Mathematics - 2019-01-24 (page 1 of 2)

@TedShifrin well its in the courtesy of our professor. The problem was "a point charge is located at the corner of cube of side a find the flux through surfaces which do not meet at q$ could be done by gauss law but professor is bit more interested in the integration that he gave it homework.

famesyasd

if I take a subset X of the real line and it is uncountable, is it necessarily dense in some ball in R?

Ted Shifrin

Oh, @Nobody, the integral does work out pretty nicely. Do it in polar coordinates over half the square. You should end up with something like $\int_0^{\pi/4}(1+\sec^2\theta)^{-1/2}d\theta$, which looks horrendous, but is doable if you use an obvious trig identity after writing in terms of $\cos$.

Mike Miller

12:09 AM

@LeakyNun Did he say it's a great project for you?

Ted Shifrin

Yes, @famesyasd. If you have an uncountable subset of $[0,1]$, can you prove it must be dense in some interval?

Mike Miller

@TedShifrin and language...

Leaky Nun

@MikeMiller he said it's the right level for me

Ted Shifrin

Yeah, that too.

Mike Miller

Ok, that's better advice than any of us can give then.

Ted Shifrin

12:09 AM

^^^^

Or you could be like Zee and ask for advice just to do the opposite ...

rolls 7 + $\pi/6$ eyes

Nobody recognizeable

@TedShifrin will try that. Thanks for the help and your time.

Ted Shifrin

You're welcome.

Leaky Nun

@TedShifrin lmfao

Mike Miller

3/8 done copyediting this monstrosity.

Leaky Nun

oh man

anyway, off I go

Karl Kronenfeld

12:24 AM

@TedShifrin No

Perturbative

I'm confused by this proof, if the set of $m \times n$ matrices of full rank is an open subset of $M(m \times n, \mathbb{R})$, how does continuity allow us to conclude the Jacobian of $F$ has full rank in some nbhd of $p$?

Karl Kronenfeld

By definition of continuity, in fact.

famesyasd

@TedShifrin but the Cantor's set on [0,1] is nowhere dense.

Karl Kronenfeld

(and uncountable)

famesyasd

so cardinality has nothing to do with density?

Karl Kronenfeld

12:34 AM

For $\mathbb R$. Indeed, the Cantor set has cardinality equal to the continuum, so we do not have to assume CH or anything of that sort.

In the other direction, finite sets are never dense ;)

Perturbative

@KarlKronenfeld But $dF_p$ is a map from $T_pM$ to $T_{F(p)}N$ and we can view the Jacobian matrix of $F$ as a map from $T_{\hat{p}}\mathbb{R}^k$ to $T_{\hat{F(p)}}\mathbb{R}^l$, the codomain of either of these maps isn't $M(m\times n, \mathbb{R})$ so when the author says continuity, which map is the author referring to?

(also $k$ is the dimension of $M$ and $l$ the dim of $N$ in the above)

Karl Kronenfeld

Use the map sending $q$ to the Jacobian at $q$.

Perturbative

Okay but then it doesn't seem obvious to me that such a map would be continuous

Can I view the map as a composition of continuous maps?

Karl Kronenfeld

Go back to definition of smooth map.

Mike Miller

12:52 AM

Or C^1 map

mick

0

Q: Is $\mathbb{Z} ((-2)^{\frac{1}{n}})$ ever a UFD?

Let $n>2$ be an integer. Consider the integral domains $\mathbb{Z} ((-2)^{\frac{1}{n}})$. For what integer values of $n$ is $\mathbb{Z} ((-2)^{\frac{1}{n}})$ a UFD ? Are there infinitely many solutions ? Same question for $\mathbb{Z} (2^{\frac{1}{n}})$.

Karl Kronenfeld

Yes, more precisely :)

Perturbative

1:17 AM

Well since $F$ is smooth all the partial derivatives of all orders exist and are continuous, but what I'm saying is that while the map $\xi : M \to M(m \times n, \mathbb{R})$ defined by $\xi(p) = \text{Jacobian matrix of $F$ at $p$}$ is a map taking $p$ to a matrix containing entries that vary continuously, I don't understand how that implies that $\xi$ is continuous. Maybe I am forgetting some basic linear algebra fact

Karl Kronenfeld

You give the space of matrices the product topology so that it is homeomorphic to $\mathbb R^{mn}$.

