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Lets consider the adjoint action

$$\mathrm{Ad}: G\rightarrow G$$ $$\mathrm{Ad}_g(g')=gg'g^{-1}$$

This can equally be written as

$$(l_g\circ r_{g^{-1}})g'=\mathrm{Ad}_g(g')$$

Lets consider a Lie group, for example the $SL(n)$ Lie group of $n\times n$ matrices $M$ with $\det M=1$. It is known that the generalized Pauli elements under matrix multiplication form a subgroup of $SL(n)$. It is also known that the generalized Pauli elements without the identity constitute a basis set for the $\mathfrak{sl}_n$ general linear Lie algebra.

The Lie algebra of a Lie group $L(G)$ is isomorphic to $T_eG$ via

$$i(A)=l_{g*} A:=L^A\in L(G)$$

being an isomorphism for $A\in T_eG$ and $L(G)$ denoting the space of left-invariant vector fields.

Crucially: $L^A_e=A$, so the vector field evaluated on $e\in G$ equals $A$.

There are elements that can be defined with both, being in the Lie group and its associated Lie algebra, and those are precisely the elements of the basis set consisting of group elements spanning the Lie algebra. It is known that

$$(h_* v)(f) = v (f\circ h)$$

Lets take a basis vector for the Lie algebra $P$ which is a group element, so $P\in SL(n)$ and $P\in \mathfrak{sl}_n$.

$$\rightarrow (l_g\circ r_{g^{-1}})_* P(f)=(l_g\circ r_{g^{-1}})_* L^P_e(f)=L^P(f\circ l_g\circ r_{g^{-1}})(e)=L^P_e (f)$$

But this does not make sense since this would indicate that the adjoint representation is always like the identity for basis vectors of the Lie algebra.

So where did I go wrong?

relativeentropy
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1 Answers1

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Firstly, I wouldn't worry about elements that are simultaneously in the Lie algebra and the Lie group. I don't think this is very helpful in general and certainly there are Lie groups where this coincidence won't occur.

Next I wouldn't call the thing you defined at the start the adjoint action. Rather, it is the conjugation action. Its derivative $\mathrm{Ad}:G\to GL(\mathfrak{g})$ is the adjoint action. In matrix groups, they both work out to be conjugation but I think you should keep them distinct nonetheless.

Then I think your issue is assuming that the vector field at the end is right invariant which it is not or mixing and matching when you are thinking of something as a vector or as a vector field. If we did have a bi-invariant vector field then the adjoint action would indeed be trivial. The adjoint action written in terms of left-invariant vector fields naturally simplifies to $r_{g^{-1}} (L^P)$ but this doesn't have the property $(r_{g^{-1}} (L^P))_e = L^P_e$

Callum
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  • Thank you for your answer. But the thing is, that I did not assume right-invariance. $L^P (f \circ l_g \circ r_{g^{-1}}) e = L^P(f(geg^{-1}))=L^P(f(e))=L_e^P(f)$ is true regardless of invariance. This is what confuses me. – relativeentropy Apr 22 '24 at 14:43
  • @relativeentropy I'm not quite following your notation but what you have written there ultimately doesn't make sense. Consider a simpler example: rotating around a point changes the directional derivatives around that point even though it doesn't move the point, whereas by your argument any diffeomorphism that keeps a point fixed doesn't change the directional derivatives there. – Callum Apr 25 '24 at 10:16
  • my notation is that we have push forwards notated with a star, $l_g$ and $r_g$ denoting left an right translation. $L^P$ denotes a left invariant vector field isomorphic to being $i(A)=L^A$ for $i: \mathrm{T}e G\rightarrow L(G)$ where $L(G)$ denotes the space of left invariant vector fields and $\mathrm{T}_e G$ being the tangent space at the identity of $G$. And $i(A)=l{g*}A$. So it makes sense to me, but I am sceptical. And I would ask for help in interpretation. – relativeentropy Apr 25 '24 at 10:44
  • @relativeentropy that isn't the bit of the notation I mean. $e$ shouldn't be moving around the expression in the way that it is. For example $L^P (f(e))$ doesn't make sense let alone being equal to $L^P_e(f)$. $L^P$ is a vector field so it acts on functions but $f(e)$ is a real number. – Callum Apr 25 '24 at 13:18
  • You are of course correct thank you for clarification and raising awareness of a bad abuse of notation here. So, if I understood you correctly, what I really have is $(L^P_e\circ f \circ r_{g^{-1}})$ as expression for an adjoint action if $P$ is an element of the Lie algebra. So $\mathrm{Ad}_{g*}(P(f))(e) \cong (L^P_e (f))(g^{-1})$ – relativeentropy Apr 25 '24 at 14:56
  • @relativeentropy You mean $(\operatorname{Ad}_g(L^P (f)))(e)$ I think. Note the $*$ shouldn't be there as $\operatorname{Ad}$ refers to the representation of the group on its Lie algebra not conjugation in the group and there's not much point saying $P(f)$ instead of $L^P (f)$ as it's only acting on functions when we think of it as a vector field anyway. – Callum Apr 26 '24 at 07:34
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    In my book that I have he uses this notation. It is however unusual as I have seen this nowhere else. Since I haven't had any other touch with Lie theory than this book, I thought this was standard notation. Thank you for raising awareness. – relativeentropy Apr 26 '24 at 12:38