Mathematics - 2023-04-29

Ted Shifrin

00:15

Seems like you totally ignored what I suggested.

shintuku

the author defined neighborhoods with balls since he started with metric spaces

but the logic wasn't working out, what if the balls went past the metric space. since it is an arbitrary metric space, this is possible

even if it isn't embedded in any other metric space

in any case i switched author and this one has his definitions worked out

properly

Ted Shifrin

@shintuku say what?

If $X$ is your world, then all points live in $X$. Period.

shintuku

oh

right heheh

Ted Shifrin

If you’re talking about $[a,b]\subset \Bbb R$, then you make it clear that you’re dealing with a subset of a metric space.

Eventually, then, you learn about subspace topology.

Silly Goose

00:44

In this example of the category $\text{Mat}_R$, so the objects are like the dimension $j$ of an $i x j$ matrix? And the morphisms $f: j \rightarrow m$ takes an $i x j$ object into say a $i x m$ object?

Or here is what I think (hopefully) more clearly: My interpretation: assign to each object $r \in R$ the collection of $\alpha \times r$ matrix where $\alpha$ can be any element of positive integers. Then, a morphism $f: r \rightarrow r'$ is the collection of all $r \times r'$ matrices. In other words, the collection of all matrices that when multiplying any $\alpha \times r$ matrix results in an $\alpha \times r'$ matrix.

Ack I mean $r$ and $r'$ are positive integers, not elements of the ring

Thorgott

no, a morphism $f\colon r\rightarrow r^{\prime}$ is the same thing as a single $r^{\prime}\times r$ matrix (note the order of $r$ and $r^{\prime}$, too)

Silly Goose

01:00

Oh I see so each such matrix is its own morphism

So in this case, the domain and codomain of all morphisms are singletons of positive integers? I.e., any morphism looks like $f: \{a\} \rightarrow \{b\}$ where $a,b$ are positive integers

Hm okay so I think this is refined: A way to think about it: assign to each object $p$ the collection of $p \times \alpha$ matrices where $\alpha$ can be any positive integer. Then, a morphism $f: p \rightarrow p'$ is a single $p' \times p$ matrix. In other words, a matrix that when left multiplying any $p \times \alpha$ matrix results in a $p' \times \alpha$ matrix.

user10478

03:40

I am trying to model the probability of winning a game like Mastermind/Wordle in a set number of turns. Intuitively, the probability of winning at the beginning is Pr(Winning), then after the first guess the player gets a clue, it becomes Pr(Winning | First Clue),

but I'm not sure if also have to condition on what round it is, like Pr(Winning | It Is Round 1), Pr(Winning | (It Is Round 2 ∩ First Clue)), etc. That looks really awkward but I don't think the probability is the same every round even if you don't look at the clues, so I'm not sure if I need to introduce two variables at every round or just one.

KZ-Spectra

03:53

Is there anyone here for conversation?

leslie townes

04:05

user: yeah, if you want to "model" wordle in a simplified sense (and not just duplicate what it does exactly), you need to make some kind of assumptions about how the player will play the game. otherwise you don't know what guesses are possible (let alone likely) in a way that you can assign probabilities. even at one guess, let alone a sequence of multiple guesses.

some of those choices in modeling mimic real life choices, e.g., do you assume the player never makes a guess that is inconsistent with revealed information (roughly speaking, wordle's "hard mode"), or do you let a player guess more generally (for example, if it allows them to more quickly rule out candidates).

and does the player want to maximize the probability of winning at a given turn, or the probability of winning within the allowable number of turns (perhaps at some cost of always making slightly more guesses on average), or something else.

if you want to assign numbers to things, there's also the modeling question of, do you use the actual dictionary that wordle uses, or do you compute with reference to a larger/different dictionary that might better reflect what people will guess.

one potato two potato

Dirichlet problem is more important than I thought. I thought it has some meaning in pde but it's used a lot in the classification of Riemann surfaces

user10478

04:30

@leslietownes Ahhh, I went ahead and posted on main site before I noticed you'd responded, so that has a bit more detail (math.stackexchange.com/questions/4688692/…) tl;dr: I'm actually modeling Mastermind rather than Wordle so the space of legal solutions is well defined at 1296 possibilities, I assume the solution is generated uniformly randomly and the player guesses uniformly randomly among only solutions that are consistent with the clues.

one potato two potato

04:58

I have one confusion: If a continuous function $h$ is subharmonic on a Riemann surface $M$, then $h$ satisfies the maximum principle. If $M$ is compact, then $h$ should obtain maximum at some point $P\in M$. Then suddenly $h$ is a constant function?

Ted Shifrin

Compact means no boundary. What does the max principle say?

one potato two potato

Oh there's nothing to confuse of

$h$ is constant. I only considered compact case

Koro

06:45

A complex valued harmonic function that is not analytic.

Why?

Harmonic function is supposed to be real valued.

But I think $f(z)=\bar z$ should answer this. f=u+iv. Both u and v satisfy Laplacian=0 is the obvious meaning of complex valued harmonic function.

Silly Goose

has anyone worked through emily riehl's category theory in context book? I was looking for an introduction to the topic and came across this text

Koro

@TedShifrin: But complex variable substitution does not seem to be working here: f is entire, then evaluating $\int_0^{2\pi} f(e^{i\theta}z) d\theta$

This integral is equal to $2\pi f(0)$ for every z in C.

Now, let's use subst. $u= e^{i\theta}z$ so that $du= i e^{i\theta} zd\theta$.

Suppose z is non zero. Then, $u$ is a simple closed curve at z. So we get: $\int_\gamma -if(u)/u du$, where $\gamma$ is $u= e^{i\theta} z$ as $\theta$ goes from 0 to 2 pi.

@robjohn: I understand now your comment about harmonic functions. Since the function is real valued on boundary, its imaginary part is 0 on the boundary hence 0 in the interior also by MMT.

I'm not sure if the subst. aftermath is correct or not.

Does $\gamma$ enclose the origin?

Koro

07:12

Why can't a non constant real valued harmonic function be in $L^1$?

robjohn

@shintuku if $X$ is the entire metric space, then each element of the open cover of $X$ is contained in $X$.

@Koro $L^1$ of what?

Koro

$L^1(\mathbb R^2)$

@shintuku I had this confusion too once. What if we take radius of a ball so large that we go beyond the metric space? But that's not possible.

-1

Q: Proving that a metric space $X$ is open and closed subset of itself.

Let $X$ be a metric space. Let $d$ be its distance function. Proving that $X$ is closed: Complement of $X$ relative to $X=\emptyset$, which is an open set (as all points of an open set are interior points and $\emptyset$ is empty and therefore vacuously $\emptyset$ is open) and hence $X$ is clos...

@shintuku please see the comments here. :-)

robjohn

@Koro Consider integrating the harmonic $u$ in balls of radius $r$ around a point $z_0$ so that $u(z_0)\ne0$. $\int_{B_r(z_0)}u(x,y)\,\mathrm{d}x\,\mathrm{d}y=\pi r^2u(z_0)$, which is unbounded as $r\to\infty$.

Taking the absolute value of $u$ possibly makes that even bigger.

Koro

@robjohn how did you get the integral?

I know mean value theorem (for integrals) holds for harmonic functions.

robjohn

The MVP of Harmonic Functions.

Koro

07:22

That is, I have for every r>0, $u(z_0)= \frac 1{2\pi} \int_0^{2\pi} u(z_0+re^{i\theta}) d\theta$.

for any $z_0\in B_r(z_0)$.

robjohn

it would be $\frac1{2\pi r}$

Koro

no?

robjohn

It is the mean value

it works for circles and disks

oh, nvm you are not using normal lebesgue measure

it is $\frac1{2\pi r}$ when the measure is $r\,\mathrm{d}\theta$, but you've already divided out the $r$.

Silly Goose

Is exercise 1.1.i.ii just asking to show that a morphism has a unique inverse (if it exists)? or something along those lines

I am confused because i didn't think of an isomorphism as something a morphism "can have" more like a morphism can be an isomorphism and if it is it can have an inverse? somethingg along those lines

robjohn

@Koro $\int_0^R\int_0^{2\pi}u\!\left(z_0+re^{i\theta}\right)r\,\mathrm{d}\theta\,\mathrm{d}r=\int_0^R2\pi ru(z_0)\,\mathrm{d}r=\pi R^2u(z_0)$ (integration in polar coordinates)

anak

07:29

@SillyGoose this exercise is poorly worded if they don't define inverse morphisms before. What they mean is that in (i) g is f's "inverse isomorphism". Then you just show that this is unique if it exists in (ii).

leslie townes

silly: i agree that "being an isomorphism" is a property of a morphism, not something that a morphism "can have," but i might feel differently about an "inverse morphism" or an "inverse isomorphism"

Silly Goose

ahh okay so the "double sided inverse" $g$ (as in gives you the identity for respective domain/codomain) is what we want to show is unique if it exists

leslie townes

see how it defines those concepts

yes

Silly Goose

okay i see

Koro

@robjohn Thanks a lot. You're so amazing :-).

5

excellent

anak

07:31

@leslietownes you are up late

Silly Goose

unfortunately such terms are not defined in the text (at least to my knowledge)

but this clears it up :D

category theory is so exciting hehe

anak

@SillyGoose yeah they are relying on the reader to be more knowledgeable on general ideas from mathematics, I expect.

Not very good pedagogically, though.

leslie townes

anak: all night every night

Koro

friday night

user 85795

How many hours do you sleep, sir?

anak

07:33

Well I am going to retire for the night. Goodnight silly goose and friends.

user 85795

cya pal

leslie townes

i'm exaggerating, but i'm often up at this hour. it's rarely as active as it is now

user 85795

Looking at your activity histogram, I would guess you average about 4 hours per day 😴

Koro

I want to adopt an Alaskan Malamute.

But I think that it won't survive weather here.

same goes for an Akita Inu.

user 85795

why?

Koro

07:46

they have double fur suited to keep them in cold temperatures.

same for a Husky and St. Bernard.

user 85795

I guess you could constantly trim down their fur; but, that seems a bit too time consuming.

Silly Goose

an identity morphisms is always an isomorphisms right? because it is (by definition) its own inverse isomorphism?

PlaceReporter99

09:19

chat.stackexchange.com/transcript/36/2023/4/28/0-20

one potato two potato

Now I see why hyperbolic geometry is important: For most Riemann surfaces (with suitable riemannain metric) $M$, its universal cover $\tilde{M}$ is conformally equivalent to upper half space (or hyperbolic space).

I always wondered why people interested in hyperbolic geometry (especially among low-dimensional topologiests)

Every Riemann surface can be classified into three types (as humans): Elliptic (i.e., compact Riemann surface), parabolic and hyperbolic. The name hyperbolic is suggestive as the surface has hyperbolic structure. But the other two cases... why?

Sonozaki

10:33

Can I deduce that $(A\cap B)\setminus C=(A\cap C)\setminus(B\cap C)$ is false because if $x \in (A\cap B)\setminus C$ then $x \notin C$ and so $x \notin (A\cap C)$, hence $x \notin (A \cap C)\setminus (B\cap C)$?

Thorgott

10:48

This works if $(A\cap B)\setminus C$ is non-empty. It can happen that both sides are empty and hence equal, though, e.g. if $C=A\cap B$.

Koro

Suppose that $f:S^1\to C$ is continuous and is such that $f(z)= f(\bar z)$ for all z in $S^1$. Can it be extended holomorphically to all of the unit disk?

user977780

11:05

Is this right time to say goodbye?

Sha Vuklia

11:20

@Thorgott This kind of (counter) example is very useful, thanks a lot. Is it ok if I put a link to your solution in my post on main, where I asked this question? Then all solutions are in one place

Thorgott

11:38

sure

one potato two potato

13:05

10

Q: The degree of a meromorphic form in a genus $g$ Riemann surface is equal to $2g-2$… is this in some way related to the Gauss-Bonnet theorem?

I’m following an introductory course on Riemann surfaces. Today, the lecturer proved the fact that the degree of a meromorphic form in a genus $g$ Riemann surface is equal to $2g-2$ (we can deduce from this, for instance, that the sum of orders of zeros of a holomorphic nontrivial form is $2g-2$ ...

differential-geometry riemannian-geometry riemann-surfaces curvature meromorphic-functions

Mad

13:33

Hello

How can i understand the concept of Degeneracy in quantum mechanics in mathematical terms?

Spesifically, regarding hte multiplicity of the eigenvalues

one potato two potato