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05:00
ah okay
so H^2 = 1.
alright
The cross-product moments are the important ones: ++ and -- both give +1, and +- and -+ give -1
so <HB> = 0.125(+1)+0.375(-1)+0.375(-1)+0.125(+1) = -0.5
and similarly for <HF> and <BF>
that's not obvious from the description of the experiment
cool
Here's the important thing: What's the expectation value of (H+F+B)^2?
well, you expand that out and evaluate it by linearity of expectation
05:04
okay
so that's <H^2>+<F^2>+<B^2> +2<HB>+2<BF>+2<FH> = 1+1+1+2(-1/2)+2(-1/2)+2(-1/2) = 3-3=0...
or, <(H+F+B)^2>=0
And that should surprise you, because that's claiming that a nonnegative quantity has an expected value of zero
is this from QM
inspired by it, yeah
the only way for this to be consistent would be that H+F+B = 0
But they're all supposed to be +1 or -1
05:07
well these aren't numbers per se, they're events
but [X] is a number thouigh
Right, and so is [X^2]
err <X> whatever
and the only way for E[X^2]=0 to be true is if P(X=0)=1
in particular, you'd need P(H+F+B=0) = 1
which can't happen since the outcomes are just +1 and -1.
what have you done to probability theory Semi
you broke it
What that tells you is that your partner can't be reporting their outcomes fairly, because they're too strongly correlated
05:10
O_O
Too many entries, will write the above in paper to better dissect it
immediately unfriends everyone on facebook to avoid probability paradoxes
that's weird
that set of correlations cannot be created by the two of you watching coin flips and reporting back accurately
lol
Now, this set of correlations can be generated in other contexts
falling asleep, ill have to hear this some other time. good night
For instance, take three coin flip variables X1,X2,X3 (outcomes {+1,-1} with equal probability) and take them to be independent. then they've all got zero mean, unit variance, and zero covariance e.g. <X1.X2> = 0
and then construct the variables Y1 = X1 - (X1+X2+X3)/3, etc.
you'll find that the new variables have zero mean, variance 1/3, and covariance -2/3
main thing now is that Y1+Y2+Y3 = 0
so you can have <(Y1+Y2+Y3)^2> = 0 just fine
@GFauxPas night
05:27
Games = {H,B,F}, Outcomes = {+,-}
Let X,Y in Games, a,b in Outcomes
Suppose: X=Y, a=b: P(ab|XY)=1
Suppose X=/=Y, a=/=b
P(+|X)=P(-|X)=1/2
P(+|X) = P(++|XY) + P(+-|XY) = 1/2
P(ab|XY)=P(ba|YX)
HB, BH, HF, FH, BF, FB: ++ (12.5%) +- (37.5%) -+ (37.5%) ++ (12.5%)
<X>=0
<X^2>=1
<XY>=0.125(+1)+0.375(-1)+0.375(-1)+0.125(+1)=-0.5
<(H+F+B)^2>=0
<X^2>=0 => P(X=0)=1
=> P(H+F+B=0) = 1
So... I am guessing, the source of the correlation is the reporting here?
> what's not obvious are again the cross-cases. suppose they all work out like this:
HB, BH, HF, FH, BF, FB: ++ (12.5%) +- (37.5%) -+ (37.5%) ++ (12.5%)
hmm..., so that means:
$$\frac{1}{2\sqrt{2}}\lvert ++\rangle + \sqrt{\frac{3}{8}}\lvert +-\rangle + \sqrt{\frac{3}{8}}\lvert -+\rangle + \frac{1}{2\sqrt{2}}\lvert --\rangle$$
But, what is this?
Y1 = X1 - (X1+X2+X3)/3
somehow the new variables Y1,Y2,Y3 ae all shifed away from the mean of all the X variables and then the correlation can be produced just fine?
is that classical?
well, the outcomes in that case are different
you'll have (Y1,Y2,Y3)=(0,0,0), (0,0,0), (4/3,-2/3,-2/3), (-4/3,2/3,2/3), (-2/3,4/3,-2/3), (2/3,-4/3,2/3), (2/3,2/3,-4/3), (-2/3,-2/3,4/3)
(0,0,0) shows up twice because (X1,X2,X3) = (+1,+1,+1) and (-1,-1,-1) both give that result.
