# 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.