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19:00
@GabrielRomon I would claim it is obvious that the determinant is $1$. Once you can see that the matrix is invertible, the claim mod 2 is of course even easier
why is it obvious that the determinant is $1$ ?
because it has enough zeroes to make it easy to calculate mentally
I think he had something simpler in mind
Expanding upon the first row, you're left with $$\det\begin{pmatrix}
1 & 1 & 1 \\
1 & 0 & 1 \\
1 & 0 & 0 \\
\end{pmatrix}$$
and then you're done by expanding wrt to the last row
ok
nods
Hi @GabrielRomon, @Tobias ... Happy holidays.
19:07
@TedShifrin Hi, you too
@GFauxPas: You have a defective copy of the book. Mine is just fine.
it has all those things in the index?what edition do you have?
Yes, second edition, printed 2000.
oh, "Indian Edition"...
I have to say that I should have done better making the index for my various textbooks. But there's a lot more important stuff about a good text than the index.
Aha ... you gets what you pays for.
19:10
>.<
right it's very good if I want to read a chapter, but if i want to look something up i needed a better index
Munkres is totally meticulous, so it surprised me to read what you'd written.
it's very good, if perhaps a bit dry
Definitely dry. On the other hand, one of the reviewers for my first book complained that a math book was no place for humo(u)r (I have more than one instance of humor).
Hi Semiclassic.
@Balarka @EricSilva: Also tangent cones at singular points of algebraic (analytic) varieties.
19:16
@GabrielRomon you can also think in terms of the Leibniz formula for the determinant: $\det(A)=\sum_{\sigma\in S_n}\text{sgn}(\sigma)\prod_{i=1}^n a_{\sigma(i)}$
@Ted That's true, the tangent cone captures local info of algebraic varieties
Hello everbody
I was telling people about how I used cone fields in hyperbolic dynamics
since the only nonzero elements from the first and fourth rows are the first and second element respectively, you need $\sigma(1)=1$ and $\sigma(4)=2$. That leaves either $\sigma(2)=3$ or $\sigma(3)=3$, but the 33 matrix element is zero so the latter isn't allowed. So you need $\sigma(2)=3$ and $\sigma(3)=4$
@TedShifrin
19:19
i.e. $(1,2,3,4)\mapsto (1,3,4,2)$. that's an even parity permutation, so you get $\det(A)=+a_1 a_3 a_4 a_2=1^4=1$
@TedShifrin Your math books are awesome
(I pay Adeek for advertising ...)
haha
blah, should've had $a_{\sigma(i),i}$ earlier
I have your all of your multi-variable analysis book definitely one of a kind
19:20
@Semiclassic: I still vote for intelligent use of cofactors.
Actually, at this point, I'm quite pleased with the undergrad diff geo notes. But they're free, so no one has to complain about how ridiculous the price is.
i'm still wrong anyways
Yeah, @Balarka, i'm not sure I know about those.
I've gotten something backwards above
@TedShifrin I have been thinking about vector bundles recently. I was discussing yesterday with Balarka about asking if one can construct invariants for complex manifolds using Grassmannians. he showed me you could do that.
very cool
@TedShifrin I mostly agree.
19:22
Indeed, that's what Chern classes are, Karim. I used Grassmannians (and flag spaces built out of Grassmannians) for everything in my thesis work.
oh awesome
@Semiclassic: I meant for this particular problem, not in general, of course.
Is there more non-linear thing we can use to construct invariants ? That is sharper than Grassmannians ?
That's one thing that's so powerful about complex geometry, Karim, is that you can prove things in a universal way using Grassmannians more effectively than you ever could for real geometry. On the other hand, that said, I like proving Gauss Bonnet in the real case using real Grassmannians as well.
But vector bundles are linear things, Karim.
@Adeek I didn't show you how to construct invariants for complex manifolds, but rather vector bundles.
