Mathematics - 2018-06-21 (page 4 of 4)

user193319

11:04 PM

If $G$ is a group acting on some set $A$, what does $S_A$ defined as? My book refers to it as the symmetric group. Does this mean that it is just the collection of all bijections from $A$ to $A$?

Leaky Nun

@MatheinBoulomenos what is Hom(Qp*, C*)?

topological group

@user193319 yes

user193319

Thanks!

Maximus

In Russel's paradox we have a set R \in R

but isn't that not allowed by the property "$\in$"?

shouldnt it be a subset symbol instead?

user193319

No, I think that's the whole point of the paradox.

MatheinBoulomenos

@LeakyNun $\Bbb Q_p^\times \cong \Bbb Z \times \Bbb{Z}_p \times \Bbb{Z}/(p-1)$ as topological groups, where $\Bbb Z$ and $\Bbb{Z}/(p-1)$ have the discrete topology. The dual of $\Bbb{Z}_p$ is $\Bbb Z[1/p]/\Bbb Z$

user193319

11:13 PM

Russell's paradox is intended to show that set theory as envisioned by Gottlob Frege doesn't work.

Leaky Nun

@MatheinBoulomenos but that isn’t the circle, that’s C star

MatheinBoulomenos

@LeakyNun doesn't matter for profinite groups

Leaky Nun

why?

so Z becomes C-star and the torsion remains right

MatheinBoulomenos

Z is not profinite

Daminark

Cue Titanic recorder music

Drew Brady

11:15 PM

Hey I'm hoping to get on the problem here: imgur.com/a/p9Sb4FW. It is part (b) (I solved (a) above it). Following the hint, it is rather straightforward to see that P(B) <= a, since P(B) = P(\cup_k H_k) = lim_k P(H_k) <= a, since 0 < P(H_k) <= a for all k. But I am having difficulty getting the lower bound P(B) >= a.

MatheinBoulomenos

every homomorphism from a profinite group to $\Bbb C^\times$ (which is a real Lie group) has finite image and is thus contained in the torsion part of $\Bbb C^\times$ which is contained in the circle

Leaky Nun

I mean, Hom(Z,C-star) = C-star

@MatheinBoulomenos the more you know

MatheinBoulomenos

So if G is profinite, we get Hom(G,C*)=Hom(G,S^1)=Hom(G,Q/Z)

Leaky Nun

cool

thanks

MatheinBoulomenos

Jun 2 at 22:00, by MatheinBoulomenos

Okay suppose we have such a $\rho$. Then the kernel of $\rho$ is a closed subgroup, so the quotient $G/\operatorname{ker}(\rho)$ is again a profinite group. By the homomorphisms theorem for topological groups, $\operatorname{im}(\rho) \cong G/\operatorname{ker}(\rho)$ is a profinite group, so in particular it is compact and totally disconnected.
By compactness of $\operatorname{im}(\rho)$ and Hausdorfness of $H$, $\operatorname{im}(\rho)$ is a closed subgroup, so by Cartan's closed subgroup theorem, $\operatorname{im}(\rho)$ is an embedded submanifold of $H$. Any topological manifold is loc…

^ The proof that a homomorphism from a profinite group to a real Lie group has finite image

Hi @Daminark

user193319

11:19 PM

@Maximus Not sure if you noticed my messages, but "$\in"$ in Frege's understanding of set theory doesn't outright deny that $R \in R$ can happen (it isn't a stipulation in his system). Because of this, the paradox arises when you take this fact together with Frege's Comprehension principle, which says that, given ANY predicate, we can form the set of things having that predicate.

Maximus

yes I saw your messages, but I was asking why is it allowed to take R \in R since we have to use the subset symbol.

MatheinBoulomenos

@LeakyNun Neukirch has the structure of the multiplicative group of a general local field somewhere in chapter 2

Daminark

How's it going?

