@Null thanks. i've been tryiing to improve the quality of my answers. if you go back a bit in time they're all non-rigorous and not very good.

So I shouldn't worry about curves that aren't homotopic to a manageable contour?

Mike has a good point.

Because I'll never use them?

That particular intuition always stuck with me

Is there a simple example of a situation in which this is a cause of problems?

@GFauxPas: But they are ... whatever "manageable" means.

Varieties.

(Projective varieties are locally isomorphic to affine varieties but not the affine space)

But what if a curve isn't in the same homotopy class as one in a simply connected set?

@Alessandro: For example, in the setting of holomorphic functions, the whole plane and a disk are not equivalent.

@ZachHauk to make it more obvious: $x=\frac{ab}{a+b}$ ;)

@GFauxPas: I don't see why the set needs to be simply connected.

Ok, that makes sense, I misread Mike's remark as emphasizing "smooth" instead of "manifold", but I couldn't see how could that be a problem with manifolds (since it's not :P)

So I can use the FTOC

So why do we even consider $\Bbb R^k$?

Yep, it's not an issue in the topological or the smooth category

You don't need simple connectivity for the FTC, @GFauxPas.

is it just because it makes examples easier?

You only need simple connectivity to guarantee that every holomorphic function has a primitive. But, for example, $z^k$ has a primitive on $\Bbb C-\{0\}$ for all $k\ne -1$.

So with the contour I'm playing with, all that matters is the beginning point and end point?

If I want to integrate some function along that contour

It will depend on the function, of course.

sin^1/2(z)

But you have to decide if the function has a well-defined integral on that curve to start with.

@ted can i interest you in a question about fourier transforms?

I have to leave, @ping. Ask @Semiclassic.

ooh thats unfortunate. thanks anyway!

@TedShifrin if we consider a linear transformation from $\Bbb R^n$ to $\Bbb R^{n-1}$,are the images of elements in $\Bbb P^{n-1}$ (1-dim subspaces of $\Bbb R^n$) points?

Actually male it simple, sin z

@GFauxPas: I don't have any clue where that function makes sense. Are you just arbitrarily making up these questions? I think you should move on and learn important things in complex analysis.

$\sin z$ is entire.

So I can integrate it on any contour

@ZachHauk P for primes?

Ted youtube.com/watch?v=HHgxOXEQaFU

@Null no, projective $n$-space

No, @Zach. Plenty of lines through the origin still map to lines through the origin.

Yes, @GFauxPas.

ok. im trying to think about what the solution to a system of $n$ equations in $m$ variables consists of, for $n < m$

LOL, thanks, @Sophie.

The dark side of the internet strikes again

@ZachHauk a bunch of m-n spaces

You can think of the solutions in the original $\Bbb P^n$, @Zach, assuming this is a homogeneous linear system. So you're intersecting hyperplanes.

@Ted Salut, comment vas-tu ?

oh, yes

Ça va, @Astyx, mais je m'en vais tout de suite.

Tant que ça va, pas de problèmes

hi chat ! :)

Thanks Ted

Sure, @GFauxPas. Hi, @Kasmir. Bye, Kasmir.

so for example, in a system of $2$ equations in $3$ variables, it is the intersection of two lines in $\Bbb P^2$?

@TedShifrin Bye! :)

Yes, @Zach. Now think about what rank tells you about the lines ...

See ya later. Off to tutor kidlets.

@BalarkaSen I just want to re-emphasis this

this is what my book has

alright, bye :)

bye @ted

the negative looks weird included there

bye @TedShifrin

@Alessandro No, it was emphasizing smooth. The problem is when you try to work with complex manifolds.

@AliCaglayan Let's say homological algebra.

Ah, ok, thanks, hopefully I'll remember that when I get there!

@BalarkaSen i read about them once before, anything interesting?

@MikeMiller Nice!

Wrong.

@MikeMiller any specific homology?

:(

why is the definition of absolutely convergent for complex summations $\sum_n |a_n|<\infty$ instead of $\sum_n |\Re a_n|<\infty$ and $\sum_n |\Im a_n|<\infty$?

Hi !

@AliCaglayan Meh. Well, I like the point of view that homotopy type of a topological space is an infty-category.

That's not what homological algebra means, @Ali. Homological algebra is a general study of chain complexes, sort of.

But Balarka was right that a more proper answer about what I've been doing is infinity categories.

yeah but you have simplical homology, intersection homology

that kind of stuff, thats what I meant

turns out that slothful is a real word in English, "the reader should be informed that the slothful authors..."

@AliCaglayan And the study of any of those would not be called homological algebra.

will we some day have figured all questions one might ask about math?

