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I saw the proof but I have no intuition for this. Please give me an intuitive picture to keep in mind alongside the proof.

I THINK this has to do with the group ITSELF being closed under group multiplication. This is because $[X,Y]$ is sort of a derivative of:

$$e^Xe^Ye^{-X}e^{-Y}$$

The above must ITSELF be a group member, being a product of group members. If $[X,Y]$ is sort of a derivative of this, then idk:

$$e^Xe^Ye^{-X}e^{-Y} \sim e^{[X,Y]}$$

If the LHS is a group member, so is the RHS. Which means $[X,Y]$ generates a group member. which means $[X,Y]$ is a linear combination of generators.

But the above feels hand wavy and probably wrong reasoning.

Ryder Rude
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  • The commutator of generators is only a linear combination of generators. The coefficients are the structure constants. They're invariant under change of representation. – calc ll Aug 11 '22 at 02:01
  • It seems that your actual question is: "Why is the Lie algebra of a Lie group closed under taking commutators?" To answer this you have to tell us how do you define the Lie algebra of a Lie group. The usual definition is in terms of left-invariant vector fields. – Moishe Kohan Aug 11 '22 at 02:04
  • @MoisheKohan I'm defining them to be the derivatives of group members at identity. – Ryder Rude Aug 11 '22 at 02:15
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    @RyderRude: This is not a meaningful mathematical definition. I suggest you read a textbook on the subject to get your definitions straight. See some recommendations here: https://math.stackexchange.com/questions/194419/whats-a-good-place-to-learn-lie-groups – Moishe Kohan Aug 11 '22 at 02:39
  • Which proof are you referring to? Please share in a [edit]. – Shaun Aug 11 '22 at 11:46
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    @Ryder If you comfortable with $e^X$ being an element of the Lie group for $X$ from the Lie algebra, then one approach that you might find the following argument convincing: for $A$ from the Lie group and $X$ from the Lie algebra, $AXA^{-1}$ is another element of the Lie algebra. With that in mind, consider the derivative at $t = 0$ of the map $t \in \Bbb R \mapsto e^{tY}Xe^{-tY}$ where $X,Y$ are elements of the Lie algebra. – Ben Grossmann Aug 11 '22 at 19:08
  • @BenGrossmann Much thanks. I saw a similar argument in my book. It said that $e^{-Xdt}e^Ye^{Xdt}$ must be a group member. For infinitesimal $dt$, this equals $e^{Y+[X,Y]dt}$. Hence, $[X,Y]$ must be a generator. – Ryder Rude Aug 12 '22 at 09:44

1 Answers1

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So the proper formal answer should be that the vector fields on a Lie group have a natural Lie bracket (as do all vector fields on a general manifold). Then the Lie algebra can be identified with the set of left invariant vector fields (which only makes sense on a Lie group) which can then be identified with just the tangent space at the identity. Then the key is that the Lie bracket preserves left invariance.

From what you've written though you seem to be only imagining matrix Lie groups and a matrix exponential. There are some problems with this approach as it implies we have already chosen a representation of both the Lie group and the Lie algebra. Different representations will behave differently so it should be less clear that what we say at this point will not depend on our choice. It turns out that matrix multiplication in the Lie algebra is not preserved under passing to different representations although the commutator is. Either way, the point is that we are not being careful enough. If we allow ourselves this looseness we can perhaps get some of the intuition you seek.

Firstly, note that the group acts on its vector fields (on the left and on the right) which indeed we are using to talk about left invariant vector fields at the start. Indeed $g \in G$ acting on the left sends the tangent space at $h \in G$ to the one at $gh\in G$. But since the Lie algebra is supposed to be the tangent space at the identity we want an action that preserves this. So we should consider conjugation instead: $e \mapsto geg^{-1} =e$. Note here we have found the adjoint action of $G$ on its Lie algebra. To find an action of the Lie group on itself we only need to differentiate. So replace $g$ by the curve $\exp(tX)$ which differentiates to $X$ at $t=0$ (Note here that using the exponential map depends a lot on already understanding the structure of the Lie algebra/group so this is somewhat loose).

So then $\exp(tX)Y\exp(-tX)$ differentiates to $XY - YX$ by the product rule and we have deduced the commutator straight from the adjoint representation.

Callum
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