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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?

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migrated from mathoverflow.net Mar 24 '15 at 11:31

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  • $\begingroup$ 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). $\endgroup$ – Ryan Budney Mar 23 '15 at 21:47
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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.

Hope this helps! Please let me know if you would like me to elaborate further.

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    $\begingroup$ Hi and thanks for your answer. Would you be able to offer an example of my main question in the context of, say, SO(3) or SU(2) or some simple similar Lie group? $\endgroup$ – user39726 Mar 23 '15 at 23:42

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