Mathematics - 2020-06-06 (page 1 of 2)

12:25 AM

Hi I have a simple question. In inner product we have a first definition that says, for u,v non-zero vector in $R^2$ or $R^3$, then the dot product is defined as: $u . v = |u||v| cos \theta$. Now, the second definition says that for u,v non-zero vectors in $R^n$, then $ u . v = u_1v_1$.
My question is: Is the first definition restricted to only $R^2$ or $R^3$? why don't we have like the first definition over $R^n$

Thorgott

your second definition is most certainly off

also tell me what $\theta$ is

skillpatrol

The angle between two vectors.

@user777

user777

12:43 AM

@Thorgott sorry, it should be as following: for u,v non-zero vectors in $R^n$, then $ u . v = u_1v_1 + u_2v_2 + .... + u_nv_n$.

Thorgott

right

the answer to your question will be that the same thing still works in arbitrary dimensions, but you have to provide a definition of the angle between two vectors in that general setting and that's slightly more annoying to do than this definition, which is more practical

AbstractAlgebraLearner

@JackOhara hey :D

Semiclassical

1:18 AM

another way to look at it is that, taken together, the two definitions together provide a perfectly good definition of "angle between two vectors"

i.e., you define the angle between two unit n-vectors as "the inverse cosine of their dot product"

Ted Shifrin

3:48 AM

@Semiclassical once you prove Cauchy-Schwarz.

user76284

Is there an example of a field other than the surreal numbers that contains the ordinals under Hessenberg arithmetic as a subsemiring?

Yuvraj

5:52 AM

morning

everybody

Knight

6:22 AM

@Semiclassical I asked a question and I don’t know how the answer that I received is in relation with my question, can you please have a look:

continued fractions should be practiced by hand; the calculations are all quite easy, but writing them out with letters and subscripts is a mess. I put an example as answer, where the two original numbers really are coprime. If I did the continued fraction for $\frac{64}{48},$ the final convergent comes out in lowest terms, here $\frac{4}{3}$ — Will Jagy 9 hours ago

math.stackexchange.com/a/3707504?noredirect=1

Yuvraj

8:11 AM

today my small cousin asked me about the value $\pi$ ,he asked me about the unqueness in the value of $\pi$ that why we have use the value 3.14..... for the circumference and area?and wether we are unaware of the circumference of the circle without the discovery of $\pi$ i do not have the answer for this .

i feel $\pi$ is unique due to its irrationality that it is longest irrational number.

A FURIOUS MIND

What does "longest" irrational number mean to you?

Yuvraj

does this irrationality cause the uniqueness or is there are some more formulas where the multiply or addition of $\pi$ changes the whole mathematics?

@AFURIOUSMIND something whose end we can see

unlike the other numbers who ends at infinity

3.1459265359

as far i know it is the longest irrational number

A FURIOUS MIND

does infinity end?

np

The three dots in 3.14... mean "and so on" without end.

Just like 1/3 = 0.333...

or 1/9 = 0.111...

infinitely many 1s^

or 3s

Yuvraj

8:26 AM

@robjohn good morning sir

A FURIOUS MIND

with the number pi there are infinitely many non-repeating digits: 3.14...

with respect to your small cousin it usually recommended to start with the irrationality of the square root of 2

here @Yuvraj

Adam L

8:57 AM

longest irrational number?

Balarka Sen

10:25 AM

@Yuvraj The value of the circumference of a unit circle is $2\pi$; you can tell your cousin to take that as definition. I don't understand what "uniqueness" means - we defined it to be something unique, namely, length of a unit semicircle.

"Longest irrational number" sounds like hot garbage to me.

There is something nice to be asked: What is the distribution of occurrence of digits in the decimal expansion of $\pi$? It is believed that it's a uniform distribution, with each digit appearing with probability $1/10$, but this has not been proven.

Such numbers are called base $10$ normal numbers, and it is known that almost every real number is normal.

(In the sense that if you pick a real number "randomly", it will be normal with probability 1)

Thorgott

10:43 AM

what's your probability distribution on R

anything absolutely continuous wrt Lebesgue measure works, I guess

Balarka Sen

yes

Thorgott

how's the proof that almost all reals are normal

I've never looked at it

Balarka Sen

I dunno

Thorgott

1

Q: What's wrong with the following proof that almost all real numbers are normal?

