Mathematics - 2018-07-23

geocalc33

00:02

what is reflection positivity?

Secret

Negative cardinality can be defined, there's a question in MSE on that. You just have to be ready to give up something

Otherwise, if I recall one of my past experiments correctly, a set of negative cardinality (defined as some S and P such that |S|+|P|=0) breaks either axiom of regularity or de Morgan's law of set operations

So they are very weird

Secret

00:29

9

A: Has negative cardinality been considered?

First you need to ask yourself what is cardinality? It's a mathematical notion which is designed to give us a way to measure the size of a set. When you want to extend cardinals to negative integers you need to ask yourself, what is the purpose of this extension? Can we measure "more sets" now?...

Ah sorry I mean axiom of extensionality

but it is much worse than that. Most of the set operation identities no longer work if you have negative cardinality

rschwieb

01:33

@Fargle Yeah, that one is cool because the operations are accessible examples of things that make ring operations, but they didn't have anything obviously to do with addition and multiplication.

Fargle

Yeah, that's exactly why I like it! It can be easy to imagine that rings only arise from numbers, but this is a case where a ring is not numerical from the outset.

rschwieb

@Fargle Have you checked out any of the examples on my site? You might find something else you like

Fargle

I didn't know you had a site D:

Leaky Nun

@Fargle my group rep lecturer would say, they might be isomorphic, but they have different personalities

Fargle

@LeakyNun That's a nice way to put it.

Leaky Nun

01:39

Also, if V is a fin.dim.v.s/K then V is isomorphic to V* = Hom(V,K) but hardly canonically

Fargle

Yeah, the isomorphism depends on a choice of basis.

Leaky Nun

exercise: find an example where there is no canonical isomorphism

Fargle

I mean, doesn't $\Bbb R^2$ work?

Leaky Nun

@rschwieb but $k$ can be finite

@Fargle well I have a canonical basis of $\Bbb R^2$ called {(1,0),(0,1)}

rschwieb

@LeakyNun Yeah, I realized that ina later comment :) I know you're getting caught up with my glut of comments though

Leaky Nun

01:42

because $\Bbb R^2$ really means $\Bbb R \times \Bbb R$

Daminark

What do you mean by canonical exactly?

Leaky Nun

you would have given me the same basis

Fargle

I might have given you (2,0),(0,2)

rschwieb

@Fargle ringtheory.herokuapp.com

Leaky Nun

@rschwieb so what do you make of my suggestions?

Daminark

01:44

Yeah I'd rather call that "natural" than "canonical"

Leaky Nun

indeed

rschwieb

@LeakyNun One of them definitely had to be a countably infinite field, and I think I must have made another countable, and I realized the last two may very well be fine as finite... so after some extra checks I'm going to make those two finite

@LeakyNun It was helpful... thanks!

Leaky Nun

exercise: find an example where there is no isomorphism that is natural

rschwieb

@LeakyNun Oh, arg, now just noticing the ones you posted on the site too :) I'll have to get caught up on thos

Leaky Nun

C(Q(i)) = Gal(Q^ab/Q(i)) is a subgroup of Gal(Q^ab/Q) = C(Q)
C(Q)/N(C(Q(i))) = group of order 2
which one am I supposed to believe

Daminark

01:50

I could see a non-natural case arising for some kinda Galois extension

Like, $\mathbb{Q}(\sqrt[3]{2},\zeta_3)$ as a $\mathbb{Q}$-vector space

Leaky Nun

hmm

Fargle

I mean, this seems like it's equivalent to the question of finding a finite dimensional vector space without a natural basis.

Daminark

Hmm, maybe that's a bit too easy but if you give some massive polynomial and just take its splitting field, then there won't be an especially obvious basis. This question also depend on how things are presented to you

Fargle

I mean, you could take the perp of $(1,1,1)$ in $\Bbb R^3$.

I mean, I mean, I mean,

Leaky Nun

@Daminark there isn't even a canonical splitting field

Daminark

01:54

I'm thinking about it in $\mathbb{C}$

Leaky Nun

I see

EnjoysMath

05:34

:R

Silent

08:13

Let $f:R^n\to R^m$ be differentiable at a point $\vec v$. The gradient at $\vec v$ points in the direction of greatest increase. Is it possible that there is other direction where $f$ increases fastest? In other words, if we know that function increases in a direction fastest at a point $\vec v$, then, can we infer from this information that gradient is pointing in that direction?

Alessandro Codenotti

@rschwieb For some reason it wasn't rendering as mathfrak on my phone, but it works fine on my computer

Silent

08:26

@AlessandroCodenotti, will you please look at my question above?

