Relations between two definitions of Lie algebra

Question

This post follows from another post What is exponential map in differential geometry about two kinds of exponential maps (of Riemannian groups and of Lie groups, separately) and Lie algebra. It is inspired by discussions following the answer, which are not repeated here.

It’s said there are two definitions of Lie algebra (tangent space, left invariant vector field).

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(The question is originally stated as ‘ Relations between two two definitions of exponential maps’, that’s something I’m also interested in, I may put another post for that if necessary.)

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(Edited to add:)

By far I guess Lie algebra is a bit like a collection $G$ of left invariant (well behaved) vector field such that from a vector at a point we can infer or generate vectors at all other points, (i.e. a well behaved vector field), for these vectors are somehow the same or homogeneous; the homogeneity and generalizability is what the invariant means. [It's, as explained below, invariant of vector fields $X$ or phase space... w.r.t. the operation $+$ of Lie group. e.g. $X_{p+q} = X_q$ for all $p, q$ in the Lie group.](Probably right invariant works too) So there is a one one correspondence between a left invariant vector field in $G$ and a vector in a tangent space $T_qM$ (it seems, according to some other posts, $q$ can be any point and we prefer identity for it’s convenient.) and so $G$ of these vector fields and $T_qM$ are isomorphic or have at least some kind of one one correspondence and so the two definitions are consistent.

The definition of Lie algebra also includes the consideration of commutability of two left invariant vector fields. For that purpose we define an unusual multiplication [,]. Why we particularly need to take care of that commutability? I guess it’s for the the expansion of log (exp(X)exp(Y)), as mentioned in the comment of the origin post. (Btw, in the tangent space definition do we need to consider commutability?) Why we do such expansion? It’s because the idea of exponential maps of Lie groups originates from exponent of matrix?

In a word, the left invariant definition seems to justify the tangent space definition (I guess there is a related proof) and if we consider tangent space at all points and carefully pick up a vector of invariant property (like of certain length and direction) from each tangent space we may well visualize ANY left invariant vector field. And it is isomorphic to a vector of tangent space at ONE point.

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(The following continues discussion, in comments on an answer, on notations in Lie group)

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About notations, using Lie group $M$ as an example, $\ell_q:M\to M$ (or in Spivak's notation, $L_a$) is adding a point $q$ to any point in $M$ (such addition is possible since we impose a Lie group structure to a manifold), while $\ell_{q*} $ (or $\ L_{a*}$) is the derived operation for the tangent space of Lie group $M$ (NOT the Lie group itself) at a point $q$, e.g. $T_pM$ or $M_p$ (it confuses me since the two denote the same thing), adding q to p (NOT adding elements in tangent space) to get tangent space $T_{q+p}M$. Using Lie group $SO(2)$ (~$S^1$) as an example $\ell_A:SO(2)\to SO(2)$ is multiplying a matrix $A$ to any matrix in $SO(2)$ , while $\ell_{A*}$ is the derived operation for the tangent space of Lie group $SO(2)$ at a point $p$, e.g. $T_pS^1$, adding q to p to get tangent space $T_{q+p}S^1$ .

Left invariant means a vector field (or a collection of vector fields, or all tangent vectors at all points or in phsical context the phase space, or in symplectic geometry and the Hamiltonian mechanics (which I know little) the similar pair of position and velocity), each element of it for any 'distance' (any element in Lie group) being transferred or moving to another point and we still get the same vector field (or vector fields, or phase space...). (Complement: considering Lie derivative of a vector field, this seems to somehow the same as saying $L_XX=0$, which in terms of Lie algebra, just $[X,X]=0$ in the definition; by seeing [ , ] as 'derivative' it seems the meaning is clearer. Put that view in the context of matrix Lie group, e.g. $SO(2)$ where $[A, A]=0, [A, B]=0$, it's like saying the two vector fields corresponding to two tangent vectors at a same point differentiated against themselves and, sometimes, even against each other equals zero.)

And Lie group basically enables us to to interpret a point at a manifold as a distance, similar to that we can treat a vector (position) in Euclidean space as a displacement (by setting the 'original point' $O$, which 'becomes' in Lie group the unit $e$). With Lie group we 'geometrify' the non-geometrical objects like a matrix set, and 'numerify' the non numerical objects like a manifold.