05:45
ah the Y variables can be greater than one
right
and can be zero as well
the point is that you can get a wider range of correlations when you don't restrict the setting as much
I recall one explanation on Bell inequality as some pressing button to light up lamp game explains how because of the richer structure of the operators that appeared in the expectation values (compared to just scalars in the classical case), it provide the wiggle room needed to achieve lager correlations thus give rise to entanglement
Actually, I wonder if your above sports example can be used to describe Bell inequality
Yeah, that's where this is coming from
In that case that is the easiest explanation of Bell's Inequality I have ever heard so far, without so many inequality symbols causing me to go BSOD
More precisely, this is a setting which Bell's inequality would forbid
05:54
yup, and thus illustrates what entanglement is without explicitly introducing quantum jargon, thus such example can reel the audience in quickly
yep
To get Bell's inequality, note that the outcomes of the games are intended to be some random vector $(X_1,X_2,X_3)=(\pm 1,\pm 1,\pm 1)$
in order for the outcomes to not show any bias, you need +++ and --- to show up equally often
and similarly for the pairs {+--, -++}, {-+-,+-+} , {--+,++-}
if you now consider the products, you see that the three pairwise products are +1 for +++ and ---
whereas for the rest you have one +1 and two -1
consequently when you add the products together you get +3 in the first two cases and -1 in the rest. as such, the expected value of X1.X2+X2.X3+X1.X3 has to be between 3 and -1
whereas in the previous case the expected value of the products worked out to be -1/2 for each of them
Is there a name for adding the covariance that way X1.X2+X2.X3+X1.X3, I don't know how is this particular expression is motivated?
well, note that p12 = <X1.X2> in this case defines the correlation coefficient of X1 and X2. (the mean values of X1,X2 are zero and their stddev's are one, so the usual definition simplifies a bit)
so the point is to look at c12+c23+c13
however, the motivation is a bit harder to explain. I guess you can partly relate it to what I was doing earlier re: (X1+X2+X3)^2
i.e. the expected values of X1^2, X2^3, X3^2 are all 1, so the expected value of their squared sum is determined by the sum of those product-moments
<H^2>+<F^2>+<B^2> +2<HB>+2<BF>+2<FH> hmmm, it feels like the expression X1.X2+X2.X3+X1.X3 is summing up all the correlations that are present in the system
3
Q: Meaning of covariance matrix row sums

user2303Say I have an $n \times n$ covariance matrix for a sample set of $n$ random variables. Is there any meaning if the sum of the rows of this matrix? Is it a meaningful measurement of the contribution of each random variable to the variance of sum of the random variables?

sounds like it, so the entrywise sum of a covariance matrix sort of keep track of the total correlations present
06:09
anyways
the main point is that, if you only work with sets of games whose outcomes are +1/-1, then the pairwise correlations are constrained more than they'd be otherwise
Right
otoh, you can instead take into account the fact that the partner may know what game you choose to watch
yup, that corresponds to extra correlations that is not taken account of (corresponds to nonlocal hidden variables in the QM case)
well, technically it's a bit stronger than that
suppose, for instance, that they doctored the correlations so that you had <HB>=<BF>=<FH> = -1
you could have any of those individually, but together they'd imply <(H+B+F)^2> = 1+1+1-2-2-2=-3
which is impossible for any set of random variables
Sounds like that is going into PR boxes territory
3
Q: Difference between PR correlations and entanglement correlations?

Secret(As pointed out by helpful users, one of the bolded claims is interpretation dependent. Therefore, unless otherwise specified, Copenhagen interpretation is used throughout) Recall that for classical correlations, the observables are already determined even before a measurement, and thus knowing ...

06:14
That's exactly what it is.
Note that, in the case of the Y1,Y2,Y3 example, that knowing the outcomes of Y1 and Y2 is enough to deduce Y3
yup because Y1+Y2+Y3=0
right
so the outcomes are not independent and therefore they can't be explained by three local (i.e. independent) hidden variables
once I know what Y1 and Y2 were, I'd instantaneously be able to deduce Y3
if you're finding this interesting, you might try tracking down Jeffrey Bub's book Bananaworld
it uses a different framing device than what I'm giving, but the approach is similar apparently
(I know this stuff more second-hand)
Hello!!

