Mathematics - 2017-11-17 (page 5 of 8)

12:33 PM

what is the non-null-homotopic involution $\Bbb S^3 \to \Bbb S^3$?

@LeakyNun What do you mean by "the"? Any map $S^n \to S^n$ given by reflecting along a codimension 1 great circle is non-null.

Leaky Nun

@BalarkaSen I read that $\pi_3(\Bbb S^3) = 2$

Balarka Sen

The hell?

Leaky Nun

so it must be "the"

Balarka Sen

$\pi_n(S^n)$ is $\Bbb Z$. I don't know what it means for a group to be equal to 2 lmao

Leaky Nun

12:39 PM

oh right, it's $\pi_4(\Bbb S^3)$, sorry

@BalarkaSen $\Bbb Z/2\Bbb Z$

Balarka Sen

Did you invent the notation "2" for the cyclic group of order 2?

1) That's a pretty shit notation 2) Don't expect people to know unofficial notation you have invented yourself

Leaky Nun

@BalarkaSen https://en.wikipedia.org/wiki/Homotopy_groups_of_spheres#Table_of_homotopy_grou‌ps

Balarka Sen

Yes, wikipedia doesn't expect people to know unofficial notation wikipedia has invented.

That's why it has included a paragraph explaining their notation lmao

In any case you were wrong.

Leaky Nun

alright

TastyRomeo

If you're lazy, the notations $\mathbb{Z}/2$ or $\mathbb{Z}_2$ are quite common. But just $2$ would be very confusing.

Balarka Sen

12:50 PM

I actually use $\Bbb Z/p$. $\Bbb Z_p$ is prone to confusion sometimes because it also denotes the p-adics

Secret

Our algebra lecturers often like to use $\Bbb{Z}/n$ for the mod n groups

TastyRomeo

Well, if the context has something number-theoretic, then using $\mathbb{Z}_p$ for the group is a bad idea

But otherwise, I just use $\mathbb{Z}_n$ for $\mathbb{Z}/n\mathbb{Z}$

Secret

I tend to reserve the subscript notation for p adics

1

Q: Contribution of non-western mathematics to modern mathematics

When I talk about the western mathematical tradition, I think roughly of the mathematics done by the Greeks, and then take up by European countries and then countries with primarily European descendants. Also, I'm not thinking of the mathematicians in particular, but the methods they used, and th...

western washed mathematics lol

luckily the chinese made their way in early in the history

aiatsis.gov.au/collections/collections-online/…

Australian Aboriginal enumeration

The Australian Aboriginal counting system was used to send messages on message sticks to neighbouring clans to alert them of, or invite them to, corroborees, set-fights, and ball games. Numbers could clarify the day the meeting was to be held (in a number of "moons") and where (the number of camps' distance away). The messenger would have a message "in his mouth" to go along with the message stick. A common misconception among non-Aboriginals is that Aboriginals did not have a way to count beyond two or three. However, Alfred Howitt, who studied the peoples of southeastern Australia, disproved...

cdu.edu.au/centres/macp/keyfindings.html

abc.net.au/news/science/2016-08-15/…

We often said mathematics is an art, but do we have ideas on how to do it the artistic way. However some will argue that without logic, we will end up with intuitive maths back in the ancient time which is misguided in all directions

but then, having only logic to guide mathematics sounds too limiting, but I don't have any reasonable solutions to this issue

Abcd

1:09 PM

@Secret How is intuitive maths "misguided in all directions"?

Secret

ooops wrong term, I should be saying "informal mathematics"

the logic room mentioned about informal mathematics were known to lead to a lot of dead ends

and it is only when logic came by that things become more steady

Abcd

@Secret was that a reply to me?

Secret

yes and no

I am also correcting the typo for that general message above

Meanwhile, it seems Periodic Table is back in peak activity again. Looks like the regulars have mostly returned

Abcd

Great! Where had they (the regulars) gone?

Secret

I have no idea. you may ask them

abenthy

1:19 PM

Let $G$ be a simple Lie group. It can have a center, but does it always arise as some quotient of a simple Lie group those center is trivial? I know that $\pi_1(G)$ can be identified with a discrete subgroup of the center $Z(\widetilde{G})$ of the universal covering group and I think the construction should be something like $\widetilde{G}/Z(\widetilde{G})$.

Tobias Kildetoft

@abenthy But $SL_n$ is not such a quotient, is it?

Mike Miller

There's a unique centerless simple lie group with a given simple lie algebra iirc

The model case of what's going on is PU(n)

Tobias Kildetoft

@MikeMiller Right, the adjoint one

abenthy

I'm a bit confused. Wikipedia states "Any simple Lie group with trivial center has a universal cover, whose center is the fundamental group of the simple Lie group. The corresponding simple Lie groups with non-trivial center can be obtained as quotients of this universal cover by a subgroup of the center."

So If I take $G$ centerless simple Lie group, then I can construct simple Lie groups with nontrivial centers out of $\widetilde{G}$

But how does the reverse process go?

Maybe take the full $\widetilde{G}/Z(\widetilde{G})$?

@TobiasKildetoft Yeah right, its $PSL_n$ which I want in that case

Tobias Kildetoft

@abenthy Right, I guess the thing is that the universal cover will not be simple, right?

(I haven't thought nearly as much about these things as I ought to. I usually just go "by usual considerations, we can assume our group to be simple and simply connected and then the results will also hold for arbitrary reductive groups")

Mike Miller

1:32 PM

@TobiasKildetoft A simple Lie group is not actually a simple group

Tobias Kildetoft

@MikeMiller That part I do know

since that is the same for algebraic groups

Mike Miller

Ah, well simple Lie groups are the same as talking about simple Lie algebras

Tobias Kildetoft

(with even more examples for algebraic groups)

Mike Miller

and the latter is preserved under covers

Tobias Kildetoft

ahh, right

abenthy

1:34 PM

Yeah, there are two different definitions for simple group and simple Lie group which are not equivalent for Lie groups

What I would like to understand is that why, given a simple Lie group, I can assume its center to be trivial.

Tobias Kildetoft

@abenthy depends on what you want to do with it

It is essentially the same as I said above that I always just state somewhere in the paper

(I did get slightly called out for it once, and had to include some more details about how this worked)

which was not too bad, because it made me have to think about it some more

abenthy

It essentially comes from my desire to understand why in working with symmetric spaces of noncompact type, one has $G/K$ with $G$ semisimple Lie group with trivial (sometimes finite) center

What goes wrong when the center is infinite? Why can I reduce to the case that its trivial?

Tobias Kildetoft

@abenthy Isn't the center of a semisimple Lie algebra always finite?

Sorry, of a semisimple Lie group

The Lie algebra obviously has trivial center

Mike Miller

It's not impossible to me that there's an infinite discrete subgroup

but maybe you can rule that out somehow

abenthy

I just googled it and found that $\widetilde{SL_2(\mathbb{R})}$ or $\mathrm{SO}(n,\mathbb{R})\times\mathbb{Z}$ are counterexamples

Tobias Kildetoft

1:42 PM

@MikeMiller For algebraic groups, it is certainly always finite (in fact, this is one way to define when a reductive group is semisimple)

Mike Miller

@TobiasKildetoft Ah, yes, there's no need for that to happen with Lie groups

Tobias Kildetoft

@MikeMiller I see. I will just stick to my nice algebraic groups then

Mike Miller

I suspect there's no naive algebraic object that models the universal cover of $SL_2$

Tobias Kildetoft

@MikeMiller I think $PGL_2$ does that

Mike Miller

ah yes, I forgot I'm working over R and you can't

Tobias Kildetoft

1:44 PM

Sure I can, I just don't

abenthy

Can I ask you an algebraic group question then Tobias? :)

Tobias Kildetoft

sure

Mike Miller

fine, then the problem is that i don't understand the relationship between real algebraic geometry and smooth things.

abenthy

Me neither

Tobias Kildetoft

(maybe this is something I can answer at least :) )

abenthy

1:46 PM

Okay, I was trying to figure out the $\mathbb{Q}$-rank of the algebraic $\mathbb{Q}$-group given by the functor which sends a $\mathbb{Q}$-algebra $A$ to the group $(A \otimes_\mathbb{Q} \mathbb{Q}(\sqrt{3}))^\times$

Is it one and a maximal torus is given by $A \mapsto A^\times \otimes_\mathbb{Q} 1$?

Tobias Kildetoft

@abenthy What is the coordinate algebra of that group?

abenthy

I have no idea

Tobias Kildetoft

I mean, doesn't this just send $A$ to $A^*\times \mathbb{Q}(\sqrt{3})^*$?

abenthy

I dont think we have $(A \otimes B)^\times \cong A^\times \times B^\times$

user193319

Does every closed subspace of $\Bbb{R}$ have a countable dense set? Let $A \subseteq \Bbb{R}$ be closed. I was thinking $A \cap \Bbb{Q}$ would do the trick, but I first need to prove $\overline{A \cap \Bbb{Q}} = \overline{A} \cap \overline{\Bbb{Q}}$, which I am unsure is true, since the intersection of the closures generally isn't contained in the closure of the intersection.

Tobias Kildetoft

1:52 PM

@abenthy Sorry, I need to go now, the bus for the retreat leaves in 10 minutes

Mike Miller

@user193319 It is true, but you'll have trouble doing it with Q. In fact you can find closed sets for which $A \cap \Bbb Q$ is empty!

abenthy

@TobiasKildetoft No problem, have a nice trip :)

user193319

@MikeMiller Really? That's pretty remarkable. It is a deep and hard to prove fact that every closed subspace of the reals has a countable dense set contained in it?

Mike Miller

No. You can prove it. But you have to build the set "by hand".

user193319

What I am working on is finding a topological $X$ that has a countable dense subset and a closed subset $A$ that doesn't have a countable dense subset. Obviously what you have said rules out $\Bbb{R}$ having any such examples...So I am not sure which space to look at for an example. Any hints on what space?

Mike Miller

2:02 PM

the proof for R actually works for any metric space with a countable dense subset

so it has to be something non-metrizable

user193319

How about $\Bbb{R}_\ell$, reals with the lower limit topology? I just proved that it isn't metrizable.

Mike Miller

Sounds like a good candidate, then! :)

user193319

Thanks!

Akiva Weinberger

2:18 PM

5 hours ago, by Balarka Sen

@AkivaWeinberger https://www.youtube.com/watch?v=qv6UVOQ0F44

@BalarkaSen I've seen that, it's cool

Alessandro Codenotti

2:29 PM

@user193319 This is kinda cheating, but given any space $X$ you can turn it into a separable space by adding a single point, by taking the open sets in $X\cup\{\infty\}$ to be those open in $X$ with the bonus point added to them

Secret

【スーパーマリオオデッセイ】#1 さっそく冒険に出かけよう！

►

Now, imagine a future where you don't even have human players

Akiva Weinberger

@MikeMiller I'm trying to figure out that proof. Does it rely on choice?

Secret

An AI doing a let's play on another AI playing a game made by an AI and with AI audience

Mike Miller

I make a countable sequence of choices. So I guess so.

Secret

(Oh, and did I mention that in those future, youtube's rating system is probably managed by a neural network?)

Alessandro Codenotti

2:32 PM

@Akiva I think dependent choice is enough

Mike Miller

what's that?

Akiva Weinberger

@Secret YouTube's monetization is run by NNs right now

(Also that video isn't loading for me for whatever reason) EDIT: never mind I got it @Secret

@MikeMiller If so, I see how to prove it, then

Mike Miller

cool

Alessandro Codenotti

nevermind, countable choice is enough

Secret

ok, did not see that coming, so I guess the rating system is already somewhat overseen by a NN.

Uh..., I have no idea, if it loads on australia, it probably is not because of geographic restrictions

(NB, this character is not actually an AI, as it is strongly suspected to be developed by some team in Japan, but close enough to give the idea)

Alessandro Codenotti

2:36 PM

@MikeMiller It says that if you have an entire relation $R$ on a set (for all $a$ in the set there is a $b$ such that $aRb$) then there is a sequence $(a_n)$ with $a_nRa_{n+1}$ for all $n$. It's a bit stronger than countable choice

Akiva Weinberger

The animated head there is creepy. Is that like Apple's animoji?

Secret

nah, it's a internet meme thingy that became very popular in korea and japan

knowyourmeme.com/memes/people/aichannel-ai-kizuna

Mike Miller

@AlessandroCodenotti Eminently plausible

Secret

As always any meme is covered by this website

Leaky Nun

So, what is the nontrivial map from S4 to S3?

Akiva Weinberger

2:42 PM

Arright, so motion capture and an animation software

Cool idea

@LeakyNun I think it's the "suspension" of the Hopf map

Just a moment

Do you know what the suspension of a space is?

https://en.wikipedia.org/wiki/Suspension_(topology)

Secret

But what *is* a Neural Network? | Deep learning, chapter 1

►

Balarka Sen

Right, it is the suspension.

Alessandro Codenotti

I'm reading some primary decomposition stuff, they say that if $\sqrt{(0)}$ is a maximal ideal then the ring is local and $\sqrt{(0)}$ is the unique maximal ideal

Balarka Sen

So that would mean the nilradical is the maximal ideal

Akiva Weinberger

> Higher homotopy groups were first defined by Eduard Čech in 1932 (Čech 1932, p. 203). (His first paper was withdrawn on the advice of Pavel Sergeyevich Alexandrov and Heinz Hopf, on the grounds that the groups were commutative so could not be the right generalizations of the fundamental group.)

Lel

Alessandro Codenotti

2:52 PM

@BalarkaSen Interesting, I didn't know $\sqrt{(0)}$ has a special name

Balarka Sen

The interesting point is it's the intersection of all the prime ideals

In particular if m_1, m_2 are two distinct maximal ideals, m_1 \cap m_2 would contain sqrt(0).

But then you're immediately rekt

as sqrt(0) is maximal by assumption

@Akiva lol poor guy

Alessandro Codenotti

@BalarkaSen Didn't know this fact either, that makes it easy

Balarka Sen

Cech spent a lot of time generalizing the fundamental group to a larger class of spaces

Akiva Weinberger

I still don't understand why the suspension of the Hopf map plus the Hopf map is nullhomotopic

Balarka Sen

$S^4 \to S^3 \to S^2$? That's not nullhomotopic, last time we decided.

Or did you mean that star itself

(i.e., \pi_4(S^2) = Z/2)

Akiva Weinberger

2:58 PM

No I mean, let $H$ be the nontrivial element of $\pi_3(S^2)$

then $\Sigma(2H)$ is $2\Sigma H$ I assume?

Alessandro Codenotti

I was thinking about picking a $b\not\in\sqrt{0}$ and showing it is invertible, consider the ideal $(\sqrt{0},b)$ which must be the whole ring by maximality, so there are some $a,c$ in the ring such that $ab+cx=1$, with $x\in\sqrt{0}$, so $cx=1-ab$ and $0=(1-ab)^k$ for some $k$ such that $x^k=0$, and so $a$ must be the inverse of $b$ (if $1-ab=0$ we get to a contradiction)

Akiva Weinberger

And $\pi_4(S^3)\simeq\Bbb Z_2$, so that's zero

$2H$ being $H+H$ where $+$ is the group operation of $\pi_3(S^2)$

Balarka Sen

@AkivaWeinberger Ah OK.

