# Kirillov orbit method for finite Lazard Lie group

## Statement

The Kirillov orbit method is a method that can be used to determine the irreducible representations of a finite Lazard Lie group. The procedure is as follows:

• Let $L$ be the Lazard Lie ring of $G$.
• Denote by $\hat{L}$ the Pontryagin dual of $L$, viewed only as an additive abelian group. Note that $\hat{L}$ is isomorphic to $L$, but there is no natural isomorphism.
• The natural action of $G$ on $L$ (called the adjoint representation) induces a natural action of $G$ on $\hat{L}$ (called the coadjoint representation). The orbits under this action correspond to the irreducible representations. Moreover, the size of any orbit is the square of the degree of the irreducible representation to which it corresponds. Note that this is consistent combinatorially with the fact that sum of squares of degrees of irreducible representations equals order of group.

## Identification of coadjoint orbits with irreducible representations

A full proof is beyond our scope here, but we can explain what is going on.

### In terms of group rings, duals, and bases

The underlying vector space of the group ring $\mathbb{C}[G]$ can be thought of as the set of all set maps $G \to \mathbb{C}$ with pointwise addition and scalar multiplication. The multiplication is defined in terms of convolution.

Note that $G$ has an element-to-element bijection with $L$, its Lazard Lie ring:

$L \stackrel{\exp}{\rightarrow} G, G \stackrel{\log}{\rightarrow} L$

this bijection induces a $\mathbb{C}$-vector space isomorphism between $\mathbb{C}[G]$ and $\mathbb{C}[L]$, where the latter is the group ring for $L$, and can be thought of as set maps from $L$ to $\mathbb{C}$.

The ring $\mathbb{C}[L]$ has another choice of basis: the one-dimensional characters of $L$. With this basis, the multiplication by convolution simply becomes coordinate-wise multiplication along with rescaling by the order of the group (this follows from the orthonormality of characters -- see linear representation theory of finite cyclic groups and linear representation theory of finite abelian groups). In other words, the ring $\mathbb{C}[L]$ is isomorphic to the ring $\mathbb{C}^{\hat{L}}$, i.e., a direct product of $\hat{L}$ many copies of the algebra $\mathbb{C}$.

It is important to note that for this analysis, $L$ and $\hat{L}$ are being treated as basis sets rather than as vector spaces or modules themselves.

The adjoint action of $G$ on $L$ induces a coadjoint action of $G$ on the set $\hat{L}$. We get orbits for this action. Each orbit is invariant under the $G$-action. Thus, the subspace of $\mathbb{C}[L]$ spanned by each orbit is a $G$-invariant subspace of $\mathbb{C}[L]$. We have thus obtained a decomposition of $\mathbb{C}[L]$ as a direct sum of a bunch of $G$-invariant subspaces.

The claim is that, under the identification of $\mathbb{C}[L]$ with $\mathbb{C}[G]$ explained above, these subspaces correspond to the minimal two-sided ideals of $\mathbb{C}[G]$. Based on knowledge of the representation theory of $\mathbb{C}[G]$, we know that the minimal two-sided ideals are the matrix algebras for the irreducible representations of $G$. We thus have:

Size of an orbit for the $G$-action on $\hat{L}$ = Dimension of corresponding matrix algebra for an irreducible representation of $G$ = Square of degree of that irreducible representation of $G$

## Explicit construction of irreducible representation in terms of coadjoint orbit: via polarizing subring

### Constructing bihomomorphism using Lie bracket and character

Suppose $\Omega$ is one of the coadjoint orbits, i.e., it is an orbit in $\hat{L}$ under the action of $G$. Pick $\omega \in \Omega$. $\omega$ is a one-dimensional character of $L$.

Define $B_\omega:L \times L \to \mathbb{C}^\ast$ as:

$B_\omega(x,y) = \omega([x,y])$

where $[x,y]$ is the Lie bracket of $x$ and $y$. Note that since the bracket itself is bilinear, and $\omega$ is a homomorphism, $B_\omega$ is a bihomomorphism (i.e., it is "bilinear" keeping in mind that the addition on the target group is what we customarily denote multiplicatively). It is also skew-symmetric due to the skew symmetry of the Lie bracket.