Say $U \sim \text{U}[0,1]$. Then $\mathbb{P}[\,U\text{ is normal}\,] = 1$. This is evident from the fact that a Bernoulli process can be used to sample the decimal digits of $U$, leading to a normal number with probability 1. But the probability measure for a uniform random variable is just the L...

Balarka Sen

11:04 AM

yeah i was thinking it has to be something obvious because you just take $X_0, X_1, \cdots \sim \text{Ber}(1/2)$ and set $U = (0.X_0X_1X_2\cdots)_2$ to simulate a uniform distribution

wasn't sure so didnt say it

Yuvraj

11:57 AM

@BalarkaSen i mean, Why is this number equal to 3.14159....? Why is it not some other irrational number?

Thorgott

cause we can calculate it

and that's what the calculations give

Yuvraj

i find it hard to explain

and i did some research and after hours of search i got

Archimedes did: using polygons inside and outside the circle!

Thorgott

I mean, it's just a number

Yuvraj

@Thorgott sorry sir ,it has some proof

Thorgott

now if you were to ask why the ratio of the circumference to the diameter turns out to be a constant that's ubiquitous in many other seemingly unrelated parts of mathematics, that is a reasonable question with potentially deep answers

but the reason why that number is equal to 3.14159265358979323... is simply cause that's what it turns out to be

polygonal approximations are simply a way to approximate pi by rational numbers

Yuvraj

12:03 PM

i got one more question

Another question to ask is that if $\pi_1r^2$ is the area of the circle and $\pi_2r$ is the circumference why $\pi_1=\pi_2$? What geometries or metrics will result in 1≠2 ?

something which more complex

Thorgott

no, what you get is $\pi_1=\pi_2/2$ in that setup

if $r$ is the radius of a circle, its area is $\pi r^2$ and its circumference is $2\pi r$

Balarka Sen

@Yuvraj You can compute it using calculus

skullpatrol

here @Yuvraj is how you calculate the digits

numberworld.org/misc_runs/pi-10t/details.html

Yuvraj

@skullpatrol hi

skullpatrol

hi pal

12.1 Trillion Digits of Pi

Satisfied? @Yuvraj

Arjun Rana

12:22 PM

Yuvraj

@BalarkaSen do you have any?

Balarka Sen

Do I have what

Arjun Rana

Hi pals anyone can give a hint?

Astyx

@BalarkaSen any

Balarka Sen

lol

Arjun Rana

12:24 PM

🤷

skullpatrol

Y'all Lives Matter

Mike Miller

skull it's much more meaningful to show up on the street to demonstrations than to comment on them online given the opportunity. if it's something you care about then you gotta show up

skullpatrol

I'm not the one posting black stereotypes in a chatroom, pal.

Balarka Sen

are you nuts? it's a joke on a line in the David Chappelle show, not a black stereotype

robjohn

12:40 PM

@Yuvraj hey there

skullpatrol

Have you ever heard the American southern accent?

robjohn

@MikeMiller hats? hoodies? hints?

Balarka Sen

Yeah what about it?

skullpatrol

Define: y'all

Balarka Sen

i can't make sense of what you are saying

seems like nonsense to me

Thorgott

12:44 PM

Y'all (/jɔːl/ yawl[2]) is a contraction of you and all (sometimes combined as you-all).

Balarka Sen

i would recommend avoiding taking every cheap opportunity to post black lives matter comments on the chat and instead showing up at the protests, like mike is doing at the moment. it only hurts the movement's reputation

skullpatrol

Y'all

Y'all ( yawl) is a contraction of you and all (sometimes combined as you-all). It is usually used as a plural second-person pronoun, but the usage of y'all as an exclusively plural pronoun is a perennial subject of discussion. Y'all is the main second-person plural pronoun in Southern American English, with which it is best associated, though it also appears in other English varieties, including African-American English and South African Indian English. == Etymology == Y'all arose as a contraction of you-all. The term first appeared in the Southern United States in the early nineteenth century...

Alessandro Codenotti

@BalarkaSen Normal is stronger right? You want $n$ digits strings to have probability $10^{-n}$ for all $n$

Thorgott

2:21 PM

if $\pi_1\colon E_1\rightarrow B,\pi_2\colon E_2\rightarrow B$ are two vector bundles and $\Delta\colon B\rightarrow B\times B$ the diagonal map, then $\Delta^{\ast}(E_1\times E_2)=E_1\oplus E_2$, right?

Mast

Flags were raised. Are room owners and/or moderators available?

M.A.R.

@Mast Chat flags are shown to all users with more than 10k rep network-wide

Mast

@M.A.R. I know, that's what brought me here and hence the comment.

Balarka Sen

@Thorgott Yes, that's correct

Mast

If RO/mods are available we can leave again, otherwise I'll keep an eye out for the next hour or so.

M.A.R.

2:27 PM

Oh you meant flags were raised, not that you raised them

Mast

:-)

M.A.R.

@Mast Meh, it's the math chat. Can't live a day without having flags here

I'm taking a nap

Mast

lol

Goodnight @M.A.R.

Thorgott

nice, that's a pretty economic of obtaining the direct sum then

Alessandro Codenotti

Thorgott is already talking like an algebraic geometer

Thorgott

2:34 PM

this isn't even my final form

Knight

2:59 PM

@MikeMiller Mike you posted something which can be called objectionable and then you and @BalarkaSen together exercised majority power over @skullpatrol. It's quite weird that you got 4 stars in all, and Balaraka called skull as "nuts" and talked in a way that he shouldn't

Balarka Sen

??? There is nothing objectionable and Mike's comment and what I told skull.

Knight

Come on Balarka, it may be because you're not in America. We can make a very big fuss about what Mike posted here

I request you to please delete it as you're one of the room owner.

And as far as going on streets sic concerned,WE DO GO THERE

Balarka Sen

I am not a room owner. I don't see what fuss is there to make.

If you think something is objectionable make a coherent argument on why it is so

But better yet, stick to math and don't talk about politics here at all. Make another room for this

feynhat

If I am given an orientable manifold $M$, and a local orientation at $x$ $\mu_x \in H_n(M, M-x)$. I should be able to find a ball $B$ (in chart around $x$) and a generator $\mu_B$ for $H_n(M, M-B)$ such that the map induced by inclusion $H_n(M, M-B) \to H_n(M, M-x)$ takes $\mu_B$ to $\mu_x$, right?

I mean this is what orientability means.

Knight

Okay. Just imagine someone is having his knees on someone's neck, and the man is crying "I cannot breathe, leave me, leave" but the man with knees on says "You're tough guy ha?"

Balarka Sen

3:05 PM

You don't need orientability for this, @feynhat. Just take a chart ball around $x$

Knight

And eventually the man who couldn't breathe dies.

Can you imagine it pal?

Balarka Sen

We are all aware of the death of George Floyd and don't support police brutality. How does that have any relevance here?

This is what I mean when I say these conversations read to me as nonsense. It's just not coherent.

Alessandro Codenotti

Can't you take any chart ball and flip it around if you get the wrong generator?

Knight

Why to post something with a black man saying something? (He posted in a matter where it was not needed)

Balarka Sen

Huh? It's a crime to post a picture of a black man saying something?

That seems like racism to me!

Absolutely ridiculous statement

Knight

3:08 PM

Well okay

IF thats the case then see what skull said

3 hours ago, by skullpatrol

Y'all Lives Matter

and how Mike replied :

3 hours ago, by Mike Miller

skull it's much more meaningful to show up on the street to demonstrations than to comment on them online given the opportunity. if it's something you care about then you gotta show up

Balarka Sen

He's right, the internet anti-racism discussions lead nowhere. You should be out on the streets, not on the internet, when protesting these things.

Knight

Well Mike posted first

Balarka Sen

Lest these protests get undervalued

Knight

What?

feynhat

My question is if I am given a generator, ofr $H_n(M, M-x)$ (say a simplex with $x$ in its interior). Is there some way to come up with a a generator for $H_n(M, M-B)$. Can I stretch this simplex so that its boundary lies in $M-B$?

Knight

3:11 PM

And don't call anyone "nuts" it is considered rude and as a misbehavior.

Balarka Sen

@feynhat Take $B$ to be a chart ball containing $x$. Then the inclusion induced map $H_n(M, M - B) \to H_n(M, M - x)$ is an isomorphism by excision. Just look at preimage of the generator you chose in $H_n(M, M - x)$.

feynhat

I want to know what this pre-image looks like.

Can I represent it by a simplex?

Balarka Sen

Yeah you just stretch the simplex. Flow outward by a radial vector field centered at $x$

Then you have stretched the simplex so that $B$ is contained in it's interior

feynhat

oh man

Balarka Sen

@Knight It's not rude, but it is meant to show contempt. I really do believe it's preposterous to think that meme had anything to do with black stereotype.

Mike Miller

3:17 PM

@Knight You live in Chicago, right? There is a march scheduled at union park at 11; I encourage you to attend if you can make it there

Knight

@MikeMiller Will you meet me up there?

Coz I have to make plans for tomorrow if I have to go there, it’s far away

Mike Miller

Stay safe; Lori hired 500 private military contractors for the weekend

Knight

Okay, I’ll send you in few minutes on your email

Mike Miller

👍 I'm out for now

DMGregory

3:41 PM

Hi @MikeMiller - your message got flagged for containing personal information. To preserve your privacy, mods have removed the message from the chat log, but please be aware that this chat is publicly visible and searchable, and be careful sharing things like emails here.

Thorgott

3:55 PM

can the product of non-trivial vector bundles be trivial?

answer is no when they're over the same base

Balarka Sen

product? you mean whitney sum?

then yes, take Mobius strip over circle

Thorgott

no, I mean product

Balarka Sen

square of Mobius strip is trivial rank 2

What do you mean by product of vector bundles?

Thorgott

if $\pi_1\colon E_1\rightarrow X_1,\pi_2\colon E_2\rightarrow X_2$ are vector budnles, the product is $\pi_1\times\pi_2\colon E_1\times E_2\rightarrow X_1\times X_2$

Balarka Sen

But take $X_1 = X_2 = S^5$, $E_1 = E_2 = TS^5$? $E_1 \times E_2$ is then $TS^5 \times TS^5 = T(S^5 \times S^5)$ which is trivial because it's parallelizable

Why is the answer no when they're over the same base?

Thorgott

4:05 PM

why is $S^5\times S^5$ parallelizable

Balarka Sen

Product of any number of spheres where one of them is odd dimensional is parallelizable. The point is $TS^{2k+1}$ has a rank 1 trivial subbundle, so you can manipulate around with that, using $T(M \times N) \cong \pi_M^* TM \oplus \pi_N^* TN$, to get it to be trivial, using the fact that $TS^n$ is 1-stably trivial (direct sum-ing with a trivial bundle makes it trivial of rank n+1)

So I am confused. How did you prove the answer is no over the same base?

Thorgott

urgh, I made a mistaken then, it seems

that's what I get for only sketching

Semiclassical

4:21 PM

so, more elementary measure stuff. was trying to prove that countable additivity (of pairwise disjoint sequences of sets) implies countable subadditivity (of a generic set sequence)

my immediate thought was to show that worked in the case of two sets, then use induction to prove in the case of finitely many sets

and then take the limit to get countably many

now, for finite sequences, i'm pretty confident that's enough

but i'm less confident re: countably many

Thorgott

how do you prove it in the case of two sets?