Yeah I don't know how to see that

Akiva Weinberger

Also, looking at the table, $\pi_{10}(S^6)$ is $0$ randomly. While $\pi_9$ and $\pi_{11}$ of that space are $\Bbb Z_{24}$ and $\Bbb Z$ respectively.

Mike Miller

Do you believe that $\pi_4(S^3)$ is framed cobordism of links in $S^4$?

Akiva Weinberger

3:01 PM

I hope that was addressed to Balarka 'cause I have no idea what that means

Mike Miller

Eh that's not an answer that will satisfy you.

Balarka Sen

@Mike yeah I don't know how to understand the Hopf map, it's suspension, or it's double in that picture

@Akiva The point is a map S^4 --> S^3 gives rise to a 1-submanifold of S^4 by taking preimage of a generic point in S^3

Mike Miller

well that's already the way people understand the Hopf map. it corresponds to the framed knot whose self-linking number is 1

(aka the preimage of two generic fibers link)

Balarka Sen

That's a bunch of circles linked togather. It also has a natural "framing on it"

which is just a trivialization of the normal bundle of the link

Mike Miller

but there's only 2 framings of $S^1$ in $S^4$ (as opposed to the $\Bbb Z$ in $S^3$, determined by self-linking number)

Balarka Sen

3:05 PM

i guess i subconsciously want an explicit way to see the nullhomotopy of twice the generator

maybe that's too ambitious

Mike Miller

the invariant that sees those framings is probably something like "push the circle off itself, pick Seifert surfaces, take intersection number" maybe

the only good way i know to see that with geometry is framed cobordism unfortunapely

Maneesh Narayanan

in Linear & Abstract algebra, 5 hours ago, by Maneesh Narayanan

2

Q: Dimension of $Image(T)$ and $Image(T^2)$

Given a $4\times4$ real matrix $A$, let $T:\mathbb R^4\to \mathbb R^4$ be the linear transformation defined by $Tv=Av$, where we think of $\mathbb R^4$ as the set of real $4\times1$ matrices. For which choices of $A$ given below, do the $Image(T)$ and $Image(T^2)$ have respective dimension 2 a...

matrices linear-transformations

Please help me.

cvsguimaraes

3:37 PM

hey folks, I'm trying to implement a parametric equation but there's something that I couldn't identify yet, perhaps someone here can tell me how its called so I can learn to interpret it myself

basically the image is a link to a function that uses "a" and "b" on x y equations but only K is given as input a k=a/b

not sure how the values of a and b are inferred from it nor how this thing is called so I'm stuck

since I can't even formalize my question properly I decide to give the chat a try, thanks for any help

Semiclassical

@cvsguimaraes isn’t there a parameterization in the image description?

Namely: “x = (a - b) * Cos(t) + b * Cos(t * ((a / b) - 1))
y = (a - b) * Sin(t) - b * Sin(t * ((a / b) - 1))

k = a/b”

cvsguimaraes

exactly

Semiclassical

Oh, so you want to infer a,b from the plots and the values of k

Thing is, suppose you divide both equations by b on both sides. That gives...

cvsguimaraes

yeah, maybe I'm missing something by since only k is given k = a/b isn't enough to figure out a and b

Semiclassical

x/b = (a/b-1) cos(t) + cos (t*(a/b-1)) = (k-1)cos(t) +cos(t(k-1))

And similarly y/b = (k-1) * sin(t)- sin(t(k-1))

cvsguimaraes

3:49 PM

makes sense thanks!

Semiclassical

so If you make a plot with a particular k and then vary b, all you’ll end up doing is rescaling the picture equally along both axes

cvsguimaraes

if x was an equation that even after dividing by b left some a or b behind that would be a hole in the function image right?

Semiclassical

The shape is otherwise unchanged

cvsguimaraes

@Semiclassical don't blow my mind just yet

Semiclassical

I’m not sure that’s true off the top of my head

The point is really that the a, b dependence of that equation is a bit overstated

In that x/b depends only on t and a/b

Not on a, b separately

for a comparison, how do the graphs of x^2 + y^2 = 4 and x^2 + y^2 = 1 relate to each other?

cvsguimaraes

3:55 PM

perhaps they both depend on x/y ?!

