# Half Haunted: Relating the 1/2's in Duflo and Harish-Chandra

This post is written together with Josh Mundinger. We seek to understand the relations between $$1/2$$’s that appear across mathematics. From the Riemann Hypothesis to the L2 norm, we aim to see the myriad and enticing ways this unfurls; each instance of $$1/2$$ connected in an anarchic network of equals. In this blog post, we examine a specific example arising in representation theory: the center of the universal enveloping algebra $$U\mathfrak g$$ of a Lie algebra $$\mathfrak g$$.

The PBW theorem implies there is an isomorphism of vector spaces $sym: \text{Sym}(\mathfrak g)^\mathfrak g \cong Z(U\mathfrak g),$ which on $$\text{Sym}^i$$ is given by the symmetrization map $sym: x_1x_2\cdots x_i \mapsto \frac{1}{i!} \sum_{\sigma \in S_i} x_{\sigma 1}x_{\sigma 2}\cdots x_{\sigma i}.$

Note that this is not the identity map because $$Sym$$ indicates coinvariants and not invariants of the symmetric group. The symmetrization map is not a ring homomorphism. Duflo showed, however, that by twisting the symmetrization map, one obtains a ring homomorphism. The Chevalley restriction theorem then identifies $$\text{Sym}(\mathfrak g)^\mathfrak g$$ with $$\text{Sym}(\mathfrak h)^W$$, where $$\mathfrak h$$ is a Cartan subalgebra and $$W$$ is the Weyl group.

On the other hand, when $$\mathfrak g$$ is semisimple, there is another way to understand $$Z(U\mathfrak g)$$. The Harish-Chandra isomorphism is a “quantized” version of the Chevalley isomorphism, $\text{Sym}(\mathfrak{g})^{\mathfrak{g}} \to \text{Sym}(\mathfrak{h})^{W}.$ It’s called quantized because we are taking taking the universal enveloping algebra, which is adding a parameter deforming the bracket of our Lie algebra (literally a deformation parameter). Then, we get back down to the classical case by taking the associated graded. Harish-Chandra proved that there is an isomorphism $$Z(U\mathfrak g) \cong (U\mathfrak h)^{(W,\cdot)}$$. Here $$W$$ acts on $$\mathfrak h$$ via the \emph{dot action} instead of the usual action: the origin is shifted by $$\rho$$, $$1/2$$ of the sum of the positive roots of $$\mathfrak g$$.

Our quest now leads us to the natural question: are the 1/2 in $$\rho$$ and the 1/2 in Duflo’s differential operator the same 1/2?

In the case of $$\mathfrak{sl}_2$$, we show via explicit computation that the answer is yes!

In a followup post, we will further see why both arise via a lovely geometric flip and adjustment. The geometric adjustment is by another $$1/2$$: a square root of the canonical bundle on a variety.

## The square with $$\rho$$ and Duflo

We investigate the compatibility between three different maps between semi-simple Lie algebras. In particular, we wish to fill in the remaining arrow. We show this commutes explicitly in the case of $$\mathfrak{sl}_2$$ by diagram chasing the generator by hand. Let’s go!

The Lie algebra $$\mathfrak{sl}_2$$ has basis $$E,F,H$$ with commutation relations $[E,F]=H, \qquad [H,E] = 2E, \qquad [H,F] = -2F.$

The matrices representing $$E, F, H$$ are as follows:

The Cartan subalgebra $$\mathfrak h$$ is spanned by $$H$$, and the Weyl group $$W = \mathbb Z/2\mathbb Z$$ acts on $$W$$ by $$H \mapsto -H$$.

## Harish-Chandra

The Harish-Chandra homorphism is defined as follows: there is the Verma module $$M_{\lambda} = U(\mathfrak{sl}_{2})/U(\mathfrak{sl}_{2}) \cdot (E, H - \lambda).$$ which we can think of as generated by an element $$1_{\lambda}$$ satisfying $$E1_{\lambda} = 0$$ and $$H1_{\lambda} = \lambda 1_{\lambda}$$. We can think of $$\lambda$$ as a basis for $$\mathfrak{h}^\ast$$ dual to the basis $$H$$ of $$\mathfrak{h}$$. The module $$M_{\lambda}$$ is indecomposable, so $$Z(U \mathfrak{sl}_{2})$$ acts by scalars. The image of $$z \in Z(U\mathfrak{sl}_{2})$$ in $$\mathbb{C}[\mathfrak{h}] = \mathbb{C}[\lambda]$$ is the function sending $$\lambda$$ to the scalar $$z|_{M_{\lambda}}$$.

In the case of $$\mathfrak{sl}_{2}$$, We know the center of $$U\mathfrak{sl}_{2}$$ is generated by the Casimir operator $$\Omega = EF + FE + \frac{1}{2} H^2$$. That is, $Z(U(\mathfrak{sl}_{2})) \simeq \mathbb{C}[\Omega].$

Let’s compute the action of Casimir $$\Omega$$ on $$1_\lambda$$:

So we have shown that under the Harish-Chandra map,

Let’s check that $$\frac{1}{2}\lambda^2 + \lambda$$ is invariant under the $$W \bullet$$ action. For $$\mathfrak{sl}_2$$, we have $$[H, E]=2E$$, so that the root associated to $$E$$ is the function $$\alpha(H) = 2.$$ Then, $$\rho$$ is $$\frac12$$ of the sum of the simple roots, in other words, $$\rho = \frac12 (2) = 1$$. The action of $$W \bullet$$ sends $$\lambda \mapsto -(\lambda+1)-1 = -\lambda - 2.$$