Perturbative

Okay but then the only result I have at hand to show that $\xi$ is continuous is the following:
Let $(X, d)$ be a metric space and let $f_1, ..., f_k$ be real valued functions on $X$ (that is $f_i : X \to \mathbb{R}$ for $i \in \{1, \dots k\}$ ) and let $\mathbf{f} : X \to \mathbb{R}^k$ be defined by $$\mathbf{f}(x) = \left( f_1(x), \dots, f_k(x)\right)$$ then $\mathbf{f}$ is continuous if and only if each of the functions $f_1, \dots, f_k$ is continuous.

So do I need to view $M$ as a metric space to show that $\xi$ is continuous

Because then since the space of matrices is homeomorphic to $\mathbb{R}^{mn}$ I can view $\xi$ as a function from $M$ to $\mathbb{R}^{mn}$ and the above result would apply

Douglas Sirk

@MikeMiller how is the orbit in Lorenz homeomorphic to $\mathbb R$? doesnt it limit onto itself?

Karl Kronenfeld

@Perturbative You're working in a coordinate chart, so you have a metric if you really need one. But, one definition of/result about continuity is that the preimage of any open set via a continuous map is open.

Perturbative

@KarlKronenfeld Okay so if the preimage of $\xi$ is open in $M$ then why can we conclude that the Jacobian of $F$ has full rank in this open set?

Douglas Sirk

1:34 AM

/././.././.

jajajajajajajaja

Karl Kronenfeld

Because, you're picking the right open set to take the preimage of using $\xi$ (the author gave this set to you). Then, consider what happens when you apply $\xi$ to anything in this preimage.

Mike Miller

@DouglasSirk Given an ODE on $\Bbb R^k$, the flowline through a point $p$ is a map $\gamma: I \to \Bbb R^k$, where $I$ is an interval, $\gamma(0) = p$, and $\gamma$ satisfies the ODE. Further we demand that $I$ is the largest possible domain of $\gamma$: there is no solution defined on a larger interval. Note that by the existence and uniqueness theorem for solutions to ODEs, in particular, the flowline is unique if eg the ODE is C^1.

For the Lorenz attractor, the largest interval $I$ is $\Bbb R$ itself.

But a flowline, by definition, is a map from $\Bbb R$, or possibly some subset (or if you like, the image of this map.)

The Lorenz continuum you are asking about is the closure of a certain flowline.

That is, as far as I know, its definition.

Perturbative

@KarlKronenfeld Okay I get it now, applying $\xi$ gives a matrix of full rank, so the Jacobian of $F$ in coordinates has full rank, thus $dF_p$ has full rank in this nbhd

Thanks for your help! @KarlKronenfeld

Karl Kronenfeld

@Perturbative Yep

no problem

mick

2:04 AM

0

Q: Is $\mathbb{Z} [(-2)^{\frac{1}{n}}]$ ever a UFD?

Let $n>2$ be an integer. Consider the integral domains $\mathbb{Z} [(-2)^{\frac{1}{n}}]$. For what integer values of $n$ is $\mathbb{Z} [(-2)^{\frac{1}{n}}]$ a UFD ? Are there infinitely many solutions ? Same question for $\mathbb{Z} [2^{\frac{1}{n}}]$.

abstract-algebra ring-theory factoring unique-factorization-domains

Silent

3:13 AM

Directrix and Dandelin Spheres: What is the directrix of hyperbola and ellipse in Dandelin sphere representation here?

user76284

3:29 AM

Is there a way to have only 1 number per align by default in LaTeX?

Tired of putting splits everywhere.

zed111

5:31 AM

Why are even rank tensors invariant under inversion?

Akiva Weinberger

7:28 AM

Secret

7:52 AM

???

Akiva Weinberger

8:14 AM

Points repelling each other

By the way

I randomly stumbled upon this

The curve hidden in plain sight...no flat Earth. Extended version

►

It's pretty cool - the simplest demonstration of the Earth's curvature that I've ever seen

(Beyond the "we see the sun go below the horizon and that would make it night everywhere at once" thing)

The comments are… disappointing

Secret

Rule number one of the internet: Do not read the comments

That video also reminds me of a concept:

Projective geometry is probably the easiest way to make use of infinite objects. Here a vanishing point at infinity actually has physically measurable results as shown

I wonder what other domains of maths can make infinity more tangible for daily uses?