Semiclassical

$A\cup B = A\cup (B\setminus A)$

Balarka Sen

Using the same trick you would get P(cup A_i) = sum P(A_i \ cup_{j \neq i} A_j) \leq sum P(A_i) by using a termwise inequality (monotonicity of P)

Semiclassical

right

Balarka Sen

well my expression isn't exactly correct but you get my point

Semiclassical

4:30 PM

yeah

so you get $P(\cup_{i=1}^n A_i)\leq \sum_{i=1}^n P(A_i)$

i.e. finite subadditivity

the question, i think, is whether it's legit to take $n\to\infty$ at that point to get countable subadditivity. what makes me hesitant is that "finite additivity" does not imply "countable additivity"

see for instance en.wikipedia.org/wiki/…

Thorgott

well, to reasonably take a limit on the LHS, you need some sort of continuity of the measure

there are such types of continuity, called continuity from above and continuity from below

for finitely additive maps, countable additivity, countable subadditivity and continuity from below are equivalent

continuity from below is that if a sequence of sets $A_n$ is increasing, then $\lim P(A_n)=P(\bigcup A_n)$

Semiclassical

increasing as in $n<m\implies A_n\subset A_m$?

Thorgott

right

Semiclassical

(should probably be $\subseteq$, since it can be a proper subset. ugh, notation)

Thorgott

so if you can show continuity from below, you can take the limit in your inequality to get the countable case

conversely, you can extend the trick you already have to a countable sequence of sets and directly obtain subaddivity and then derive continuity from below

then you will have in fact established their equivalence for finitely additive maps

Semiclassical

4:41 PM

the text i'm using has us prove that $A\subseteq B\implies P(B)=P(B\setminus A)+P(A)\geq P(A)$ prior to the "countable additivity" part

and the codomain of $P$ in this case is $[0,1]$

so if you've got an increasing sequence of $A_n$, you also have an increasing sequence of $P(A_n)$ which is bounded above by $1$

is that enough to establish that $\lim_{n\to\infty} P(A_n)$ exists?

Thorgott

sure, a bounded, monotonically increasing sequence converges to its supremum

Semiclassical

huzzah, monotone convergence

one proof attempt I'd made went like this. since $P$ is subadditive, we have $$P(\cup_{k=1}^\infty A_k)=P(A_1\cup[\cup_{k=2}^\infty A_k])\leq P(A_1)+P(\cup_{k=2}^\infty A_k)$$

and more generally $P(\cup_{k=1}^\infty A_k)\leq \sum_{k=1}^n P(A_n)+P(\cup_{k=n+1}^\infty A_k)$

with the task seeming to be: does $P(\cup_{k=n+1}^\infty A_k)\to 0$ as $n\to\infty$

in that case, though, we're taking fewer and fewer sets, and therefore the sequence of sets is decreasing rather than increasing

so now would this be an argument via continuity from above?

my main suspicion being that, if this route worked, then it should work too for the case of $=$ rather than just $\leq $

Balarka Sen

5:07 PM

Oh you're doing something else, nevermind

Wasnt following the discussion

I don't see why this is turning out to be a long discussion though. Let $X_n = \cup_{k = 1}^n A_k$. Then $\{X_n\}$ is an increasing sequence of sets whose union is $\cup_{k = 1}^\infty A_k$.

Thorgott

the argument does and would work

continuity from above is slightly less well-behaved than continuity from below

you need a decreasing sequence of sets whose first element has finite measure

Balarka Sen

Write $Y_n = X_n \setminus X_{n-1}$. Now $\{Y_n\}$ is a disjoint sequence of sets whose union is all of $\cup_{k = 1}^\infty A_k$ once again

Thorgott

of course, for probability measures that's always a given

in the general case, you can have the sequence $[n,\infty)$, which decreases to the empty set, but each element has infinite measure

of course, this reasoning works for = rather than just <= too

but you're not proving anything new in that case

Semiclassical

hmm

Thorgott

cause continuity from above follows from countable additivity, but not necessarily from finite additivity

in fact, continuity from above will also be equivalent to countable additivity, at least for probability measures

I'm not sure whether that holds for general measures tbh, I'd assume it fails

Semiclassical

5:14 PM

so: both continuity from above and below work here, but the former has to be made more carefully?

Thorgott

nah, both work just fine here (and so does Balarka's argument)

Balarka Sen

They are both true for probability measures

Thorgott

continuity from above behaves worse only in more general settings

Semiclassical

hmm

well, by "has to be made more carefully" I include having to invoke properties specific to probability measures

Balarka Sen

Yeah you need that the whole space has finite measure

Thorgott

5:17 PM

@Balarka do you know if continuity from above for a content implies countable additivity in general

Balarka Sen

Otherwise starting to decrease from an infinite measure set will be complicated

Semiclassical

but I guess the argument from below does invoke that $P$ is bounded above by 1

Alessandro Codenotti

In the general setting you need to assume that there is a set with finite measure in your sequence

Semiclassical

so either way you do use the fact that $P$ assigns finite measure

Balarka Sen

@Thorgott Not off the top of my head, nope.

Alessandro Codenotti

5:19 PM

What is a content? @Thorgott

Thorgott

finitely additive map on a $\sigma$-ring

and such that empty set has $0$ content

codomain $[0,\infty]$

Alessandro Codenotti

that was probably wrong

math.stackexchange.com/questions/448605/… I don't know about $\sigma$-rings, but it seems true for finitely additive measures

Thorgott

that's continuity from below

this equivalence holds in full generality too via the same proof

feynhat

5:55 PM

@BalarkaSen Does it work if I work in charts and stretch the simplex so that it contains the ball? Lets say, I have a simplex containing x in its interior (suppose this simplex lies inside a chart around p). Suppose B is the chart ball on which the local consistency holds. Look at the image of our simplex in the chart at stretch it so that it contains the image of B. Then, bring it back to the manifold.

feynhat

6:21 PM

No way he slept this early.

Thorgott

yeah, I don't believe that lol

anakhro

6:35 PM

Hi all.

Balarka Sen

6:53 PM

@feynhat Yeah for sure, that's a simpler way to say flowing radially outward by some field centered at x

lol

sorry for writing "multiplication by c" in a complicated way

Hi Balarka.

Hey

What's up?

Nothing really

Things been going okay?

Been keeping busy, I hope.

Balarka Sen

6:55 PM

more or less

wbu

anakhro

Learned anything interesting lately?

I have been busy, but not with what I want to be busy with. :P

Just marking stuff.

Balarka Sen

yeah a bunch

gotcha

anakhro

Any bite-sized morsel you can share?

Alessandro Codenotti

Hi Anakhro

anakhro

Hey Alessandro! Perhaps you have some mathematical morsel to share.

Thorgott

7:06 PM

morsel theory

Alessandro Codenotti

Depends on how interested you are in dimension theory and/or descriptive set theory lol

anakhro

I am open to absolutely anything.

I am not picky when it comes to mathematics.

Alessandro Codenotti

I'm trying to figure out why $\mathrm{Homeo}(X)$ is $G_\delta$ in $C(X,X)$ now for compact metrizable $X$ now

Balarka Sen

@anakhro Let $f : \Bbb R \to \Bbb R$ be a smooth function such that $f(x) = f(-x)$. Prove there is a smooth function $g : \Bbb R \to \Bbb R$ such that $f(x) = g(x^2)$

Pig

hi all

Thorgott

7:10 PM

ok, who wnats to hear about colax-slice 2-categories

Balarka Sen

Hi @Pig

Pig

just show continuity of derivatives at 0 of the natural candidate @BalarkaSen?

what's colax slice lol

Balarka Sen

yeah

well, and you have to extend beyond 0

Pig

but there shouldn't be any issue at all outside of 0 for the natural candidate

at 0 you may need to worry about left/right

Balarka Sen

what is the natural candidate

yeah left/right is what i mean

anakhro

7:12 PM

@AlessandroCodenotti after doing that, was there something you wanted to use it for in particular?

Pig

oh hm i was thinking about $g(x) = f(sgn(x) \sqrt{|x|})$ or something

maybe there is some subtlety here

Alessandro Codenotti

I just want to say that Homeo(X) is a Polish group

Clarinetist

Hello all. I haven't been here in months. Just glad my masters degree is finally done with everything that's happening.

Thorgott

it's like the slice 2-category, but you don't require the natural trafos that constitute your 1-morphisms to be invertible and it's lax/colax depending on the direction in which it goes

Pig

i don't even know what a slice 2-category is lol

Thorgott

7:15 PM

like a slice-category, but the defining triangle for 1-morphisms only commutes up to a 2-morphism

anakhro

@AlessandroCodenotti ah explains the descriptive set theory premise.

Pig

i have to take it back, i don't know what a slice category and what a 2-category is =_=

let me take a look lol

alternatively, any typical example i can keep in mind?

Balarka Sen

@Pig it seems doubtful that that will in fact work at 0

one of them will

but I could be wrong

Alessandro Codenotti

@anakhro Fair enough

Do you know what the Lebesgue covering dimension is? @Anakhro

anakhro

Yup!

Alessandro Codenotti

7:18 PM

Nice

Thorgott

take a category $\mathbf{C}$ and consider the new category $\mathbf{C}^{\prime}$ whose objects are tuples $(X_i)_{i\in I}$ where each $X_i$ is an object in $\mathbf{C}$ and $I$ is some set and the morphisms from $(X_i)_{i\in I}$ to $(Y_j)_{j\in J}$ consist of a function $f\colon I\rightarrow J$ and a morphism $X_i\rightarrow Y_{f(i)}$ in $\mathbf{C}$ for each $i\in I$

anakhro

Is this related, @AlessandroCodenotti?

Leaky Nun

@Thorgott so the functor category

Thorgott

this is a certain subcategory of a certain colax-slice 2-category

no, it's not

Alessandro Codenotti

Recently I learned that every space separable metric space $X$ with $\dim X\leq n$ embeds into $\Bbb R^{2n+1}$ (in fact it embeds into $N^{2n+1}_n$, the Nöbeling subspace of $\Bbb R^{2n+1}$ of points with at most $n$ rational coordinates)

Thorgott

7:20 PM

it's almost (a subcategory of) the slice category $\mathbf{Cat}/\mathbf{C}$ except the the triangles only commute up to a natural transformation

Alessandro Codenotti

@anakhro No, it's just a cool fact that I learned recently

Well the proof first deals with the case in which $X$ is compact and it does use that $C(X,\Bbb R)$ is Polish (in particular Baire)

anakhro

@AlessandroCodenotti there was some result in measure theory that uses it prolifically. I don't recall what it is now.

Alessandro Codenotti

Hm, I'm not aware of any such result, let me know if you remember what that is, sounds interesting!

anakhro

The Lebesgue covering dimension, that is.

The Polish space stuff I have not heard much about.

I had a friend who was into Polish space stuff but at the time I was not very good with analysis so I didn't quite follow most of what he liked about it.

Pig

@BalarkaSen ah you are right, hm

user462942

7:29 PM

Hello

Clarinetist

7:54 PM

Hi @Semiclassical, long time no chat. How's it going?

Ted Shifrin

8:26 PM

Well, there's that erstwhile statistician, @Clarinet.

Clarinetist

Hi @Ted, long time. Just finished my M.S. this semester, and as of last November, I now work at the U of Minnesota

as a data analyst

Ted Shifrin

Very cool.

Clarinetist

I'm self-learning measure theory before the fall. Haven't committed to the PhD program yet; they're willing to let me try it part-time if I do well on the qualifying exams.

Ted Shifrin

Oh oh ... sounds scary :)

Clarinetist

Oh, and I'm now a faculty member at a community college as well. I teach one 3-credit class, now for a year - an intro data science class

Ted Shifrin

8:33 PM

Oh oh ... Clarinet is growing up :D

Clarinetist

Life's been pretty good to me, even despite what's happening out here in Minneapolis and with COVID-19. My data science students - keep in mind, this is the first semester of data science they've taken - recently competed in the "advanced" level against 4-year institutions. We were the only 2-year school in the contest, and we made 3rd place. I'm so proud of them.

Ted Shifrin

That's quite awesome.

Clarinetist

What's been happening here lately? I've just been out of the loop

Ted Shifrin

Some of the same old characters. Some of the fixtures from a few years ago have totally disappeared. (Except for Balarka and Ted.)

Clarinetist

Are you still part-time teaching in CA?

Ted Shifrin

8:38 PM

Nah, after the second year I decided it wasn't rewarding enough (the kids weren't putting in much time outside of our 1 hr 45 min classes — other than doing some of the on-line homework, which I wasn't that thrilled with).

So now I'm a full-time bum. I am only recently venturing out at all to do grocery shopping (including farmers markets). It's going to stay scary for a few years, I think. So proud of our government.

