Given a Lie group $G$ there is a natural representation of $G$ on the dual of its Lie algebra $\mathfrak{g}^*$ given by the coadjoint representation. This representation is obtained by differentiating the diffeomorphisms given by conjugation $\text{con}_g(h) = g h g^{-1}$ and then taking the adjoint. From this one gets two important sets: for $\mu \in \mathfrak{g}^*$ the coadjoint orbits $G \cdot \mu$ and the coadjoint isotropy subgroups $G_\mu = \{ g \in G \ | \ g \cdot \mu = \mu \}$.

From a purely symbolic perspective, this is a very natural construction: the representation is an intrinsic property of the Lie group itself, since no external objects are needed to define it. However I don't quite see why this representation is interesting from an intuitive point of view - it actually seems quite arcane. How can one imagine elements in $\mathfrak{g}^*$ and the action of $G$ on them? What is the meaning of the coadjoint orbits and the coadjoint isotropy subgroups, how can they help to understand the Lie group $G$ better?

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    $\begingroup$ One can study irreducible unitary representations of certain Lie groups by looking at their coadjoint orbits, this was first discovered by Kirillov: en.m.wikipedia.org/wiki/Orbit_method $\endgroup$ Aug 8, 2021 at 18:03
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    $\begingroup$ I can't say anything as comprehensive as @‍NinjaDarth's answer, but, in the spirit of @IvanSolonenko's comment, I think one reason that it's hard to see the significance of the coadjoint representation is that, in characteristic $0$, it's the same as the adjoint representation (at least for semisimple groups)! In positive characteristic they can differ, and the coadjoint representations can be more useful; see, for example, Kac and Weisfeiler - Coadjoint …. $\endgroup$
    – LSpice
    Dec 30, 2022 at 21:59

2 Answers 2


Actually, you have it backwards. It's more intuitive and less arcane and (as Kirillov himself noted) it may have been how Lie originally tried to frame his formalism. This is best seen with an example.

$\newcommand\so{so}\DeclareMathOperator\SO{SO}$The Lie algebra $\so(3)$ for the group $\SO(3)$ may be presented as being generated by the basis $(X, Y, Z)$, subject to the identities $$[Y, Z] = X, \quad [Z, X] = Y, \quad [X, Y] = Z.$$ In the dual $\so(3)^*$ the role of basis elements and coordinates is reversed, so that $(X, Y, Z)$ are now thought of as coordinates (with respect to the dual basis). The Lie bracket becomes a Poisson bracket $$\{Y, Z\} = X, \hspace 1em \{Z, X\} = Y, \hspace 1em \{X, Y\} = Z,$$ over the function space $C^\infty\left(so(3)^*\right)$, which contains the linear functions, i.e. $\so(3)^{**}$, which is equivalent to $\so(3)$, itself, so that the Poisson bracket is an extension of the Lie bracket to non-linear functions.

Associated with the Lie group $\SO(3)$ is a family of symmetry transformations generated by infinitesimal transformations that correspond to the elements of $\so(3)$. Over this Poisson manifold, each $\Lambda = xX + yY + zZ$ gives rise to an infinitesimal transformation $\Delta = \{\cdots, \Lambda\}$, whose actions on the basis/coordinate-functions is: $$\Delta X = yZ - zY, \hspace 1em \Delta Y = xZ - zX, \hspace 1em \Delta Z = xY - yX.$$ In the language of mathematicians, $\Delta$ is a vector field and its exponential $e^{s\Delta}$, its flow, is the action of the identity component of the Lie group (here: the Lie group itself, since it's connected) on the space. Here, it can be tabulated as follows: $$\Delta^2(X, Y, Z) = (x, y, z) (xX + yY + zZ) + \lambda (X, Y, Z),$$ where $\lambda = -\left(x^2 + y^2 + z^2\right)$, $$\Delta (xX + yY + zZ) = 0, \quad \Delta^3(X, Y, Z) = \lambda\Delta(X, Y, Z),$$ so $$\begin{align} e^{s\Delta}(X, Y, Z) &{}= C(X, Y, Z) \\ &{}+ S(yZ - yZ, xZ - zX, xY - yX) \\ &{}+ D(x, y, z)(xX + yY + zZ) + D\lambda (X, Y, Z), \end{align}$$ where the functions $C(\lambda, s)$, $S(\lambda, s)$ and $D(\lambda, s)$ are given by $$(C, S, D)_{s=0} = (1, 0, 0),\hspace1 em\Delta(C, S, D) = \frac{\partial}{\partial s}(C, S, D) = (\lambda S, C, S),$$ which, for $\lambda < 0$, work out to: $$(C, S, D) = \left( \cos\left(\sqrt{-\lambda}s\right), \frac{\sin\left(\sqrt{-\lambda}s\right)}{\sqrt{-\lambda}}, \frac{1 - \cos\left(\sqrt{-\lambda}s\right)}{\lambda}\right).$$

Since this action is actually taking place over the function space of $\so(3)^*$, what is denoted by the coordinates $(X, Y, Z)$, in these expressions, are actually linear functions, i.e. elements of $\so(3)^{**}$ — effectively the basis elements of $\so(3)$, itself. The result is that the adjoint action of $\so(3)$ by the group $\SO(3)$ is extended to non-linear functions.