Secret

I had a feeling this will be downvoted to oblivion...

0

Q: Event suggestion: A history of mathematics and future direction panel discussion?

So recently, I was pondering about historical connections of subjects, such as how concepts, ideas and experiments done in history change the course of humanity as they open new doors to exploration and technologies. Having been to many panel discussions that covered topics from quantum computin...

discussion

Semiclassical

@cvsguimaraes what?

Secret

Now to report back to Martin and see if he had further advices...

Semiclassical

Well, it hasn’t been downvoted yet

So maybe wait for that first :)

cvsguimaraes

@Semiclassical sorry just brain farted, I see now they're the same equation with a different constant that changes the circle radius

Dattier

4:22 PM

Hi,

Determine the prime integers p, such that $\ mathbb F_p ^ *$ have 2 subgroups, one of order 17, G and order 19, H, such that there exists g in G with g + 1 in H.

NV-US

Give an example of a function$ f : [a, b] → R$ which is continously
differentiable and such that $|f(x) − f(y)| < |x − y|$ for all distinct
$x, y ∈ [a, b]$, but such that $|f'(x)| = 1$ for at least one value of $x ∈ [a, b]$.

can someone give me an example

i cant think of one

Ted Shifrin

@Liad You need to reread the theorem. You only need the primitive along the curve, as this is just the usual Fundamental Theorem of Calculus.

NV-US

hi @TedShifrin

Ted Shifrin

hi @NV-US.

You want a hint for your question?

NV-US

f(x) = |x|, in [-2,-1]

this will do, right?

Ted Shifrin

4:34 PM

Nope.

For that function $|f(x)-f(y)| = |x-y|$ for all $x,y$.

NV-US

ofcourse, sorry

hint please

Ted Shifrin

The Mean Value Theorem has $f(x)-f(y)= f'(c)(x-y)$ for $c$ where precisely?

NV-US

between x and y

Ted Shifrin

Right. So suppose you made $|f'(x)|=1$ happen at an endpoint of your interval and $|f'(x)|<1$ otherwise?

Antonios-Alexandros Robotis

hello @all

Ted Shifrin

4:37 PM

hi @Antonios

Antonios-Alexandros Robotis

how are you @ted

Ted Shifrin

Doing fine, and you?

Antonios-Alexandros Robotis

Not bad. Probably going to do some algebraic geometry problems.

Ted Shifrin

Cool.

Antonios-Alexandros Robotis

Trying to get a proper grasp on sheaves.

Also, sitting in a café which is playing the Allman brothers which is pretty great

NV-US

4:39 PM

i get your point. By MVT for all points inside the interval i have |f'(x)|<1, so i can only work on the endpoints. @TedShifrin

studrayght5

Out of curiosity, apart from commutative algebra, what else does one need to learn algebraic geometry?

Antonios-Alexandros Robotis

An affinity for pain.

jk, seems like a good breadth of knowledge really helps

studrayght5

Haha!

MatheinBoulomenos

algebraic geometry can be learned on a lot of different levels

Antonios-Alexandros Robotis

but I think comm. algebra is the brunt of it. that said, I'm not an expert.

MatheinBoulomenos

4:41 PM

If you want to jump straight into scheme theory, apart from commutative algebra, category theory and homological algebra can be helpful

Ted Shifrin

@NV-US: No, you aren't obliged to have $|f'|<1$ everywhere inside just because of the inequality, but what I'm suggesting is certainly easiest.

Note that the Mean Value Theorem does not say that for every $c$ you must have $x,y$ with $f'(c) = \frac{f(x)-f(y)}{x-y}$.

MatheinBoulomenos

If you want to do something over $\Bbb C$, complex analysis, Riemann surfaces or even higher dimensional complex manifolds can be useful as motivation (but @Ted is the expert on that)

It depends on the approach

Ted Shifrin

Everyone should learn about Riemann surfaces and algebraic curves ;P

3

user104729

@studrayght5 What do you want to learn?

