When constructing the Lie algebra $L(G)$ of a Lie group $G$, one usually uses the identification of the tangent space $T_1 G$ with left invariant vector fields $\mathcal{V}^l(G)$ to construct the Lie bracket on $L(G)=T_1 G$. However, in the literature on Lie groupoids and Lie algebroids some authors use left-invariant and some use right-invariant vector fields (and I think that right-invariant vector fields are more natural here).

There are several reasons for using one convention or the other, for instance using the right invariant vector fields yields the usual Lie bracket on the vector fields, which are the sections of the Lie algebroid of the pair groupoid. This can be rephrased by saying that the natural action of the bisections (which are just diffeomorphisms in the case of the pair groupoid) yield a Lie algebra homomorphism from the right invariant vector fields on the Lie groupoid to the vector fields on the manifold. However, in finite-dimensional Lie theory one often wants to make Lie algebra elements act on functions on the Lie group, which then involves a $^{-1}$ in the left regular representation and thus $\mathcal{V}^l(G)$ does act naturally on $C^\infty(G,\mathbb{R})$.

This has the unfortunate effect, that if we consider $G$ as a Lie groupoid with one object $(G\rightrightarrows *)$, then the Lie bracket on its Lie algebroid $L(G\rightrightarrows *)$ is not the same as the Lie bracket on the Lie algebra $L(G)$, considered as a Lie algebroid $L(G)\to *$.

Now I have two questions on this:

  1. What are further reasons for using one or the other convention (for instance, calculations that are substantially easier in one or the other)?
  2. In your opinion, is this is a historical accident that one should stick to and endure the resulting signs or do you see good reasons for breaking with the conventions in favour of a unified construction of the Lie functor on Lie groups and Lie groupoids?

    Note: One cannot simply use left-invariant vector fields on Lie algebroids, what is frequently done in the literature. One would also have to change the way how these objects act naturally, for instance that diffeomorphisms act on a manifold naturally from the left by evaluation.

  • $\begingroup$ Perhaps the answer to (2) is "yes". I get the impression from books by Jerry Marsden on mechannics that much of Lie's work was motivated by studying the Rigid body. Here the phase space ends up being the left invariant vector-fields, and this may have motivated Lie's convention. Just a guess, so I'm hesitant to post it as an answer. $\endgroup$
    – hoj201
    Aug 19, 2014 at 0:12

1 Answer 1


My view is that one needs both, left and right invariant vector fields. Some reasons: For $X\in\mathfrak g=T_eG$ the left invariant vector field $L_X$ has a flow consisting of right translations: $$ \text{Fl}^{L_X}_t(x) = x.\exp(tX) $$ and conversely. For a left action $\ell:G\times M\to M$ we need the right invariant vector field $R_X$ so that $R_X\times 0_M$ on $G\times M$ is $\ell$-related to the fundamental vector field (infinitesimal left action by $X$) $\zeta_X$ on $M$. Etc.

The situation is very transparent for the left action of the diffeomorphism group $\text{Diff}(M)$ on a compact manifold $M$ itself. The Lie algebra of the diffeomorphism group is the space $\mathfrak X(M)$ of smooth vector fields on $M$ with the negative of the usual Lie bracket as bracket. Here one sees, that both left and right make their appearance at the same time.

As for mechanics of a rigid body with configuration space a Lie group, as well as for continuum mechanics of a fluid, say, with configuration space the group of (volume preserving) diffeomorphisms, one uses both trivializations: Let $G$ be the group in both cases, with $\mu:G\times G\to G$ the multiplication, $\mu(x,y)=x.y=\mu_x(y)=\mu^y(x)$. Let $g(t)$ be a curve in the group acting on a material point $x$ in space by $g(t).x$. Then:

  • The material or Lagrangian velocity is in the tangent bundle of group itself: $\partial_t g(t)\in TG$

  • The spatial or Eulerian velocity is right trivialized in the Lie algebra, the right logarithmic derivative: $\partial_tg(t).g(t)^{-1} = T(\mu^{g(t)^{-1}})(\partial_tg(t))$.

  • The body or convected velocity is left trivialized in the Lie algebra, the left logarithmic derivative: $g(t)^{-1}.\partial_tg(t) = T(\mu_{g(t)^{-1}})(\partial_tg(t))$.

All 3 are used: see 15.2 of `Marsden, Ratiu: Introduction to Mechanics and Symmetry'


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .