf{k}}^{-1}(0)//G_{\mathbf{k}})$. Now applying Theorem 5 to the action of $G$ on $\mathcal{O}(\mu^{-1}(0))$, we conclude that $\mathcal{O}(\mu^{-1}(0)//G) \otimes_S \mathbf{k}$ surjects onto $\mathcal{O}(\mu_{\mathbf{k}}^{-1}(0)//G_{\mathbf{k}})$. On the other hand, since $\mathcal{O}(\mu^{-1}(0)) / \mathcal{O}(\mu^{-1}(0)//G)$ is flat over $S$, we get that

$$ \mathcal{O}(\mu^{-1}(0)//G) \otimes_S \mathbf{k} \to \mathcal{O}(\mu^{-1}(0)) \otimes_S \mathbf{k} $$

is injective. So, the restriction map

$$ \mathcal{O}(\mu^{-1}(0)//G) \otimes_S \mathbf{k} \to \mathcal{O}(\mu_{\mathbf{k}}^{-1}(0)//G_{\mathbf{k}}) $$

is an isomorphism. Therefore, $\mathrm{gr}(R) \otimes_S \mathbf{k} \cong \mathrm{gr}(U_{\tilde{\chi}}(G_{\mathbf{k}}, X_{\mathbf{k}}))$. Now flatnest of $\mathrm{gr}(R)$ over $S$ implies that $\mathrm{gr}(R_{\mathbf{k}}) = \mathrm{gr}(R) \otimes_S \mathbf{k}$. Hence we conclude that $R_{\mathbf{k}} \cong U_{\tilde{\chi}}(G_{\mathbf{k}}, X_{\mathbf{k}})$.

Since $\chi(e_1), \dots, \chi(e_n)$ are algebraically independent over $S$, we get that $\tilde{f}(\chi^{[1]}) \ne 0$ for all $p \gg 0$ and an appropriate base change $S \to \mathbf{k}$. Hence $\tilde{\chi} \in U_{\mathbf{k}}$. As $G_S$ acts freely on $\mu^{-1}(U)$, we conclude that $G_{\mathbf{k}}$ acts freely on $\mu^{-1}(\tilde{\chi})$. So Lemma 3 implies that $U_{\tilde{\chi}}(G_{\mathbf{k}}, X_{\mathbf{k}})$ is an Azumaya algebra over a symplectic variety under the reduction modulo $p$ Poisson bracket. So, conditions of Lemma 4 are met. Hence we have shown that algebra $U_\chi(G, X)$ is simple for very generic values of $\chi$.

Now suppose there exists a nonzero $f \in g_Z/[g_Z, g_Z]$ such that $G$ acts freely on $\mu^{-1}(\chi)$ when $\chi(f) \ne 0$. Let $S$ be a finitely generated subring containing $\chi(e_i)$ satisfying conditions as above. Write $f = \sum_i f_i e_i$, $f_i \in \mathbb{Z}$. Then given a base change $S \to \mathbf{k}$, we have

$$ \tilde{\chi}^{[1]}(\tilde{f}) = \sum_i \tilde{f}_i ((\tilde{\chi}(\tilde{e}_i))^p - \tilde{\chi}(\tilde{e}_i)) = \tilde{\chi}(\tilde{f})^p - \tilde{\chi}(\tilde{f}). $$

Let $\chi$ be so that $\chi(f)$ is irrational. Then it follows from the Chebotarev density theorem that there are arbitrarily large primes $p$ and a base change $S \to \mathbf{k}$ to an algebraically closed field $\mathbf{k}$ of characteristic $p$, such that $\tilde{\chi}(\tilde{f}) \notin \mathbb{F}_p$. Hence $\chi^{[1]}(\tilde{f})$ is nonzero in $\mathbf{k}$. So, $G_{\mathbf{k}}$ acts freely on $\mu^{-1}(\tilde{\chi})$, and arguing just as above we may conclude that the algebra $U_\chi(G, X)$ is simple. □
We may apply the above result to certain filtered quantizations of quiver varieties as follows. Let $\mathcal{Q}$ be a quiver with $n$ vertices, let $\alpha$ be a its positive root. Then $G = \prod GL_{\alpha_i}/\mathbb{C}^*$ acts on the space of $\alpha$-dimensional representations $\mathrm{Rep}(\mathcal{Q}, \alpha)$ giving rise to the moment map $m_{\alpha}: T^*(\mathrm{Rep}(\mathcal{Q}, \alpha)) \to \mathfrak{g}^*$. We will identify $(\mathfrak{g}^*)^G$ with $\lambda \in \mathbb{C}^n$ such that $\lambda \cdot \alpha = 0$. From now on we assume that the moment map $m_{\alpha}$ is flat. The set of such dimension vectors $\alpha$ was fully described by Crawly-Boevey in [2, Theorem 1.1]. Denote by $A_{\lambda}(\mathcal{Q}, \alpha)$ the corresponding quantum Hamiltonian reduction of the ring of differential operators $\mathcal{D}(\mathrm{Rep}(\mathcal{Q}, \alpha))$ with respect to the character $\lambda$.

We have the following direct corollary of Theorem 1. Remark that stronger results on generic simplicity follows from the works of Losev on quantizations of quiver varieties (see for example [4, Theorem 1.4.2].)

**Theorem 6.** Let $\alpha$ be a positive root as above. Let $\lambda \cdot \alpha = 0$ be such that $\lambda \cdot \beta \notin \mathbb{Q}$ for any positive root $\beta < \alpha$. Then $A_{\lambda}(\mathcal{Q}, \alpha)$ is simple.

## References

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[2] W. Crawley-Boevey, “Geometry of the moment map for representations of quivers”, *Compos. Math.* **126** (2001), no. 3, p. 257-293.

[3] V. Franjou, W. van der Kallen, “Power reductivity over an arbitrary base of change”, *Doc. Math. Extra Vol.*, **Andrei A. Suslin's Sixtieth Birthday** (2010), p. 171-195.

[4] I. Losev, “Completions of symplectic reflection algebras”, *Sel. Math., New Ser.* **18** (2012), no. 1, p. 179-251.

[5] A. Tikaradze, “Ideals in deformation quantizations over $\mathbb{Z}/p^n\mathbb{Z}$”, *J. Pure Appl. Algebra* **221** (2017), no. 1, p. 229-236.