Mathematics - 2016-06-19 (page 2 of 2)

Ted Shifrin

17:20

Shure is kwyet in hear.

Semiclassical

ya shure, you betcha it is

Ted Shifrin

Hi @Semiclassic

Semiclassical

(in my best stereotypically minnesotan voice)

morning, @ted

Ted Shifrin

Now that Garrison Keillor is done, you'll be taking over ...

Semiclassical

ya shure, you betcha

Owatch

17:24

Googling Trinomial theorem, get exclusively binomial :(

Mike Miller

it's not a super pleasant theorem

Ted Shifrin

What's the big deal, @Owatch?

Semiclassical

I don't actually listen to him myself. I just find it annouying

Ted Shifrin

G'night, @MikeM

Owatch

I guess I can use this multinomial theorem.

Mike Miller

17:24

morning

Semiclassical

trinomial is just multinomial for n=3

Ted Shifrin

What are you trying to do, @Owatch?

Owatch

@TedShifrin I have a program that is designed to solve a quadratic equation.

However.

An expression given might not be in the right form, so I need to build the coefficients from the expression tree.

One problem right now is: (expression)^n power

Ted Shifrin

We're talking quadratic ?

Owatch

Yes, it's a bit confusing. But I decided that the best way to preserve the [ax^2 +bx + c] components is pass the expression being raised to a power to a function that will perform a multinomial expansion on it.

Then just use the last three elements of that. IDK.

Semiclassical

17:27

Can you give a concrete example? I'm a bit perplexed by that description.

Ted Shifrin

I'm not sure I understand, but if you're going to have higher powers, why should they disappear?

Owatch

They shouldn't necessarily disappear. But given the function is supposed to find the roots for a quadratic equation they aren't going to be used. I still want to keep the multinomial expansion function though, because if someone wants to actually do a multinomial expansion they should still be able to use it for that.

Ted Shifrin

I'm not sure this makes sense.

Owatch

If you're trying to solve a quadratic which is raised to any power higher than one then you can't use the quadratic formula.

Ted Shifrin

Depends on the situation.

Owatch

17:30

Because it's not a quadratic. Yeah I know I'm not. Let me try to recollect myself.

anon

a quadratic function raised to a power is an expression, not an equation...

Ted Shifrin

I could have $(ax^2+bx+c)^2 + e(ax^2+bx+c) + f = 0$ and solve it by using the quadratic formula twice.

Well, greetings, gracious @anon.

Owatch

Well I'm assuming all roots need to be found, so it is an equation then.

anon

(ax^2+bx+c)^n=0 is what you solve to find the roots of (ax^2+bx+c)^n. do you mean a polynomial applied to a quadratic?

(as Ted wrote)

Owatch

My initial problem was this: I may be given a polynomial that when expanded, is actually a quadratic.

Like: (x+2)^2

My function tries to fill out a static array of coefficients: [a, b, c] for ax^2 + bx + c.

anon

17:33

IOW you want to build a computer algebra system that simplifies polynomial expressions into standard form

Owatch

I can't do anything with an unexpanded binomial, so I'm trying to find a way to adapt to situations where an expression to a power is encountered.

The idea was to pass the existing coefficient array: [a, b, c] to a function that could perform a multinomial expansion on it to the power I need and then return it with the correct coefficients.

anon

what does computing (ax^2+bx+c)^n have to do with your "initial problem"?

Owatch

You mean, why do I care for a trinomial expansion specifically?

You know what, this is a bad idea.

anon

I mean what is the connection between computing (ax^2+bx+c)^n and your "initial problem" of simplifying a polynomial expression to standard form.

Owatch

Well. The connection is that (ax^2 + bx + c)^n isn't in standard form.

So I need to expand it to get it there. But I'm being confusing and stupid so I'm going to try something different.

I will focus on converting input to standard form before I try to do any computations with it..

anon

17:41

well, neither are (ax+b)(cx+d)(ex+f) or (rx+s)^3+(ux+v)^2+(px+q) or an infinite other types of expressions in standard form. do you just feel especially curious about how to expand (ax^2+bx+c)^n?

Owatch

I can't provide a rational explanation for why I settled on wanting to know that.

Just kind of became a priority in my train of thought.

In any case, I'm no longer pursuing it for now. I've got to rethink how I work with user input.