### Definition of degenerate subgroup

As before, $\Omega$ is a coadjoint orbit and $\omega \in \Omega$.

Let $N$ be the subset of $L$ that is degenerate with respect to $B_\omega$, i.e.:

$N = \{ x \in L \mid B_\omega(x,y) = 1 \ \forall \ y \in L \}$

In other words, $N$ is the largest subset of $L$ such that $[N,L] \subseteq \operatorname{ker} (\omega)$.

Clearly, $N$ is an additive subgroup of $L$, and moreover, $N$ contains the center of $L$. Does $N$ have to be a subring or an ideal? This is unclear.

### Definition of stabilizer

For any fixed $\omega \in \Omega$, the stabilizer of $\omega$ under the action of $G$ is a subgroup of $G$, which we call $\operatorname{Stab}_G(\omega)$. By the fundamental theorem of group actions, the coset space of this subgroup corresponds to $\Omega$, so in particular, the subgroup has index $|\Omega|$ in $G$.

It seems to be the case (proof pending) that, under the Lazard correspondence, the subgroup $\operatorname{Stab}_G(\omega)$ of $G$ corresponds to the degenerate subring $N$ of $L$.

### Definition of polarizing subring or polarizing subalgebra

A Lie subring $P \subseteq L$ is termed a polarizing subring for $\omega$ if it is a subring maximal with respect to the restriction of $B_\omega$ to $P \times P$ being trivial. Another way of putting this is that $P$ is maximal among subrings of $L$ with respect to $[P,P]$ being in the kernel of $\omega$ (which itself need not be a subring, but is a subgroup with cyclic quotient group). Note that if the Lie ring has an algebra structure over a commutative unital ring, then any polarizing subring is a subalgebra and is termed a polarizing subalgebra.

It turns out that all polarizing subrings for all $\omega \in \Omega$ have the same order as each other. More specifically, they all have index equal to $\sqrt{|\Omega|}$.

### Construction of irreducible representation in terms of polarizing subring

We are now in a position to construct the irreducible representation of $G$ corresponding to a coadjoint orbit $\Omega$. Let $\omega \in \Omega$ and let $P$ be a polarizing subring for $\omega$. (Both the choice of $\omega$ and the choice of $P$ are non-unique, but it will turn out that the irreducible representation that we finally construct is unique up to equivalence of linear representations).

Let $H$ be the subgroup of $G$ obtained from $P$ under the Lazard correspondence. Pulling $\omega$ back along the Lazard correspondence, we see that $\omega$ also defines an equivalent one-dimensional character of $H$, say $\omega'$. Note that this is justified as follows: the fact that $[P,P]$ is in the kernel of $\omega$. So, $\omega$ descends to a character of $P/[P,P]$, which, when pulled back along the Lazard correspondence, corresponds to $H/[H,H]$. Sincewe are now doing a Lazard correspondence for an abelian group, the correspondence is a group isomorphism, so we get a character of $H/[H,H]$. Composing with the quotient map gives a one-dimensional character $\omega'$ of $H$.

The irreducible representation we are looking for is the induced representation $\operatorname{Ind}_H^G(\omega')$. We have:

Degree of this irreducible representation = $[G:H]$ times degree of $\omega'$ = $\sqrt{|\Omega|} \cdot 1 = \sqrt{|\Omega|}$

### Kirillov character formula

For any coadjoint orbit $\Omega$ and any element $g \in G$, the character of the irreducible representation corresponding to $\Omega$, evaluated at $g$, is given by:

$\frac{1}{\sqrt{|\Omega|}} \sum_{\omega \in \Omega} \omega(\log g)$

Here, $\log g$ is the element of $L$ corresponding to $G$. If we identify $G$ and $L$ as sets under the Lazard correspondence, we would simply write it as:

$\frac{1}{\sqrt{|\Omega|}} \sum_{\omega \in \Omega} \omega(g)$

Note that this character formula follows pretty directly from the explicit construction of the irreducible representation as an induced representation.