Akiva Weinberger

Incidentally

Cool geometry thing

Reconfigurable Materials

►

A metamaterial is a material whose function is determined by its structure rather than its composition. An example is butterfly wings. There's no pigment there - the texture at the nanoscale determines what light is reflected and in what way!

Nature is really good at nanotechnology, much better than we are. So most of our attempts at metamaterials are at a much larger scale. Still, they're pretty cool.

Secret

8:32 AM

Not surprising given it has so much time to experiment with many things

Akiva Weinberger

8:56 AM

Snow gets its color from its microstructure (rather than nanostructure), I think

Snowflakes are a good example of natural complexity that doesn't come from Darwinian evolution

Leaky Nun

10:12 AM

@MatheinBoulomenos!

Secret

9

Q: Does the concept of infinity have any practical applications?

I know what you're thinking: "of course it has, for example, it can be used to tell you how many times you can go around a circle". But that isn't really true, now is it? You'd be dead or the world would go under long before an infinite amount of loops had been reached. Are there any practical a...

infinity applications

Actually, I found that infinity often comes from extrapolating some regular process or pattern. But why do we always believe it will eventually lead to something?

Akiva Weinberger

11:13 AM

> Absolutely, infinity has countless (:P) practical applications.

rolls countless eyes

vimeo.com/49081053/recommended

^More wibbly wibbly wobbly

mick

12:10 PM

1

Q: Prove $ \mathbb{Z}[\sqrt -5] $ is not a UFD without factoring?

We know the ring $ \mathbb{Z}[\sqrt -5] $ is not a UFD. The typical proof is showing that $6$ factors in $2$ ways. But there must be a better way to show this than to do trial and error factoring of some random element of the ring. So how does one prove $ \mathbb{Z}[\sqrt -5] $ is not a UFD wi...

ring-theory factoring alternative-proof unique-factorization-domains

ÍgjøgnumMeg

@Mick compute the Minkowski bound

Hey @Alessandro

@Mick sorry I misread your question. In any case, the machinery your commentors are probably referring to is computation of the ideal class group of $\Bbb Z[\sqrt{-5}]$, which you'd do well to learn about if you're interested in this sort of thing

(and which also requires you to compute the Minkowski bound anyway, if you wanna save yourself a chunk of work)

Alessandro Codenotti

12:27 PM

Hi @ÍgjøgnumMeg

ÍgjøgnumMeg

Hi :)

How's it going?

Alessandro Codenotti

Exams are getting very close

ÍgjøgnumMeg

The pit of doom

Alessandro Codenotti

But the next semester is going to be great, I'll do a lot of set theory

ÍgjøgnumMeg

Sounds just like what you'd enjoy!

Alessandro Codenotti

12:29 PM

Indeed! I signed up for the graduate seminar in set theory just yesterday, I got a cool topic for my talk

user237650

12:39 PM

Hello! Can someone give the formula to solve (1/2 + 1/12 + 1/30 + .......till infinity) Thanks

ÍgjøgnumMeg

@Alessandro Nice, what's your topic (bearing in mind I have absolutely no idea about set theory)

Alessandro Codenotti

Kunen's inconsistency theorem and cardinals on the edge of inconsistency

ÍgjøgnumMeg

lol

sounds quite arcane

Alessandro Codenotti

Basically there is a hierarchy of large cardinals (which will be topic of the whole seminar), cardinals with particular properties whose existence cannot be proved in ZFC, simple examples are inaccessible cardinals: uncountable cardinals $\kappa$ such that $2^\lambda<\kappa$ for every $\lambda<\kappa$

ÍgjøgnumMeg

@Alessandro one of my old students is doing his undergrad dissertation on elliptic curves, which I was happy to hear

@Alessandro I see, sounds cool still

Alessandro Codenotti

12:44 PM

Less simple examples are measurable cardinals, cardinals $\kappa$ with a finitely additive nontrivial 2-valued measure

ÍgjøgnumMeg

bleh

Alessandro Codenotti

But there's plenty of large cardinals, weakly compact, supercompact, indescribable, extendable, huge, etc.

ÍgjøgnumMeg

lmfao

sounds so silly

Definition: Huge

@Alessandro I didn't get the job I applied for, so now I'm sending yet another application for a Master in Frankfurt...

lol

Alessandro Codenotti

Kunen's inconsistency refers to a particular type of extremely strong large cardinals, namely Reinhardt cardinals, and it shows that this is in fact too strong and inconsistent with ZFC

So it is in some sense an upper bound on how far this hierarchy of cardinals can grow

ÍgjøgnumMeg

Nice

Alessandro Codenotti

12:47 PM

And there are some cardinals (rank into rank) which are slight weakenings of Reinhardt cardinals whose consistency with ZFC is still open

@ÍgjøgnumMeg Oh, I'm sorry to hear that :/ But hopefully you won't have to wait to long before starting your Master then

ÍgjøgnumMeg

@Alessandro yeah.. it's all gone a bit panicked because Heidelberg didn't work

which was my fault anyway but yeah, hopefully I can just start in April as planned

Alessandro Codenotti

That'd be the best

ÍgjøgnumMeg

that's what I want anyway lol

Might have to call upon the formidable finances of my aunt and uncle though

I saved about 5000 GBP working for 6 months

but that aint enough to finance an entire master

Alessandro Codenotti

No but it should be a good chunk of it

Convert it to real money before brexit though :P

Tobias Kildetoft

@ÍgjøgnumMeg I was censor on oral exams in elliptic curves the past two days. Meant I had some reading to do beforehand, as I have not done anything with elliptic curves since my second year of undergrad.

ÍgjøgnumMeg

12:57 PM

@Tobias I haven't done anything on elliptic curves, but he said my dissertation inspired him to do something number theory related so he chose that

which was really nice

Tobias Kildetoft

neat

Two of the students were slightly stumped when I asked them if they could name an example of a group where the discrete log problem is very easy (since they has presented the MOV attack on it for elliptic curves, which reduces it to multiplicative groups of finite fields which are somewhat easier, but still potentially hard when the embedding degree is high).

Rithaniel

3:08 PM

Does there exist a single group satisfying $\langle x,y\mid y^2=x^2=1\neq (xy)^n, \forall n\in\mathbb{N}\rangle$ or would that be a family of groups? (Or perhaps does no such group exist?)

Mike Miller

I mean if you're asking I'd that's a presentation it's not

Rithaniel

presentation?

Does that refer to the group generator I tried to type up?

Mike Miller

Google group presentation

It's a way to specify a group

You have not specified a group, just stated a condition: it is 2-generated by x and y, degree 2 elements, so that (xy)^n is never the identity

Rithaniel

Okay, so any number of different groups could satisfy that condition?

Akiva Weinberger

Continuing my series of "random geometry stuff that I link in the chat randomly"

https://youtu.be/H-CJyzL3WPY

Mike Miller

3:14 PM

@Rithaniel yes, why not?

You haven't had any reason to believe otherwise yet as far as I can see

Rithaniel

I also have no reason to believe that there would be multiples. I might suspect that multiple groups exist which satisfy that, but I've not put in the work yet, and for all I know, there could be some proof out there that only one such group exists.

(Thank you for the info, by the way)

Tobias Kildetoft

@Rithaniel But there is of course a group with presentation $\langle x,y\mid x^2,y^2\rangle$, which does satisfy the additional condition you have there.

Actually, any group satisfying the conditions will be a quotient of that, and as far as I recall, it only has finite quotients.

Rithaniel

I'm still fairly early in my abstract algebra course. We've been touching on morphisms this week and we haven't yet gotten to quotient groups.

Hmmm, so this would necessarily also be a normal subgroup? I would have expected that. Also, the line $xGx^{-1}=G$ is kind of confusing me. Does that mean that the relation holds for every element in $G$?