MatheinBoulomenos

I think I wrote that down on MSE somewhere

@Daminark pretty well, thanks. I gave a talk today in our projective geometry seminar and it went really well

user193319

@Maximus Again, the point of Russell's paradox is that we can, according to Frege's theory, write $R \in R$; it isn't prohibited. The fact that $R \subseteq R$ is allowed doesn't generate any problems/paradoxes, so it is pointless to consider it with respect to Russell's paradox.

Maximus

11:23 PM

I see, so how is this connected with a set of all sets

the notation {x : x $\notin$ x}

Daminark

Nice!

Leaky Nun

@Maximus that's free comprehension

i.e. the x ranges through every set

which leads to contradiction

Maximus

because it also ranges with itself

?

Leaky Nun

ZFC solves this by not allowing set comprehension to range through every set

in ZFC, set comprehension must only range through elements of a set

user193319

@Maximus Well, it's actually connected with "is a member of itself". Because of Frege's comprehension principle, the predicate "is a member of itself" gives rise to a set (the one you just wrote down).

Leaky Nun

11:24 PM

i.e. {x in A : proposition of x}

this set comprehension ranges through elements of A

MatheinBoulomenos

@Daminark people asked for counterexamples and I had prepared a lot, including a single variable polynomial over the quaternions such that the zero set is a 2-sphere (so not quite finite) and some weird stuff with finite fields where a "plane" is a "curve"

Maximus

so it limits it then for not being able to choose x \in x

?

MatheinBoulomenos

How's it going for you? @Daminark

Leaky Nun

you can still construct the set { x in A : x notin x } in ZFC

and it won't lead to any contradiction

Maximus

I see, so the contradiction comes when we dont restrict subsets and we can choose x \in x correct? @LeakyNun

@user193319 I see, thank you

MatheinBoulomenos

11:27 PM

unrestricted comprehension implies the existence of a set of all sets
and set of all sets with restricted comprehension implies unrestricted comprehension (well, it implies anything, really), so there's a connection

Leaky Nun

@MatheinBoulomenos hmm, I feel kinda frustrated how the proof of classification of finite fields flows very naturally for me (in fact I can prove it myself), but that when I tell my friends I realize that it's based on quite a handful of non-trivial stuff (that has internalized in me)

@Maximus right

MatheinBoulomenos

@LeakyNun we're doing a lot of finite fields stuff in our "elementary" number theory course for which I'm the sole TA

so far we had 3 exercises on finite fields that nobody got correct

Leaky Nun

test me :P

Drew Brady

all you need to know is the Frobenius map lol

and that (a -b)^p = a^p - b^p

Leaky Nun

what you said are equivalent

Drew Brady

11:29 PM

i agree

Leaky Nun

i.e. the Frobenius being a homomorphism

Drew Brady

that pretty much sums up finite fields.

(kidding, kinda)

MatheinBoulomenos

@LeakyNun Let $p$ be a prime and $D \in \Bbb{F}_p^\times \setminus (\Bbb{F}_p^\times)^2$, then how many solutions does the "reduced Pell's equation" $x^2-Dy^2=1$ have with $(x,y) \in \Bbb{F}_p^2$?

that equation is to be read mod $p$ of course

Leaky Nun

I think you have mentioned this before

not that I know the answer

MatheinBoulomenos

but I thought all you need to know is that $(a+b)^p=a^p+b^p$? :P

The solution is cool, but not easy to come up with

Maximus

11:35 PM

@LeakyNun what confused me is that in my book it says the following:

But this is before Comprehension axiom is introduced. That confused me I guess.

MatheinBoulomenos

@DrewBrady can you solve that? if finite fields are that easy

Leaky Nun

@Maximus I don't find any problem with that

Maximus

So after comprehension is introduced you can have any set with any property since you're taking subsets. What they are talking about is some properties don't describe sets if we don't take subsets, correct?