We have the ring $R=\mathbb{Z}_5[x]$ and the element $b=(x-4)(x-3)^2$ of $R$. Let $A$ the set of $a$ in $R$ such that $at=b$ has a solution $t\in R$. How can we determine the number of elements of $A$ ? Could you give me a hint?
06:20
okay, now I should sleeeeeeep
hmm, so I am guessing, the extra correlation beyond entanglement in the PR box is that equation AB=(a+b) mod 2 that forces both inputs and outputs to correlate with each other even though the inputs can be randomly chosen between 0 or 1
Polynomial rings sounds like a Mathien problem to me. I don't have enough understanding of irreducible polynomials to solve that
But... can I bruteforce that somehow...
o wait no, if one of the polynomial is e.g. $x^2+k$ for $k \in \{0,1,2,3,4\}$, then the number of cases to try will skyrocket
But maybe...
$(x-4)(x-3)^2 = (x-4)(x^2-6x+9)=x^3-10x^2+33x-36$
So...
$x^3-10x^2+33x-36 = x^3-0x^2+3x-1$ mod 5
$(ax^n+b)(cx^m+d)=acx^{n+m}+bcx^m+adx^n+bd$
06:37
@MaryStar Can you use that the ring is a UFD?
Jun 8 at 4:58, by Secret
@Zee http://www.mast.queensu.ca/~andrew/notes/pdf/2007c.pdf
To be drawn with pretty pictures in a few minutes
06:52
Does it make sense to talk about: "The set of stabilisers of vectors of an irreducible subspace corresponding to a certain irreducible representation." Or is there a better way to express that?
@Rudi_Birnbaum It is not quite clear what it would mean
What is an irreducible subspace corresponding to an irreducible representation?
@TobiasKildetoft: Can you specify were the uncertainty comes about exactly?
OK
@TobiasKildetoft: The point is, that the subspace should be spanned only one irrducible representation. So is it that already implied in "irreducible subspace"?
"only by"
well, an irreducible representation is already a vector space
OK, but it should be spanned only by one irrep. How to say that?
Is the setup that you have some larger representation, and inside that you have a subspace which is an irreducible representation?
usually, we only speak of "spanned by" when we have some set of vectors that do not already form a vector space
06:58
I have group and want to investigate stabilizers comming from one irrep and not from more. But the one irrep maybe higher-dimensional.
sure, from the point of view of the mathematics, being higher dimensional does not change much
OK, I thought thats the "vector space" part.
all representations are vector spaces
But then I guess 1D also gives a vector space.
right, just a 1-dimensional one
07:00
So how to address this set of stabilisers then properly?
By stabilizer, you mean those that stabilize everything in the representation, right?
Ok, these form a (normal) subgroup, usually called the kernel of the representation
As far as I know, they from a set of subgroups called epikernel, which have an intersection called kernel.
Not familiar with the term epikernel, but from that it looks like it means the stabilizer of a single vector
07:06
In case of higher-than one dimensional irreps, there are different possibilites which elements of $G$ are stabilized, which leads to different subgroups of $G$ their intersection is the kernel.
Does that make sense?
Right, since for a $1$-dimensional one, we can just pick any non-zero vector
But for higher dimensional ones, we can have different stabilizers of different vectors
Yes, sounds right!
So when you originally said "the set of stabilizers", you meant this set of subgroups?
Great. Let me think of how I would phrase it in a concise way
07:09
@TobiasKildetoft: :-) yes! :-)
Do you only want stabilizers of a single vector, or should this also contain stabilizers of sets of vectors?
What are the vectors here again? This is always the bit I am struggling about. Is it some "object" expressed in terms of a basis?
Or is it only the irreps?
I think it must be "sets of vectors" so in general.
So the vectors are the elements of the irrep
Remember that to a mathematician, a representation consists of two objects: A vector space (this is where the vectors I am talking about are) and an action of the group on this space.
the collection of characters for all group elements?