NV-US

yes, right. @TedShifrin

studrayght5

4:46 PM

@user104729 I want to eventually understand some algebraic number theory/ class field theory.

MatheinBoulomenos

You can learn algebraic number theory without learning algebraic geometry first

Ted Shifrin

But you still need serious commutative algebra.

user104729

@studrayght5 Say from the perspective of Serre - Algebraic Groups and Class Fields?

studrayght5

@user104729 Yes, that's right.

MatheinBoulomenos

Yeah, everyone should learn commutative algebra

Antonios-Alexandros Robotis

4:48 PM

Atiyah-Macdonald is nice.

user104729

I would say, read some Matsumura, and Atiyah-Macdonald before looking at say, Gathmann and Ravi-Vakil's notes.

Antonios-Alexandros Robotis

I actually found a hardcover copy of Atiyah-Macdonald in a local bookstore from way back. It's awesome haha

Ted Shifrin

I still like Eisenbud's book, even though it's not short.

studrayght5

I'm learning commutative algebra -- using online notes, mostly. I'm thinking of using Eisenbud's book.

Ted Shifrin

Yeah, I took a course based mostly on Atiyah-Macdonald, but it's too terse.

MatheinBoulomenos

4:49 PM

Not surprised @Ted.

user104729

If you want to learn up to a second year graduate student level though, in only 24 hours, you can save years of your life by just doing "Twenty-four hours of local cohomology".

(The context of my joke is here: bookstore.ams.org/gsm-87 )

Ted Shifrin

@NV-US: Interestingly, the converse of the Mean Value Theorem holds "most" of the time, actually. If you're interested, you can find some questions about it on MSE and also get some results by googling.

NV-US

f(x)=sinx, in [0,pi/2], @TedShifrin

oh

Antonios-Alexandros Robotis

I've heard only good things about Eisenbud's book @TedShifrin . Had the pleasure of meeting him at the MSRI which was cool :)

Ted Shifrin

There you go, @NV-US. That's a good example.

NV-US

4:52 PM

thank you. searching for converse of mvt.

MatheinBoulomenos

I liked Matsumura's book more than Eisenbud's, but I wouldn't want to do without either of those, you should look into both and see what suits you more

studrayght5

I just want to eventually understand things like Falting's theorem etc, but there are so many things to learn before one gets there, it seems.

Ted Shifrin

@Antonios: I heard him give a lecture in Paris back in the 80's on work he and Joe Harris were doing back. It was great.

user104729

@TedShifrin Did you ever meet Grothendieck?

Ted Shifrin

Nope.

studrayght5

4:54 PM

@user104729 How much algebraic geometry does that Serre book require?

user104729

@studrayght5 Depending on where you look in the book, not much, or much.

Antonios-Alexandros Robotis

@TedShifrin that's so cool.

MatheinBoulomenos

@TedShifrin in our uni, algebraic curves are treated after at least one year of algebra and another 3/4 year of scheme theory and sheaf cohomology as an application

user104729

@studrayght5 To me, it feels like one of those books that you read over a long period of time. Reading a little, finding something you lack, and then correcting that - then returning to the text.

ALannister

Last one. I swear it.

0

Q: Probability density and expected value

Consider for one car owner the insurance policy with the following clauses: Deductible: If the loss $X>d$, then the insurer pays only for loss above $d>0$. Coverage Limit: If the loss $X>l$, then the insurer pays only for loss below $l>d$. Distorted Distribution: The insurer may base the premiu...

probability probability-distributions expectation density-function

Ted Shifrin

4:57 PM

@Mathei: That's nuts. And there's nothing wrong with doing Riemann-Roch more geometrically. Griffiths and Harris tell a beautiful story.

Semiclassical

@ALannister setting aside the measure theory context, it seems like there's three cases of interest (the same as in your piecewisie definition)

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$Mathematics$ Mathematics