# Half Haunted: The 1/2 in Harish-Chandra via the Fourier Transform

This post is written together with Josh Mundinger. Last time, we compared the Harish-Chandra isomorphism $$Z(U\mathfrak g) \cong (\text{Sym} \mathfrak h)^{W,\cdot}$$ for $$\mathfrak g= \mathfrak{sl}_2$$ to the Duflo isomorphism $$Z(U\mathfrak g) \cong (\text{Sym } \mathfrak g)^{\mathfrak g} \cong (\text{Sym} \mathfrak h)^W$$, and found that they differ exactly by a translation by $$\rho$$. In this blog post, we study just the Harish-Chandra map $$Z(U\mathfrak g) \to \mathbb C[\mathfrak h]$$, using the Fourier transform to explain why the image is invariant under the dot action $$(W,\cdot)$$. Recall that the Harish-Chandra map sends $$z \in Z(U\mathfrak g)$$ to the action of $$z$$ on the Verma module $$M_\lambda$$. The dot action of $$W$$ is defined by $$w \cdot \lambda = w(\lambda + \rho) - \rho$$. Thus, for $$\mathfrak sl_2$$, we need to show that the center of $$U\mathfrak sl_2$$ acts by the same scalar on $$M_{\lambda}$$and $$M_{-\lambda - 2}$$.

## Twisted differential operators

The Beilinson-Bernstein localization theorem provides a geometric way to understand the universal enveloping algebra of a semisimple Lie algebra $$\mathfrak g$$ through differential operators. For a smooth variety $$X/\mathbb C$$, the sheaf of differential operators $$D_X$$ on $$X$$ is the sheaf of $$\mathbb C$$-linear operators $$\mathcal{O}_X \to \mathcal{O}_X$$ which locally look like $\sum_{i_1,\ldots,i_n} f_{i_1,\ldots, i_n} \frac{\partial^{i_1}}{\partial x_1^{i_1}}\frac{\partial^{i_2}}{\partial x_2^{i_2}}\cdots \frac{\partial^{i_n}}{\partial x_n^{i_n}},$ where $$f_{i_1,\ldots, i_n}$$ are regular functions and $$x_1,\ldots, x_n$$are local coordinates. More generally, if $$\mathcal L$$ is a line bundle on $$X$$, then the sheaf of differential operators $$D^{\mathcal L}$$ on $$\mathcal L$$ is the sheaf of operators $$\mathcal L \to \mathcal L$$ which look locally as above. This ring may be expressed in terms of $$D_X$$ by the formula $D^{\mathcal L} = \mathcal L \otimes D_X \otimes \mathcal L^{-1}$ In our case, we are interested in $$X = \mathbb P^1 = SL_2/B$$, the flag variety for $$SL_2$$. Line bundles on $$\mathbb P^1$$ are exactly $$\mathcal O(\lambda)$$ for $$\lambda \in \mathbb Z$$, and we will write $$D^{\lambda} = D^{\mathcal O(\lambda)}$$.

It turns out that the algebra $$D^\lambda$$ makes sense even when $$\lambda$$ is not an integer! We have a map $\pi: \mathbb{C}^2\setminus 0 \to \mathbb{P}^1.$ This is a quotient map for the action of $$\mathbb G_m$$ by dilation on $$\mathbb C^2$$. The derivative of the $$\mathbb G_m$$-action is given by the Euler vector field $eu = x_1 \frac{\partial}{\partial x_1} + x_2 \frac{\partial}{\partial x_2}$ where $$x_1$$ and $$x_2$$ are linear coordinates on $$\mathbb C^2$$. In these terms, $D^\lambda := (\pi_*D_{\mathbb C^2 \backslash 0}/(eu - \lambda))^{\mathbb G_m}.$ The above formula makes sense for any complex number $$\lambda$$.The sheaf of rings $$D^\lambda$$ is a sheaf of twisted differential operators on $$\mathbb P^1$$.

## Infinitesimal action and a Lemma of Beilinson-Bernstein

The theorem of Beilinson and Bernstein relates $$\mathfrak g$$ to the flag variety $$X = G/B$$. The line bundles on $$G/B$$ are $$\mathcal O(\lambda)$$ for $$\lambda$$ a character of $$B/[B,B]$$. In case $$G = SL_2$$, then $$G/B = \mathbb P^1$$, and $$\mathcal O(\lambda)$$ for integer $$\lambda$$ are the usual line bundles on $$\mathbb P^1$$.

Lemma (Global Sections of D) If $$G$$ is a semisimple complex algebraic group and $$X = G/B$$ is the flag variety of $$G$$, then $\Gamma(X,D^\lambda) \cong U\mathfrak g/I_\lambda$ where $$I_\lambda$$ is the ideal generated by the kernel of $$Z\mathfrak g$$ on the Verma module $$M_\lambda$$.