### Tying together the ideas

Here is the overall picture. Fix a coadjoint orbit $\Omega$ and an element $\omega \in \Omega$. Then, under the Lazard correspondence:

• Degenerate subring for $\omega$ in $L$ $\leftrightarrow$ Stabilizer $\operatorname{Stab}_G(\omega)$ for $\omega$ in $G$, and both have index $|\Omega|$ (in $L$ and $G$ respectively)
• Polarizing subring for $\omega$ in $L$ $\leftrightarrow$ Its exponential in $G$, and both have index $\sqrt{|\Omega|}$ (in $L$ and $G$ respectively).

## Action of automorphism group

The automorphism group of $G$ naturally becomes the automorphism group of its Lazard Lie ring, again via the natural identification of $G$ with $L$ under the logarithm and exponential maps. Thus, it is a subgroup of the automorphism group of $L$ as an abelian group, and hence also a subgroup of the automorphism group of the Pontryagin dual $\hat{L}$.

The action of $G$ on $\hat{L}$ itself lives inside this action. Recall that group acts as automorphisms by conjugation, so there is a natural map:

$G \to \operatorname{Aut}(G)$

whose kernel is the center $Z(G)$ and whose image is the inner automorphism group $\operatorname{Inn}(G)$. Composing, we get a composite map:

$G \to \operatorname{Aut}(G) \to \operatorname{Aut}_{\mathbb{Z}}(\hat{L})$

The claim is that the action of $G$ on $\hat{L}$ via this composition is the same as the coadjoint representation. Thus, the action of $\operatorname{Aut}(G)$ on $\hat{L}$ extends the action of $G$.

This means that the orbits under this action are unions of coadjoint orbits. Hence, the action of $\operatorname{Aut}(G)$ on $\hat{L}$ can be reduced to an action of $\operatorname{Aut}(G)$ on the set of coadjoint orbits in $\hat{L}$. Going further, since $\operatorname{Inn}(G)$ stabilizes each coadjoint orbit, we get an action of the outer automorphism group $\operatorname{Out}(G)$ on the set of coadjoint orbits.

The claim is that under the identification of coadjoint orbits with irreducible representations, this action coincides with the usual action of the outer automorphism group on the set of irreducible representations.

## Relation with isoclinism

Suppose two Lazard Lie groups $G_1,G_2$ are isoclinic groups. Then, their corresponding Lazard Lie rings, which we denote $L_1$ and $L_2$, are isoclinic Lie rings. The Kirillov orbit methods for these two groups proceed in a very similar fashion.

• Even though $G_1$ and $G_2$ need not be isomorphic, the inner automorphism groups $\operatorname{Inn}(G_1)$ and $\operatorname{Inn}(G_2)$ are isomorphic as part of the isoclinism data. This is significant because the actions we use for the Kirillov orbit method all descend to the inner automorphism group.
• Under the isomorphic identification of $\operatorname{Inn}(G_1)$ and $\operatorname{Inn}(G_2)$, the actions on $\hat{L_1}$ and $\hat{L_2}$ are equivalent as group actions on sets (up to size scaling -- they are precisely equivalent if $G_1$ and $G_2$ have the same size). Note that $\hat{L_1}$ and $\hat{L_2}$ may or may not be isomorphic as additive groups, so it does not make sense in general make sense to ask whether these are equivalent as group actions on groups.
• When we combine the similar orbit structure with the fact that degrees of irreducible representations are square roots of orbit sizes, this comports with the already know fact that isoclinic groups have same proportions of degrees of irreducible representations.

## Application

The Kirillov orbit method can be used both for particular groups and for group families with enough common features that a joint analysis can be done. Below are some examples where the Kirillov orbit method can be used: