Mathematics - 2018-07-23

A: Has negative cardinality been considered?

12:29 AM

9

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

1:33 AM

@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.

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.

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

I didn't know you had a site D:

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

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

1:39 AM

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

Yeah, the isomorphism depends on a choice of basis.

exercise: find an example where there is no canonical isomorphism

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

@rschwieb but $k$ can be finite

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

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

1:42 AM

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

What do you mean by canonical exactly?

you would have given me the same basis

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

@Fargle ringtheory.herokuapp.com

@rschwieb so what do you make of my suggestions?

1:44 AM

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

indeed

@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!

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

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

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

1:50 AM

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

hmm

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

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

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

I mean, I mean, I mean,

@Daminark there isn't even a canonical splitting field

1:54 AM

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

I see

EnjoysMath

5:34 AM

:R

8:13 AM

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?

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

8:26 AM

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

What about a constant function

Q: Why is gradient the direction of steepest ascent?

42

$$f(x_1,x_2,...x_n):\mathbb{R}^n \rightarrow \mathbb{R}$$ 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)

@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

8:50 AM

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

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

o..o

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

sorry

@Secret this is not a function!

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)

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

9:00 AM

because all the other directions are worse

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

@mercio ok, reading.

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

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

oh!

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

Thank you very much!

This was super clarifying.

9:11 AM

Im glad I could be of help

mathjack51

9:26 AM

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.

draw a picture

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

9:56 AM

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

mathjack51

@mercio No I am not integrating f(0)

10:14 AM

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

12:46 PM

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.

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

@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.

consider the torsion of both sides

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

1:02 PM

quotient the torsion

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?

then quotient by the torsion

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

Ah, very nice! I see. Thanks

@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.

@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 $M_1/N_1 \simeq M_2/N_2$.

1:12 PM

@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

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

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?

Oh, that's good to know. Thanks!

@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

ok

1:15 PM

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

Thanks again for those suggestions

@Secret Do you know about bravais lattices?

a little bit

depends on what context

Hey y'all

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

1:30 PM

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

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

yup, it's working for me

With proxy it works for me as well, cool

Btw, what are you up to now @AlessandroCodenotti

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!

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

1:35 PM

I see, that sounds cool

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

@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

@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.

not for me

oh

1:42 PM

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

2:22 PM

@Secret which one?

theres no perfect square there only the diamond is clear

2:39 PM

the parallelgrams, which are the squares viewed from an angle

2:50 PM

@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

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

3:07 PM

No. Dont look identical to me.

i have even rotated it and seen.

5:29 PM

@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.

@rschwieb yes

@rschwieb yes, yes

OK, great

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

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?

5:33 PM

@rschwieb compact infinite topological groups are uncountable

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

There are, I suppose, finite completions, right?

hmm

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

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

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

5:36 PM

eh I don't think it is a subgroup

I think it is a quotient

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

it's the subset of a product

countable product of continuum sets

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

ok

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

5:39 PM

Direct sum of rings

are the second/third homemorphism theorems important ?

$R\oplus S$ 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 $S$)

@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?

sure

Know any more pithy explanation?

5:41 PM

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

aha

of course.. density

or separability rather

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

@AlessandroCodenotti Thanks!

@AlexKChen yep

The wikipedia page is bit confusing though on Direct sums

Tobias Kildetoft

5:43 PM

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

hi @TobiasKildetoft

@LeakyNun Hi

Hi @Tobias

it depends on whether your rings have unity or not

although all my rings are commutative and unital

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

5:49 PM

...

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.)

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

6:32 PM

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

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

@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.

Yeah, sometimes you don't have any nonconstants.

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 :/

6:48 PM

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

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

The norm topology?

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.

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

7:16 PM

For any Diff Geometers out there, I'd just like to check and see if my answer to my question is correct: math.stackexchange.com/a/2860684/266135

If my assumption is correct, are there any other (important) instances like this in diff geom where authors call something, something else without telling the reader?

7:33 PM

The operator d/dx_j is not a "renaming" of anything - that is the name you learned it under in calculus many years ago!

9:27 PM

Is a transformation always a function?

11:05 PM

@rschwieb the source of the craziness comes from the fact that holomorphic functions extend across codim 2 subsets (hartog's theorem) :p

11:15 PM

Hey @loch

Hi @geocalc33

how are you

good

and you?

not bad

@loch are you familiar with comlplex analysis

I know a little bit of complex analysis sure

11:20 PM

could i post my question here and could you give me some help/

cant guarantee ill be helpful but sure

Q: Conformal mapping - known points?

2

I have a hopefully rather simple question: I want to experiment with different geometries of flowlines and equipotential lines in a 2-Dimensional space in order to fit experimental data. Flow lines and equipotential lines are always orthogonal to each other, and hence form a basic rectangular gri...

transformation conformal-geometry

this isn't my question but it's very similar

here's mine

0

Q: Finding a suitable transformation function for the picture

I haven't taken complex analysis yet, so there may be many words to sift through and not much concrete mathematical notation. I do know a little math though. Anyway, to prepare for my complex analysis course I'm trying to find the conformal mapping that corresponds to the picture. Given a local...

complex-analysis transformation conformal-geometry quasiconformal-maps

11:35 PM

conformal transformations preserve angles, which is clearly not happening

can it happen?

well you could use the identity map

just map the same thing to the same thing?

oh that actually would work

but consider a non euclidean geometry

even if it doesn't look conformal

it still could be

and as more gridlines are added i would think the geometry would flatline

and approach euclidean geometry, where the angles of triangles add up to 180 deg

do you know what a quasi-conformal map is ?

without having taken complex analysis ?

i don't think that tag applies actually

all i know

is that it's almost conformal

so i assume some conditions are being relaxed

11:44 PM

but you don't even have defined an it to talk about