# Which Lie algebras are realised as vector fields on a group?

Fix a Lie group $G$. The vector fields on $G$ form an (infinite-dimensional) Lie algebra with the commutator of vector fields as Lie bracket. What (finite-dimensional) Lie algebras can I realise as subalgebras of this? (Let's suppose also that the vector fields must be non-vanishing, although I am happy to ignore this assumption if it helps).

I know that the set of left-invariant (or right-invariant) vector fields form a Lie algebra isomorphic to $\mathfrak{g}$, the Lie algebra of the group $G$. We can also get subalgebras of $\mathfrak{g}$ this way. Are there other Lie algebras (i.e. not isomorphic to $\mathfrak{g}$ or its subalgebras) that we can get by taking some appropriate subset of vector fields on $G$?

As a concrete example: Consider the group $G = SU(2)$. The left-invariant vector fields will form an $\mathfrak{su}(2)$ Lie algebra. Any single vector field on $SU(2)$ will form the abelian Lie algebra $\mathfrak{u}(1)$, and two commuting sets of vector fields will form the $\mathfrak{u}(1) \times \mathfrak{u}(1)$ Lie algebra. Can we find three vector fields which give us, say, the Heisenberg Lie algebra?

$[v_1,v_2]=v_3, \quad [v_2,v_3] = 0, \quad [v_1,v_3]=0$

If we were able to find three (globally defined) commuting vector fields, then these would form a three dimensional abelian Lie algebra $\mathfrak{u}(1) \times \mathfrak{u}(1) \times \mathfrak{u}(1)$, however there are topological restrictions to this, so it seems we can't simply obtain every finite dimensional Lie algebra.

• If you don't care about invariance, there's no point in considering a Lie group, or at least in mentioning its group structure, and it would be more natural to refer to $SU(2)$ as the 3-sphere. Similarly if you work on $SL_2(\mathbf{R})$, then it would be more natural to refer to it as the manifold product of a real plane and a circle. – YCor Apr 14 '17 at 1:58
• This question arose naturally in the context of Lie groups, but of course you are right, I could look at vector fields on any manifold. I restrict to Lie groups because I think (and hope) that one can say more about which algebras are possible. For example, we know that for a Lie group there are always at least two algebras we can obtain ($\mathfrak{g}$ and $\mathfrak{u}(1)$). For an arbitrary manifold, we can't even get $\mathfrak{u}(1)$ unless it is parallelisable. – Mark B Apr 14 '17 at 2:03
• @Mark: that is not true; any nonzero vector field gives a copy of $\mathfrak{u}(1)$, and every positive-dimensional manifold admits a nonzero vector field. – Qiaochu Yuan Apr 14 '17 at 2:06
• Sorry, brain fart! – Mark B Apr 14 '17 at 2:08
• On any manifold $M$, pick nonoverlapping open sets, and for each open set, pick one vector field vanishing outside that open set. These vector fields then commute, forming an abelian Lie algebra. So even the abelian Lie algebras on the real number line cannot be classified up to diffeomorphism. If a Lie group is diffeomorphic to a ball, then the same construction works: you pack lots of balls, extend vector fields to vanish outside balls, get lots of Lie algebras. – Ben McKay Apr 14 '17 at 9:10

If you fix a specific Lie group and look for finite dimensional Lie subalgebras of the algebra of vector fields then the answer is not trivial. Take an easy Lie group, e.g. $G=\mathbb R^2$. Already in this case the answer is quite involved, you can find it here: http://www-users.math.umn.edu/~olver/q_/lavf.pdf. (Here it is resonable to "classify" under change of local coordinates, i.e. local diffeomorphisms)
• The paper you mention discusses only the real analytic Lie algebras. There are many more $C^{\infty]$ examples. On any manifold $M$, pick nonoverlapping open sets, and for each open set, pick one vector field vanishing outside that open set. These vector fields then commute. So even the abelian Lie algebras on the real number line cannot be classified. – Ben McKay Apr 14 '17 at 9:06