Charlie

My god I didn't realise that was a vsauce video and the start made me jump

At some point someone hanging around in a deep enough cave would accidentally connect to their wifi

I do appreciate your answers and help though

I remember the last time I touched it maybe a year ago it made virtually no sense but now I'm a bit older and wiser it makes moderate sense

Because I'm watching my favourite series on geometric physics by Frederic Schuller and I was reading the wiki page about PFB's at work while some stuff was loading and I got to the bit about connections and it reminded me about this lie algebra valued form stuff lol

The thing is I'm not really super interested in how exactly your make it super formal to the point that I would go out of my way to read at length about the topic :P

I will continue to think about it periodically, I'm only looking at principal fibre bundles tangentially

I need a return tram ticket to Heidelberg :P

It's almost there for me, all of the words make sense but I just can't put it all together just yet

Yeah we had the discussion on the tensor product stuff when someone asked about Dirac spinors a while back

@Slereah Well but because the $\mathrm dx^i$ live in a vector space they come with a notion of scalar multiplication, but if you "let the components be $V$-valued" you'd need some definition of $V\cdot \mathrm dx^i$ is all I mean

Like a real number

When I said it's very formal I just mean that $f(x)\mathrm dx$ makes sense but only if $f(x)$ is something that can be "multiplied" by a form

Ok I follow that

Uh actually maybe that does make sense

In terms of just figuring out the concept

Though I don't think it matters

I'm not distinguishing clearly between working with a field and just working at a point

I'm not sure I follow the equivalence after thinking about it for a minute, if we just deal with 1-forms then in what sense are $\omega_i\mathrm dx^i$ and $\omega(\vec v)$ related? if $\vec v$ is just some vector in $T_pM$

@ACuriousMind At this point the expansion in the $dx^i$'s is very formal now right

I was specifically reading this line

Well that's what I was thinking, because I was reading today and it reminded me of your description of the "operator valued vectors" in QFT, namely thinking of the Dirac operator as the tensor product of the endomorphisms over $\mathcal H$ and the spinor space.

Just the "matrix elements" of a linear vector/operator @ACuriousMind

Or the kernal of the pushforward of the projection

I thought the definition of the vertical subspace involved it being the kernal of the projection or something

Awesome

Oh yeah cdot is better I mistyped it

So the fact that I can construct a map $g(p):=p\circ g^{-1}$ and that $f(g(p))=(p\circ g^{-1})\circ g=p$ proves that the map $f:p\mapsto p\circ g$ is surjective

I do see what you mean though

That's cool that the requirements are kind of dual to each other like that

Ahhhhh

Surjectivity of $f$ that is

If we only had $f\circ g=\mathrm{id}_Y$ alone that would only show surjectivity and not injectivity right?

Ohhh

Damn

Am I right so say that if $f:X\rightarrow Y$ and I can construct a map $g:Y\rightarrow X$ such that $g\circ f=\mathrm{id}_X$ then $f$ is necessarily surjective?

I think I might just need to think about it for a minute, I think Jakobians comment above makes it clearer

@leslietownes $x\cdot g^{-1}$?

@Jakobian I think this answers my question

Hmm ok

Or maybe I'm semantically wrong and it "is invertible"

But a map that is just injective has an inverse

Couldn't it only be defined on a subset of the codomain of the initial map?

Is it not possible for me to construct an inverse of a map without the inital map being surjective?

@leslietownes Yes and also how to show it

Yes

I get that $g^{-1}$ is the map

Oh I hadn't considered that that was an equivalent statement

I can start with assuming that no such point exists, but I'm not really sure where to go from there

Could someone give me a hand with this, I don't really feel like I understand how to prove surjectivity properly. The specific case I'm considering is proving that for a topological space $X$ and topological group $G$ that for fixed $g \in G$, the right action on $X$, namely $p\mapsto p\cdot g$ with the standard axioms is a homeomorphism. I don't really understand how you show that every point in $X$ is mapped to by at least other point in $X$?