I think the best process should be to convert an expression to standard form right away.

Owatch

17:58

Converting an expression to standard form looks to be a massive undertaking on its own...

user139655

m

user139655

18:14

I typed in "m" and pressed "enter" to return to the main site. It turned out that I didn't click on the mouse properly and it got typed into the chat.

Agawa001

18:44

hello dear mathers

Akiva Weinberger

19:03

hello deer mathers

user 1618033

I expect Romania will win EURO 2016. We can be the best in the world at anything (we wanna be the best).

@Agawa001 o/

Romania-Albania (right now on TV)

Andrew Thompson

@MikeMiller Nothing beyond the definition.

Mike Miller

19:25

Is there a more sane definition than what's on Wikipedia?

Andrew Thompson

Are you familiar with the Ext and Tor-functors?

anon

@Mike So I'm back to yesterday of trying to figure out how a principal $G$-bundle on nice $M$ yields a group homo $\pi_1(M,m)\to G$ (with $G$ finite). Say we pick a lift $\bar{m}$ of $m$. Then $\pi_1(M,m)$ acts on the fiber $p^{-1}(m)$. For every $\gamma\in \pi_1(M,m)$, we have a $\bar{m}\gamma$ from the monodromy action, for which $\exists ! g\in G$ s.t. $g\bar{m}=\bar{m}\gamma$, call it $g(\gamma)$.

This defines a map $\pi_1(M,m)\to G$, $\gamma\mapsto g(\gamma)$. But why would this be a homomorphism? For one thing, $\gamma$ and $g(\gamma)$ won't necessarily act the same way on $p^{-1}(m)$, which might be an indication this isn't the right road to go down.

looks like I'm being @balarka for now

cap

19:42

Hi, why are there different definitions of direct sum? One definition requires the components $U,V$ of $U \oplus V$ to intersect trivially, but for abelian groups the direct sum and direct product are the same, even if they don't intersect trivially.

anon

@cap there is a distinction between internal and external direct sums of groups

when speaking internally, if the two subgroups intersect nontrivially, then their sum is not direct

so one would write $U+V$ rather than $U\oplus V$ if $U\cap V\ne0$

cap

@anon is there a distinction between all structures (modules, rings, etc) or just for groups? When I see this symbol for modules, which definition should I assume?

anon

when $U$ and $V$ are not part of a common group, it doesn't even make sense to speak of their intersection, and that is the context one takes the external direct sum

cap

this helps, thanks

anon

@cap are you asking when to tell if the notation $U\oplus V$ is referring to exernal or internal direct sum? If $U$ and $V$ are subgroups of a common group it's internal, otherwise it's external.

cap

19:45

yes, I am trying to learn more about these but I was confused by the competing notions.

anon

Note that even with external direct sums, $U$ and $V$ canonically embed into $U\oplus V$, and their embedded copies (namely $U\oplus 0$ and $0\oplus V$) within intersect trivially.

cap

Will that intersection of embedded copies always be trivial?

anon

yes, by construction

Say $(u,v)$ is in both $U\oplus 0$ and $0\oplus V$. By being in the first we get $v=0$, and by being in the second we get $u=0$, hence $(u,v)=(0,0)$ is the trivial thing in $U\oplus V$.

cap

Does this mean that if I want to decompose a vector space $V$ into a direct sum I can find any two (nontrivial) subspaces $U,W$ such that $V=U+W$ and this will be isomorphic to some $U' \oplus W'$?

anon

the meaning of the phrase "decompose a vector space V into a direct sum" means find $U$ and $W$ such that $V=U\oplus W$ internally. thus, if $U'\cong U$ and $W'\cong W$ then $V\cong U'\oplus W'$ (external direct sum)

this is true for R-modules, or direct products of (not necessarily abelian) groups, or whatever

Mike Miller

19:54

@anon I can respond in a bit

@AndrewThompson I am

anon

k

cap

got it!

Andrew Thompson

@MikeMiller Alright, so let me explain it from my perspective, which I hope is better than that of Wikipedia.

For any algebra $A$ over a commutative ring $R$ (i.e. the group ring $\Bbb Z G$ considered as an algebra over the integers), define the stable module category as the usual module category, but do the following trick to the hom-sets: demand that morphisms that factor through a projective is zero.