19:24
But in this particular case I'd say one can notice that the only way to 'go from the top of the matrix to the bottom', if you take my point, is to do $a_{11}a_{23}a_{34}a_{42}$
and I like that.
I was wondering if there is invariants which are built from non-linear thing. Because, in my mind such invariants will capture more information about the space @TedShifrin
Invariants for complex manifolds come either from algebraic topology, from the tangent bundle, or from various holomorphic sheaf cohomologies in some form or other.
Well, you'd be surprised that you're mostly wrong :P
(though to match the version of the Leibniz formula given by Wikipedia I should really be doing left-to-right. oh well)
@TedShifrin why ?
Because how are you going to capture the complex structure of the complex manifold?
19:26
Everything is linear algebra
oh yeah
Manifolds are locally linear
Shrug
There is way less extrinsic geometry in the complex setting than in the real setting. Way more stuff is intrinsic.
Yeah, @Balarka, but there's more going on because of the smooth/holomorphic difference.
ugh my combinatorics isnt strong, how do I count the set of possible combinations of subsets of a three element set $\{x,y\},\{\alpha,\beta\},\{z\}$ where $\{x,y\}\cap\{\alpha, \beta \} = \{z\}$?
@TedShifrin That's an interesting remark, which I don't really understand.
19:28
oh ok
Right, the almost complex structure.
I don't understand the obstructions from almost complex to complex, however
Witness how mean I was in my final comment here‌​.
that is why in the right settings(i.e for the right complex manifolds) things are given by polynomials @TedShifrin
I'm not even thinking about such things, @Balarka.
That's surprisingly unuseful, Karim.
I guess we have that holomorphicity is very very rigid
19:29
It just tells algebraic geometers they should care about complex geometry.
@mlaci thanks million
hi @TedShifrin
But there's complex geometry beyond projective varieties.
hi Leaky
yeah
And a lot of my work was local geometry, not global.
I like the more analytic side of complex geometry @TedShifrin
19:29
The whole point of Kodaira embedding is it tells which complex manifolds are projective, I think
compact
@GFauxPas $2^3$?
Would that be non-hyperelliptic surface stuff?
@Adeek Projective manifolds are automatically compact, so that's a superfluous adjective
yeah sure
19:30
It gives a sufficient condition, @Balarka, but it's not intrinsic. There are things like Hodge-Riemann bilinear relations that tell you which tori are projective.
all these elements are picked from a 3 element set Leaky
theyre variables
@TedShifrin the joys of the period mapping
@TedShifrin Oh wow, there are non-projective torii? HMMMM
@Balarka: I can look at non-compact submanifolds of $\Bbb P^n$. In particular, I did that for my thesis.
Yup, there are non-projective tori.
If I knew this fact I have forgotten it
19:31
(Not in dimension 1, of course.)
cool @TedShifrin
Of course.
a relevant remark from Wikipedia: "All curves of genus 2 are hyperelliptic, but for genus ≥ 3 the generic curve is not hyperelliptic. This is seen heuristically by a moduli space dimension check. Counting constants, with n = 2g + 2, the collection of n points subject to the action of the automorphisms of the projective line has (2g + 2) − 3 degrees of freedom, which is less than 3g − 3, the number of moduli of a curve of genus g, unless g is 2."
You guys aren't going to confirm that I was mean?
19:32
@BalarkaSen all the facts you've learned, you've forgotten?
oh, wait. hyperelliptic is stronger than projective.
deeerp
i guess i'll just enumerate them all
@TedShifrin how is deformation theory btw I am planning to learn that at some point.
@Semiclassic: Hyperelliptic curves are a very special subset of curves of genus $\ge 3$. Most curves are not. Hyperelliptic curves are ones that admit a degree 2 mapping to $\Bbb P^1$.
19:33
Deformation theory is an important part of geometry/algebra. Kodaira developed it in the beginning.
So the fact that there are non-projective tori is a good deal more special than there being non-hyperelliptic curves.
