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Given a lie group $G$ and an action $G \curvearrowright M$ where $M$ is a manifold, how can I get a homomorphism $\mathfrak g \to Diff(M)$. Here $Diff(M)$ is the universal differential operators on $M$.

My professor wrote down this step on the board.

The only case in which I know how to do this is the multiplication action of $G$ on $M=G$.

My main problem is that I don't know the definition of $Diff(M)$ for arbitrary $M$ - I only know the definition for a lie group.

Do you have any ideas?

De Yang
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2 Answers2

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I will use the notation of $\mathcal{D}(M)$ for global differential operators on $M$. This is a filtered $\mathbb{R}$-algebra, defined as follows: $$\mathcal{D}(M)_0=C^\infty(M)$$ and for $n>0$, define $\mathcal{D}(M)_i$ inductively as: $$\mathcal{D}(M)_i=\{T\in End_\mathbb{R}(C^\infty(M)):[T,f]\in\mathcal{D}(M)_{i-1}\text{ for all }f\in C^\infty(M)\}$$ where $[T,f]$ means commutator as linear operators on $C^\infty(M)$. Then set: $$\mathcal{D}(M)=\bigcup\limits_i\mathcal{D}(M)_i$$ A good exercise is to check that $\mathcal{D}_1(M)=C^\infty(M)\oplus Vec(M)$, where $Vec(M)$ is the Lie algebra of smooth vector fields on $M$.

Another, equivalent way of defining it (although these aren't the only ones) is to define $\mathcal{D}(M)$ to be the set of linear operators $T\in End_\mathbb{R}(C^\infty(M))$ such that in local coordinates $(t_1,\dots,t_n)$ on $M$, $T$ is given by a finite sum $\sum\limits_{\alpha}f_\alpha\partial^\alpha$, where the sum is over multi-indices $\alpha$ (I hope the notation is clear).

Anyhow, now suppose that a Lie group $G$ acts $M$. Then $G$ acts on $C^\infty(M)$ via pullback. This is now a representation of $G$ on an algebra (it preserves the multiplicative structure of $C^\infty(M)$), so we can differentiate the action and get an action of the Lie algebra $\mathfrak{g}$ on $C^\infty(M)$ acting by derivations, i.e. a map of Lie algebras $\mathfrak{g}\to Vec(M)$.

This map $\mathfrak{g}\to Vec(M)$ of Lie algebras induces, by the universal property, a map of $\mathbb{R}$-algebras $\mathfrak{U}(\mathfrak{g})\to\mathcal{D}(M)$. This, I believe, is the desired map.

freeRmodule
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  • Thanks for your answer. This enabled me to use the same steps that are used in showing that $ad(X)(Y)=[X,Y]$ to show that the action on $f \in C^\infty(M)$ is just the derivative of the pullback you mention $\frac d {dt} |_{t=0} exp(tX)^* f$. This in turn gave me the simple description in my answer. – De Yang Nov 30 '16 at 04:31
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Short answer: Just extend and lift the assignment sending $X \in \text{Lie } G$ to the vector field $\widetilde{X} \in \mathfrak{X}(M)$ defined by $\widetilde X(y)= \frac d {dt}|_{t=0} exp(tX) y$.

Here is some more if you like:

It is obvious from this description that this is a left invariant vector field on $M$. Thus I have defined a map from $\mathfrak{g} \to \mathfrak{X}(M)^G$. By universal properties, this extends and lifts to a map from $U(\mathfrak{g})$ to invariant differential operators on $M$.

De Yang
  • 353