So we have shown that under the Harish-Chandra map,

## Duflo

The Duflo map is defined on the level of vector spaces by acting by a linear operator $$\partial_q: S\mathfrak g \to S\mathfrak g$$ followed by the symmetrization map $$S\mathfrak{g} \to U\mathfrak{g}$$. We start with expanding a function $$q:\mathfrak{g} \to \mathbb C$$ into a power series. A power series expansion of $$q: \mathfrak{g} \to \mathbb{C}$$ in terms of the coordinate basis of $$\mathfrak{g}$$ is a power series in $$\mathfrak{g}^\ast$$. To get $$\partial_q$$ from $$q$$, we replace each element of $$\mathfrak{g}^\ast$$ in the power series with the corresponding partial derivative $$\text{Sym}\mathfrak{g} \to \text{Sym}\mathfrak{g}$$. Then, we define the map $$\text{Sym}\mathfrak{g} \to \text{Sym}\mathfrak{g}$$ by applying the differential operator to the polynomials that comprise $$\text{Sym}\mathfrak{g}$$.

We now define the power series $$q(x)$$. Let $$p(x)$$ be the function $$\mathfrak g \to \mathbb C$$ given by $p(x) = \det\left( \frac{sinh(ad(x)/2)}{ad(x)/2}\right) = 1+ \sum_{n=1}^\infty \frac{1}{2^{2n}(2n+1)!}tr(ad(x)^{2n}).$ and let $$q(x)$$ be a square root of $$p(x)$$ at $$0$$.

We now explicitly express the process of computing the Duflo map with our favorite example $$\mathfrak{sl}_2$$, using the basis $${E,F,H}$$. We know the center of $$U\mathfrak{sl}_2$$ is generated by the Casimir operator $$\Omega = EF + FE + \frac{1}{2} H^2$$. Hence, it suffices to compute the action of the Duflo operator on $$\frac{1}{2}H^2 + 2EF \in (S\mathfrak{g})^{\mathfrak g}$$. Since the Casimir is quadratic, we only need to know $$q$$ up to second order: we have $$q(x) = 1 + \frac{1}{2} \left(\frac{1}{4(6)}tr(ad(x)^2)\right) + O(x^4) = 1 + \frac{1}{48}tr(ad(x)^2) + O(x^4)$$. Let’s compute $$tr(ad(x)^2)$$ in our coordinates $${E,F,H}$$. If $x = \begin{pmatrix} a & b \ c & -a \end{pmatrix} = aH + bE + cF ,$ then $$tr(ad(x)^2) = 8a^2 + 8bc$$. Hence our differential operator $$\partial_q$$ is $\partial_q = 1 + \frac{1}{48} (8 \partial_H^2 + 8 \partial_E \partial_F) + h.o.t. = 1 + \frac{1}{6}(\partial_H^2 + \partial_E\partial_F) + h.o.t.$

Applying this differential operator to $$\frac{1}{2}H^2 + 2 EF \in (\text{Sym}\mathfrak g)^{\mathfrak g}$$ gives $\partial_q(\frac{1}{2}H^2 + 2EF) = \frac{1}{2}H^2 + 2EF + \frac{1}{6}(1 + 2).$ So applying the symmetrization map gives $Duflo(\frac{1}{2}H^2 + 2EF) = \Omega + \frac{1}{2},$ where $$\Omega$$ is the Casimir.

## Chevalley

Now we compare with the Chevalley isomorphism. The Chevalley map is defined by viewing an element of $$\text{Sym}(\mathfrak g)^{\mathfrak g}$$ as an element of $$\mathbb C[\mathfrak g]$$ via the Killing form, restricting to $$\mathfrak h$$, then again using the Killing form to get an element of $$U\mathfrak h$$.

We now express this explicitly in the case of $$\mathfrak{sl}_2$$. Under the Killing form, the dual basis to $$\langle H,E,F\rangle$$ is $$\langle \frac{1}{8}H, \frac{1}{4}F, \frac{1}{4}E\rangle$$. Hence, dualizing, restricting to $$\mathfrak h$$, and dualizing again sends $$H$$ to $$H$$,$$E$$ to $$0$$, and $$F$$ to $$0$$, thus sending our element $$\frac{1}{2}H^2 + 2EF \in \text{Sym}(\mathfrak g)^{\mathfrak g}$$ to $$\frac{1}{2}H^2$$. This is a $$W$$-invariant element of $$U\mathfrak h$$ which acts on a highest weight vector $$1_{\lambda}$$ by $$\frac{1}{2}\lambda^2$$.

## Filling in the Square

We know the center of $$U\mathfrak{sl}_2$$ is generated by the Casimir operator $$\Omega = EF + FE + \frac{1}{2} H^2$$. Hence, it suffices to diagram chase this generator through our square to conclude that our square commutes. We showed that the Duflo map sends the symmetrized Casimir operator $$\frac{1}{2}H^2 + 2EF$$ in $$\mathfrak{g})^{\mathfrak g}$$ to $$\Omega + \frac{1}{2}.$$ Further composing with the Harish-Chandra map gives that $$\frac{1}{2}H^2 + 2EF \in (S\mathfrak g)^{\mathfrak g}$$ is sent to $$\frac{1}{2}\lambda^2 + \lambda + \frac{1}{2} = \frac{1}{2}\left(\lambda+ 1 \right)^2$$.

We now diagram chase the other side. We showed that the Chevalley map sends the symmetrized Casimir operator $$\frac{1}{2}H^2 + 2EF$$ to $$\frac{1}{2} \lambda^2$$ where $$1_\lambda$$ is the eigenvalue of $$H$$. The $$\rho$$-shift map sends $$\lambda$$ to $$\lambda + 1$$, which sends the image of the Chevalley map $$\frac{1}{2} \lambda^2$$ to $$\frac{1}{2} (\lambda + 1)^2$$. Therefore, the diagram commutes. We have filled in the square victoriously!

Written on September 16, 2023