(Ah, nope, it refers to cosets)

ÍgjøgnumMeg

3:36 PM

@Rithaniel it means that there exist $g, h \in G$ st $xgx^{-1} = h$

user131753

In The Joy of Cats it is written that (slightly paraphrased), "[If $F:(\mathbf{A},U)\to (\mathbf{B},V)$ be a concrete isomorphism between the concrete categories $(\mathbf{A},U)$ and $(\mathbf{B},V)$ then the fact] that such a concrete isomorphism exists means, informally, that each structure in $\mathbf{A}$, i.e., each object $A$ of $\mathbf{A}$, can be completely substituted by a structure in $\mathbf{B}$,

user131753

namely $F(A)$ (keeping, of course, the same morphisms).

user131753

What does it mean to say that each $A$ object $\mathbf{A}$ is "can be completely substituted by a structure in $\mathbf{B}$"? What is the informal difference between a concrete isomorphism and an isomorphism?

Mike Miller

@TobiasKildetoft Ah, you are right! I am really embarassed. The group $\langle x, y \mid x^2 = y^2 = 1\rangle$ is better known as the infinite Dihedral group, whose quotients are all itself, finite dihedral, or finite cyclic.

user131753

Anyone? Any idea?

Mike Miller

3:50 PM

@Rithaniel To be explicit, Tobias has made the point that this does specify one group. However, I will repeat the point I was making: normally you cannot specify a group by specifying some generators, some relations, and some non-relations.

Rithaniel

Alright, well that is still a good thing for me to keep in mind. I need to be careful about how I specify things.

Mike Miller

Writing some number of generators, and then some equalities those generators must satisfy, does indeed specify a group.

Karl Kronenfeld

@user170039 Concrete isomorphisms are stronger than isomorphisms in general, since concrete isomorphisms establish an equivalence of structures on the same sets, while an arbitrary isomorphism does not come with any reference to the category of sets. If one concrete category is isomorphic to another, it means the structures (on the corresponding sets) given by the one do not mean anything different from the structures given by the other -- you can swap out the structures and nothing changes.

Rithaniel

So, in this case, if I were to say, add another requirement, assuming for a moment that this added requirement is possible, does that necessarily result in a different group?

Mike Miller

I do not understand your question

What do you mean by "added requirement"?

An equality of words in the generators?

(That is the interpretation for which you are still giving me a group presentation, which specifies a group.)

In that case, you can clearly make formulas whose truth follows from the formulas you already have. The stupidest example is $\langle x \mid x = 1\rangle$, which is the same as... $\langle x \mid x = 1, x= 1\rangle$. I've added the same relation twice ;)

user131753

4:00 PM

@KarlKronenfeld "If one concrete category is isomorphic to another, it means the structures (on the corresponding sets) given by the one do not mean anything different from the structures given by the other -- you can swap out the structures and nothing changes." - why so?

Mike Miller

Less pedantic is eg $\langle x \mid x = 1, x^2 = 1\rangle$

user131753

Essentially this was the part that I find difficult to understand. I have no problem with the forma definition itself, @KarlKronenfeld.

Karl Kronenfeld

This comes down to the premise of category theory: the morphisms tell you everything important about a structure.

Rithaniel

Okay, so what about allowing that you have an additional requirement whose truth does not necessarily follow from the formulas already presented, such as "$xy^{-1}=x^n$ for some $n\in\Bbb{N}$"?

Mike Miller

@Rithaniel how do you know that it doesn't already follow from the formulas presented?

Karl Kronenfeld

4:04 PM

@user170039 To back that up a little, often structures (especially algebraic structures) can be restated using morphisms alone, so it makes sense that morphisms would characterize such structures. Extending beyond this context is a matter of hypothesis and, really, ymmv.

Rithaniel

Honestly, I don't, but if it does, we can just replace the $=$ with $\neq$ and the "for some" with "for any"

user131753

@KarlKronenfeld Indeed. But that would mean that the morphisms of the concrete category $(\mathbf{A},U)$ (and similarly of $(\mathbf{B},V)$) tells me everything important about a structure.

Karl Kronenfeld

@user170039 Yep

That is the point.

Mike Miller

@Rithaniel No, I tried to make clear above that you cannot specify a group by adding inequalities.

The fact that it worked in the one case you cooked up was unfortunately a fluke.

Rithaniel

Ah, right, right, sorry.

Mike Miller

4:06 PM

no need to be sorry

This is the issue. If you have a group presentation, and you write down some word in the generators (note that every formula $w_1 = w_2$ is equivalent to $w_1 w_2^{-1} = 1$; we may as well just be adding the relation $w = 1$ for some word $w$), then there is no algorithm to determine whether or not $w = 0$ follows from your other relations.

