Hey guys!

I have a mathematics question

Could anyone help me?

Its here if u could help me

Thanks

@mathvc_ Your question is currently not comprehensible.

@Danu why? I suppose it is

For example, consider the union of two solid tori, and identify their boundary (intuitively their circles) via the identity map to obtain a new space

But a torus has one hole. My question is about doing the same thing with a solid space with 2 holes

1. There is no "boundary" on a torus. 2. You never made it clear that you are talking about surfaces, or rather the solids that fill them out, in the first place.

I reworded the question a little bit, hope you don't mind @mathvc_

Yes! Thats great

@Danu Dont you have any idea about my solution? I guess I have made a mistake, I am not sure whether I can use the relative Mayer-Vietoris sequence in this case

@DavidZ halp physics.stackexchange.com/posts/338790/revisions

or any other mod

A lie algebra is the tangent space to the identity of a lie group, but the commutator $[A,B]$ involves second derivatives of the group commutator, $[A,B] = \frac{\partial^2}{\partial s \partial t} g_1(t) g_2(s) g_1^{-1}(t) g_2^{-1}(s)$, hmm...

@dmckee are you around?

or even @ACuriousMind

@bolbteppa This is not so hard to see

$\operatorname{Ad}_g(X)=D_e c_g(X)$ where $c_g$ is conjugation by $g\in G$, and this gives a map $\operatorname{Ad}:G\to \operatorname{Aut}(\mathfrak g)$. The adjoint action $\operatorname{ad}:\mathfrak g\to \operatorname{End}\mathfrak g$ is its derivative, hence the two derivatives in total.

physics.stackexchange.com/review/suggested-edits/178177

:-)

lol

who upvoted it though

physics.stackexchange.com/posts/338790/revisions

:D

You can get the commutator in two ways, one way by the adjoint, the other way using the group commutator as above, if you do it this way, two derivatives, i.e. defining two derivatives on a manifold, which involves a bunch of extra structure no? math.stackexchange.com/q/961922/82615

@AccidentalFourierTransform What the hell is going on xD

this is fun

oh no

@bolbteppa Nothing beyond everything that is immediately there when you have a smooth manifold. I just used the fact that a map induces a map on tangent bundles, its "derivative" as some may call it (though not everybody does).

@ACuriousMind thanks

^

nah they are all good people

specially when they are young

no matter how kind you are, German children are kinder

Just intrinsically, without adjoints, taking a second derivative of a group element, it still lives in the tangent space, is that not odd?

@bolbteppa What do you mean "just intrinsically"? All the structure I use is intrinsic to Lie groups. I'm literally only using the conjugation map.

The point is perhaps that $\mathfrak g$ is in itself already an object that always "comes with derivatives", so it's not so strange to have a second derivative pop up; maybe that's a more intuitive explanation for you.

I mean, I just took a second derivative of an element of a lie group, not it's first derivative but a second derivative, yet the second derivative is a tangent vector

@bolbteppa You didn't take a derivative "of an element". You took the derivative of one map, then took another derivative of a related map.

But in principle, your wonder about why the Lie bracket on vector fields manages to produce another vector field rather than a "second order object" is not so strange---it's just a surprising feature. I don't think I can do more to make it appear natural than I have so far :)

But $g_1(t)g_2(s)g_1^{-1}(t)g_2^{-1}(s)$ is just an element of the group? I just took two derivatives of it?

(but this has nothing to do with Lie groups)

@bolbteppa It's rather a map $I^2\to G$ that you are taking derivatives of, but maybe you don't like to make that distinction.

That might explain it

In any case, I think that the point of view that I outlined makes it appear quite natural that you have 2 derivatives, yet produce an element of $\mathfrak g$.

Take the derivative of conjugation, and then take the derivative of the map that that induces (which is not the same map, but it is defined in terms of the old derivative!!)

I gotta go now, bye!

I will think about that, that might explain it in general, but if you play dumb and treat the group commutator as just an element of the Lie group, it's still confusing

Ok cool thanks

@AccidentalFourierTransform sorry I was AFK, thanks for the ping though

I see that ACM took care of things

It's plausible that defining the lie algebra commutator involving two derivatives this way may not be an element of the lie algebra, but an element of the enveloping algebra giving the same element that you get from the adjoint, I just find the two derivatives on a group element odd

@Mostafa Not always but also I experienced they somehow don't find funny what I do

For weak gravitational fields where $g_{\mu \nu} = \eta_{\mu \nu} + h_{\mu \nu}$, is it okay to call $\eta_{\mu \nu}$ the baseline metric tensor?

As opposed to Minkowski metric

If you say that you're saying it could be any baseline metric tensor but you've chosen a specific one

What if I say the baseline metric tensor due to special relativity?

I think it's clear what you mean

So it makes sense I hope

Baseline Minkowski metric tensor?

Sure, but I basically want a synonym for that