Clarinetist

I think now, more than ever, is the perfect time to not have any commitments

Ted Shifrin

So, congrats on the competition. Are you doing any good learning (I suppose I have to consider your learning Lebesgue analysis good)?

Clarinetist

Yeah, I'm learning measure theory through Yeh's book before I take the slightly-less-detailed version that PhD stats people are required to take in the fall

Ted Shifrin

Never heard of that author.

Clarinetist

It's the only measure theory text that I've found with a solutions manual. amazon.com/Real-Analysis-Theory-Measure-Integration/dp/…

Ted Shifrin

8:44 PM

People keep emailing me (often rudely) asking for solutions manuals for my various books. In two cases, they are available only to instructors through the publisher. But in the other two cases, they don't exist.

Clarinetist

I do think that solutions manuals are nice when one doesn't have an instructor to guide you through material, but even then, they need to be used sparingly

Ted Shifrin

Yes, based on what I see on MSE, students generally abuse answers.

That said, I did just write out a very detailed answer for a differential geometry/Lie groups calculation, but the OP put in extreme good effort, and I wanted to correct it a little and add some insight.

Clarinetist

I took an algorithms class taught by a colleague at the community college this last spring. My colleague mentioned throughout the class to the students that he found a ton of cheating and/or copying-and-pasting of answers from Chegg. Quite unfortunate, since algorithms is probably one of the most important topics for a CS curriculum.

Ted Shifrin

It was actually interesting. I did the computation two ways and was off by a sign. It took me a few minutes to realize what caused that.

Yeah, I don't have any knowledge of Chegg, but I hear it's a big cheat source.

Clarinetist

He pays $5 a month to catch people cheating on there. I might have to do that myself... but honestly, it's usually pretty easy for me to catch cheating, since I make original homework assignments and quizzes every semester (yes, it's exhausting), and I make my students do exams orally

Ted Shifrin

8:50 PM

Oral exams? Wow.

For me, making students write mathematics and grading their actual written work was essential.

Clarinetist

Yes, because I did not want to at all deal with the possible complications of cheating this semester with COVID happening and my students were online already, I just moved all proctored quizzes and exams to all being oral

This class is very different from a typical math class

Ted Shifrin

Yeah, I'm sure.

Still, there's always an issue with fairness and having exams be "comparable" or "the same."

Tanner Swett

Hi everyone.

So I've got a question, but it's a rather complex question and I don't know how to write it up. :D

Ted Shifrin

Well, then we can just answer the non-question, @Tanner.

Here's your answer .... * * *

Tanner Swett

The question is, essentially: I've come up with a certain concept, and I'm sure it's a concept that other people have studied before; what is this concept actually called and where can I read about it?

Clarinetist

8:53 PM

Right, so what I do there is for quizzes, because (as you might imagine from the results that I've gotten from the contest) I throw so much material at them, I give them a question list beforehand and select questions randomly depending on a number that the students give me before they take the quiz. No two students are allowed to have the same number. For the final exam, though, it's similar but they are allowed to use notes but will not see the questions beforehand.

Ted Shifrin

Well, that may be impossible to answer or very easy to answer. Who knows.

That's a lot of hard work for you, @Clarinet. But kudos to you.

@Tanner: So what is the concept, or what does it concern?

Clarinetist

The programming skills I've developed over my jobs have been really useful for things like this. I have some code that I've written in R which will automatically generate a weekly calendar given a start date and end date in LaTeX. I can't imagine going straight into teaching from grad school without being able to do things this way.

Ted Shifrin

I'm still an old fuddy-duddy type of teacher.

Tanner Swett

Well, the concept is actually identical to the concept of a variety of algebras (en.wikipedia.org/wiki/Variety_(universal_algebra)), except I'm only interested in the initial algebra of the variety.

Ted Shifrin

OK, @Tanner, I know absolutely nothing about such things, but there are some folks who hang out here who might ...

Tanner Swett

8:58 PM

I'm especially interested in, giving two varieties-of-algebras V and W, proving that the initial algebra of V is essentially isomorphic to the initial algebra of W.

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$Mathematics$ Mathematics