Drew Brady

@MatheinBoulomenos haven't looked at it, and I was certainly joking about finite fields. I'm waiting to get help on my probability theory problem...

MatheinBoulomenos

@LeakyNun I can give you a hint if you want

Leaky Nun

11:37 PM

@MatheinBoulomenos I don't want

So $[\Bbb F_p(\sqrt {D}) : \Bbb F_p] = 2$, and then consider $(x+\sqrt D y) (x-\sqrt D y) = 1$, i.e. the elements in $\Bbb F_p(\sqrt {D})$ with norm 1. By local artin reciprocity, $N : \Bbb F_p(\sqrt D)^\times \to \Bbb F_p^\times$ sends $p^2-1$ elements to $(p-1)/2$ elements, so the kernel has size $2(p+1)$.

MatheinBoulomenos

@LeakyNun that's wrong actually. $\Bbb F_p$ is not a local field

Leaky Nun

ok I may need to calculate the order of the image of the norm

give me a moment

MatheinBoulomenos

there's CFT for finite fields, but it doesn't work as it does for local fields

Leaky Nun

:(

Jake S

a newbie linear algebra question, generally if I am determining whether a set of vectors is linearly independent, should I write the matrix using the vector coefficients as rows or columns?

for example if I am determining whether or not the following polynomials span the set of all polynomials of grade equal to or lesser than 2: $\{x^2, x^2-x+1, 2x-2, 3\}$ should the first row be (1 0 0) or (1 1 0 0), or is this indifferent as one is just the transpose of the other?

Drew Brady

11:43 PM

usually columns, because you're looking for nontrivial elements in the nullspace

Ax is just a linear combination of the columns of A weighted by the entries of x.

Maximus

@user193319 can you have $R \in R$ in ZFC?

Leaky Nun

@MatheinBoulomenos ok do you have hints?

@Maximus no

MatheinBoulomenos

is there another way to write the norm?

Leaky Nun

$N(z) = z \overline z$

MatheinBoulomenos

what is $\overline{z}$?

Leaky Nun

11:49 PM

the non-trivial element of the galois group

MatheinBoulomenos

yeah, I think you know something about that element

Maximus

@LeakyNun Which axiom disallows $R \in R$ in ZFC?

Leaky Nun

@Maximus foundation (consider {R})

Daminark

Sorry I was out but yeah not too much, just kinda going through stuff somewhat randomly

@MatheinBoulomenos

MatheinBoulomenos

@Daminark what kind of stuff?

is this related to your REU?

I'm also taking a course on modular forms right now

Daminark

11:56 PM

Yup, the candidate topics at the moment seem to be elliptic curves, modular forms, "raw" ANT, and group/Galois cohomology

Maximus

@LeakyNun What do you mean by consider {R}

MatheinBoulomenos

so much good stuff to choose from

Leaky Nun

use foundation on {R} @Maximus

Maximus

I see

MatheinBoulomenos

@Daminark just do all of that

Drew Brady

11:58 PM

Hey I'm hoping to get on the problem here: imgur.com/a/p9Sb4FW. It is part (b) (I solved (a) above it). Following the hint, it is rather straightforward to see that P(B) <= a, since P(B) = P(\cup_k H_k) = lim_k P(H_k) <= a, since 0 < P(H_k) <= a for all k. But I am having difficulty getting the lower bound P(B) >= a.

Maximus

@LeakyNun makes sense thank you so much

Daminark

Yeah, for sure. It's tricky to say, the NT talks at the moment are building to modular forms and I think we have a standard ANT class here doing stuff like class groups and all so I've considered zooming in a bit on either cohomology or curves, but we'll see. Probably gonna talk to my mentors some more, read some more, and get a feeling

Lmao I wish I could, though the REU is just 2 months, so that's probably not enough time to go deep enough for that

MatheinBoulomenos

yeah if you can just take ANT as a regular class, then it would be a waste to miss the other topics

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$Mathematics$ Mathematics