This is called the undecidability of the word problem.

user131753

Under this viewpoint one case if two categories are isomorphic we can treat them as being "essentially same". Right @KarlKronenfeld?

Mike Miller

If you can ever prove that adding this new relation actually changes the group, you have proved a theorem.

It does not follow trivially.

Rithaniel

Okay, I think I understand, then.

So, there is no easy way to evaluate, at least at a glance, if adding a requirement to the infinite dihedral group will necessarily change it?

Karl Kronenfeld

@user170039 Yeah.

Rithaniel

Or any group, for that matter

Karl Kronenfeld

4:10 PM

@user170039 Isomorphic objects of a category are usually regarded as "essentially the same", though you often want the/an explicit isomorphism. (Unique up to unique isomorphism is a common phrase)

user131753

@KarlKronenfeld My question then simply is this: How will two concretely isomoprhic categories be viewed keeping in mind the analogy of isomorphism between "abstract" categories?

Mike Miller

that does not seem to be a question

@Rithaniel Correct; you can certainly prove results, since the infinite dihedral group is reasonably tractable.

But they do not come from just looking.

user131753

@MikeMiller Sorry.

Mike Miller

@user170039 now your equivalence preserves the set-level description of objects and morphisms. that's all.

Karl Kronenfeld

Same idea as having rings that are isomorphic as groups under addition, or isomorphic as rings.

user131753

4:14 PM

@MikeMiller Why sets? A concrete category is just a pair $(\mathbf{A},U)$ such that $U:\mathbf{A}\to\mathbf{X}$ is a faithful functor for some category $\mathbf{X}$.

Karl Kronenfeld

You have extra structure on you categories, namely the functors to sets, and it is preserved by a concrete isomorphism.

Mike Miller

the only definition of concrete category I have ever seem in my life has $\mathbf{X} = \mathsf{Set}$

user616128

Is there a place where one can find a decent collection of example chern polynomials?

Karl Kronenfeld

You could also go to another concrete category $X$, I guess.

Mike Miller

@user616128 no, what more precisely are you after?

Rithaniel

4:16 PM

And I assume that you could also potentially prove that an added requirement would result in contradiction?

Mike Miller

@Rithaniel No, you can never find a group that does not satisfy some list of generators and equalities between words in the generators. That always specifies a group.

user131753

@MikeMiller I read the definition from here pp 58 Def. 5.1(1).

Mike Miller

If you write down a bunch of dumb shit you might just specify the trivial group, eg $\langle x, y \mid x = y, xy^2 = 1, xy = yx, xy = y^3x\rangle$

user616128

More precisely, examples of nonvanishing third chern classes

Mike Miller

There are the universal examples, there are examples on projective spaces.

Rithaniel

4:18 PM

Okay, so you always have such a group, but you don't know if one group is necessarily different from another, hence why isomorphisms are so important.

Mike Miller

@user170039 what they call a construct is what everybody else calls a concrete category

user616128

I've seen $X$-concrete used when they mean X\ne Set

Mike Miller

when Freyd wrote the paper "homotopy is not concrete" I assure you he was not claiming that the category $\mathsf{Ho}(\mathsf{Top})$ does not admit a faithful functor to any category whatsoever. :)

anyway if you want to replace by $\mathbf{X}$ above then you just take my sentence "now your equivalence preserves the set-level description..." and replace set with $\mathbf{X}$.

there is no functional difference, but now X is some random thing that we do not usually imagine we have a picture of.

nbro

4:31 PM

Érico Melo Silva

6:30 PM

@Daminark i heard through the grapevine that ur jumping into the Neves-reading

Daminark

Maybe, I know someone wanted to do a reading course with Neves next quarter and he was like yeah if there's enough people sure, and he said that since he's gonna be doing Danny's Riemannian geo he might ask to have a course on extending stuff we'll do this quarter, maybe Morse theory, maybe characteristic classes, etc

That I'd be down for at least attending, even if I don't take it as a class

Ted Shifrin

hi Demonark @Eric

Érico Melo Silva

hlo hlo

Daminark

Also today in difftop I have found a newfound appreciation for PDE since we talked a bit about how Brouwer gives you the Yamabe problem

Hey @Ted!

