Mathematics - 2023-02-02

ペガサス Seiya

12:30 AM

@TedShifrin You wouldn't get it

Ted Shifrin

Not the first time.

ペガサス Seiya

@TedShifrin Only gourmet cooks would understand that

Xander Henderson

Ugh... I don't feel well. Three years ago, I would have just canceled class and told everyone to do the reading. But now I can teach from home when I don't feel well. :(

But I just finished my class, and only made one truly dumb mistake.

(Did you all know that if $\frac{1}{2} > |2| |x-3|$, then $1 > |x-3|$? TRUE FACT!

Ted Shifrin

Sigh. I hope it’s not you-know-what. Two of my friends (one a doctor) have just come down with it after escaping 3 yrs.

Well, $4=1$, so sure!

Xander Henderson

@TedShifrin Home test says no.

Ted Shifrin

12:45 AM

Try several days.

Xander Henderson

It feels very much like allergies, and I took a Claritin at noon, which seems to be helping some.

Ted Shifrin

I have had allergies and runny nose for many weeks.

Xander Henderson

But every cold I've ever had starts with a day of "Man, my allergies are really bad today!", followed by a terrible sore throat the next day.

Ted Shifrin

I take Claritin daily.

Xander Henderson

So we'll see how tomorrow goes.

@TedShifrin I do that for about 6 weeks in the late spring / early summer.

But try to avoid it most of the rest of the year.

Ted Shifrin

12:46 AM

I’ll cross 5 fingers for you.

Xander Henderson

Thanks.

In any event, I'm done for the day, and going away now.

Laters.

Ted Shifrin

Night!

ペガサス Seiya

You wear a mask in Japan regardless, especially during allergy season

D.C. the III

1:03 AM

I've been summoned from the depths of the Mariana Trench, to whom am I to answer?

ペガサス Seiya

Lupin the III

D.C. the III

@ZaWarudo I could perhaps help you along to give advice to move in the right direction, but what Ted and Leslie have said sum up the most important parts. You're going to need a damn near obsessed work ethic and even then the most important facet is feedback.

Answer manuals are not of much help. They provide the end goal, but math is everything in between the start and end goal....that is "math".

Ted Shifrin

1:19 AM

Lupin or Lapin? Lupin was a French detective. Lapin is a French rabbit.

robjohn

2:28 AM

Lupin is also a flower

Ted Shifrin

Right, it is. I was in French mode.

Purplish flower, quite beautiful.

robjohn

Lupin was also a werewolf in Harry Potter.

Koro

How do we show that O(n) is homeomorphic to $SO(n)\times Z_2$?

I can show this for n= odd.

I consider the map $f: O(n)\to SO(n)\times Z_2: f(A)=(|A| A, |A|)$, which is continuous because its projection maps are continuous (the projection onto the first coordinate is continuous, continuity onto the second is continuous as the following shows)

$\pi_2\circ f:O(n)\to Z_2$ is to be shown continuous. Z_2={0,1} with discrete topology and $(\pi_2\circ f)^{-1}(X)$ is open in O(n), where X is either emptyset, {0,1} or {0}.

$(\pi_2\circ f)^{-1}({1})= SO(n)=det^{-1}((1/2,\infty))\cap O(n)$.

which is open in O(n), hence $\pi_2\circ f$ is continuous.

O(n) is compact, $SO(n)\times Z_2$ is Hausdorff, f is a bijection, hence a homeomorphism. This proves the desired homeomorphism.

But how to prove this for n= even? In this case |A|A is not in SO(n) anymore.

Ted Shifrin

2:44 AM

@Koro homeo, not group iso. Be cleverer with your map. Clearly $O(nj$ has two connected components, so done.

Koro

ok so for n=even, I define f(A)=(|A| A', |A'|), where A'= matrix obtained by multiplying the first row of A by -1. So det(|A|A')=$ -|A|^n |A|=-|A|^{n+1}$

Ted Shifrin

Even easier, just do cases so it’s clearer.

Remember, the space is disconnected, so continuity is easy.

Koro

you mean cases for n=odd, n=even or using connected components and then pasting lemma for continuity? This f defined for n=odd is also a homoemorphism I think.

Ted Shifrin

No pasting!

Koro

because everything is disjoint?

Ted Shifrin

2:55 AM

Actually, you can do a map independent of parity.

Right. Two components.

Koro

@TedShifrin ohh nice.

Koro

4:45 AM

Am I right in saying that Q with distance metric is a meagre set because Q=U_{q\in Q} {q} and each {q} is closed with empty interior?

leslie townes

yes

same for any countable subset of R (or R^n) with the distance metric, or indeed any countable subset of any topological space where points are both (1) closed and (2) not open

Ted Shifrin

5:13 AM

If there are isolated points, just exclude them from eligibility.

copper.hat

6:21 AM

@SineoftheTime It was a joke.

Koro

hi @copper.hat!!

copper.hat

@Koro Hi Koro! Hope you are well!

Koro

Thanks. I wish you the same.

@leslietownes yes, I just wanted to confirm just in case.

For learning open mapping theorem, closed graph theorem and their applications does one need to know 'weak and strong convergences'?

Momo (food)

Momo are bite-size dumplings made with a spoonful of stuffing wrapped in dough with origins from Tibet. Momo are usually steamed, though they are sometimes fried or steam-fried. Meat or vegetables fillings becomes succulent as it produces an intensively flavored broth sealed inside the wrappers. Variants first developed in Tibet, after it became popular among Asians. Eating dumplings on the first day of the new year was a widely spread custom in northern China. Written records show that dumplings became popular during the Southern and Northern dynasties (420–589 AD), the earliest unearthed real...

This tastes good sometimes :).

copper.hat

6:41 AM

i have had many momos!

i had a lovely dosa a few nights ago.

Koro

ohh nice :)

leslie townes

@Koro if you were ordering material in a course, i can't think of any formal need to put those forms of convergence ahead of those theorems (although i also can't think of any strong reason not to).

Koro

I see. Thanks.

So one can straightway go from Uniform boundedness principal to open mapping theorem.

copper.hat

my neck, my back, ...

leslie townes

yeah, you can probably go from any one of the big theorems to any other one of the big theorems.

its like the ABCs, man, there's no inherent reason why B is after A except someone decided to do it that way, man

Koro

6:58 AM

:-)

Sine of the Time

@copper.hat what are you referring to?

one potato two potato

manifold with corners... manifold with corners... manifold with corners...

copper.hat

@SineoftheTime i made a silly remark about identity theft

Sine of the Time

@copper.hat you're right, totally forgot about it. I'm getting old

copper.hat

we are getting old at the same rate...

one potato two potato

7:14 AM

@copper.hat really?

copper.hat

well, it's all relative, i was assuming we were at the same altitude...

gn folks

Cotton Headed Ninnymuggins

7:44 AM

Is there any reason behind the naming of trig functions? cotangent is 1/(tangent) but cosine isn't 1/(sine). I'm filing a complaint.

Cosinux

8:02 AM

Hi. According to my literature a "perfectly" reconstructed sampled signal can be represented as
$$ y{\left ( t \right )}=\sum_{k=-\infty }^{\infty }x\left ( kT \right ) sinc{\left ( \frac{\pi t}{T}-k\pi \right )} $$
Since the sinc factor is the only one that depends on t, I figured that the Fourier transform must be:
$$ Y{\left ( \omega \right )}=\sum_{k=-\infty }^{\infty }x\left ( kT \right )\pi rect{\left ( \frac{1}{2}\left ( \frac{\pi t}{T}-k\pi \right ) \right )} $$
But I can't see how these pi/2 spaced rectangles can reconstruct the signal in the frequency domain. What am I missing?

PNDas

8:42 AM

I'm reading Evans and in the proof of weak maximum principles of elliptic PDE, they said that since $A$ is symmetric and positive definite so there exists an orthogonal matrix such that $OAO^t=$ a diagonal matrix with positive diagonal elements. And then they defined $y=x_0+O(x-x_0)$ and used change of variables to write partial derivatives w.r.t x in terms of partial derivative w.r.t y.

I want to see it geometrically what is happening.

Can you please help me?

