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A note for self-reference: this post continues but differs from another post: Lie derivative of a function (of a point) with respect to a vector field

Lie derivative $L_XY$ of vector field $Y$ at a point $p$ with respect to another vector field $X$, which is also a vector field. We can deduce that $L_XY=XY-YX$, or in more complete notation (to avoid confusing $(L_XY)f=(XY)_pf$ with $X_p(Yf)$), $(L_XY)_p(f)=X_p(Yf)-Y_p(Xf)$, namely:

  • 'a vector (of the Lie derivative vector field) acting on/differentiating any proper function $f$
  • equals a linear combination of a vector (of a vector field) acting on/differentiating another function 'of similar domains and image' (the Lie derivative of the function $f$).'

(Note that the above stuff can be abbreviated as (XY)f=X(Yf)-Y(Xf), the associate law doesn't exist here. About vector field as differentiating, as seen in differential equations, here is a post containing my relevant puzzle Vector field on a manifold and its basis)

So we can define a bracket operation [ , ]. It's easy to prove

  1. $[X,Y]=-[Y,X]$,
  2. $[X,X]=0$
  3. 'Jacobian identity': $[X,[Y,Z]]+[Z,[X,Y]]+[Y,[Z,X]]=0$.
  4. linearity of [ , ] with respect to its arguments.

In Lie algebra, we define a similar operation $[\ ,\ ]_\frak{g}$, ($\frak{g}$ is unnecessary, here I just use it to distinguish the two bracket operations), with 2nd and 3rd properties.

So my questions are

A. what's the relations between [ , ] and $[\ ,\ ]_\frak{g}$? (main question)

[B. why we choose 2nd, 3rd properties, not others, to define $[\ ,\ ]_\frak{g}$ in Lie algebra.] [Just find Question B is invalid since all mentioned properties are used to define $[\ ,\ ]_\frak{g}$.)

C. What motivates definition of [ , ], is it similar to the motivation for definition of $[\ ,\ ]_\frak{g}$? (Seems not, motivation for $[\ ,\ ]_\frak{g}$ involves exponential maps of Lie groups, which means not much for Lie derivatives.)

ps: I am only familiar with notations in Spivak's book, which is what I mainly use above.


(Edited to add:)

An extra minor note for self reference. I see what causes my confusion, namely, $[\ ,\ ]_{\frak{g}}$ has more kinds of arguments (e.g. matrix) than those of [ , ], (i.e.the vector fields). That implies operations and actions (e.g. matrix product and its action on vectors, $AB, Av$) involved in Lie algebra may not follow the laws those of vector fields ($ Xf, XY=L_xY$) follow, (Since these laws more depend on specific type of elements used as arguments of bracket operation, and are 'independent' of bracket operation itself.), e.g. non-commutativity law of vector field is shared by matrices, but non-association law is not: $(XY)f\neq X(Yf)$ (implying possibly $(XY)f-(YX)f\neq X(Yf)- Y(Xf)$, but actually we have '=' (correct?) for perhaps 'increment' cancels out), while $(AB)v=A(Bv)$.

Summarize it out, we see, particularly if we don't think of meanings in these notations, then apparently $(XY)f=(L_xY)f=(XY-YX)f$ (distribution law fails), $XY-YX\neq 0$ (commutativity law fails), $(YX)f\neq Y(Xf)$ (association laws fail). If we carefully think of meanings in these notations, we see these laws may not fail; but we need to be careful when we 'apply these laws' in Lie derivatives (and in Lie algebra too).

This is very rough understanding. I guess I need to go to many details of calculation to see what Lie derivatives and Lie algebra mean and how they work.

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2 Answers 2

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For A, the first is a special case of the second. If $\mathscr X$ is the space of vector fields, you can check that the bracket of vector fields gives $\mathscr X$ the structure of a Lie algebra.

For B, the bracket in any Lie algebra is required to have all four properties you listed.

For C, the definition of a Lie algebra (and hence the Lie bracket $[{-},{-}]_{\mathfrak g}$) is actually inspired by the behaviour of the algebras of vector fields on differentiable manifolds.

That is the general overview, but your question is actually very broad and we could expand upon every one of your points.