(I.e., the homsets are $\text{Hom}_A(M, N)/P(M, N)$, with $P(M, N)$ the ideal consisting of morphisms factoring through any projective $P$.

Mike Miller

Why?

Andrew Thompson

I can define a first syzygy $\Omega^1(M)$ as the kernel of a chosen surjection $P \to M$. This construction is unique up to direct projective factors (for two choices, consider the pullback and compute the kernel of the projections) and hence defines a functor on the stable module category. (As projective modules are isomorphic to zero here.)

Now I can define Tate cohomology as follows:

$\hat{H}^n(G, M)$ is the colimit of the system $\underline\text{Hom}(\Omega^{n + m}(\Bbb Z), \Omega^m(M)$ with $\Omega^m(M)$ the syzygyconstruction iterated $m$ times,

Mike Miller

20:09

oh somehow I thought we were doing cyclic homology

Andrew Thompson

and \underline{Hom} is the morphisms on the stable module category

Mike Miller

but this is still helpful

Andrew Thompson

I am glad.

Mike Miller

Why am I doing this?

Andrew Thompson

So I know a bit more in the direction of stable module categories, nothing more in the direction of Tate cohomology.

I am curious about that as well.

Agawa001

20:15

@user1618033 soccer ?

user243301

Very thanks much @LeakyNun and also good nights to you, Semiclassical and all users.

Andrew Thompson

(And I can't edit at this point, but the Hom is over the group ring ${\Bbb Z} G$, so no, all Tate cohomology groups are not 0.)

Mike Miller

This should have some sort of relationship to group cohomogy and a "dual" theory that I don't know how to define properly. Eg for $G=A= \Bbb Z/2$, the group cohomology ring should be $\Bbb Z/2[U]$, $|U|=1$; Tate cohomology $\Bbb Z/2[U,U^{-1}]$, and the "dual theory" $\Bbb Z/2[U,U^{-1}]/U\Bbb Z/2[U]$

Danu

Heyo people

Andrew Thompson

@MikeMiller As the characters in Dragonball should patiently wait for Goku, I believe we should wait for @TobiasKildetoft.

Mike Miller

20:17

so the first is nonzero precisely in nonnegative degrees, the second nonzero in all degrees, the third nonzero in nonpositive degrees. I don't know how to get these objects and these relationships group theoretically.

RIP Krillin

Danu

While you are waiting, perhaps entertain yourselves with something easier? ;)

I'm trying to write down the second homology group (cellular) of the 2-skeleton of $T^3$.

Mike Miller

where did you learn what you know?

Andrew Thompson

My Bscthesis essentially.

Danu

And I've sorta convinced my self by looking at it (it's the quotient of the empty cube in $\Bbb R^3$ that it's generated by the three planes $x=0$, $y=0$, $z=0$ in the cube.

Andrew Thompson

(Did you mean stable cat-stuff or Tate cohomology? Tate cohomology I don't remember where I learned.)

Danu

20:20

I'm not completely sure how to actually go about proving that, though.

Mike Miller

Tate cohomology. I bet the references to your thesis would be enlightening. Can I see a copy?

Andrew Thompson

@MikeMiller sciencedirect.com/science/article/pii/002240499290116W This article defines Tate cohomology the same way I did. (And is written by Benson, yey!)

Sure, I can send it to you. No bullying for silly typos.

Mike Miller

Might be too new-fashioned for me.

Thanks. I won't.

Ted Shifrin

@AndrewT: Ah, two of my former colleagues.

Andrew Thompson

Sent. So there is no Tate cohomology there, but a bit on stable stuff which seems to be the modern approach to it.

@TedShifrin G'day.

Ted Shifrin

20:25

G'day. And g'night @MikeM. Hi @Danu

Danu

Hi @TedShifrin

(also night here; 10:25 PM, in case you're wondering)

Mike Miller

Hi @Ted. Want to explain Tate cohomology to me?

Ted Shifrin

It's not night where Mike is. It's just our perennial argument.

I know nothing, Mike.

Semiclassical

where are you, Danu, somewhere in Europe?

Mike Miller

@AndrewThompson The problem is that modern is too modern for me. I have modern definitions for the things I'm thinking about, dual spectra blah blah. I want to write down a damn chain complex and take its damn homology.