And exponential maps basically links (though not necessarily one one) a tangent vector to a point at a manifold (geometrical manifolds like surface or more abstract manifold like a matrix set, the two corresponding to the two kinds of exponential maps I guess) interpreted as a 'distance'/displacement. With exponential maps we link tangent space (a vector space) to the manifold (now made a Lie group).

But here comes another question, which I states in another post: why we need to, with exponential maps, make a link between a tangent space and the manifold?

Answer 1

Let $G$ be a Lie group. We say $X\in \mathfrak{X}(G)$ is left invariant if $\ell_{g,*}X=X$, where $\ell_g:G\to G$ is the left multiplication map. More precisely, for any $p\in G$, $\ell_{g,*,p}X_p=X_{gp}.$ There is a simple bijection between left invariant vector fields on $G$ and $T_eG$, given by sending a left invariant vector field $Y$ to $Y_e\in T_eG$. This defines an isomorphism of vector spaces.

Let's denote the left invariant vector fields ${}^G\mathfrak{X}(G)$. The left invariant vector fields $Y\in {}^G\mathfrak{X}(G)$ have the benefit that they come with a natural operation, given by $[Y,Z]=YZ-ZY$, defined on functions by $[Y,Z](f)=Y(Z(f))-Z(Y(f))$. You can check that this defines a vector field (e.g. a derivation of $\mathscr{C}^\infty(G)$). Note that defining $Y\cdot Z$ by $(Y\cdot Z)(f)=Y(Z(f))$ does not result in a vector field in general as it will not satisfy the Leibniz rule. The bottom line is that ${}^G\mathfrak{X}(G)$ has a natural bracket product structure making it into a Lie algebra. That is, $[\:,\:]$ is $\mathbb{R}-$bilinear, alternating, and satisfies the Jacobi identity.

We use this bracket $[\:,\:]$ on ${}^G\mathfrak{X}(G)$ coupled with the isomorphism $T_eG\cong {}^G\mathfrak{X}(G)$ to define an analogous operation on $T_eG$ by $[X_e,Y_e]=[\widetilde{X},\widetilde{Y}]_e$ where $\widetilde{X}$ is the unique left invariant vector field on $G$ with $\widetilde{X}_e=X_e$ defined by $$ \widetilde{X}_g:=\ell_{g,*,e}X_e.$$ $T_eG$ has the useful concrete interpretation as being the tangent space to $G$ at $e$, and as mentioned in your other post this allows us to define $\exp:T_eG\to G$ in concrete geometric terms as geodesics with respect to a bi-invariant metric on $G$. Anyway, using this natural identification we call the resulting algebra $\mathfrak{g}$, the Lie algebra of the group $G$. In addition, using $\mathfrak{g}=T_eG$ makes the functoriality of $G\mapsto \operatorname{Lie}(G)=\frak{g}$ perhaps more transparent. That is, given a smooth map $f:G\to H$ of Lie groups, there is a map $df_e:T_eG\to T_eH$ which is inducted. This gives a map $\mathfrak{g}\to \mathfrak{h}$.

The bottom line is that both interpretations of the Lie algebra of $G$ are fruitful and there is a canonical identification of the two.

Perhaps I have not really answered your question regarding commutativity. My interpretation is that $X\cdot Y=XY$ will not define a vector field, so the "natural" multiplication is not actually natural at all. Instead, the vanishing of $[X,Y]=\mathcal{L}_XY$ (Lie derivative) implies that the flows associated to $X$ and $Y$ commute which is of geometric significance and was a historically relevant quantity prior to the "invention" of abstract Lie groups.

Answer 2

Question: What is the commutability you are referring to?

Remark: Given a manifold $M$, the set of all tangent vector fields on $M$ is same as the set of global $C^{\infty}$-derivations. Which comes with a natural lie algebra structure. In particular when one considers a Lie Group, you are looking at a specific sub-algebra which captures the action of the group.

Also, if you start with a group representation $\phi$ then for a connected group understanding this representation reduces to understanding $d\phi$ which is a representation of Lie Algebras.

I hope this would provide some insights to the definitions of Lie Algebra of a Lie Group.