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I believe that all of the axioms you listed are to be satisfied in order for a mapping to be called a Lie bracket $[\;,\; ]_{\mathfrak{g}}$.

To elaborate, perhaps as a tid bit of information - it is fruitful to look at the Lie algebras as tangent spaces to Lie groups at the identity, $\mathfrak{g}=T_{e}G$ (a Lie group is a group endowed with the structure of a smooth manifold - multiplication and taking the inverse are smooth maps). With this interpretation, the Lie bracket is a second derivative of a certain mapping.

This article will perhaps make a better job at explaining the technicalities than I could:

Adjoint representation

The Lie derivative of smooth vector fields on a Lie group (actually, on any smooth manifold, but let's stick to this case) $\Gamma(G;TG) = \{X:G\rightarrow TG, X \text{ is smooth} \}$ is defined in the standard way by taking the pushforward of one vector field (say, $Y$) backwards along the flow of the second vector field $X$. $$ (\mathcal{L}_{X}Y)(p) =\frac{d}{dt}\bigg{(} \big{(}\mathcal{G}^{X}_{-t\,*}Y\big{)} (p) \bigg{)}\bigg{\rvert}_{t=0}. $$

I denote the one-parameter diffeomorphism group associated with the X vector field flow by $\mathcal{G}^{X}_{\bullet} : [-\epsilon,\epsilon]\times G \rightarrow G$ and by $\mathcal{G}^{X}_{s \,*} : \Gamma(G;TG) \rightarrow \Gamma(G;TG) $ the pushforward of a vector field along this map.

Within the standard framework of differential geometry (no need for Lie groups, just any smooth manifold), it indeed holds that the Lie derivative of a vector field is the Lie bracket: $\mathcal{L}_{X}Y = [X,Y]$.

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  • $\begingroup$ I guess my confusion is about that in $[\ , \ ]$ about Lie derivatives, it's defined w.r.t. its action on a function while in definition of Lie algebra, $[\ , \ ]_\frak{g}$ we don't mention the function. $\endgroup$ Jul 29, 2020 at 8:04
  • $\begingroup$ {This puzzle can be solved by considering the 'natural order of operations'/law of association, for example, for matrix, it's natural order of calculation goes from right to left, more specifically $[A , B ]_{\frak{g}}=AB-BA$ and $[A , B ]_{\frak{g}}(v)=(AB-BA)(v)$ (where v is of similar role to that of f above)) $=A(Bv)-B(Av)$ which, which we compare it to its analog in [ , ] of vector fields, is supposed to $\neq (AB)v-(BA)v$ (but this is obviously incorrect.*} $\endgroup$ Jul 29, 2020 at 8:04
  • $\begingroup$ What's in the {} in my last comment is incorrect somewhere, but it does let me see what causes my confusion, namely, $[\ ,\ ]_{\frak{g}}$ has more kinds of arguments (e.g. matrix) than those of [ , ], (i.e.the vector fields). That implies operations and actions ($AB, Av$) involved in Lie algebra may not follow the laws those of vector fields ($ Xf, XY=L_xY$) follow, e.g. non-commutativity law of vector field is shared by matrices, but non-association law is not: $(XY)f\neq X(Yf)$ (implying possibly $(XY)f-(YX)f\neq X(Yf)- Y(Xf)$, but it's '=' (correct?)), while $(AB)v=A(Bv)$. $\endgroup$ Jul 29, 2020 at 8:32
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    $\begingroup$ i must admit I don't quite follow what the problem is. We cannot compose vector fields with vector fields, since a vector field acting on a function is a mapping: $X: \mathcal{C}^{\infty}(M) \rightarrow \mathcal{C}^{\infty}(M)$. Therefore the only way is to act on the function on the output again, that is, by $XYf$ we implicitly always mean $X(Y(f))$. Likewise, $[X,Y](f) \equiv X(Y(f)) - Y(X(f))$. $\endgroup$
    – K.T.
    Jul 29, 2020 at 12:15
  • $\begingroup$ Sorry, I will correct the second statement I gave. The composing of vector fields is of course possible via their action on a function. $\endgroup$
    – K.T.
    Jul 29, 2020 at 12:34

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