Alessandro Codenotti

What about a constant function

Secret

42

Q: Why is gradient the direction of steepest ascent?

$$f(x_1,x_2,...x_n):\mathbb{R}^n \rightarrow \mathbb{R}$$ The definition of the gradient is $$ \frac{\partial f}{\partial x_1}\hat{e}_1 +\ ... +\frac{\partial f}{\partial x_n}\hat{e}_n$$ which is a vector. Reading this definition makes me consider that each component of the gradient correspon...

The direction is always unique since there can only be one direction of greatest increase on an affine space

and the gradient is the tangent vector at a point of the function f

(and higher order tensors as well)

Silent

@AlessandroCodenotti so, gradient there is zero vector.

so, i think it says that there is no way to reach steeper place

Is there no direction of greatest increase or all the directions are of greast increase? @AlessandroCodenotti

Secret

08:50

There are no directions of greatest increase. To see that, consider e.g. $2x^2+y^2=z^2$ and compute the gradient at $(0,0)$

Silent

@Secret What will be gradient vector? i am sorry, but first i thought that it will be (4x, 2y)

mercio

o..o

are you saying functions into $\Bbb R^m$ have gradient vectors ?

Silent

sorry

@Secret this is not a function!

mercio

if the gradient is nonzero then it points to the direction of fastest increase

if it is zero then all directions are equally bad at increasing (looking at first order only)

Silent

@mercio yes, and how do we know that this direction is unique?

mercio

09:00

because all the other directions are worse

have you read some of the answers in the linked question ?

Silent

@mercio ok, reading.

by the way I saw a proof in prof. Ted's lectures, which uses $x\cdot y=||x||||y||\cos \theta$

mercio

so many bars

use \cdot

also if you have 2 directions $u,v$ where the function is equally good at increasing, then by looking at $u+v$ you get a better direction

Silent

oh!

mercio

maybe you are not realizing that "differentiable in several variables" is much stronger than "differentiable along all directions"

if you have several directions of (nonzero) fastest increase, then your function is not differentiable at that point

Silent

Thank you very much!

This was super clarifying.

mercio

09:11

Im glad I could be of help

mathjack51

09:26

I had a confusion while exchanging summation and integral. If f is a number-theoretic function then $$\int_1^x \sum_{n\le y} f(n) dy = \sum_{n\le x} \int_n^x f(n) dy$$ how did the change of limit in integral take place? I am a bit confused. Please help.

mercio

draw a picture

for which $y \in [1,x]$ are you integrating $f(0)dy$ ?

Abcd

09:56

$$\int_1^e\dfrac{1+x^2\ln x}{x+x^2 \ln x} dx$$

mathjack51

@mercio No I am not integrating f(0)

mercio

10:14

well you are integrating a bunch of $f(n)$s

maybe you would have better luck turning it into a sum whose limits are independent of $y$ and compensating that with an indicator function ?

then it's easier to switch the two things

user193319

12:46

Let $R$ be some integral domain, $N$ a nonzero $R$-module, and $R^n$ free module of rank $n$. I am trying to show that $R^n \oplus N \simeq N$ implies $n=0$. I could use a hint.

Leaky Nun

@user193319 $N = \bigoplus_\Bbb N R$, $n = 1$

user193319

@LeakyNun Ah, a counterexample? Hmm...perhaps I need to be more specific. I am trying to show that $R^n \oplus R/(a_1) \oplus ... \oplus R/(a_m) \simeq R/(a_1) \oplus ... \oplus R/(a_m)$ implies $n=0$, where $a_i \in R - \{0\}$ are not units and $a_1 | a_2 |...|a_m$.

Also, $R$ is actually a PID.

Leaky Nun

consider the torsion of both sides

user193319

The torsions are isomorphic because $Tor(R^n) = 0$, right? That doesn't seem to give a contradiction.

Leaky Nun

13:02

quotient the torsion

user193319

By what? If I take the torsion of both sides of $R^n \oplus R/(a_1) \oplus ... \oplus R/(a_m) \simeq R/(a_1) \oplus ... \oplus R/(a_m)$, both sides are $ R/(a_1) \oplus ... \oplus R/(a_m)$, right?

Leaky Nun

then quotient by the torsion

you get $R^n$ on the left and $0$ on the right

user193319

Ah, very nice! I see. Thanks

rschwieb

@AlessandroCodenotti oh! That explains it :) . But yeah, if you ever see something that looks wrong please let me know. I do want to make sure there aren't serious mistakes.

user193319

@LeakyNun Actually, what permits me to quotient by sides by their respective torsions? Isn't the following generally false, which we are implicitly making use of: if $M_1 \simeq M_2$ and $N_1 \subseteq M_1$ is isomorphic to $N_2 \subset M_2$, then $M_1/N_1 \simeq M_2/N_2$.

Leaky Nun

13:12

@user193319 if $M_1 \cong M_2$ then $M_1/\operatorname{Tor}(M_1) \cong M_2/\operatorname{Tor}(M_2)$

because you're doing the same thing to both sides

user193319

Okay, so the statement does hold for this particular submodule.