When we talk about the elements of a representation, we mean the elements of the above mentioned vector space
So in the case we are considering, there is just a single representation (i.e. a single character)
(but when viewed as a character, there are no longer any vectors to be seen)
Do you prefer that we consider these in terms of characters instead? It doesn't change the theory really, but it changes what we call things.
07:21
Yes, I want to focus on one certain specific representation, like e.g. the trivial representation. But then we get only the set containing $G$ as the only element. No please no characters. One motivation for this is to get independent from characters in the formulation.
ok
Right, for the trivial representation, we just get $G$ whatever we do
So maybe we should take another concrete example.
The set with $G$ as the only element ;-)
right
So if we want the set to have more than one element, we need the rep to not be $1$-dimensional
(or rather, to make it not just consist of the kernel and the group itself, since the group will be the stabilizer of $0$)
07:24
One example would be $C_3$
Which group is that to you? (to me it is of order $3$)
The group generated by 120° roation.
Ok, and we want real representations of this?
The irreps are $A$ and $E$.
(since the only irreps over the complex numbers are $1$-dimensional for this group)
07:26
Well (I personally think out of continous usage only over the reals, I know that it is an issue in communicating, sorry for that)
It's fine. So we have the irrep on $\mathbb{R}^2$ given by rotation by a third
So the non-trivial irrep in this group is in my nomenclature 2-dimensional, ok?
Unfortunately, this will be a very uninteresting example
right
You're right. Lets go to a better one, one moment please.
There is a group of order 6 we call $D_3$.
Great, we call it that too
So far our names for groups have at least coincided
07:30
$C_3$ is subroup, but it contains in addition a perpendicular rotation by 180°.
Our notation is called "Schönflies"
Yeah, we can see this group as the symmetries of a triangle (including flips)
You mean in 2D?
In 3D its actually a bit more complicated, there it belongs to "chiral" objects.
which are not congruent with their inflection images.
The reason I mentioned the triangle, is that it gives us a nice description of a non-trivial irrep
07:32
OK fine we use that!
It has three irreps $A_1, A_2$ and $E$. OK?
$E$ is the two dimensional one.
8)
Now we want to what we call "descend in symmetry" along "E". That is in your words ...
No clue actually :)
(we will get \{C_1,C_2\}) come that is what we talking about ;-)
I am now lost
07:36
.. we want the stabilisers of the irreducible subspace spanned by $E$, don't we?
Thats the same thing!
Or in my words, we want the stabilizers of vectors in $E$
Ahh - OK. (We call it "descend in symmetry").
interesting
07:39
Well I think then we kind of mutally understand each others!
We are at least getting there
So for each vector in $E$, we want to know what group elements stabilize it
But unless that vector is $0$, we can see that none of the rotations can stabilize it (hence why the example of $C_3$ was not so interesting)
Hmm, I don't know, I think thats not that important right now.
Was that not what we were looking for?
I am not sure, I am just looking for how to express that map $\phi(\rho_k): G \to \{H_j\}_{j\in S}$ in a mathematically concise way. With $\rho_k$ being a specific irrep of $G$ (like $E$) and the set of $H_j$ the stabilizers.
I need a way to define $\phi(\rho_k)$ so that every mathematician immediately understands what is meant.
Ahh, ok. So finding the subgroups is not important right now, it is just about expressions precisely that these are the subgroups
07:44
Yes!
Finding them is no issue - thats specific computations we don routinely. But I have conjecture which I like to express in a waterproof way (mathematically).
And we wanted not just those that stabilize a single vector, but also those that stabilize some set of vectors?
The whole concept should be formulated without recurring to any specific representations (by that I mean objects in Euklidian space and stuff ).
Ok, so what sort of object is a representation to you?
The an element of the result of a group homomorphism?
It can be seen as a group homomorphism (not the result of one, but as the homomorphism itself)
specifically one which takes values in matrices
07:52
OK, but the matrix thing is just like a concrete example, isn't it?
Not necessarily. It is quite common to define a representation to be a homomorphism to the group of $n\times n$ matrices
I know that its common, but I always thought there would be some more general principles "behind it" that can be formulated without recurring to linear algebra
or at least without $n\times n$ matrices.
