# Clarification on notation of “left invariant fields” (Lie groups)

In these notes in Definition 1.4 we learn that

A vector field $X$ on a Lie group $G$ is called left invariant if $d(L_g)_h(X(h))=X(g(h))$ for all $g,h \in G$, or for short $(L_g)_*(X)=X$.

where $L_g : G \to G$ are the left invariant fields. I am very confused on both the quoted text as well as on what the left invariant fields are. Please note I am a physicist and if the above are trivial I still cannot quite understand them. What is this $d$? Is it some form of a derivative? Could you provide some easy example?

## migrated from mathoverflow.netMar 24 '15 at 11:31

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• If $G$ is a matrix group, $X(A) = AB$ is left-invariant for any matrix $B$ fixed. Similarly, $Y(A) = BA$ is right-invariant for any $B$ fixed. As Amitsh mentions $d$ is a way of writing the derivative (as a linear transformation). – Ryan Budney Mar 23 '15 at 21:47

The formal answer is that $d$ is the differential of the left translation map $L_g:G\to G$ given by the rule $L_g(h)=gh$ for $h\in G$. The map is smooth by the definition of "Lie group" and so there is a notion of the differential of this map. In particular, $d(L_g)_h$ denotes the differential of $L_g$ at $h\in G$; it is a map of tangent spaces $d(L_g)_h:T_{h}G\to T_{gh}G$.
A more intuitive way of thinking about this is to think of vector fields as differential operators (on $C^{\infty}(G)$) and left-invariant vector fields as left-invariant differential operators. Let me define what this means. The group $G$ acts on $C^{\infty}(G)$ by left-translations: if $g\in G$ and $f\in C^{\infty}(G)$, then $L_gf(x)=f(gx)$. A (first order) differential operator is a derivation $C^{\infty}(G)\to C^{\infty}(G)$ (that is, a linear map satisfying the Leibniz rule: $D(fg)=D(f)g+fD(g)$ for $f,g\in C^{\infty}(G)$) and we define a differential operator $D:C^{\infty}(G)\to C^{\infty}(G)$ to be left-invariant if $D(L_gf)=L_gDf$ for all $f\in C^{\infty}(G)$. A vector field on $G$ is the same thing as a (first order) differential operator and a left invariant vector field on $G$ is the same thing as a (first order) left invariant differential operator.
A tangential note: in my answer I refer to "first-order left-invariant differential operators on $G$" which should be thought of as being precisely the elements of the Lie algebra of $G$. Higher order left-invariant differential operators correspond to elements of the universal enveloping algebra of the Lie algebra, in case you have encountered this construction before.