ペガサス Seiya

10:47 AM

Both cats and hamsters have whiskers

Some either all cats are hamsters or all hamsters are cats

shintuku

11:17 AM

apparently, $\sum_{n=1}^\infty \frac{1}{|x|^n} = \frac{1}{|x|-1}$, for $\frac{1}{|x|} < 1$ but I'm getting $\frac{|x|}{|x|-1}$

any tips?

Peter

11:42 AM

@shintuku The hint is that we have $$\sum_{n=0}^{\infty} q^n=\frac{1}{1-q}$$ for $|q|<1$

shintuku

oh the index! then we get our desired result with $\frac{1}{|x|-1} + 1$

thanks!!

PM 2Ring

12:02 PM

@CottonHeadedNinnymuggins Think of the fundamental right triangle definitions. The cosine is the sine of the complementary angle. I.e., $\cos(\theta)=\sin(\frac\pi2-\theta)$. The cotangent is the tangent of the complementary angle. Similarly with secant and cosecant.

ペガサス Seiya

12:31 PM

I think secant should've been 1/sine and cosecant would then be 1/cosine

Makes much more sense

Gwyn

1:03 PM

Consider a function $g:\mathbb{R}\to\mathbb{R}$ continuous at $x_0$. Can I say that $\int_{x_0}^x g(t)dt=f(x_0)|x-x_0|+o(|x-x_0|)$ as $x\to x_0$ because of the mean value theorem for integrals? I am not sure because it requires the continuity in all the interval of endpoints $x_0$ and $x$, while in my hypothesis I only have continuity at $x_0$.

Sorry, I wrote $f(x_0)|x-x_0|$ once but I meant $g(x_0)|x-x_0|$.

user 726941

1:32 PM

@AkivaWeinberger are you subscribed to Prof Hamkins' substack?

ILikeMathematics

Does anyone have a recommendation of a real analysis book put up by the author on the internet?

Akiva Weinberger

1:52 PM

@user726941 I am not

Costs money, I gather

user 726941

Yup.

Akiva Weinberger

@ILikeMathematics I think Terrance Tao's An Epsilon of Room is online? Not sure though

(also I think it's more advanced topics than an introductory sequence, though I could be misremembering)

ILikeMathematics

2:11 PM

@AkivaWeinberger Thank you. That seems to be mostly about his articles on the blog

Mad

2:24 PM

Can you guys construct me a matrix which has charahcteristic polynomial of x^5 + 1

Preferablly it is orthogonal

Ajay

2:41 PM

There's no way that $9x^6 + 432x^4 - 2304x^2 - 49152x - 28672$ can be solved using algebra-precalc techniques.

I've been attacking this problem for 2 hours and at this point i'm just trying to guess divisors.

Sine of the Time

@Ajay what do you have to solve? is the expression equal to $0$?

Franklin

Can anyone please explain the proof of this :Find the locus of the point of intersection of three mutually perpendicular tangent planes to the paraboloid $ax^2+by^2=2cz.$ in this [post][math.stackexchange.com/questions/4631003/…

Actually I had a pic of the proof associated with it...so I couldn't post the picture here ...

Mad

My dudes

Can anyone give me a matrix with that polynomial ?

should be orthogonal

Other than the obvious method of finding the solutions, how to prove that the value of the solutions (the norm) is equal to one? using linear algebra shinanegans

i am thinking finding an orthogonal matrix with that polynomial then its good

Mats Granvik

@Mad It is possible to list the first few matrices in Mathematica of which one or more has that characteristic polynomial and is orthogonal, but it takes a lot of computation time.

Mad

what should i write in mathematica?

Mats Granvik

2:52 PM

Wait I while and I will give it a try. I will try to answer in a few minutes.

Franklin

Anyone please help me with this proof explanation 😅 , I tried it for 2hrs...and I am 😫 😩 😫

Koro

$\begin{bmatrix}0&0&0&0&-1\\
1&0&0&0&0 \\
0&1&0&0&0\\
0&0&1&0&0\\
0&0&0&1&0\end{bmatrix}$

@Mad

Mad

You the man koro

Franklin

Anyone helping me with this math.stackexchange.com/questions/4631003/… pleaaaaaassssse 😫 😩 😫😫 😩 😫😫 😩 😫

geocalc33

3:17 PM

what is the difference between complex vs. real zonotopes aside from using complex vectors vs real vectors?

complex zonoids are approximated by complex zonotopes

are complex zonoids studied as complex manifolds?

and is there a complex analogue of this?: As n->infty, the polar zonohedron of order n approaches a solid of revolution created by rotation of a sine curve

Franklin

Anyone interested to help me with this math.stackexchange.com/questions/4631003/…

Please help me...

Thorgott

3:38 PM

consider helping us by not spamming instead

Franklin

What is the meaning of spamming? Sorry, I am not a native

geocalc33

it's sending the same message indiscriminately to (large numbers of recipients) on the internet

Franklin

I am feeling helpless man!😂😂😂😂😂😂😂😂😂 maybe you can help me please 🥺

Hey! Truly unfair thorott .... you could have helped me instead of using facy words. Ughhh I am tired...I will again chat tomorrow..besides if you can me pleeeeeeeeese do....and yeees this is my last message for this day as my fingers are tired from typing....

PNDas

4:05 PM

The proof of weak maximum principle ($c=0$) shows that If $Lu<0$ in $U$ then $u$ can't attain maximum in interior, right?

mathcat

Hey guys, quick probability question, my calculations seem fine, but something just isn't adding up,

If I have a prism with the base of an equilateral triangle, the probability of rolling the base is 1/(sqrt(3)a²+12ab), right? And rolling it again in that order would be ^ squared? (a is the length of the base)

Koro

@Franklin: Hi, I looked at your post. Yes, $l_i, m_i, n_i$ are direction cosines.

@Franklin: Do you know matrices? If yes, then there is a very short answer to your question(s).

mathcat

4:21 PM

@mathcat The sub problem was to solve for a if the probability of rolling the base <= the probability of rolling the face twice.

Franklin

@Koro Yes!!! I know matrices only the basics...not a pro in that...but I can still give it a try to understand it in that approach! Finally, someone responds!!!!! This means a lot! Thank you!

mathcat

4:41 PM

@mathcat Anyone? Basic algebra isn’t helping me.

PNDas

Does the uniform ellipticity condition imply every coefficient is positive?

It's not true as laplacian satisfies this condition but $a_{ij}=\delta_{ij}$. But if I take $\xi=(1,1,0,\ldots)$ then $a_{1,2}(x)\geq \theta>0$ so it must be positive.

What is happening here?

Got it.

Koro

5:13 PM

@Franklin consider a 3 by 3 matrix B, whose ith row is ($l_i,m_i, n_i$).

Then note that $B^T B=I$.

So $BB^T=I$ (because $B^T$ is the inverse of B).

Look at the entries of $BB^T$ and compare them with the entires of I. This will answer your question.

robjohn

Because the rows (or columns) are orthogonal.

If they are unit length, then they are inverses

Koro

in general one has the following result: For n by n square matrices A and B, AB=I implies and is implied by BA=I.

The Substitute

5:37 PM

Good day. I heard ChatGPT has passed portions of the MCAT and bar exams. Can the AI pass a math qual?

Ted Shifrin

If the qual is just regurgitation of memorized stuff, sure.

user193319

$\mathbf{p}(t) = [p_1(t),...,p_n(t)]^{T}$ and $\mathbf{q}(t) = [q_1(t),...,q_n(t)]^{T}$, how exactly is $\frac{\partial}{\partial \mathbf{p}}(\mathbf{p})$ defined?

Whoops...meant to write $\frac{\partial}{\partial \mathbf{q}}(\mathbf{p})$

I know $\frac{\partial}{\partial \mathbf{q}}(\mathbf{p})$ should be a matrix...would it be the matrix whose $(i,j)$-th entry is $\frac{\partial}{\partial q_j(t)} (p_i(t))$? But how does $\frac{\partial}{\partial q_j(t)}$ make sense?

I am trying to follow along with this: math.stackexchange.com/questions/2197249/…

Koro

5:53 PM

@leslietownes @Thorgott: I thought about characterising the sets in R^n which are unit ball of some norm. But I don't see how to do it if the norm is not coming from an inner product.