The wikipedia page on higher category theory is really unclear

So how does infty categories come into play?

@Null No.

@Alessandro The mental images formed by that are all variants of this.

how would you know when all questions have been asked, you can't know what you don't know

A common cause of death in sloths is over eating

@Balarka LOL, that's a quote from the Guillemin Pollack book

Rawr

On a more serious note I think I heard it before multiple times

@usukidoll hello

@BalarkaSen in books or being said to you?

Hi

@AliCaglayan Chain complexes form an infinity-category. When G is a topological group, the natural notion of a G-action on a chain complex (up to homotopy) is best phrased in terms of functors between infinity categories. I need some of the basic theory of complexes equipped with G-action.

I had never seen it before neither in spoken nor in written English but I guess is not such a common word

So we won't even ever know what we can't answer.

Hey ! I have to use the the expansion series $ \dfrac {1} {1-x} $ to represent $ ln(x) $ in power of (x-4) , what does that mean ? O.o

@Maks integrate

Hell hath no fury like Ted's projective geometry exercises. ;)

@MikeMiller hmm interesting

Any test taking tips for gre?

@usukidoll hi

But actually I need to work with something slightly more general than the category of chain complexes and working that out has been annoying.

Hello

People in this part of the world doesn't talk much in English. We just use some common phrases and proverbs, as @Krijn saw in one of the movies I recommended him (which he still hasn't finished, I think).

@Sophie Yeah, I get $ \dfrac {1} {x} $ but I dont know what do I need to achieve

@BalarkaSen Converging towards finishing it

@Krijn Fair enough :)

Have you read Dostoyevski in the meantime?

hey guys can i ask you guys one thing? is dx/dt = rx(1-e^-bx) a linear or nonlinear?

I am not forcing ya, tastes do vary.

@Maks taylor series expansion around the point a = 4

@Allie I don't think this looks very linear

if we switch the columns of a matrix in $GL(2,\Bbb R)$, is it equivalent to a mirror over $y = x$?

i assume so.

@Krijn Been too pressed with exams. Not much Dostoyevsky on my plan yet, but let's see (I want to read Brothers Karamazov at one point). I hope to get a poetry collection by Arseny Tarkovsky near Christmas, and watch Bela Tarr and Angelopoulos.

@Maks I think he wants you to express $\ln(x)$ as a power series like $\sum_{n=0}^\infty a_n(x-4)^n$

@BalarkaSen That's the father, right?

Yep.

@Allie what is your definition of linear

Also Brothers Karamazov has incredibly good parts, but you have to plough through some dirt to get there

what do you mean @Zach?

@Sophie And why does it give me formula $ \dfrac {1} {1 - x} $ ?

There's just too much to do and read, and there's just too little time.

@Alessandro so we have a matrix in $GL(2,\Bbb R)$

say, $M$.

because differentiating ln x gives 1/x

differentiating a series

etc..

does switching the two column vectors of $M$ create some $M'$ such that

@AliCaglayan In any case, I'm not doing infinity categories because I like the flavor, I'm doing them because I need to.

$M'\boldsymbol{x}$ is $M\boldsymbol{x}$, flipped over the line $y = x$?

so, you want to compose the transformation represented by $M$ with a reflection

what do you know about composition of linear functions?

@MikeMiller what do you need them for? As in why are you looking at commutative algebra

also can you write a matrix in $GL_2(\mathbb{R})$ representing the reflection?

yes

arxiv.org/abs/1609.09132

"There will be time, there will be time / To prepare a face to meet the faces that you meet / There will be time to murder and create /And time for all the works and days of hands / That lift and drop a question on your plate / Time for you and time for me / And time yet for a hundred indecisions / And for a hundred visions and revisions / Before the taking of a toast and tea." yet "And indeed there will be time / To wonder, “Do I dare?” and, “Do I dare?”"

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

Which speaking of, I have also been reading some of Eliot, @Krijn

i know why now, @Alessandro

don't worry :)

if you have 2 linear transformation represented by the matrices $A$ and $B$ can you write their composition?

ok then :)

just because i don't want to leave that question hanging; as the product $AB$

why is it simplicial rather than simplical? the latter is stuck in my head for some reason

yep, and you already noticed what multiplication by the matrix you wrote above does to the columns

I guess in my mind a simplex is a pretty rigid object so it should sound rigid

@MikeMiller yes seems interesting, I am confused to why you would be interested in knowing how it is constructed however?

because I need to construct something similar

gotchya

wow, a lot of Ted's exercises are coming pretty quickly to me. it feels good to know i understand the content :P

@MikeMiller Yes

@MikeMiller This paper has some really pretty pictures

mhm