Ted Shifrin

20:26

In other words, Mike is old-fashioned.

Danu

@Semiclassical I study in Munich.

@MikeMiller Hear, hear! :D

Krijn

@Danu Erasmus?

Semiclassical

so you want something semiclassical?

Krijn

I thought you were in Amsterdam

Danu

I want to understand what you're talking about!

@Krijn No

Andrew Thompson

20:27

@MikeMiller That was how I felt my whole bsc-thesis. The final section you see me try to employ an all-or-nothing strategy to compute the Hochschild homology of group algebras. It only worked in the trivial case.

user10607

What can nAr mean in the context of probabilty? (its something similar to nCr for combinations and nPr for permutations)

Ted Shifrin

@Semiclassic: Don't let your ego get in the way :P

Danu

@Krijn I'm from Amsterdam. But I study in Munich (it's my MSc.)

Semiclassical

lol

Krijn

Ah!

Danu

20:28

I'm in LMU Munich's TMP (Theoretical & Mathematical Physics) program.

Semiclassical

@danu I figured it might be somewhere in Germany, mostly because I know from recent experience that Germany is 7 hours ahead of my time zone (3:30 pm here)

Danu

...as is most of West-Europe, but OK.

Semiclassical

hey, i didn't say it was -foolproof- logic

hence why i generalized it to Europe in general

Danu

:)

Semiclassical

(my advisor was in cologne for a few weeks this month, so i have that on the brain)

Mike Miller

20:29

@anon: I don't agree that $\gamma$ and $g(\gamma)$ act differently on the fiber. That they act the same way is more or less true by the way you defined $g$.

This whole definition is predicated on the fact that it doesn't matter what $\bar m$ you chose upstairs.

If that's true (it is), then you just follow $\gamma_1$, starting with lift $\bar m$, then follow $\gamma_2$, starting with lift $g\bar m$.

That's how you see it's a group homomorphism.

@Danu Those are the only three 2-cells!

Do you mean you don't see how to show that they're all cycles?

Danu

I see that they're cycles. I want to see why any 2-cycle must be a sum of them.

I guess it follows from the fact that the boundary operators vanish and hence the generators of the cellular chain groups are the generators of the homology groups?

Mike Miller

What, literally, is $C_2$?

Danu

$\Bbb Z^3$

So my above message is correct?

Mike Miller

More explicitly. Why cubed?

Danu

There are three cells in dimension 2

Mike Miller

20:36

What are they?

Danu

Those three planes

Mike Miller

So $C_2$ is defined to be freely generated by those three planes.

Danu

I'm talking about everything in terms the cube before the projection---then they're the planes $x,y,z$ integer (one for each)

@MikeMiller Yes.

And because $d=0$ $H_2$ is, too?

Mike Miller

yes

(and because $C_3$ is zero, of course)

Danu

Wait, $C_3$ is $\Bbb Z$ (but the map from it is zero)

Mike Miller

20:38

I thought you said you were calculating the homology of the 2-skeleton

Danu

Woops

:D

Mike Miller

Interestingly I don't know a name for the 2-complex of the 3-torus other than "2-complex of the 3-torus"

Danu

The exercise is about distinguishing spaces with identical groups by the ring structure on cohomology, so we're doing $T^3$ and its 2-skeleton wedged with $S^3$.

Mike Miller

I guess you can describe it by taking three 2-tori and gluing them together along certain circles

Yeah, that works

Danu

@MikeMiller The empty cube in $\Bbb R^3$ after opposite face identifications? Not very nice.

@MikeMiller Wut

Mike Miller

20:40

Nah it's three tori, written as $S^1 \times S^1$, (let's label them with elements of $\Bbb Z/3$) where you cyclically identify $(\{0\} \times S^1)_a$ with $(S^1 \times \{0\})_{a+1}$

Each of the planes is one of the tori

Danu

But my description also works, no?

Mike Miller

Sure.

I do not find that an inspirational use of the cup product. But nobody likes me, so don't worry about that.

4

Danu

To me, talking in terms of tori is much more mind-boggling. What is the advantage?

@MikeMiller Do you know a nice example?

Mike Miller

The advantage is that I understand one alright and not the other.

Danu

My TA is very much running out of inspiration for our exercises.

Mike Miller

20:42

A nice example of what?