Leaky Nun

actually

if $M_1 \simeq M_2$ and $N_1 \subseteq M_1$ is isomorphic to $N_2 \subset M_2$ via the same isomorphism, then $M_1/N_1 \simeq M_2/N_2$

@rschwieb how goes my cardinality suggestions?

user193319

Oh, that's good to know. Thanks!

rschwieb

@LeakyNun I haven't had any time to work on them between late last night and now, when I'm just arriving at my job :P

Leaky Nun

ok

rschwieb

13:15

@LeakyNun . I think you can probably expect a 24-48 hour turnaround, depending on complexity of things

Thanks again for those suggestions

Abcd

@Secret Do you know about bravais lattices?

Secret

a little bit

depends on what context

Krijn

Hey y'all

Anyone know of a alternative to gen.lib? Seems down

Abcd

13:30

@Secret I am just not understanding how CCP and FCC are the same thing

Krijn

@AlessandroCodenotti Really? Can' t reach it from here, will try a VPN

Alessandro Codenotti

yup, it's working for me

Krijn

With proxy it works for me as well, cool

Btw, what are you up to now @AlessandroCodenotti

Alessandro Codenotti

I completed my bachelor last week and I'll move to Germany for my master, in october

What about you? It's been a while since you last were in chat!

Krijn

I graduated, took some time off and started a job now in july

Which is mostly non-math consultancy, but I get some time to look into (Post quantum) cryptography

Alessandro Codenotti

13:35

I see, that sounds cool

Krijn

Have to say that research there is very interesting, so I'll have to discuss how to exactly spend my time...

Secret

@Abcd For both structures, the atoms are in the same positions in the unit cell. It's a bit hard to see, thus consider this diagram:

seas.upenn.edu/~chem101/sschem/metallicsolids.html

Crystallographically speaking, the length of the unit cell for each direction is the same and all atoms are as close they can be possibly backed

Abcd

@Secret Its still not clear to me :/ I can't spot FCC in it.

@Secret Has Google ruined google images through a new update? Basically whenever I am clicking a picture in google images the entire page is reloading

Please try google images once on your pc and let me know if you face the same thing.

Secret

not for me

Abcd

oh

Secret

13:42

As for your question: Concentrate on the bottom right dotted square, see how there are 4 atoms at the corner and one at the center

Now movve this plane one step to the top left, now in the second slice, you see 4 atoms arranged in a diamond shape. Now move the plane again so it is at the top left, there are now 5 atoms in it.

This 4 5 4 arrangement is exactly what you expect for a FCC

Abcd

14:22

@Secret which one?

theres no perfect square there only the diamond is clear

Secret

14:39

the parallelgrams, which are the squares viewed from an angle

Abcd

14:50

@Secret chemtube3d.com/solidstate/_ccp(final).htm

I found this^

but they have literally placed the rows one over the other

without any distinction of voids :(

click "ABCA" layers

Secret

well, look at the atoms being enclosed by that box and compare that to the FCC structure. Are they identical?

Abcd

15:07

No. Dont look identical to me.

i have even rotated it and seen.

rschwieb

17:29

@LeakyNun I'm struggling to remind myself about this: if $F$ is a finite field, then the cardinality of a $\kappa\geq\aleph_0$ dimensional $F$ vector space should be $\kappa$ exactly right? I think I remember learning that a union of a chain of $\kappa$ many sets with cardinality strictly less than $\kappa$ cannot exceed $\kappa$, but that's only a fuzzy memory.

Alessandro Codenotti

@rschwieb yes

Leaky Nun

@rschwieb yes, yes

rschwieb

OK, great

Alessandro Codenotti

If $\dim_k(V)\geq\aleph_0$ then $|V|=|k|\cdot\dim_k(V)$

rschwieb

What's an easy way of seeing ringtheory.herokuapp.com/rings/ring/92 is uncountable? I am not good with completions, but I seem to remember them as being described as a subring of an infinite product.. or maybe it isn't. Can I see it that way somehow?

Leaky Nun

17:33

@rschwieb compact infinite topological groups are uncountable

rschwieb

OK, but can you convince me in another way or two?

There are, I suppose, finite completions, right?

Leaky Nun

hmm

rschwieb

What group are you referring to? Compact in what topology?

Leaky Nun

let's say R[X,Y]/I is already uncountable and is a subgroup

rschwieb

ok, sure, I guess that is the case here.

Leaky Nun

17:36

eh I don't think it is a subgroup

I think it is a quotient

rschwieb

How do we know the completion doesn't have cardinality more than $\mathfrak c$?