Érico Melo Silva

yamabe problem is sick as hell

Alessandro Codenotti

6:36 PM

Hi @Ted @Dami @Érico

Ted Shifrin

Hi demonic @Alessaandro

Alessandro Codenotti

It looks more demonic with all those vowels

Érico Melo Silva

Aaleessaandroo

Alessandro Codenotti

lol

For some reason a lot of non Italians tend to remove an s or double the l in my name

Ted Shifrin

I was creative.

Daminark

6:39 PM

For sure. We didn't exactly go through the details of the estimates and all, just said that the inverse Laplacian can be seen as a compact operator $C^{0,\alpha} \to C^{0,\alpha}$ and then if you take the set of $u \le 0$ such that $e^u \le R$ for some appropriate $R$, then that satisfies hypotheses of Schauder fixed point theorem, so $\Lambda^{-1}(e^u) = u$, so yeah.

Ted Shifrin

Nice stuff, Demonark.

Érico Melo Silva

that's weird cuz like who spells that name w one s

i think a bunch of the romance languages spells it w x and spanish w maybe a j?

Daminark

Then he mentioned that this implies that if you have a manifold with metric $e^u$ times the Euclidean metric, then you had constant scalar curvature. Didn't go through the details and I'm not exactly the master of diffgeo to follow totally what this means but still seems pretty sweet

Ted Shifrin

Well, that's not very interesting diff geo. That's just a simple formula.

Alessandro Codenotti

@ÉricoMeloSilva it has a single x in all the Germanic languages I think

Érico Melo Silva

6:42 PM

fixed point theorems all over PDE

@AlessandroCodenotti yeah this sounds right

Alessandro Codenotti

But then it becomes a ks if you further east

Ted Shifrin

or a zh

no, I guess not.

Érico Melo Silva

my middle name is spelled like the french way

alexandre

we pronounce the x like "sh" tho

Daminark

Oh wait I misremembered, apparently scalar curvature is actually $-1$, not just constant

Ted Shifrin

No, not from what you said.

I don't think ...

Eric should know this in his sleep, but I'd need to compute.

Érico Melo Silva

6:46 PM

idr the scalar curvature transformation law but it's not that hard to compute

Ted Shifrin

Nope, but I'm not gonna crank up moving frames just yet.

Érico Melo Silva

the big daddy yamabe for positive yamabe invariant i think u need to use some big daddy theorem like PMT or something

it's hard

Ted Shifrin

what's PMT?

Érico Melo Silva

positive mass

schoen-whoever

yau?

Ted Shifrin

Yau

yeah

Érico Melo Silva

6:48 PM

classic

@Daminark this reminds me, I was talking to him about complex and he said something like "if i taught complex id accidentally end up ranting about how awesome the yamabe problem is and no one would learn complex analysis"

and naturally in my diff geo class second year he talked about it a bunch and seems hes done it in diff top too

man likes what he likes

Daminark

Amazing

To be fair about complex analysis, it's only ever 2/3 complex analysis

At most

(Though I hear Lawler's probability stuff when he taught actually linked back to the complex analysis)

Érico Melo Silva

i feel like i got screwed over w all the number theory last year

i did not like it at all

Daminark

Yeah I sorta wish we did Riemann surfaces or something

Érico Melo Silva

that's what i wouldve preferred

Daminark

I feel like that's definitely something that falls under the "general interest" category moreso than analytic number theory

Smart's doing grad complex this year and I might just sit in tbh

Érico Melo Silva

7:00 PM

oh word if he's doing it i might show up too

Daminark

Like I guess there's definitely good content in analytic number theory and I might try to revisit it at some point in the future just because, but somehow it didn't excite me much

Érico Melo Silva

it kinda killed my desire to ever look at it again tbh

Analytic NT and foundations are probably the things im least likely to ever look at in my life

Alessandro Codenotti

Analytic NT is very understandable

But foundations are the best

Érico Melo Silva

i can understand ppl liking foundations it's just not my taste

but with analytic nt actually i just dont know why ppl do it

i just could not stand it

Daminark

Lol, some stuff in analytic NT like modular forms (which looks absolutely nothing like what we did in class) I definitely still wanna try out and they seem like things I may very likely end up thinking about later. Prime number theorem type of stuff... I can't really get behind somehow

Alessandro Codenotti

7:05 PM

You need an unusual combination of interests in maths to like ANT I think

Daminark

I think the path we took toward doing it was a bit more convoluted than usual though

Érico Melo Silva

@Daminark totally fair

user76284

7:23 PM

Can we bound the rank of the wreath product $G \wr H$ in terms of the ranks of $G$ and $H$?