Ted Shifrin

@Danu That's supposed to be the professor's job :D

Danu

Distinguishing spaces with isomorphic (co)homology via the ring structure

@TedShifrin Ha-ha-ha.

Professors do not give any fucks

(here)

Ted Shifrin

Such language!

Danu

I'm sorry.

Andrew Thompson

$S^2$ wedge $S^4$ and $\Bbb C P^2$, I think

Mike Miller

20:44

@Danu I just said I don't find that very inspirational. But the standard one is $S^2 \vee S^4$ and $\Bbb{CP}^2$. Or "any space with nontrivial cup product" and "a wedge of Moore spaces" eg $T^2$ and $S^2 \vee S^1 \vee S^1$.

Danu

Last semester, one of the professors (Mukhanov) discovered that his course (General Relativity) had been given an open-book exam for the past 5 years or so. He was very pissed off ;)

@MikeMiller What are Moore spaces?

Andrew Thompson

Something one uses to construct more spaces.

Mike Miller

@Danu Spaces with only one nonzero homology group.

Andrew Thompson

I didn't know how to TeX $\vee$ until now. Maybe I shouldn't aim to become a topologist after all.

Danu

Ah, those CW complexes with prescribed homology?

@AndrewThompson lol

Mike Miller

20:46

There are many, many different Moore spaces. You are describing one way of finding some.

Danu

I really liked that exercise---finding a space with prescribed homology groups.

Mike Miller

@AndrewThompson Brown's group cohomology book seems to do what I want.

Andrew Thompson

@MikeMiller Oh, that has come very highly recommended by several friends of mine. Probably a good choice.

Mike Miller

Looks like it even does it for spaces, which is the ultimate goal. It'd be cool if this just works. But I suspect it will only work as stated for finite groups.

Andrew Thompson

What exactly are you trying to do?

Mike Miller

20:51

There are four flavors of equivariant homology. I want a chain complex (for each) that calculates them for $G$-spaces where $G$ is a compact Lie group.

I can find $G = S^1, S^3$ in the literature, but the constructions I've seen only work because $BG$ is very simple in those cases.

Andrew Thompson

Hmm. I think I know of a paper of Segal that might be of interest, let me check.

Ah, found it. Don't think it's relevant but who knows: maths.ed.ac.uk/~aar/papers/atiyahsegal1.pdf

Mike Miller

That was my guess. Not so helpful, unfortunately.

Thanks for checking!

Andrew Thompson

No problem!

Do you have any friends going to the conference in Denmark, by the way?

Mike Miller

Which one?

Andrew Thompson

math.ku.dk/english/research/conferences/2016/ytm2016

Krijn

21:02

Why are elliptic curves so nice?

Mike Miller

Oh, YTM. I don't, maybe Chris Ohrt will. I forgot to apply this year.

Adeek

@TedShifrin hi

Mike Miller

@Krijn Why not?

Adeek

Do you think I should take differential topology or modular forms ?

Krijn

@MikeMiller Most other algebraic curves are less nice

Adeek

21:03

@TedShifrin

Krijn

Why would $g = 1$ make a curve nice?

Andrew Thompson

@Adeek You didn't ask me, but differential topology.

Danu

^

Adeek

Yeah I was leaning towards it.

Mike Miller

Depends what you mean by nice. Adding more topological complication makes an object less likely to have automorphisms of any sort, in particular say holomorphic automorphisms.

Like if you can quantify what nice means maybe I can say something.

user 1618033

21:08

Romania lost, unfortunately. Every war is first won in the mental plan, which is the case that didn't happen today.

Krijn

@MikeMiller Is genus a good measure of topological complication?

Mike Miller

It's the only measure for oriented surfaces.

Krijn

Ell. C. are nice in th Mordell-Weil theorem sense

Really the fact that you can define addition for points makes them quite nice already

Mike Miller

There's not a group for other curves, so it's harder to say anything. I would actually say that if you're looking for a finiteness theorem, higher genus curves are extremely nice: there are only finitely many rational points!

Krijn

That's Falting's, right?

