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21:01
Hey, if $\sigma$ is the permutation $(1~2~3)\in S_3$, can I write $\sigma(a,b,c)=(b,c,a)$? If not, what's the proper notation for applying a permutation to an ordered tuple?
And let's say $A\subseteq S_n$ is a permutation group, and $X,Y$ are $n$-tuples. Is there an accepted notation to signify that $\exists \sigma\in A:\sigma X=Y$? I would write $X\sim_A Y$ if I had to make something up, but I wonder if there's something standard.
$Y\in AX$? @AkivaWeinberger
forgot for a moment that you meant $X,Y$ as elements rather than subgroups
I guess that would work. $AX=AY$ could be better, as it looks more equivalence-relation-y.
Assuming that the answer to my first question is "yes"
I think that that function notation is fine
I was going to object that I wasn't sure if $AX=AY$ was the same thing, but $e\in A$
21:19
@PedroTamaroff Hello
Who is this?
Just a member
I am a Masters student
Sorry to bother. Do you have any idea about Hochschild cohomology? @PedroTamaroff
A little.
@MikeMiller I made this mistake in teaching linear algebra before. (not including the 3rd axiom because I didn't think about the case where the subset is empty).
In what kind of situation do we need that cohomology? @PedroTamaroff
21:25
That's kind of a broad question.
I am studying cohomology of Banach algebras. They came up.
Could you try and be more specific?
the starred comment on the right seems applicable here
Oh, Hochschild cohomology of Banach algebras?
Yes
21:26
Aren't you considering continuous maps?
Not just continuous maps.
I mean, Hochschild cohomology for Banach algebras perhaps takes into account the topological nature of a Banach algebra.
They are called derivations.
Right, those represent elements of $H^1$.
Can you explain on that point "topological nature"?
21:29
@Mambo: Your maps are demanded to be continuous.
They may be better than continuous, but they are in particular continuous.
If you have a Banach algebra, I would guess your cocycles are continuous.
That is, you're using the topological algebraic Hom, not the algebraic Hom alone.
Sorry I don't know what a cocycle is.
Elements in the kernel of the Hochschild (co)boundary.
@MikeMiller They are bounded linear maps.
@PedroTamaroff OK
Right, bounded linear since you're taking into account the Banach structure.
21:33
Which is the same as being a continuous linear map.
@PedroTamaroff They are even certain algebras which does not demand the maps to be continuous. But automatic continuity is available there.
So what is your question?
My question is Why does one study Hochschild cohomology?
21:37
That's a broad question. I am not sure I can answer it.
Hochschild cohomology, as many other cohomology theories, classifies things.
For the first time, I am reading some cohomology theory.
For example, $H^1$ classifies derivations, and this has geometric relevance when algebras are nice enough.
$H^2$ classifies infinitesimal extensions, which also has geometric content in good cases.
It is also relevant in purely algebraic situations, like Wedderburn's principal theorem.
Oh good!
Have you come across amenability of Banach algebras?
Hochschild homology happens to be really geometric when dealing with smooth algebras, and in this case there is a famous theorem by Konstant Rosenberg and Hochschild that says that Hochschild homology coincides with the homology of differential forms on $A$.
And you have Künneth-like theorems for HH, so you can interpret cohomology too.
@Mambo No, not really.
I haven't studied homology theory too.
@PedroTamaroff There are some special kind of groups called amenable groups. Inspired from their group algebras, amenability of Banach algebras was developed which extensively uses Hochschild cohomology.
21:47
amenable=?
Well Thank you very much for the relevances.
@PedroTamaroff
@Semiclassical has a "mean" operator.
riiiiight
@Semiclassical Do you know what a Haar measure is?
more or less. it's the integration measure you use when doing averages over Lie groups, I think?
(perhaps more general than Lie groups, but that's the one i'd know)
21:52
It's simple. Every locally compact topological group admits an rotation invariant measure. That is Haar measure.
right, right
Like Lebesgue measure for reals
@Semiclassical: Consider the algebra $L^\infty(G)$ of functions that are bounded except at a measure 0 set mod functions of measue zero, on $G$. A mean is a continuous nonnegative functional on this of norm 1. The norm should be inconsequential, since you can just scale.
A group is amenable if it has a left-invariant mean.
oh. i thought you were making a joke about a-'mean'-able
21:54
well, there we are
the Haar measure on a compact group defines a mean
actually, that makes it sound like that functional serves as a probability measure
Yes it does
being positive and normalizable
It is a finitely additive measure
21:56
mmkay. then one can almost surely translate that into quantum mechanics language, somehow :)
In fact Haar measure on a compact group defines an invariant mean. So every compact group is amenable.
That joke really exists in the literature @Semiclassical
The pun is literally the etymology.
Exactly !
...i'd say that's terrible, except I'm pretty sure physicists have no ground to bitch about terrible naming puns
i mean, particle physicists have things called 'squarks' and 'sneutrinos' from super-symmetric theories
so yeah :/
22:03
hi
if i have a triangle and a line that passes through the triangle and i want to reflect it to i reflect on both sides of the line?
*do
@mambo as a place where haar measures show up in physics, see this wiki section
now I find myself wondering what Hochschild (co)homology would describe in the physics context
Keep wondering @Semiclassical
though probably things are nice enough for lie groups that the theorem @pedro cited guarantees that one might as well think about differential forms
Bye
22:08
brutal
22:25
i created a math question can anyone tell me if it makes sense or not
A triangle $T$ is given in the plane with sides $3,4,5$. A second triangle $R(T)$ is created by reflecting $T$ about a line. Find $\min\{[R(T) \cup T]\}$ where $[\cdot]$ denotes area.
does it make sense?
23:00
23:12
what about my question
23:27
Hello guys...i have a very quick question. When x<<d can i eliminate the x from (d/2)^2+x^2 ?
@GeorgeSmyridis Um, if $x\ll d$ just means that $x$ is much smaller than $d$… we could still have $x\ne0$, no? So we'd just have $(\frac d2)^2+x^2\approx(\frac d2)^2$.
(I think I'm misunderstanding the question)
@MikeMiller I thought you just study homology to distinguish things. xd
Let $H = \{\phi\in A(S) : x_0\phi = x_0 ~\text{or}~ x_1 \phi = x_1\}$. Is $H$ a subgroup of $A(S)$? ($A(S)$ is the set of all one-to-one mappings from $S$ onto itself).
What group is $b_+$?
@user19405892 Yes, it makes sense, though I'm not sure if that's the best notation
I think the notation for reflection over a line $l$ is $r_l(T)$, though I'm not sure.
So perhaps state that you're looking for $\min\limits_l[r_l(T)\cup T]$? I dunno
23:51
You're basically putting two 3-4-5 triangles of opposite orientation on top of each other and trying to minimize the total area, yeah?
i mean that when x is much smaller than d then is this equality (approximation) true?
(d/2)^2+x^2=(d/2)^2
seems wrong to me
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