Right, the matrices are there as placeholders for linear maps
But it is rarely important to distinguish between the two
There are several equivalent ways to view representations, and we just need to pick one in order to make things precise (though there can also be merit in using different ones depending on what one wishes to talk about)
07:58
So what would you suggest?
A question:
For speaking about stabilizers, it seems the most convenient to consider it as a vector space together with an action of the group
When you use this vector space $V$ is it meant that all groups can act at this same space?
Or maybe even to split it into even more data: It is a vector space $V$, and for each group element $g$ it has a linear map $\varphi_g: V\to V$
This is what confuses me often.
Hmm, not sure what you mean by all groups here.
Certainly all groups could act on the same space (for example all acting trivially)
08:03
Not just the peculiar $G$ we are talking about but also then also its subgroups, I suppose yes, but that is always a little bit hard to follow for me, since I cannot clearly see the connections between different groups, so to say.
So for subgroups, this is just a matter of only taking some of the symmetries. But for other groups that are unrelated to the given one, it just becomes a matter of pure abstract stuff
Of course its clear when there is subgroup relation, but that is not always trivially seen. I don't know if that is comprehensible, don't matter.
OK, so then my feeling was not so wrong :-)
OK lets continue.
Of course it should stabilise all possible vectors.
I mean we don't want to overlook things because we choose our $V$ too small, is it that you wanted to know?
No, I meant given $V$, the question is whether we only want subgroups stabilizing a single vector in $V$, or if we also want those that stabilize some set of vectors
(these latter will just be intersections of the former though)
I see, when we have the formulation of for the single vector we say afterwards the intersection of this set is the kernel, right?
08:10
So we go for the single vector.
Ok, then a short formulation that I am fairly certain should be unambiguous to mathematicians would be that this is the set of point-stabilizers for $G$ acting on $V$
Can we somehow implement the specific irrep named $\rho_k$ in that expression?
@TobiasKildetoft how do I make the map $G/H \times G/H \to (G \times G)/(H \times H)$ topologically?
that is $\rho_k$ an irrep of $G$
product doesn't have ump...
08:13
@Rudi_Birnbaum So when you say that $\rho_k$ is an irrep, what sort of object is it? The name $\rho$ suggests that it is a homomorphism?
@LeakyNun What do you mean by make it topologically?
it means in the category Top
@TobiasKildetoft: Its the $E$ in our $D_3$.
and what is ump?
universal mapping property
@TobiasKildetoft for example, one specific irrep from all the irreps of $G$.
08:14
@Rudi_Birnbaum Right, but I don't think you actually defined that (other than as the irrep of dimension more than $1$)
And how to make mention of it here depends on what sort of object it is mathematically
No I didnt. But we need that to specify the our beautiful set of our point-stabilizers.
@TobiasKildetoft: Because for each $k$ we get a different set of point-stabilizers for $G$ acting on $V$.
@Rudi_Birnbaum But when we change $k$, we also change $V$
Example: $\phi_E(D_3)=\{C_1,C_2\}$, $\phi_A1(D_3)=\{D_3\}$, ... here the $\rho_k$ has been E and A, so that were different maps $\phi_k$.
@TobiasKildetoft: And I need to address these maps specifically, as well.
@LeakyNun what happens if you just write down the obvious map
@loch you need to prove that it is continuous
the one thing every topologist hates doing
I could do it if I unfolded enough definitions
but that's ugly
08:27
@TobiasKildetoft: Maybe it helps when I shortly try to explain what would be the next step.
@TobiasKildetoft: The next step would be to map the ireps in a stabilizer to the irrep(s) in $G$. And that will depend strongly on $k$, when I am not mistaken.
@TobiasKildetoft: In the sense of $\{A,B\}$ in $C_2$ came from $E$ in $D_3$. My conjecture is about some functional connection between these irreps.
@LeakyNun oh yes i don't know a clean way of doing it.. does (finite) direct product coincide with (finite) direct sum in the category of top groups? or maybe more generally do they coincide when we're looking at group objects in some category?
direct sum... free product??
my groups aren't abelian
oops ignore me
@Rudi_Birnbaum Ohh, now we seem to be getting into differences in terminology again. The stabilizers so far have been subgroups of $G$, so there are no irreps "in" them
>my groups aren't abelian

The horror
08:41
@TobiasKildetoft: Yes, yes, but the "old" representations can be reduced.