Mike Miller

21:12

Yes.

user 1618033

I didn't see any of your solutions to this one

$$\sum _{n=1}^{\infty } \frac{\psi (n) \psi \left(n+\frac{1}{2}\right) \psi \left(n+\frac{1}{3}\right)\cdots\psi \left(n+\frac{1}{m}\right)}{n \left(n+\frac{1}{2}\right) \left(n+\frac{1}{3}\right)\cdots \left(n+\frac{1}{m}\right)}, \ m\ge2$$

@Semiclassical ^^^

Krijn

The fact that there are only finitely many rational points is mind boggling as well

It does not make any sense to me why :(

user 1618033

@BalarkaSen related to your message I think also a middle school textbook might get you more tired than being sick.

Ted Shifrin

Sorry, Kareem, was on the phone.

user174558

I am drinking coffee, folks.

Krijn

21:14

Make theorems, then

Ted Shifrin

OMG, @Jasper. What sin is next? :P

user 1618033

@JasperLoy Magnificent JASPER!

user174558

@TedShifrin Smoking weed.

user174558

@user1618033 What did Romania lose?

Ted Shifrin

Ah, seems appropriate ... :D

user 1618033

21:15

@JasperLoy Well, footbal ...

Adeek

It is okay my advisor @ University told me two good choices would be either modular forms or differential topology. I think I will go with differential topology. @TedShifrin

user174558

@user1618033 Oh I don't watch football, unless it's the world cup. I enjoyed watching Germany in 2014.

Ted Shifrin

IMHO, differential topology is one of the most beautiful and important courses for anyone serious about mathematics. You definitely need to know the inverse/implicit function theorems very well before you start.

Adeek

yeah

Krijn

"most beautiful" is subjective but important, sure

Adeek

21:17

I am studying your courses and reading a very nice written book by some guy named pugh.

Ted Shifrin

Is the professor a good one? Do you know what book?

Adeek

yeah

John lee

user174558

@Krijn He said one of the.

Adeek

the book is by John lee called differentiable manifolds.

Krijn

@JasperLoy Still

Ted Shifrin

21:17

Oh, that's not really differential topology. That's a basic graduate course on manifolds.

user 1618033

@JasperLoy Well, it was a matter of spirit and attitude here, and when you start anything, you wanna be the best, and do things as if you go right there, on the very top.

user174558

@Adeek Wrong, Introduction to Smooth Manifolds.

Ted Shifrin

Differential topology would be more on transversality, degree, intersection theory.

user174558

@TedShifrin Yeah, though there are universities which give it that name for lack of a better name.

Adeek

oh

Ted Shifrin

21:18

How 'bout differentiable manifolds? :D

Adeek

I guess it is called differentiable topology 1 LOL.

user174558

Smooth manifolds is short and sweet.

user 1618033

@JasperLoy it's essentially impossible to get on the very top without being very powerful mentally, without having the spirit of a great warrior, that is to be able to meet and face the impossible anytime.

Mike Miller

@Krijn I think the justification for Faltings theorem comes long after one has discovered and proved Mordell-Weil for abelian varieties in higher dimensions. I learned an argument to this effect from MO at some point but I don't remember where.

Ted Shifrin

Kareem: Depends whether they're getting to the sort of material I mentioned or not. But everyone needs to learn about manifolds, vector fields, bundles, differential forms, etc. Just a different course.

Adeek

21:20

This is my professor. @TedShifrin math.ualberta.ca/Lewis_JD.html

Mike Miller

If you pick a basepoint $x_0$ on an algebraic curve $X$ you have a map $X \to \text{Jac}(X)$, the Jacobian variety of let's say degree 0 divisors, by sending $x \mapsto [x-x_0]$. Since the dimension of $\text{Jac}(X)$ is $2g$, when $g>1$ this embeds $X$ as a positive codimension subvariety of a finitely generated abelian group.

user174558

@Adeek D J Lewis is another professor who died recently. I have one of his books.

user 1618033

@JasperLoy it's about dragging yourself completely out from the confort zone and preparing to meet anything for the very top performance.

Ted Shifrin

Hmm, an arithmetic algebraic geometer. I have no idea what he'll teach.

Mike Miller

Somehow from here it would be pretty damn weird if there were lots of points on the image of this embedding.

user174558

21:21

@user1618033 Good good. Will your book be out by the end of the year?

