Mathematics

2:11 PM

under what conditions can we guarantee that the entries of a character table integers

Hi! :-)

@robjohn Yesterday my cell phone went off and I could not thank you for your kind answer to my question. Thank you very much for your very nice answer!

robjohn

@Alex The answer to the question I answered implies the answer to the question you were asking about.

but not vice-versa

@Alex Of course, in the answer to the question you linked, $\frac{x^{2n}}{(1+x^{2n})^{(n-1)/n}+\cdots+1}\lt\frac1n$, for $0\le x\le1$ which gives convergence.

UnderMathUate

2:27 PM

@leslietownes If you're still interested, I ended up with $f(x) = {4}\frac{1-x)$ for $(-\infty, 0] \rightarrow (0, 4]$

Alex

@robjohn the user284331's answer should not be corrected?

@robjohn Thank you :-)

UnderMathUate

why is the fraction not working ;-;

f(x) = $\frac{4}{1-x}$ for f: (-00, 0] -> (0, 4]

robjohn

@Alex The part after what I just mentioned doesn't seem to make much sense, either.

@UnderMathUate it works for me. what are you trying to make it do?

UnderMathUate

Show that |(-00, 0]| = |(0, 4]|

Semiclassical

@mathsresearcher this seems relevant : mathoverflow.net/questions/10635/…

robjohn

2:31 PM

@UnderMathUate what is the absolute value of an interval? its measure? its cardinality?

UnderMathUate

Oh, sorry. I'm referring to cardinality. I want to prove that the intervals are numerically equivalent.

robjohn

and is $-00$ supposed to be $-\infty$?

UnderMathUate

yeah

I just got frustrated with chatJax lol

robjohn

@UnderMathUate In what way is the fraction not working?

Alex

@UnderMathUate Some time ago, I was also partially defeated :-(

Semiclassical

2:34 PM

In particular the second-rated answer references a theorem of Serre which says that “characters are rational iff characters are integers iff conjugacy classes are rational”

UnderMathUate

@robjohn Oh, I was referring to the message I sent above that, where it wouldn't display what I was trying to type. As for the function itself, I believe this works for this mapping.

@Alex Wdym?

Semiclassical

What’s not clear is whether there’s examples of such besides the symmetric groups

robjohn

@UnderMathUate Oh, I thought I fixed that message

Your last edit was $A = \{\frac{m}{2^n)\}$ $: m, n \in \mathbb N}$

user193319

I read somewhere that one can always define a nontrivial homomorphism from an HNN extension to $\Bbb{Z}$, thus proving an HNN always has a proper normal subgroup. How does one define this homomorphism? Let $G = \langle S \mid R \rangle$ and let $\alpha : H \to K$ be an isomorphism of subgroups of $G$. Then $G \ast_{\alpha}$ denotes the HNN extension of $G$ relative to $\alpha$. I was thinking the homomorphism could be defined by $s \mapsto 0$ for $s \in S$ and $t \mapsto 1$...

robjohn

@UnderMathUate I think you wanted $A = \{\frac{m}{2^n}: m, n \in \mathbb N\}$ . I also added resized braces.

user193319

2:47 PM

...where $t$ is the stable letter of the HNN extension. If $t$ has infinite order, this would work...Does the stable letter always have infinite order?

UnderMathUate

@robjohn Oh yeah, thanks

I always forget the syntax for that

Alessandro Codenotti

@user193319 doesn't this follow from Britton's lemma?

Alex

@robjohn Ok, I understand.

robjohn

@Alex: I have added a comment

user193319

@AlessandroCodenotti Does what follow from it? That $t$ has infinite order or that we can define a map to $\Bbb{Z}$?

Alessandro Codenotti

2:57 PM

Infinite order

Wolgwang

@robjohn What is a triangle in the wedge? This ?

robjohn

@Wolgwang Do you see the circular wedge and the triangle inside it?

