Given a complete vector field $Z$ on a smooth manifold $M$, let us write $\exp(Z)$ for the time-$1$ map of the flow of $Z$. Thus, $\exp(Z)$ is a diffeomorphism of $M$. On the other hand, if $G$ is a Lie group and $X$ is in the Lie algebra $\mathfrak{g} = T_e(G)$ then $\exp(X)$ is an element of $G$. So, $\exp(Z)$ and $\exp(X)$ are different types of objects.

Of course the two notions of exponential map are not unrelated. Given $X \in \mathfrak{g}$, there is precisely one left-invariant vector field $X_\ell$ and precisely one right-invariant vector field $X_r$ on $G$ with $X_\ell(e)=X_r(e)=X$. The vector fields $X_\ell$ and $X_r$ are generally different, but the integral curve through $e$ is the same for each and we can use either of them to define the exponential map of $G$.

$$ \exp(X_\ell)(e) = \exp(X_r)(e) = \exp(X)$$

So it would seem there is not much reason to prefer left-invariant vector fields or right-invariant vector fields, and it is more standard to work with left-invariant vector fields. However, I can think of one reason one might slightly prefer to work with the right-invariant vector fields and that is that, for right-invariant vector fields, one has \begin{align*} \exp(X_r)\circ\exp(Y_r)(e) = \exp(X) \exp(Y) && X,Y \in \mathfrak{g} \end{align*} whereas for left-invariant vector fields, one has \begin{align*} \exp(X_\ell) \circ \exp(Y_\ell)(e) = \exp(Y) \exp(X) && X,Y \in \mathfrak{g}, \end{align*} the order of multiplication getting turned around. This is not an extremely big deal, but it still leaves me slightly preferring to work with right-invariant vector fields, absent any argument for the alternative.

I guess the real issue is that, if we define a left-invariant (resp. right-invariant) diffeomorphism $\phi :G \to G$ to be one satisfying $\phi(xy)=x\phi(y)$ (resp. $\phi(xy)=\phi(x)y$) for all $x,y \in G$, then all of them arise uniquely as right-multiplications (resp. left multiplications) and $\phi \mapsto \phi(1)$ defines an anti-isomorphism (resp. isomorphism) from $\{\text{right-invariant diffeos}\} \to G$ (resp. $\{ \text{left-invariant diffeos}\} \to G$.)

Question: Is their any advantage to using left-invariant vector fields over right-invariant vector fields? Or, is this just an inconsequential convention?

  • 1
    $\begingroup$ It really doesn't matter, it's just convention $\endgroup$ – user408856 Jun 7 '18 at 5:16

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