For example $\text{rank}(G \wr H) \geq \text{rank}(H)$ I think

And of course $\text{rank}(G \wr H) \leq \text{rank}(G) + \text{rank}(H)$. Is there a better upper bound?

Akiva Weinberger

Not too long ago I had an epiphany about why $\sin x+\cos x=\sqrt2\cos(x-\frac\pi4)$

Consider a unit segment with one end fixed to the origin and the other swinging around it. The $x$-coordinate of the free end is described by $\cos x$.

ÍgjøgnumMeg

@Alessandro Danish and Norwegian use a "ks" instead of an "x", while Swedish usually uses an "x"

err

Alessandro Codenotti

Ah I see

Akiva Weinberger

Now consider the same setup but rotated ninety degrees. This gives us another swinging arm whose $x$-coordinate is $\sin x$.

Now perform the vector sum of the arms by gluing the second one to the edge of the first.

ÍgjøgnumMeg

@Alessandro for example the word for "trousers" in Norwegian is "buksa" while it is "Byxa" in Swedish

Akiva Weinberger

7:36 PM

You end up with a bent L-shape swinging around the origin.

The distance from the origin to the swinging tip of that L-shape is $\sqrt2$.

Alessandro Codenotti

Interesting

Akiva Weinberger

And the vector from the origin to the tip is rotated 45 degrees from the vectors making up the L-shape.

So we end up with $\sqrt2\cos(x-\frac\pi4)$.

Alessandro Codenotti

Also just for completeness I messed up the definition of measurable cardinal earlier, I want a proper measure, not a finitely additive one (finitely additive nontrivial 2-valued measures exist on every cardinal, assuming AC)

Akiva Weinberger

This also explains why $A\sin(x)+B\cos(y)$ has amplitude $\sqrt{A^2+B^2}$.

/end

user10478

8:10 PM

Is this identity legal? i.imgur.com/RKButP2.png

Akiva Weinberger

8:26 PM

@user10478 The first and last are equivalent but the middle isn't because $\int_a^bf(x,y)dy$ is still a function in $x$

You can write $\int_c^d\int_a^bg(x)f(x,y)dydx$ in the middle instead

and then swap the integrals like $\int_a^b\int_c^df(x,y)g(x)dxdy$

and separate it out into the last expression in the pic

user10478

8:44 PM

@AkivaWeinberger Okay, so to get the middle result, both f and g have to be functions only of their respective variable of integration, to move between the outer results and the expression you added, at least one must be a single variable function, and if both are functions of x and y, you can't do anything, correct?

Or even if you have g(x) matched up with dy and f(x) matched up with dx, you can't do anything?

user10478

9:11 PM

I'm having a hard time seeing why the second and third expressions in my image aren't equal.

Since the definite integral of g(x) dx resolves to a constant.

Akiva Weinberger

The definite integral of f(x,y) dy does not resolve to a constant.

user10478

But isn't $\int_a^b f(x,\ y) k dy$ identical to $k * \int_a^b f(x,\ y) dy$?

Akiva Weinberger

9:28 PM

You didn't take $\int g(x)dx$ out of the integral of $f\ dy$. You tried to take $\int f(x,y)dy$ out of the integral of $g\ dx$. Notice which one is the outer integral in the first expression.

user10478

True, but now I'm wondering if we can go from the first expression to your expression, to the third expression, to the second expression, making them all interchangeable after all.

user193319

10:42 PM

Theorem from Rudin's FA book: If $\mathcal{B}$ is a local base for a topological vector space, then every member of $\mathcal{B}$ contains the closure of some member of $\mathcal{B}$. I'm screwing up my quantifiers. Is this statement saying that, given $B \in \mathcal{B}$, there is some $B' \in \mathcal{B}$ such that $\overline{B'} \subseteq B$?

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$Mathematics$ Mathematics