# What is the physical meaning of a Lie algebra symmetry?

The physical meaning of a Lie group symmetry is clear: for example, if you have a quantum system whose states have values in some Hilbert space $H$, then a Lie group symmetry of the system means that $H$ is a representation of some Lie group $G$. So you want to understand this Lie group $G$, and generally you do it by looking at its Lie algebra. At least initially, this is understood as a way to make the problem of classifying Lie groups and their representations easier.

But there are Lie algebras which are not the Lie algebra of a Lie group, and people are still interested in them. One possible way to justify this perspective from a physical point of view is that Lie algebras might still be viewable as ("infinitesimal"?) symmetries of physical systems in some sense. However, I have never seen a precise statement of how this works (and maybe I just haven't read carefully enough). Can anyone enlighten me?

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One perspective is: if you have a Lie algebra of symmetries, pretend it is the Lie algebra of a group, and try to do as much as you can without ever touching the (nonexistent) group. Surprisingly, this will get you quite far lots of examples. –  Mariano Suárez-Alvarez May 17 '10 at 8:35
"But there are Lie algebras which are not the Lie algebra of a Lie group". Given the physics context, I assume you mean perhaps infinite-dimensional Lie algebras over R: every finite-dimensional Lie algebra over R integrates to a group. –  Theo Johnson-Freyd May 17 '10 at 16:51
Incidentally, not every Lie group action has a corresponding Lie algebra action. For example, consider R acting on L^2(R) by translation. The infinitesimal generator of translation is the operator d/dx --- i.e. the Lie algebra action on any function space corresponding to the group action by translation is necessarily: the basis vector of the Lie algebra acts by d/dx. But, of course, d/dx does not act on L^2(R): the derivative of an L^2 function is not L^2. –  Theo Johnson-Freyd May 17 '10 at 16:53
Theo, there is, though, a very well developed of partially defined infinitesimal generators of groups. In the case of the translation action on L^2(R), the infinitesimal generator d/dx is closed, and that will take you very very far provided you are determined enough. –  Mariano Suárez-Alvarez May 18 '10 at 13:47

I'm not sure which sort of examples of Lie algebras without the corresponding groups you have in mind, but here is a typical example from Physics.

Many physical systems can be described in a hamiltonian formalism. The geometric data is usually a symplectic manifold $(M,\omega)$ and a smooth function $H: M \to \mathbb{R}$ called the hamiltonian. If $f \in C^\infty(M)$ is any smooth function, let $X_f$ denote the vector field such that $i_{X_f}\omega = df$. If $f,g \in C^\infty(M)$ we define their Poisson bracket $$\lbrace f, g\rbrace = X_f(g).$$ It defines a Lie algebra structure on $C^\infty(M)$. (In fact, a Poisson algebra structure once we take the commutative multiplication of functions into account.)

In this context one works with the Lie algebra $C^\infty(M)$ (or particular Lie subalgebras thereof) and not with the corresponding Lie groups, should they even exist.

Symmetries in this context are functions which Poisson commute with the hamiltonian, hence the centraliser of $H$ in $C^\infty(M)$. They define a Lie subalgebra of $C^\infty(M)$.

Another famous example occurs in two-dimensional conformal field theory. For example, the Lie algebra of conformal transformations of the Riemann sphere is infinite-dimensional: any holomorphic or antiholomorphic function defines an infinitesimal conformal transformation. On the other hand, the group of conformal transformations is finite-dimensional and isomorphic to $\mathrm{PSL}(2,\mathbb{C})$.

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The second example looks more like what I want. What do you mean by a "Lie algebra of conformal transformations"? Something like locally invertible conformal maps? –  Qiaochu Yuan May 17 '10 at 21:23
Oops. I didn't see this comment before now. A pair of holomorphic functions $f(z),g(z)$ defines an infinitesimal conformal transformation $z \mapsto z + f(z) + \overline{g(z)}$. They generate the Lie algebra of conformal transformations. It is in fact isomorphic to two copies of the algebra of diffeomorphism of the circle. –  José Figueroa-O'Farrill May 31 '10 at 16:43
Can you explain further? As I understand it (as a topologist not a physicist), smooth tangent vector fields on a manifold correspond to local flows = "infinitesimal actions" of the group of real numbers. The Lie algebra of all vector fields corresponds in a sort of formal sense to the group of all diffeomorphisms. One can speak of the Lie subalgebra preserving some tensor field (meaning that the Lie derivative vanishes), and this structure corresponds in the same way to local diffeomorphisms preserving it. This fits with your discussion of the Hamiltonian setup, but ... –  Tom Goodwillie Jun 23 '10 at 4:14
... in case of conformal structure I would think that the vector fields preserving that structure correspond naturally to the holomorphic sections of the complex tangent bundle, so that locally such a field corresponds to one holomorphic function -- where's the antiholomorphic one? -- and globally on the Riemann sphere they form a 3-d complex Lie algebra, that of the (finite-dimensional) global conformal symmetry group. What am I miss(understand)ing? –  Tom Goodwillie Jun 23 '10 at 4:20