It's natural if you're going to study different complex structures on the same underlying smooth manifold, for example.
Yeah I have his book planning to read it after I learn more complex geometry.
@Semiclassic: Totally unrelated.
Every curve that's a torus is projective.
Alright, I'll stop pretending I know what I'm talking about :P
19:34
I'm getting 3 possible sets
@GFauxPas which 3?
There's only a tiny bit on deformation theory in Morrow-Kodaira, Karim. He wrote volumes of papers on it.
I had also forgotten that the 180 noscope around an axis of the genus g surface is not in general a holomorphic double cover over the sphere
let the constants be $a,b,c$
and i'm omitting the set symbols and commas
@GFauxPas: Are we supposed to know wth you're talking about?
19:35
@BalarkaSen have you had any progress on the holomorphic cover?
I started a while ago Ted
@TedShifrin Does that fit in with variation of Hodge structure stuff, or am I just talking garbage
@LeakyNun I haven't thought about it. That's why I gave it to you to think about :P
$\{ab,ac,a\}$, $\{ab,bc,b\}$,$\{ac,bc,c\}$
Balarka I want to see you write a textbook at some point
19:36
@BalarkaSen Faith in your new apprentice, misplaced may be.
@GFauxPas I don't quite understand your question though
I think people usually construct it using a theta function of some sort. Read up on it and teach me about it :)
@Semiclassic, yes, it fits somewhat, but VHS is even more general.
cool
huh, neat: "In other words, deformations are regulated by holomorphic quadratic differentials on a Riemann surface, again something known classically. "
take a 3 element set. pick two distinct 2-element sets
19:36
anyway going back to work cya guys in a bit
@TedShifrin I buy that.
have their intersection be a singleto
See you!
how many ways can I do that?
@Daminark Indeed
19:37
Bye, Karim.
cya everyone
oh, wait, its always going to be a singleton, isnt it
It'll be titled "An Introduction To Inter-Galactic Teichmuller Theory: A Geometric Approach", coauthored with Neil deGrasse Tyson
though the most I ever understood was stuff like "consider a family of riemann surfaces $y^2=p(x,t)$ parameterized by $t$, and the periods as a function of $t$"
and then Picard-Fuchs mumbo jumo
smacks Balarka
19:38
And it'll have truly beautiful exposition. "We slurp this little fucker of a manifold..."
Anyway enough memeing for now I'm trying to actually be productive
Do you wanna snek lemma?
I'm drowned with statistics right now. Got the last lab work for my school year tomorrow
Need to get this done
Ah, I see, good luck with that!
Get it done.
19:39
If somebody here likes functional analysis: Is Schwarz space on $\Bbb R^n$ isomorphic (as a topological vector space) to the space $C^\infty$ on $\Bbb R^n$ where all derivatives should vanish at $\infty$?
@TedShifrin silly question. A (real) matrix is Gramian if it's of the form $V^TV$ for another matrix $V$.
...nope, question is too silly
@s.harp: I'm puzzled by your question. Obviously, there are functions in the latter that are not Schwartz. So then you say isomorphic. How is that isomorphism supposed to work?
OK @Semiclassic. I'm fine with "too silly."
nice hat btw
I got tired of the menorah. Although I noticed I have a new secret hat, but Balarka probably has it too.
Nope
I stopped bothering
I am stuck with 6
19:43
@Ted for example $L^2[0,1]$ and $L^2(\Bbb R)$ are isomorphic as Hilbert spaces, the obvious map $L^2[0,1]\to L^2(\Bbb R)$ is however not an isomorphism. Here also the obvious inclusion $\mathcal S\to C^\infty_0$ is not an isomorphism, but maybe there is another map so that they are the same TVS.
I have 6 also, @Balarka. I just put on the new secret one. Do I need to leave and come back?
Hmm, that didn't work.
I've got zero :P
I see the jester hat right now
@s.harp: I don't think about this sort of thing, admittedly. But I don't know how you'd distinguish things like $1/x$ from the tamest element of the Schwartz space under the new identification.