Adeek

• Preliminary Material: Differentiability, inverse and implicit function theorems.
• Basic Definitions: Partitions of unity, manifolds and submanifolds, tangent spaces, differentials, vector
fields, vector bundles, and metrics.
• Tensor and exterior algebras: Differential forms, tensor fields.
• Lie groups: Exponential map, homogeneous manifolds.
• Integration on Manifolds: Orientation, Stokes’ theorem, Poincar´e and Dolbeault lemmas, de Rham and
Dolbeault cohomologies.
• Axiomatic Sheaf Cohomology: Sheaves and presheaves, cohomology theory and the de Rham isomorphism

this is course outline

Mike Miller

Time will not permit.

Ted Shifrin

OK .. This is definitely a standard manifolds course, not what I consider differential topology material. You should learn it anyway, of course, but I retract my "one of the most beautiful ..."

user174558

@Adeek The Hodge theorem is seldom proven in books.

user 1618033

@JasperLoy At this point good news on my book don't depend on me anymore. I cannot tell you more details, but things that were related to me have always been done in time.

Krijn

21:22

@MikeMiller This is sort of what my presentation tomorrow is about

Ted Shifrin

Not true, @Jasper. Many books do it.

Krijn

When $X$ is an ell. curve

Ted Shifrin

Warner being the most well-known.

Adeek

oh

user174558

@user1618033 OK. They must be having a hard time understanding your theorems which are too deep!

Ted Shifrin

21:22

The topic list looks like it comes from Warner, pretty much.

Mike Miller

@Krijn Mordell-Weil? Fair enough. I don't know enough algebraic geometry to give you higher-dimensional philosophy.

user174558

@TedShifrin Warner is one out of many. =)

Krijn

Anyways, I should be studying for that exam right now

Mike Miller

Good luck.

user174558

@user1618033 I do not know any of Rama's work, too deep for me...

user174558

21:25

@Krijn May the Force be with you.

Mike Miller

@Krijn Here is the argument I was thinking of. Posted by the inimitable Matt Emerton.

user 1618033

@JasperLoy That was a pretty private info. We'll see how things are at the right moment.

user174558

@user1618033 OK.

Krijn

@MikeMiller Thanks, I'll have a look

user174558

It is interesting that both Chern and Yau have a book called 'Lectures on Differential Geometry'.

Adeek

21:27

Anyone here tried to use the new ipad to take notes instead of paper ?

user174558

@Adeek I will never buy iProducts.

Adeek

I am considering to buy the new iPad and use it to take notes and write ideas on instead of paper.

@JasperLoy I get really messy with papers lying everywhere

user174558

First I don't have the money.

Adeek

actually If ipad works well, then you would be saving up money in the long run.

user174558

Paper works best.

Adeek

21:29

because you can just store things in it instead of having million paper lying everywhere!

user174558

I have nothing to store. It's all in my brain.

user 1618033

@JasperLoy It was pretty much a kind of mistake to express my attitude toward math here, on this chat (during the time). One needs much maturity to understand what I tried to express so far related to mathematics and that most of the time was wrongly catalogued.

user174558

@user1618033 Well, sometimes we are misunderstood by others.

user174558

@Adeek Lee is my god. I like all his books.

user 1618033

@JasperLoy Only by the right attitude and spirit one can calculate such problem series$$\sum _{n=1}^{\infty } \frac{\psi (n) \psi \left(n+\frac{1}{2}\right) \psi \left(n+\frac{1}{3}\right)\cdots\psi \left(n+\frac{1}{m}\right)}{n \left(n+\frac{1}{2}\right) \left(n+\frac{1}{3}\right)\cdots \left(n+\frac{1}{m}\right)}, \ m\ge2$$

Please test my serie and see who can do any tiny step to it ... say, for $m=2016$

Adeek

21:32

Yeah my topology Prof here likes him a lot @JasperLoy

user174558

@Adeek There is a book by Hirsch called Differential Topology, which might be closer to what differential topology is. But it's just a different name for a different subject.

Ted Shifrin

Hirsch is very sophisticated. Guillemin & Pollack would be a good, accessible place to start, Kareem.

Adeek

Yeah I am currently reading Guilleman & Pollack very nice book.

@TedShifrin

user174558

@ted Now that you are retired, you can follow Spivak and write The Great American Book on Differential Geometry.

user 1618033

@JasperLoy Maybe Ted would be able to go near, sometimes he seems to me like Ramanujan.

