Hi, looking at the text that you have linked, you get (1) because of lemma 4.3. $x exp(X) x^{-1} = exp(Ad(x)X)$. Remember that $exp(X)$ is the value at $t=1$ of $exp(tX)$ the curve that by definition has differential $X$ at $e$ when $t=0$. If we take the adjoint representation of $x$ and we want it to act on $X$, i.e. $Ad(x)X$ we will get another tangent vector in $T_e$. What is this vector? It will have to be the differential of the coniugation on my curve, $xexp(tX)x^{-1}$ but this is equivalent to the differential of a curve which is directly $exp(tAd(x)X)$. The two curves have same value

at $t=0$ so they must coincide I think. Then you just take $t=1$ and you get lemma 4.3. Your method seems a bit harder as the differentiation gets a bit more complex in that case, but I haven't tried to complete it. Hope this helps!

@l4teLearner Thank you now (1) is clear. To go from (1) to (2). I think my method was incorrect, and that in the LHS $\frac{d}{dt}|_{t=0}\exp(tZ)=Z$ just because of the definition of integral curve. Now I still need to justify the RHS

well think of $exp(sZ)$ as an element of the group. and $Ad(exp(sZ))$ is just its adjoint representation, that acts on elements of $T_e$. So in (1) the RHS just specified a curve with desired velocity $Ad(exp(sZ)Z$ makes sense? what is a bit involving here is that the argument of $Ad$ is an element of the group, but $Ad(something)$ as output of this process is again just a function on elements of $T_e$.

@l4teLearner Why does this curve $t\mapsto tAd(exp(sZ))$ has velocity Ad(exp(sZ)). I think I cannot differentiate as if it were a calculus 1 function

$s$ is fixed here, $t$ moves. the curve is not the one you wrote but is the RHS. I am not differentiating, just applied the definition of $exp$ as in 3.3 in your text.

@l4teLearner oh I see. Can you take a look at equation (4) that I just added. I am scratching my head here. I don't know how that $T_0(exp) $ shows up

and I don't know if I can just differentiate the LHS of (2) as if it were a calculus 1 constant function to get 0

I am not sure about that notation. but I would just skip that passage and look at the second equation of definition 4.5

@l4teLearner ok so (2) becomes. $\frac{dZ}{ds}|_{s=0}=\frac{d}{ds}|_{s=0}Ad(exp(sZ))Z$. Now the LHS shuch be 0, but I am not sure why, is not like I can use the defition of integral curve here. And what is also strange is that in the RHS that $Z$ seems to be just there unaffected by the 2 previous derivations

Hi

hi. we are not in calculus 1 but a Lie group is a manifold! so here the differential is well defined, and it is a linear map from tangent spaces to tangent spaces. the LHS stays constant for any s, because is not a function of s, which means that the differential brings any vector to 0. hence it is 0

for the same reason Z is unaffected because it is not a function of s or t.

But when you say differential, I understand "(T_(some point)(smoothfunction))(some vector)". In this case we are doing d/ds(...)_{s=0} , there is no differential here?

one nice way of defining the differential is using curves. it has always been very helpful to me but if you think to the curve in the source manifold, f transforms it in a curve in the target manifold. the differential of f brings velocities to velocities. let me see if in your book this is explained somehow

well this is used at the beginning of chapter 3

I have taken a screenshot from another book, how do I send it to you?

I think I am understanding something. In the proof (lemma 4.9) it says "Differentiating this relation with respect to t at t = 0", and we differentiated exp(tZ) in the LHS which is a curve. So that was the derivative of a curve, so a calculus 1 derivative. The second differentiation read "Differentiating this with respect to s at s = 0 we obtain:" But now we are not going to differentiate a curve but a vector, since the equation to differentiate is now "Z=...", so this is not a calcu

calculus 1 derivative, but a differential.

there is an upload button at the right of the chat

it seems like you can upload images saved in your pc

hm, i am on mobile, let me go on chrome desk

well it is a curve in the tangent space, which is a manifold after all..

so my point is that d/dt and df are not the same, right?, so to differentiate exp(tZ) we used the first one, but to differentiate Z we have to use df

what book is it by the way?

d/dt is a special case of differential, where the manifold from where you start is R. it maps vectors in R to vectors in R (if it is a function from R to R). the difference is that d/dt is a number while df is a linear function, but in the calculus 1 case for single variables is is just a proportionality coefficient from the tangent space of x and the tangent space in y, loosely speaking

the book is from Arvanitoyeorgos, but I do not recommend it, it is full of typos (I just studied the first two chapters)

I highly recommend Loring Tu Introduction to Manifolds. but you should have already taken a manifold course if you are studying Lie groups

well actually i took a differential geometry course without manifolds , like in R^3 and R^2 ,because topology was a parallel course, then I read some 100 pages manifolds lecture notes for a thesis, so no exercises there

so I have little experience in actually using the stuff

I am reading Lee's smooth manifolds but it's taking me forever

I highly recommend that book, it has also a good section on topology. I spent a lot of time on it, because I am self taught (I am not a math student, but an engineer) but it gives you very strong foundations.

Lee's one covers more stuff. I think Loring Tu is simpler to approach. yet it is very clear and full of exercises

give it a try! a little bird me told me that you can find a copy online (I have a printed copy, though)

Loring Tu chapter on lie groups is too basic. just gives you the definitions and has nothing on Ad ad and company. unfortunately no, Arvanitoyeorgos has some exercises but as I told you is not a goos introduction. I have Daniel Bump one but it is uncomprehensible to me! very hard

but I suggest you to make some clarity on d/dt and df. took me some time to get familiar with it

yeah I will

I gotta go. I hope you clarified your doubts on the ad representation. let me send you how Arvanitoyeorgos explains it

cheers, good luck!

thank you!!