Right, there's a sun one that's a new secret hat, @Semiclassic.
It's showing up on my profile.
19:46
if I had a hat I'd put it in the center of my swirl
Oh, it's too much work to move hats.
Especially for those of us with little hair.
it's too much work for me to bother to earn any hats
I haven't bothered. They've just shown up.
19:47
hey there guys.
I never have gotten into the bothering-for-hats game like some people.
heya @Lucas.
@Ted Thanks^^ I think Daniel Fischer knows almost everything there is to know about these kinds of questions, would it be rude for me to ping him in the chat?
Daniel Fischer is quite the analyst. I don't think it's rude at all.
I miss having him around to learn from.
(Maybe I taught him one thing or two ... I dunno.)
BBIAB
20:03
@DanielFischer do you know whether or not Schwarz space is isomorphic as a topological vector space to $C^\infty_0 = \{ f: \Bbb R^n\to\Bbb C\mid f\in C^\infty, \lim_{x\to\infty}|D^\alpha f(x)|=0\text{ for all }\alpha\}$?
20:15
7
Q: Limit finding without L'Hospital's rule

Leos KotropI have to find: $$\lim_{x\to0}{\frac{\ln(1+e^x)-\ln2}{x}}$$ and I want to calculate it without using L'Hospital's rule. With L'Hospital's I know that it gives $1/2$. Any ideas?

sigh
yet another "without" question
@LeakyNun That reminds me of Do X without Y xP
@Mr.Xcoder right
Hello!
I'm back!
I passed all of my math classes!
20:53
Nice @orbit!
What courses are you taking next semester @Daminark
Definitely doing functional analysis, algebra 2 (used to be (multi)linear algebra, rings, and modules, though now it may change since we now have a linear algebra course), and a civilization class, likely ancient empires
Ancient empires? Is that some kind of domain?
You?
Lol Alessandro, it's just a class about ancient empires :P
I know, I was joking
20:56
3 courses? Are you doing extracurricular stuff?
I still need to figure out what my 4th class is
Originally I wanted to do algorithms, but that might not be able to happen
And the problem is there aren't too many suitable classes going on right now
Hmm, if you're interested in computer science beyond algorithms, I'd recommend a systems class
I found it incredibly enlightening.
And a lot of fun.
I don't think I'd be that into ODEs or point-set topology, commutative algebra is beyond my reach, and I don't think I'd be able to do computability theory this quarter, so... shrug
So, the thing about the compsci department here is that it does everything in C for the programming classes
Intro ODE's are soo boring.
Interesting. We have Racket in first year, then Java for the next programming class, and C++ for the algorithms class.
We don't code in algorithms at all
There's an intro sequence which I took, a quarter of Racket followed by a quarter of C
21:01
How about a reading course?
Then there's the "Intro to Computer Systems" which is a ridiculously hard class, uses a lot of different stuff like C, Assembly, sometimes you even have to be able to read machine code
Yeah! Do that
Sounds like a lot of fun!
@Dami point-set?
That's not offered this quarter
Wasn't there a descriptive set theory thing too?
21:02
Oh hey, your course uses the same textbook as my course
Also, I've had a nasty experience with C in the past, generally I don't like low-level stuff, so Systems would be a nightmare
@Alessandro that's in the spring, unfortunately
If it were now I'd totally jump on it
Hmm, I see. What about something from the physics department?
I may do a different reading course instead, but at that point the question is, who would I do it with?
Or statistics?
Stats, I could consider actually
21:03
@Daminark ah, that sucks. Will you take it then?
Physics, I also had a very bad experience with, so I'm inclined to avoid it now
@Alessandro at minimum I'll audit it
Though I'm juggling quite a lot of classes this spring
Print out all the available courses and throw a dart to pick one
"Phonology and semantics of medieval French"
Do any of you know the calculus of variations?