Ted Shifrin

21:34

No, thanks.

user174558

Volume 1-10, LOL

Ted Shifrin

OK, Kareem, good.

Adeek

@JasperLoy Did you see some of ted's videos they are excellent !

user174558

@Adeek Yes. He looks cool.

user174558

@user1618033 I am curious how many university level math books you have at home...

Mike Miller

21:37

@AndrewThompson Brown was nice but still a little complicated, and it won't work for compact Lie groups. I've forgotten anything about what's salvageable in the compact Lie setting; whatever is salvageable is only salvageable through extreme violence. Probably whatever I'm thinking of right now is the wrong approach.

user 1618033

@JasperLoy I have some (university level math) books, but I didn't count them (I might though).

user174558

@user1618033 Are you still doing the same job?

user 1618033

@JasperLoy These days I'm not really dedicated to learning (from books), but to creating, developing new tools in mathematics for the stuff I like.

user174558

@user1618033 OK. I am not at the creator level yet, only a learner.

Andrew Thompson

@MikeMiller Violence sometimes works. But alright. If I come over something seemingly relevant this summer I'll tell you.

Adeek

21:40

I am currently reading a book of history of math by stephen hawking.

Very nice

user174558

Bourbaki has a book on the history of mathematics.

Mike Miller

@AndrewThompson I should probably figure out the answer in the next week or so.

Adeek

It is called God created the integers. (I guess the title refers to Kronecker quote)

Andrew Thompson

@MikeMiller Why the timelimit?

user174558

@Adeek I don't like these popular layman books though.

user 1618033

21:41

@JasperLoy I want to bring a new spirit for the approaches of the problems I like, and I think I already managed to do it by the little publishing activity I had so far. I have much stuff to publish but things go very slowly.

Adeek

@JasperLoy Before I sleep I used to watch stupid things. So instead of doing that I thought I should read layman books it would be more productive.

Mike Miller

@AndrewThompson Got other stuff to do.

user174558

@user1618033 Good good. Now all you need is to do a degree and a PhD and then you will become famous!

user174558

@Adeek I am wondering what stupid things... Hmm, LOL.

Andrew Thompson

The struggles of the PhD.

Adeek

21:42

Youtube videos @JasperLoy don't jump to conclusions haha

user174558

@Adeek OK. I watch stupid things all the time, LOLLOLLOL

Adeek

I noticed it decreases your attention span

user174558

My attention span is only changed by my own willpower.

user 1618033

@JasperLoy Well, here is a point: becoming famous won't help that much, I don't see how. My needs cannot be satisfied with this title. :-)

Adeek

No, because your mind get used to certain things.

user174558

21:45

@user1618033 I think I will be satisfied if I can get well and marry someone like Laura.

user 1618033

@JasperLoy simple things are the most precious ones, like playing with my dogs ... ;)

Mike Miller

@AndrewThompson It's not a major point of what I'm doing, and I think I have another way of doing it (that's just unfortunately sort of gross).

user 1618033

@JasperLoy :-))))))))

user174558

@user1618033 Well, I don't really like animals. I don't understand why people do.

user 1618033

@JasperLoy never had a dog?

user174558

21:47

@user1618033 Never had any pet.

user 1618033

@JasperLoy Hard to describe, but having a dog is such a experience ... :-)

(well, I have more)

user174558

@user1618033 I think I can solve my mental problems by the end of this year. I think I can be completely well then.

user 1618033

@JasperLoy Maybe a pet might help you much.

user174558

@user1618033 Besides Laura Ramsey movies, I also like Agnes Bruckner.

user 1618033

@JasperLoy :D

Krijn

21:51

I was googling stuff, and now I have found mathematically named songs on mathematically named albums, wtf

Dopplereffekt - Hyperelliptic Surfaces

►

user 1618033

@JasperLoy Monica Bellucci is my favourite one. ;)

user174558

@adeek Besides Lee's book on smooth manifolds, there is also Conlon's Differentiable Manifolds, which is more sophisticated.

user174558

@Krijn That looks German to me.

Krijn

@JasperLoy Or archaic Dutch

Adeek

oh no I would like to start with something less sophisticated.

Semiclassical

22:15

stack exchange looks to be offline

Transcript for

$Mathematics$ Mathematics