Can you please direct me to a reference which discusses this?

I didn't find this on mse.

Thorgott

I don't know a reference (leslie might), but I can tell you the answer: they are precisely the convex, compact, symmetric sets with non-empty interior

(if you drop the non-empty interior condition, you get the precisely the empty set and the unit balls of some norm on some linear subspace of R^n)

Koro

thanks :-).

But how do I prove this? What are the steps?

Leslie or you said that this is easy to see.

leslie townes

i definitely wouldn't say it is easy to see. but it is "low tech" - you don't need any big tools to prove it, or really any tools other than the definitions. and knowing the answer certainly makes it simpler than it would be otherwise.

a good start would be to prove that any unit ball has to have those properties.

then at least you know the answer makes sense as an answer.

Thorgott

the necessity is standard stuff. to see sufficiency, you check that the Minkowski functional (or its inverse? I forget the convention) for this subset defines the desired norm.

Koro

yes, the unit ball in R^n indeed is convex, compact and has non empty interior. Compactness is true because all norms on R^n are equivalent, R^n is finite dimensional so closed and bounded = compact.

Thorgott

6:03 PM

the geometric point is that if you have any point, its norm is determined by checking how much you can scale until you enter/leave the unit ball

the conditions on the set are precisely the necessary ones so that this geometric idea behaves as you would expect it to

Koro

Thorgott and Leslie: you should teach at my college.

@Thorgott ohh thanks.

But for Minkowski functional to be defined, we also need the underlying set to be 'absorbing' in R^n.

ohh I think non empty interior takes care of that.

Take an $x$ in R^n. Scalar multiplication is continous so there is some scalar c such that cx is in the set that we're trying to show to be the unit ball of some norm.

thanks a lot :-)

Thorgott

yeah, it's a bit technical to check all the details, but the geometric idea guides the proof

a related fact using similar ideas is that any compact, convex subset of $\mathbb{R}^n$ is homeomorphic to $D^k$ for some $k\le n$

Koro

I think Minkowski functional should get me the result except that Minkowski functional may not be a 'norm'.

Minkowski functional is subadditive and positively homogeneous though. But under the given conditions, it may be a norm. I'll try to show this.

smth

6:32 PM

Let $A_n, A$ be closed and bounded, all $A_n$ connected and $A$ compact, if $d_H (A_n, A) \to 0$ then $A$ is connected, d_H - Hausdorff metrics. How to prove this? I have tried something by assuming that $A$ is not connected, then there exists a surjection from A to {0,1} but I don't know how to get a contradiction. Any help is welcome.

Ted Shifrin

6:45 PM

What's the Minkowski functional?

@smth Of course you need to say that all these are subsets of some fixed metric space.

Can you extend your continuous surjection to a continuous function on the whole space, or at least a neighborhood, mapping to $[0,1]$?

smth

@TedShifrin metric d_H - Hausdorff (definition with open balls and \epsillon) on F= { A \subset X | A is non empty, closed and bounded

Ted Shifrin

I know what the Hausdorff metric on subspaces is, but you need an ambient metric on the whole space to start.

In particular, you need to visualize these sets all living in one fixed space.

smth

That fixed space is some arbitrary metric space (X,d)? I think that is given in my problem

@TedShifrin

Ted Shifrin

Right.

smth

What with that? Sorry I am stuck

Ted Shifrin

6:58 PM

So I asked you a question above about extending the function. But what happens if you think about this in the simple-minded way? Suppose you had disjoint open sets $U,V\subset X$ separating $A$.

smth

Okay so maybe if I prove that that sets also making disconnection of some A_n I will get a contradiction?

but no

That does not have to be case

Ted Shifrin

Why not?

smth

If A_n tends to something they do not contain for example?

What disconnection of A has to do with that?

Ted Shifrin

I don't know what you mean. The $A_n$ tend to $A$.

Can the distance from the boundary of $U$ to $A$ be $0$? Or must it be positive?

smth

A is compact, that distance can be zero?