Wolgwang

@robjohn Shape below AB is wedge?

user193319

3:12 PM

@AlessandroCodenotti Hmm...I don't quite see it. I'm reading the wiki page on HNN extensions and apparently every element of $G \ast_{\alpha}$ of finite order is conjugate to an element of $G$. Hence, if $t$ were of finite order, then $t = x^{-1} gx$ for some $x \in G \ast_{\alpha}$ and $g \in G$. My guess is that this implies $t \in G$...? That would certainly be a contradiction, but I don't see it.

Alessandro Codenotti

@user193319 Look at the "alternate form" on wiki. Does the string $ttt...ttt$ satisfy its hypothesis?

user193319

@AlessandroCodenotti Yes, it looks like the second condition involving $n > 0$ is satisfied, so $t^n \neq 1$ for every $n \in \Bbb{N}$.

Alessandro Codenotti

Which is the same as saying that $t$ has infinite order

user193319

Indeed!

maths researcher

3:41 PM

@Semiclassical sure, but that's for only symmetric groups. Whereas most group im encountering have character table with integer entries, so I assume there is a more general reason

jay

3:52 PM

Can anyone help me understand this ^, considering I know near to 0 differential geometry. e.g what is a local norm

here $d$ is the wasserstein distance,

leslie townes

5:21 PM

jay, this looks similar to the recipe by which a riemannian metric (= way of measuring lengths of tangent vectors to curves in a space) relates to a metric (= way of measuring distances between points in space). here, they start with the latter kind of metric and show it can be recovered by minimizing 'lengths' of curves as measured by (2.2) and (2.3). give or take some squaring stuff this is the usual recipe from riemannian geometry

the 'local norm' is a way of measuring the lengths of tangent vectors (if you like, 'velocities' of curves) in your space. the use of the word 'formally' and reference to a 'rigorous definition' suggests that there may be some mathematical details under the hood here that are not express in the formula

for what it's worth, it looks like PDF copies of both [4] and a draft of [6] are available if you google

although they look very technical and maybe won't help. i couldn't get much out of them

i think by [4, 10] they mean [4], [10]. [4] has no section named 10

amWhy

@leslietownes That recipe is just scrumptious, when adhered to correctly!

leslie townes

i inherited my family recipe of relating riemannian metrics to metric space metrics from my grandmother. her secret is to use cream or half-and-half instead of milk.

amWhy

@leslietownes ;-) Just came from reading comments above, about overcooked meat, and spinach, etc.! I mean just look @robjohn. Seems to look overcooked, unless he's a square orange!

leslie townes

spinach is really hard to cook. i like it blended in palak paneer, or added very late in a vegetable soup with other elements. ideally after the heat has been turned off and it's just cooking from the heat of the soup.

i think spinach somehow violates conservation of mass. i've definitely put like 5 bags of spinach into one pot of soup before.

nobody can explain how this works to me.

amWhy

I get why he a mean squared (orange), 'cuz more spherical oranges probably tormented as a child orange mean square.

Semiclassical

5:36 PM

@mathsresearcher found a more direct Q&A, though with the same citation to Serre: math.stackexchange.com/q/2792741/137524

jay

when $J$ is a functional and $\rho$ is probability density what do authors usually mean when they write $\frac{\delta J}{\delta \rho}$?

amWhy

@leslietownes transforms into slime in green water, some of which evaporates.

Semiclassical

Functional derivative

In the calculus of variations, a field of mathematical analysis, the functional derivative (or variational derivative) relates a change in a Functional (a functional in this sense is a function that acts on functions) to a change in a function on which the functional depends. In the calculus of variations, functionals are usually expressed in terms of an integral of functions, their arguments, and their derivatives. In an integral L of a functional, if a function f is varied by adding to it another function δf that is arbitrarily small, and the resulting integrand is expanded in powers of δf...

jay

^ Thanks

Semiclassical

in particular, see the "Coulomb potential energy functional" application: taking $J$ to be the Coulomb electron-electron potential energy, they compute $\delta J/\delta \rho$ explicitly

Wolgwang

6:03 PM

@LeakyNun $\int_a^b \sqrt{1+(f'(x))^2 } \mathrm dx$ ?