@Daminark help me
and congratulations you made it
Woohoo, thanks!
And it seems you're trying to prove that if $G$ and $H$ are topological groups, you want to find a continuous map $(G/H)^2 \to G^2/H^2$?
@TobiasKildetoft: And clearly $E$ from $D_{3h}$ can be assigned to $A,B$ in $C_2$, in this "descent in symmetry". It seems in some cases there can be ambiguities but in point groups these can be resolved, that means you know that you end up with a set of possibilities and can name all these.
actually I'm trying to prove that (G/H)^2 -> G/H is continuous
maybe there isn't a neat proof afterall
Ah, so you're showing that $G/H$ is a topological group
08:44
yes
Have you already proven that the projection $\pi:G\to G/H$ is a quotient map in the sense of spaces?
(In particular, it's an open map)
let's say we have
@TobiasKildetoft: A good example would be that the trivial irrep in $G$ will always stay the trivial irrep in the stabiliser but the trivial irrep in the stabilizer might come from other irreps from $G$ depending on which irrep we choose to get to our stabiliser.
@Rudi_Birnbaum Ok, I think I see what you are saying. We take our representation and restrict it to one of these stabilizers. Then we decompose this as a representation for the stabilizer?
@TobiasKildetoft: That makes sense.
08:50
And you can see why the map $\pi\times\pi:G^2 \to (G/H)^2$ is also continuous and open?
So in that case, we will always get at least one copy of the trivial representation
@Daminark let's say we can
Okay so let's say $m:G^2 \to G$ is our product normally, which is assumed to be continuous, and $\mu:(G/H)^2\to G/H$ is our new guy
I claim that $\pi \circ m = \mu \circ (\pi\times \pi)$
@TobiasKildetoft: The only bit I am still missing is that you do not mention the irrep which we used to get to the stabiliser. Again in our example we use $E$ in $D_3h$ to get to $C_1$ and $C_2$ and then we check what happened to $E$. But we could also have used $A_2$ that would have led to $C_3$ as stabilizer. And then $E$ from $D_3h$ would have gotten/or stayed the irrep called $E$ in the new group $C_3$.
@Daminark sure
08:56
Well, $m$ and $\pi$ are continuous, so $\pi\circ m$ is. But then $\mu \circ (\pi \times \pi)$ is continuous and $\pi\times \pi$ is a quotient map. The universal property of the quotient thus implies that $\mu$ is continuous
(Universal property of the quotient here says that $gf$ is continuous iff $g$ is continuous if $f$ is a quotient map)
@Rudi_Birnbaum Well, once we go to a subgroup we will in general need to completely rename the irreps to make sure not to confuse them
because even though we can see $E$ as an irrep both for $D_3$ and $C_3$ this is a bit of c coincidence
@Daminark :o
@Daminark what do you mean by a quotient map
A surjective map $f:X\to Y$ is a quotient map if $U\subset Y$ is open iff $f^{-1}(U) \subset X$ is open
Basically that $Y$ has the quotient topology wrt this map
hmm
let's get more abstract
in Optimi(s|z)ing optimi(s|z)ing, 5 mins ago, by Secret
That head bang moment when you need to freeze a dihedral in 256 mlecules where the last atom in a sequence of atom numbers will vary depending on which molecule it is
Ugh, someone divide me by zero so I can stop thinking about this hellhole
09:01
forget quotient and subspace
do everything in terms of filter and complete lattice
@TobiasKildetoft:That is when we choose the subgroup just like that. But according to my understanding there is (at least in point groups) a possibility to get certain informations on a connection between the irreps when you do determine the stabiliser with respect to some irrep of the group, so this is not just some subgroup selection.
I guess we defined the topology on $G/H$ to be the quotient topology so really the content here was that if $f_1:X_1\to Y_1$ and $f_2:X_2\to Y_2$ are quotient maps, then $f_1\times f_2 : X_1\times X_2 \to Y_1\times Y_2$ is a quotient map, as well as the categorical characterization of the quotient
@Daminark so a surjection $f:X \to Y$ is a quotient map iff $Y$ has the coinduced topology
@LeakyNun what are these exactly?