This allows you, for instance, to talk about invariants in an intrinsic way; i.e. without recourse to matrix or other representations. For example, you have: $$X^2 + Y^2 + Z^2$$ which you can now directly show is an invariant, since, under transformation: $$\begin{align} \Delta\left(X^2 + Y^2 + Z^2\right) &{}= 2X\Delta X + 2Y\Delta Y + 2Z\Delta Z \\ &{}= 2(X(yZ - zY) + Y(zX - xZ) + Z(xY - yZ)) \\ &{}= 0. \end{align}$$

The locus of all transforms $\Delta = \{\cdots, \Lambda\}$ corresponding to all $\Lambda$ that $\so(3)$ gives rise to is one and the same as the symplectic leaf generated from the point $(X, Y, Z)$ that these transforms act on. Here: we see that it's a sphere with radius $X^2 + Y^2 + Z^2 > 0$ if $(X, Y, Z) \ne 0$ and the fixed point $(0, 0, 0)$ if $(X, Y, Z) = (0, 0, 0)$.

All of this involves non-linear functions. You can't directly address any of it with the Lie algebra $\so(3)$, since that's a linear space.

So, the coadjoint orbits — and the related Poisson manifold machinery — together provide a way to talk about the Lie algebra $\so(3)$ as a non-linear space. As Kirillov noted, it appears that Lie was originally striving for a non-linear theory. So, it ties off a long-standing loose end.

It also provides a way to by-pass much of the machinery associated with Lie algebra contractions. Consider, for instance, a 3-parameter family of Lie groups whose basis elements comprise the following 11 generators (with 9 of them arranged together as 3-vectors, for convenience):$\newcommand\bJ{\mathbf J}\newcommand\bK{\mathbf K}\newcommand\bP{\mathbf P}\newcommand\bW{\mathbf W}\newcommand\bomega{\boldsymbol\omega}\newcommand\bupsilon{\boldsymbol\upsilon}\newcommand\bepsilon{\boldsymbol\epsilon}$ $$\bJ = \left(J_0, J_1, J_2\right), \quad \bK = \left(K_0, K_1, K_2\right), \quad \bP = \left(P_0, P_1, P_2\right), \quad H, \quad M,$$ whose Lie brackets give rise to a 3-parameter family of Poisson brackets identified by the following transform law: $$\begin{align} \Delta\bJ &{}= \bomega\times\bJ + \bupsilon\times\bK + \bepsilon\times\bP, \\ \Delta\bK &{}= \bomega\times\bK - \gamma\bupsilon\times\bJ + \bepsilon M - \beta\tau\bP, \\ \Delta\bP &{}= \bomega\times\bP - \bupsilon M + \lambda\bepsilon\times\bJ - \kappa\tau\bK, \\ \Delta H &{}= -\beta\bupsilon\cdot\bP - \kappa\bepsilon\cdot\bK, \\ \Delta M &{}= -\gamma\bupsilon\cdot\bP - \lambda\bepsilon\cdot\bK \end{align}$$ corresponding to $\Delta = \{\cdots,\Lambda\}$, with $\Lambda = \bomega\cdot\bJ + \bupsilon\cdot\bK + \bepsilon\cdot\bP - \tau H + \psi(M - \alpha H)$. The parameters are $(\alpha, \beta, \kappa)$, with $(\gamma, \lambda) = (\alpha\beta, \alpha\kappa)$, with its invariants including the fundamental invariants: $$\begin{align} \mu &{}= M - \alpha H, \\ \nu &{}= W^2 - \gamma W_0^2 + \lambda W_4^2, \\ \rho &{}= \beta P^2 - 2HM + \alpha H^2 - \kappa K^2 + \alpha\beta\kappa J^2, \end{align}$$ as well as derived invariants, such as $$\mu^2 - \alpha\rho = M^2 - \gamma P^2 + \lambda K^2 - \gamma\lambda J^2,$$ where $$W_0 = \bJ\cdot\bP, \quad \bW = \left(W_1, W_2, W_3\right) = M\bJ + \bP\times\bK, \quad W_4 = \bJ\cdot\bK.$$

They can actually all be combined into a single Poisson manifold — by just throwing in $(\alpha, \beta, \kappa)$ as extra coordinates. Together, this comprises the coadjoint orbits of all the members of the Bacry Lévy–Leblond classification of possible kinematic groups, with $\gamma > 0$ and $\kappa = 0$ being the Poincaré group, centrally extended with a trivial central extension by $\mu$ and with $\alpha = 0$, $\beta \ne 0$ and $\kappa = 0$ being the Bargmann group, which is the (non-trivial) central extension of the Galilei group.

What was originally Lie algebra contractions, involving the parameters $(\alpha, \beta, \kappa)$, are now just continuous transitions in the coordinates $(\alpha, \beta, \kappa)$ of the combined Poisson manifold.