1
Q: Constrained optimization using calculus of variations (entropy maximization)

ClarinetistPlease note that I don't know (almost) anything about the calculus of variations, and I'm familiar with analysis only at an undergraduate level. Let $p: \mathbb{R} \to \mathbb{R}_{\geq 0}$ be such that $\int_{\mathbb{R}}p = 1$. Let $$H(p) = -\int_{\mathbb{R}}p\log p\text{.}$$ We define $p(x)\log...

@Alessandro lmao
Like, I'm definitely doing algebra 3 (should be field/Galois theory), combinatorics, and 99% gonna do complex analysis.
Then there's the reading course in descriptive set theory, there's harmonic analysis, intro to formal languages
Harmonic analysis is hard stuff
Descriptive set theory and formal languages both sound cool to me but you might not like logic as much as I do so...
21:11
You can never know until you try
Keepin' bio on the back burner?
And the other 3 classes I have may be quite hard. Algebra should be alright but combinatorics and complex analysis will be heavy
Harmonic is a second year grad course, and while the prof said I have the background, it might just be too much to do in one quarter
Formal languages is my inclination
And then auditing the other two
Also @orbit what are you gonna do next quarter?
Also @skull I'm done with bio
I see, ancient civilizations are coming :-)
@Clarinetist Robjohn does
21:24
Konnichiwa
Hi all
hey ya
Here's a question Ive been struggling with for hours: suppose I want to generate a random subset of ${1,...,n}$ proportional to $n \choose k$ where $k$ is the size of the subset. I figured out that my probability function should be $ \frac{n \choose k}{2n \choose n}$
What does this expression hint for, I've been asking myself for hours.
It is obvious I should generate a subset of $1....,2n$ of size $n$ and then do something with it
what does it mean for a set to be proportional to a number?
21:34
With probability proportional to something, I meant
@Daminark, real analysis 2, graph theory, intro to complex variables, intro to statistical inference, regression, time series and forecasting
What we require here is that:
$1 = \sum_{S} P(S) = \sum_{S} \beta \cdot {n \choose |S|}$, where S is a subset of {1,...,n}
I get why you're dividing by 2nCn now
Yes. Now this expression is hinting for something which I am trying to figure out
I am certain that my algorithm should first generate a subset of {1,...,2n} of length n (assume I can do that)
then doing something with it to yield a subset of {1...,n}
BTW how can I compile math here?
see "LaTeX in chat" in room description above the starboard -->
@TheNotMe you can intersect it with {1,...,n} to get a subset of {1,...,n}
after all, summing (n choose k)^2 over k amounts to summing (n choose k)(n choose n-k) which amounts to picking k things in {1,...,n} and n-k things in {n+1,...,2n} which amounts to picking n things out of {1,...,2n}
21:38
is the hyperreals homeomorphic to the reals?
Thanks, rendered
@anon
@anon, how does this yield each subset of {1,...n} with probability proportional to its (n choose k) ?
err
the number of ways of getting a given k-subset of {1,...,n} from this procedure equals the number of (n-k)-substes of {n+1,...,2n}, hence the probability of getting a given k-subset of {1,...,n} equals (n choose n-k)/(2n choose n)
@LeakyNun No. The hyperreals are not connected
@MatheinBoulomenos how?
the set of infinitesimals and the set of non-infinitesimals are both open
21:44
why is the set of non-infinitestimals open?
if a point is a infinitesimal distance from a non-infinitesimal, it is itself non-infininitesimal
I'm using the order topology here
I thought 0+ε=ε
@anon, I apologize, I don't get it. Let me try and see this in more details. Fix a k-subset received from the procedure. What is the probability we receive it? It is the probability we pick these k numbers out of 2n, and then to pick the other numbers (that we don't care about, except that they must be out of {n+1,...,2n})
@LeakyNun oh yeah I meant 0 + infinitesimals and non-zero non-ininitesimals
???/
how is the latter open
21:47
pick a point
@anon I think I see your point. The nubmer of ways to actually get a given k-subset is actually equal to (equivalent to) the number of ways to get n-k subsets of {n+1,...,2n} sinbce the other n-k numbers must be out of {n+1,...,2n}, correct?