Ted Shifrin

7:08 PM

Give me an example of that.

user1294729

is here someone who knows about algebraic geometry?

Ted Shifrin

That is a vast subject, @user1294729. Depends what in particular you're curious about.

user1294729

I have posted a question about the sheaf of meromorphic functions and how it is used in the definition of a Cartier divisor.

Ted Shifrin

Well, OK, either ask the question or link to it.

user1294729

1

Q: How to think about the sheaf of meromorphic functions?

I am currently reading a book about divisors. In the beginning they are speaking about the sheaf of meromorphic functions, and they are using this notation then when they introduce Cartier divisors. Unfortunately I don't get what this definition really want to tell me. Let $(X,\mathcal{O}_X)$ be...

smth

7:11 PM

@TedShifrin maybe actually that distance cannot be zero because if it is such a case other set in disconnection should be empty which is a contradiction?

user1294729

This would be the question. Sorry I thought that maybe it will not help since no-one feels comfortable about this

Ted Shifrin

Hold on, @user1294729. I'll look at it. Also talking to smth.

user1294729

no problem there is no hurry end your talk first!

Ted Shifrin

I'm not following, smth. Why not try to prove some stuff here instead of just wondering?

Thorgott

hmm, the idea should be that regularity is some sort of improved non-vanishing condition, so the presheaf you're looking at are well-defined quotients of usual morphisms. the sheafification will produce functions that are locally of this form.

Ted Shifrin

7:14 PM

@user1294729 My first suggestion is to think about this all much more concretely on a Riemann surface or algebraic curve before you worry about general schemes.

smth

I have tried but I am a beginner and I need a little help

user1294729

@TedShifrin the problem is that our lecture is so horrible that I we have not introduced algebraic curves. We have started directly with general schemes or more precisely affine schemes but never mind.

Ted Shifrin

A beginner should start with concrete stuff and not with advanced stuff, but ...

Oh, great.

More horrid pedagogy.

user1294729

yes it is horrible

Ted Shifrin

I would suggest you look at Griffiths and Harris, for example.

They discuss this precise question in the setting of complex manifolds.

user1294729

7:20 PM

thanks.

Ted Shifrin

Locally, any meromorphic function is a quotient of holomorphic (or regular) functions. Sheaves see things that are built locally. Whether you can globalize is one of the whole points of sheaf cohomology.

user1294729

And what do Cartier divisors?

smth

@TedShifrin I have just asked for some hint, If you are willing to say, If not, that is also okay, no one forcing you to help. I don't think that problem is so advanced because I got that in my course, but I haven't succeeded, so I wrote for help... that is it @TedShifrin

Ted Shifrin

@user1294729 Start with divisors on a curve. These are $\sum n_P P$, where $n_P \in\Bbb Z$, $P\in X$, and $n_P\ne 0$ for only finitely many points $P$.

Rithaniel

So, I was looking at info on Bernoulli numbers, and apparently their generating function plus $\frac{x}{2}$ is equal to $\frac{x}{2}\text{coth}(\frac{x}{2})$. Is there any immediate geometric interpretation of this?

Ted Shifrin

7:26 PM

@smth I have given you two separate hints. I'm suggesting that if $A\subset U\cup V$, then $A_n\subset U\cup V$ for large enough $n$. I'm not going to do the work for you.

user1294729

So isn't this a weil divisor?

Ted Shifrin

The question is this: If $A$ is contained in an open set, is some $\epsilon$-neighborhood of $A$ contained in that set?

Thorgott

the difference appears only in higher dimensions iirc

Ted Shifrin

Yes, that's true.

Start by understanding that case, @user1294729. For smooth varieties, they're the same notion.

user1294729

@TedShifrin sorry I don't get what you mean by understanding that case? So we only had the definition of a weil divisor and we have defined the cycle map which is a map between Cartier divisors an weil divisors

Ted Shifrin

7:30 PM

Anyhow, the definition you posted fits into this in terms of thinking of meromorphic functions. Think of sections of $\mathcal K^\times/\mathcal O^\times$ by patching together local sections.

You need to think about zeroes - poles regardless.