The formalism is more intuitively grounded, as well. The components of $\Lambda$ in the transform $\Delta = \{\cdots, \Lambda\}$ are:

  • infinitesimal rotations $\bomega = (x, y, z)$, which we've already discussed,
  • infinitesimal boosts by an infinitesimal change $\bupsilon$ in velocity,
  • infinitesimal spatial translations $\bepsilon$,
  • infinitesimal time translations $\tau$ (with the negative on $H$ being a legacy convention),
  • infinitesimal actions $\psi$ for the central charge $\mu = M - \alpha H$.
The transforms give you the transform laws for the generators themselves under the respective actions, e.g. $\Delta\bK = \bepsilon M$ under a spatial translation $\Lambda = \bepsilon\cdot\bP$ (which shows that $\bK$ is an "$M$-moment"), or $\Delta\bP = -\bupsilon M$ under a boost $\Lambda = \bupsilon\cdot\bK$ (which shows that $\bP$ is an "$M$-momentum").

It only seems "arcane" because it's usually being presented only in the abstract; but when get down into the details to see what it's actually all about, without the abstraction, you see that it's all quite intuitive, and intuitively grounded.

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    $\begingroup$ I love the historical background of this answer, but perhaps your opening could be phrased more objectively as, for example, "this perspective is historically backwards"? I don't think one can objectively argue about whether one perspective is more or less intuitive. For example, for me, writing things out explicitly in the way that you do makes things much harder, not easier, to hang a geometric intuition on—but I understand that there are plenty of people who reason otherwise, and I definitely feel it's important to have both viewpoints available. $\endgroup$
    – LSpice
    Dec 30, 2022 at 21:53
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    $\begingroup$ Ok. I'll remove the editorial, if you wish. $\endgroup$
    – NinjaDarth
    Oct 25, 2023 at 21:39

It is fair question probably encountered by everyone who begin to study these things.

The modern line of thought put everything into bigger picture, and that relation is some particular case of more general phenomena. But for that one needs to be familiar with certain ideas from several areas, which are kind of far from another, so probably it is some source of difficulties (at least for me when I started to study these things many years ago).

Let me try to sketch the ideas along that line of thought:

  1. Representations of Lie group — more or less same as representations of Lie algebra.

  2. Representations of Lie algebra — same as representations of universal enveloping algebra.

  3. Universal enveloping algebra — is an associative algebra, so as for any algebra looking for representation we should pay attention on ideals, (which are related to kernels of representations).
    So look on principal ideals $\hat C_i = 0$ — for some elements. But for non-commutative algebras $\hat C_i$ should be central elements for the ideal to be two-sided ideal and everything works smoothly.

So we arrive to the simple idea that: "Central Elements = Constants" — gives us some (actually main) source of the representations of any non-commutative algebra.

  1. The final step is to relate "Central Elements = Constants" to coadjoint orbits. That is better to understand in the general framework of quantization. But let us try to be a simple as possible.

  2. From $U(g)$ to $\operatorname{Fun}(g^*) = S(g)$ — by isomorphism of modules over $G$. The point is that universal enveloping algebra is kind of special algebra which is not far from commutative algebra of functions on the $g^*$. (It is its quantization, but let us try to avoid that term). Central elements in universal enveloping algebra (Casimir elements) they are, of course, invariant with respect to action of the Lie group — but as representation of the Lie group universal enveloping algebra is isomorphic to $\operatorname{Fun}(g^*)=S(g)$ — just the commutative algebra.

Point: So central elements in $U(g)$ are exactly the same as invariant elements in $\operatorname{Fun}(g^*)=S(g)$.

  1. That is more or less all what we need. From central elements $\hat C_i$ in $U(g)$ we got just functions $C_i$ on $g^*$.

Point: $ C_i = c_i$ — define precisely the coadjoint orbits. Just because invariance of $C_i$ and so invariance of defining functions is more or less the same as the manifold to be an orbit. Since it is an orbit in $g^*$, it is a coadjoint orbit.

So I hope the story above at least puts some ideas on the table and might be helpful. It is not the end of the story — one needs to explain how to construct the representations. That story is somewhat open-ended — there is clear philosophy why that might work, but it is not that much clear how to go from philosophy to constructions and when these constructions are possible, when not. Also the line of thought above should be better understood in light of general ideas of quantization — where a coadjoint orbit is just an example of a Poisson leaf, and it is expected that Poisson leaves should give rise to representations — it is a part of general "classical" to "quantum" dictionary.

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    $\begingroup$ Although the post seems clearly interested in characteristic $0$, and anyway one has to start the story somewhere and this is a good place to start it, I would like to re-beat the positive-characteristic drum and point out that your very first step, ‘nearly’ identifying representations of (connected) groups and their centralisers, is much more tenuous in positive characteristic—and yet the connection to coadjoint orbits still stands in many cases, thus indicating that we've surely got the right end of the stick. $\endgroup$
    – LSpice
    Dec 31, 2022 at 17:55

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