the curves are functions

bleh

@ÍgjøgnumMeg here's group 1

these are nameless coloured lines to me

what's the composition law ?

in reality there should be infinite lines

@Arrow maybe this stuff is not as unnatural as it seems: any functor from R-mod to S-mod induced a functor from R,T-bimodules to S,T-bimodules

okay so you have an infinite set of nameless coloured lines

@mercio multiplication is the operation

how does one compose them?

Yes, but viewing invariants and coinvariants as $R$-modules seems "right" because they're limits of the representation viewed as a functor $\mathbf BG\to \!\!\;_R\mathsf{Mod}$

that is why I'm so reluctant to move away

@ÍgjøgnumMeg you mean you want to know the explicity functions?

@geocalc33 do you have any concrete description of the functions?

@geocalc33 right, it might be useful to know what functions you're composing

@Arrow then think of my argument as an adjunction between $\varepsilon^* \circ (-)_G$ and $\varepsilon^* \circ (-)^G$

or

multiplying I should say

@ÍgjøgnumMeg the functions are all $ (e)^{s/\text{log}(x)} $ for different values of $s$

well at least all the functions for group 1

@Arrow I think from a module perspective, viewing them as $R[G]$-modules is also natural: one is the largest submodule with a trivial action and the other is the largest quotient with a trivial action. In representation theory, it is important to work with the trivial representation as a representation and not just a vector space

@geocalc33 Right, and what happens if you multiply two such functions?

it's closex

closed

@geocalc33 say you have $f_1 = e^{\frac{s_1}{\log x}}$ and $f_2 = e^{\frac{s_2}{\log x}}$

@MatheinBoulomenos actually it seems that plugging $Y=X^G$ into here might give what's needed: $\!\!\;_{R}\mathsf{Mod}(X_{G},X^G) \cong \!\!\;_{R[G]}\mathsf{Mod}(X, \varepsilon ^\ast R\otimes _R X^G)) .$

@ÍgjøgnumMeg yep

@geocalc33 tell me what $f_1 \cdot f_2$ is

I think the arrows you $X\to X^G$ you covered in your answer are secretly arrows $X\to \varepsilon ^\ast R\otimes _R X^G$, no?

$ = e^{\frac{s_2+s_1}{\log x}} $

@Arrow yeah

Man, this stuff is confusing.

@geocalc33 okay, what is the inverse of $f_1$?

itself

What makes you say that?

I calculated it

What is the identity in the group (let's call it $G_1$ to use your notation)

the identity element is 1

@geocalc33 okay nice! So how do we get $1$ by composing functions of the form $e^{\frac{s_1}{\log x}}$?

(I'm assuming $x$ is fixed?)

x is a variable

@MatheinBoulomenos I'm going to sleep. Thank you very much for your patience!

but @ÍgjøgnumMeg you do $ e^{\frac{s_1}{\log x}} \times e^{\frac{-s_1}{\log x}} $

@geocalc33 right! So.. $1$ corresponds to $s = 0$, and the inverse of an element corresponds to $-s$ right?

yeah

So it's almost like

the whole

$e^{\frac{\cdot}{\log x}}$ bit doesn't matter that much

lol someone stop me if I'm talking out of my ass

@Arrow good night! No problem, I like this stuff

@geocalc33 what I mean is, it looks like all you're doing is adding real numbers

or

wherever $s$ is

well maybe that's the wrong identity element

No

for what you're doing

(multiplying functions)

you have $1 \cdot e^{\frac{s}{\log x}} = e^{\frac{s}{\log x}} \cdot 1 = e^{\frac{s}{\log x}}$

because this is the same as multiplying by $e^{\frac{0}{\log x}}$

yeah

so all I'm doing is adding real numbers?

Right, if you wrote down everything you were doing you might notice that the only thing that's changing is the $s$ in the numerator of the exponent

you might save ink by omitting the $e^{\frac{\cdot}{\log x}}$ and instead using the operation $+$

but then your identity has to change with this and your inverses are now just $-s$ etc. etc.

I'm disappointed

aw

But it's good! This is an interesting thing

why

because

you have a correspondence between elements of the form $e^{\frac{s}{\log x}}$ and real numbers $s$

in such a way that

if you take two real numbers $s_1$ and $s_2$ and add them in $\Bbb R$ you get $s_1 + s_2$

and then using the map $e^{\frac{\cdot }{\log x}}$ you get $e^{\frac{s_1 + s_2}{\log x}}$

but you could've started with $s_1$ and $s_2$ separately

used the map $e^{\frac{\cdot}{\log x}}$ and got $e^{\frac{s_1}{\log x}}$ and $e^{\frac{s_2}{\log x}}$ and then composed these and get the same answer $e^{\frac{s_1 + s_2}{\log x}}$

@geocalc33 proofwiki.org/wiki/…

look at this

lol

I looked at it

so it is a group?

Yus, you have a group $G_1 = \lbrace e^{\frac{s}{\log x}} : s \in \Bbb R \rbrace$ where the composition law is given by multiplication, the identity is $1$, the inverses are the ones you mentioned earlier

and associativity holds because associativity has never not held

but the point I'm making is that this is basically the same as $(\Bbb R, +)$ with different notation

(at least, depending on the value of $x$)

Hmm it seems to be of no use

I should go to bed, hopefully my explanation wasn't too bad. I'd recommend reading up on group theory if you're interested; I was talking about group isomorphisms here.

A real mathematician might be able to help you some more!

Night all

Night

@MikeMiller

@KasmirKhaan

@geocalc33 Gnight , are you testing chat pager system ?

Yeah

okay that is all fine I guess

"If a covariance function is isotropic then it is invariant to all rigid motions", what does rigid motions mean here?