Thorgott

speaking of AG, @Ted, I had some fun earlier today calculating the dimensions of various incidence correspondences

Ted Shifrin

Oooh, incidence correspondences were my life :P

My thesis was all about those.

user1294729

@TedShifrin ah okey

Ted Shifrin

@user1294729 I assume your course has taught you (or assumes you know) how to think about Cech cohomology and patching local things to global.

user1294729

no I don't know the cech cohomology. So I know about gluing schemes together

Ted Shifrin

7:38 PM

Oy.

Time to go back and learn some basics.

Even @Thor will agree with me on this, and he loves fancy nonsense :D

Thorgott

yeah, I checked that you would heuristically expect a generic degree $d$ hypersurface in $\mathbb{P}^n$ to contain finitely many lines is precisely when $d=2n-3$, cute stuff

@TedShifrin I will

I never got into higher-dimensional divisors myself

Ted Shifrin

Oh, you need higher-dimensional divisors if you live beyond curves :P

Hard to study surfaces without divisors :)

By finitely many lines you mean a nonzero number?

Thorgott

hmm, I don't know if it's non-zero

Ted Shifrin

So you mean not an infinite number? Ruled varieties are very rare ...

Thorgott

there's an obvious projection from the incidence correspondence to the moduli space of degree $d$ hypersurfaces. if $d=2n-3$, these have the same dimension. there are two cases. either the projection is not dominant, in which case the fiber is generically empty, or the map is dominant, in which case the fiber is generically a non-empty finite set

I don't know which of these cases actually is the case

Ted Shifrin

7:47 PM

And then enumerative geometry gets you to count them :)

I think you want to distinguish between 0 and positive :)

Thorgott

sure, I just don't know the answer

we'll finish counting the infamous $n=3$ case in lecture next week

Ted Shifrin

I know how to do that by doing divisors/linear systems and blowing up $\Bbb P^2$ at the appropriate number of points. :)

Thorgott

ah, we didn't do any divisors in this lecture, I think we'll do some magic by passing to the analytic topology (only doing it over $\mathbb{C}$)

actually, I think there's a lapse in my logic above. I'm not convinced I can say that either the fiber is generically empty or generically non-empty finite. rather, I should say the fiber is generically empty or non-empty finite, which is of course less restrictive. damn order of logical operations.

Ted Shifrin

8:05 PM

Of course I've only ever done this over $\Bbb C$. I went through it carefully when I taught complex geometry back in 1980. I have never taught it since.

geocalc33

8:19 PM

$g(x)=\log (\sum_{k=1}^\infty \prod_{n=1}^\infty f(\cdot)) $ for $f(\cdot)$ some expression involving a variable $x$ and $k,n$ can you switch the order of the product and sum?

or is there any other simplification to make?

Not Tfue

9:32 PM

dude Ted's thicc

Shifrin Math 3510 Day 24: Differential Forms (complete)

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Odestheory12

Is this true? $P(\bar{A} / H)=\frac{P(H) \cdot P(H / \bar{A})}{P(H)}$

Shouldnt be $P(\bar{A} / H)=\frac{P(\bar{A}) \cdot P(H / \bar{A})}{P(H)}$?

Thorgott

9:46 PM

@Koro @leslie I thought about it slightly more and realized there's an (in my opinion) slicker and geometrically clearer proof (at least compared to what I had in mind originally)

Ted Shifrin

10:03 PM

@NotTfue huh?

Rithaniel

"Thicc" in this context is probably meant as "robust" (if you haven't seen the term "thicc" before, it's usually meant as an attractive quality, but perhaps more often it is used in offhand comments that are meant to be humorous and casual)

Ted Shifrin

10:35 PM

Thanks for translating — I think.

robjohn

@Odestheory12 yes, that looks correct. The other just cancels out.

Odestheory12

I see, I am trying to understand why my notes state that

Ted Shifrin

11:50 PM

@Odestheory12 You probably wrote down a wrong letter ... or the professor did. The way to think about this is simple. $P(A|B) = \dfrac{P(A\cap B)}{P(B)}$. Similarly, $P(A\cap B) = P(B|A)P(A)$.

leslie townes

@Thorgott i love it

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$Mathematics$ Mathematics