Leaky Nun

it's the subset of a product

countable product of continuum sets

rschwieb

ah, the product is countable... and we're back to the thing I mentioned earlier

OK, that filtration is countable and makes a countable product

got it

thanks

Leaky Nun

ok

Alex K Chen

What does this symbol $\oplus$ means w.r.t Ring theory ? For example, I'm seeing $R/I \simeq \mathbb{Z}_5 \oplus \mathbb{Z}_5$ ?

Alessandro Codenotti

17:39

Direct sum of rings

Alex K Chen

are the second/third homemorphism theorems important ?

Alessandro Codenotti

$R\oplus S$ is a ring whose underlying set is $R\times S$ and the operations are defined coordinate-wise, meaning $(r,s)+(r',s')=(r+r',s+s')$ (note that the first addition is in $R\oplus S$, the second in $R$ and the third in $S$)

rschwieb

@LeakyNun I know the cardinality of continuous functions on $\mathbb R$ is $\mathfrak c$, and I guess a similar argument does the same thing for $\mathbb C$, and hence the ring of holomorphic functions?

Leaky Nun

sure

rschwieb

Know any more pithy explanation?

Alessandro Codenotti

17:41

@rschwieb Yes (the argument is that they are determined by their value on $\Bbb Q+i\Bbb Q$)

rschwieb

aha

of course.. density

or separability rather

Alex K Chen

@AlessandroCodenotti Thanks ! So $(r,s) \cdot_{R \times S} (r',s') = (r \cdot_R r', s \cdot_S s')$ ?

rschwieb

@AlessandroCodenotti Thanks!

Alessandro Codenotti

@AlexKChen yep

Alex K Chen

The wikipedia page is bit confusing though on Direct sums

Tobias Kildetoft

17:43

@AlexKChen Though I prefer to use $\times$ instead, as one often uses $\oplus$ for a coproduct

Leaky Nun

hi @TobiasKildetoft

Tobias Kildetoft

@LeakyNun Hi

Alessandro Codenotti

Hi @Tobias

Leaky Nun

it depends on whether your rings have unity or not

although all my rings are commutative and unital

rschwieb

I was just looking into the ring of holomorphic functions the other day. I found out here msp.org/pjm/1953/3-4/pjm-v3-n4-p.pdf#page=41 that Schilling had claimed here ams.org/journals/bull/1946-52-11/S0002-9904-1946-08669-3/… that every prime ideal of the ring of entire functions is maximal.

I thought it was a little weird they didn't specify "nonzero" in either paper. Surely neither of them thought the ring was a field

Leaky Nun

17:49

...

rschwieb

But anyhow, I think I read elsewhere that Henriksen's result shows the Krull dimension fo that ring is actually $\mathfrak c$, far from being $0$.

Anyone know much about that?

(In case it wasn't clear in my earlier comment, of course that claim of Schilling's was wrong.)

Mike Miller

18:32

I know nothing about that, but I believe it's a theorem that the Krull dimension of the Ring of holomorphic functions on an affine complex variety is either 0 or uncountably infinite.

39

Q: What is the Krull dimension of the ring of holomorphic functions on a complex manifold?

Consider a connected holomorphic manifold $X$ and its ring of holomorphic functions $\mathcal O(X).$ My general question is simply: in which cases is the Krull dimension $\dim \mathcal O(X)$ known? Of course if $X$ is compact $\mathcal O(X)=\mathbb C$ and that dimension is $0$. There are also ...

cv.complex-variables ac.commutative-algebra

rschwieb

@MikeMiller That's pretty interesting. I believe it, but how can it become $0$ if it's even $\mathfrak c$ on $\mathbb C$? I guess sometimes it's just exactly $\mathbb C$?

Not sure if I"m reading it right.

Mike Miller

Yeah, sometimes you don't have any nonconstants.

rschwieb

Does the case of holomorphic functions from $\mathbb C\to \mathbb C$ not a case where the domain is compact? I guess I'm not sure what topology is going on

That post is also implying the Krull dimension is $0$, if $\mathbb C$ is compact. But as I understand it, the ring is a nonfield domain @MikeMiller

Clearly I don't have a grasp on the whole context :/

Mike Miller

18:48

I am really confused what you are talking about? $\Bbb C$ is not compact

rschwieb

@MikeMiller OK, good. I'm not sure what topology is intended.

The norm topology?

Mike Miller

Yes, I wouldn't talk to you about something that wasn't a manifold :)

If $M$ is a compact complex manifold without boundary such as the Riemann sphere, there are no nonconstants holomorphic functions. This is because they must have compact image in $\Bbb C$, which is closed; however the open mapping theorem says that nonconstant functions have open image. Thus we are constant.

rschwieb

I think it's a little crazy that you can have a noncompact X that still has only constant entire functiosn

the complex numbers are just too nice

THey're like the Mr. Rogers of fields.

Perturbative