@TobiasKildetoft: Maybe there is some kind information on an orientation of the elements to be imposed, first.
09:03
filter is a generalization of sequences and nets and things like that
you get things like
X is a Hausdorff space if and only if every filter base on X has at most one limit.
X is compact if and only if every filter base on X clusters or has a cluster point.
X is compact if and only if every filter base on X is a subset of a convergent filter base.
X is compact if and only if every ultrafilter on X converges.
and complete lattice is saying that the possible topologies on a set S forms a complete lattice
i.e. it is closed under arbitrary supremum and arbitrary infimum
@Daminark you have the neighbourhood filter at each point, and you can express continuity in terms of filter
Ah logic style filters
@TobiasKildetoft: Lets assume we have only one-dimensional real irreps with entries "-1" and "1". Then we select one irrep, and remove all $g$ with entries $-1" from the table. And we are there. Then we exactly know which irrep came from which.
if you have an order topology, you can have a filter that represents a neighbourhood of "infinity", and you can express limit in terms of filter
right
Actually next week there are some talks about links between these ultrafilter guys and combinatorics, looking forward to it
09:08
Finally gonna learn what those things are, I keep hearing about them floating around but never actually figured them out
Once you learn nets, you will never go back to sequences again because they are uncountably more awesome
And I believe once you learn ultrafilters, you won't want to go back to nets again
Usually the more general and abstract the object is, the more expressive it is
(Disclaimer: I have not seriously look at ultrafilters yet, my order theory background still need some spicing up to understand them)
Someone said ultrafilters?
I am not ready to comprehend those yet, but Leaky say they behave like nets generalised
supermegaultrafilters
So I'm vaguely familiar with nets and I definitely do think they get some good stuff done over sequences by virtue of generality, though as it stands it feels to me like if you're in a context where nets do things that sequences don't, you're in that side of topology where it's a bit more set pushing than I'd like
For the most part I've neglected point-set however in functional analysis there was a context where it mattered
It seems to me that you have sorta two main ways in which your spaces begin to go to shit
09:18
@TobiasKildetoft: So some irreps are already reduced, and some have to reduced again.
In analysis, especially functional analysis, the axioms of countability begin to matter
And in algebra it seems like separation is more what hurts you
We had a pset problem one time which said to prove that if an infinite-dimensional Banach space is reflexive or has separable dual, then you can find a sequence on the unit sphere converging weakly to 0
Now in class we already proved that the weak closure of the unit sphere is the closed unit ball so we were like, how is this not trivial, and why do you have these assumptions in particular?
The point was that we were talking about a topology which a priori wasn't first countable. The content of the problem was to say that if a space has separable dual, then its unit ball is weakly metrizable, so you can actually do things with sequences
And if a space is reflexive, you take the closure of the span of some sequence of linearly independent vectors, then that gives a separable subspace, which is reflexive as well by virtue of living in a larger reflexive space
Now, if a space is separable and reflexive, then its dual is separable and reflexive (in general separable doesn't imply separable dual), so you can apply the above business there
yeah, when I first learnt about nets from some conversation in this chat, I was discussing with user21820 and leaky about the difference between countablity and uncountability, which then the discussion leads into topologies that are not first countable, which is why I found nets more appealing since the context I am interested in are those topological spaces which cannot be adequately converged with sequences
Nets being based on directed sets which in turn are preorders imposed on sets are also mroe generalised than well orderings, thus doing problems involving this help me to build intuition on how to deal with posets that are not well ordered (and possibly don't always have sup for every subset)
so that when I get used to them enough, problems involving sequences will be easier since well orders are "nicer" than generic posets
One of the things being discussed is the relationship between uncountability and completeness of the space, that is something I want to understand more
because e.g. you cannot have a complete vector space of infinite dimensions with a hamel basis of countable cardinality (and other related things)
and for non separable hilbert spaces, even a countable orthgonal basis is not enough to complete it
Topology will thus help me to understand and comprehend the relevant proofs on how the failure of completion occurs and why ramping up the cardinality and then suddenly you can complete the space
thus give a better idea on what properties of an object is an uncountable cardinal is controlling
10:18
$$\lim_{x\to \infty} o..o$$

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