l4teLearner

@Michael developer here.

folks. what are your thougts about the "Kerr vs Penrose" thing? I am particularly curious if Penrose has ever commented last Kerr article...

for me it still is a remote future :)

a lot of formula references are scrumbled and there are some obvious typos here and there but it seems a good book to me

ciao :) nice to meet you guys. a presto

right! they wrote an english version, by the way

anyways, thanks for having had a look. cheers

I must have done a dumb computation problem somewhere or a mistake in the modeling

it's from Fasano, Marmi Analytical Mechanics

there must be a dumb error somewhere or something I can't see

oh, nothing quantum. it's just plain old boring springs and masses...

hi I guess that people visiting this chat to promote their question on stackexchange are seen as annoying and unpolite, right?

well, here is the question I was talking about. if anyone has ideas, it would be great: math.stackexchange.com/questions/4547149/… thanks!!

@MaryStar note that that the max value I was taking about is the max value of $f$ but I am not sure what you are indicating with | and || in your question. Is that a measure of some kind?

@MaryStar I would be tempted to say yes, it seems to me a segment from (0,1) to (1,0) for k=1 whose right end gradually moves towards (0,0) as k grows.. but maybe I am missing some catch, seems too simple

Hi all! How poorly received is spamming over here to ask consideration for a question? Or, equivalently, are bounties usually taken into consideration over math.SE or are useless? I am poor!! :)

cheers, good luck!

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

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

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

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

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

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.

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

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

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

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

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

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

well this is used at the beginning of chapter 3

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

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

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

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

$s$ 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.

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 $T_e$.

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!

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

you are welcome. everyone needs to figure out things so that they click for them :) ( by the way the answer mentioned that the manifold has to be connected, so maybe it is my fault to have cited it a bit out of context). cheers!

i am not sure who closed your answer. i didnt.

yes you are right, the answer I cited only holds for connected manifolds. this is also what Moishe said yesterday

yeah I think that we agree on this. hope this is a bit more clear now.

postimg.cc/phrV1r7x indeed.

yeah, it will be nicely different from zero. i take it as the manifold equivalent of the zero theorem

yep sure. all this holds if M is connected!

so if $\omega$ determines an orientation, all the $f \omega$ forms with $f>0$ will determine the samd orientation. while the others, $f<0$ will disagree. these are all and only the possibilies.

postimg.cc/9r6rsXXj this is better, the other got cropped