The map in this Lemma is induced by the infinitesimal action of $$G$$on $$X = G/B$$. Given a path $$\gamma: (-\epsilon, \epsilon) \to G$$ with $$\gamma(0)=1$$, we obtain a family of automorphisms of $$X$$; differentiating them gives a vector field on $$X$$.This vector field depends only on $$\gamma’(0)$$, and so we obtain a Lie algebra homomorphism $\mathfrak g \to \text{Vect}(X).$ In case $$G = SL_2$$, $$X = \mathbb P^1$$, we can calculate these operators on $$\mathbb C^2 \setminus 0$$, then descend using the map $$\pi: \mathbb C^2 \setminus 0 \to \mathbb P^1$$. Here are formulas for the infinitesimal action on $$\mathbb C^2 \setminus 0$$:

These formulas give a Lie algebra homomorphism $$\mathfrak{sl}_2\to \text{Vect}(\mathbb C^2 \setminus 0) \to D_{\mathbb C^2 \setminus 0}$$, inducing a homomorphism of associative algebras $$U\mathfrak{sl_2} \to D_{\mathbb C^2 \setminus 0}$$. The image of $$\mathfrak{sl_2}$$ is $$\mathbb G_m$$-invariant since the action of $$SL_2$$ commutes with dilation. So we get a ring homomorphism $\mathrm{act}: U\mathfrak{sl}_2 \to D^\lambda = \pi_\ast(D_{\mathbb C^2 \setminus 0}/(eu - \lambda))^{\mathbb G_m}$ for all $$\lambda \in \mathbb C$$. We have now given explicit formulas for the map in the Lemma above.

## Fourier

In this section we show that the Fourier transform descends to an equivalence of categories between the category of $$D^\lambda$$-modules to the category of $$D^{-\lambda - 2}$$-modules for generic $$\lambda$$. The Fourier transform naturally transforms differential operators on a vector space $$V$$to differential operators on the dual space $$V^*$$.We show how this induces a map on twisted differential operators on $$\mathbb P^1$$, where the $$\rho$$-shift will naturally appear.

Let $$V = \mathbb{C}^2$$ with coordinates $$x_1,x_2$$. On $$V^*$$, we will use dual coordinates $$(\frac{\partial}{\partial x_1}, \frac{\partial}{\partial x_2}) =: (y_1, y_2)$$.Note that for $$\mathfrak g$$ acting on a vector space $$V$$, the action of $$x\in \mathfrak g$$ on $$V^*$$ is defined as $$x(v^*) = v^* \circ(-x)$$. Thus, on the dual basis, we get the following action of $$\mathfrak{sl}_2$$:

Note that these matrices are different than the ones we had before. In this basis, $$E$$ acts by $$-F$$ on the dual, $$F$$ acts by $$-E$$ on the dual, and $$H$$acts by $$-F$$ on $$V^*$$

The Fourier transform $$\phi: D_V \to D_{V^*}$$ is defined by $\phi(x_i) = \frac{\partial}{\partial y_i}, \quad \phi\left(\frac{\partial}{\partial x_i}\right) = - y_i.$ The goal of this section is to show that the Fourier transform $$\phi$$ induces the following commutative triangle.

where $$\mathrm{act}$$ is the infinitesimal action of the last section. This commutative triangle implies that the central character of $$\lambda$$ is the same as the central character of $$-\lambda - 2$$, which gives us that the Harish-Chandra homomorphism is invariant under the $$(W,\cdot)$$ action.

We begin by showing the vertical arrow is well defined by an explicit calculation with the Euler fields on $$V$$and $$V^*$$: $\phi (eu) = \sum_i (-\frac{\partial}{\partial y_i}y_i) = \sum_i (-y_i \frac{\partial}{\partial y_i}-1)= -eu - 2.$ Hence $$\varphi( (eu - \lambda)D_{V}) = \varphi( (eu - (-\lambda - 2))D_{V^*})$$. Now $$D_V$$and $$D_{V \setminus 0}$$ are not the same, as the latter is a sheaf on the non-affine scheme $$V \setminus 0$$. Nonetheless we get a map $\varphi: D_{V \setminus 0}/ (eu - \lambda)D_{V\setminus 0} \to D_{V^* \setminus 0}/ (eu - (-\lambda - 2)D_{V^* \setminus 0}$ and thus $$D^\lambda_{\mathbb P V} \to D^{-\lambda - 2}_{\mathbb PV^*}$$.

To fill in the triangle, we need to check that $$\varphi$$ intertwines the infinitesimal action of $$\mathfrak{sl}_2$$ on $$V$$ and $$V^*$$. We calculated both of these earlier! The Fourier transform sends infinitesimal actions defined by equations (1), (2), (3) to (4), (5), (6) respectively. Thus we conclude the triangle commutes.

## Central Characters

By the Lemma (Global Sections of D), this commutative triangle induces the following triangle on global sections:

where $$I_\lambda$$ is the ideal of the central character of $$\lambda$$ under the Harish-Chandra homomorphism.

This shows that the Harish-Chandra homomorphism is invariant under $$\lambda \mapsto -\lambda - 2$$, as desired.

The reason that the shift $$\lambda \mapsto -\lambda - 2$$ shows up in today’s story comes from the Fourier transform of the Euler field. When the Fourier transform is treated in a more coordinate-free manner, the canonical bundle shows up. The canonical bundle of $$\mathbb P^1$$ is $$\mathcal O(-2)$$, and that $$-2$$ is the same $$-2$$ as in $$-\lambda - 2$$. We will elaborate on this in the future.

Written on December 31, 2023