then it contains the interval $(x-\varepsilon,x+\varepsilon)$
it doesn't
As I said, I'm using the order topology here
yes, it does
21:47
@TheNotMe yes
2+ε/2 isn't there
why not? 2+ε/2 is non-zero and non-infinitesimal
oh that's what you meant by non-infinitesimal
But why does that mean that the probability of getting a given k subset of {1,..,n} equal to (n choose n-k)/(2n choose n)?
yeah, infinitesimal is smaller than 1/n for all n
21:49
fair enough
@TheNotMe probability of an event is equal to the number of outcomes in that event divided by the total number of possible outcomes. and there are (n choose n-k) ways of getting a given k-subset of {1,...,n}, as that's the number of (n-k)-subsets of {n+1,...,2n}
I thought you were saying: the number of ways of getting a n-k subset out of {n+1,...,2n} is (n choose n-k)/(2n choose n)
22:07
@robjohn nice to see that you and Daniel Fischer still have the same taste in hats; it's our first winter bash without you guys in this chat room....I miss the good old days of arriving and leaving the room on a sled. Anyway, Merry Christmas and have a Happy New Year!
Merry Christmas and happy new year :-)
Quick question: suppose you have two curtains infront of you, one is red, and the other is purple. Behind one of them is a million dollars, and behind the other is nothing. You can't see behind which curtain the million dollars are. You are allowed to open one curtain, and if it is the one hiding the money, you get to have the money.
Which curtain do you choose?
(not a math question btw -- just for fun -- assembling statistics :D)
do you work for derren brown or something?
no, just a theory that crossed my mind
I would pick the red one
I would assume most people pick the purple one
I disagree
I think that most, like you, will pick red.
the reason is simple: red is more common to see in daily life. much more common than purple.
22:18
If it's completely random I'll probably go for the classic "Eenie meenie mynee moe"
and people tend to non-conciously reside in things they see daily
@Daminark - the colours of the curtains do not shift your pick towards one of them?
@orbit-stabilizer - you sound like you disagree with what I said
I associate purple with royalty. And I associate royalty with riches. So I would've picked purple if I wasn't trying to be contrarian.
Good point :-)
you picked red -- some way or the other.
Ill see what others pick
my gf picked red as well... she said it is a "more comfortable" color
my parents also picked red
only one friend picked purple, I asked why, and he said because red is comfortable -- and comfortable is fishy
22:31
@TheNotMe it's possible that if I come across this situation IRL I'd be different
But reading it, I just about ignored the colors
I see... thanks for the input
Last night dream has a few more maths than previous ones:
1. The notion of a "duel correlation": There are two people X,Y on a square grid with some probability distribution to find them in each tile determined by the correlation matrix of X,Y, but the correlation matrix is not symmetric.
The result is that in order to find the tile which minimise the distance between X and Y, you need to pick the coordinates of the tile which has the highest probability of X and Y appearing respectively, and that tile also corresponds to the largest value in the correlation matrix of X,Y
2. This is the more interesting one, which I am not sure how can one even approach to prove it or give a counterexample: There's a weird conjecture (real life counterpart unsure) that is shown by some frequency plot, showing how the number of irrationals are bounded until a certain critical value of e^{-1087e + (some long polynomial)}, where the number of irrationals are no longer being bounded and started to shoot up faster than exponentially.
1. "which has the highest probability of X and Y appearing respectively" --a more precise definition is needed. The average between the two probabilities? the sum? the...?
22:51
I think the dream is using the wrong terminologies as usual. From the scenes that is shown, which has some kind of heat map on the grid, I think what is actually shown is the probability distribution of positions of X (independent of Y) $P(X)$, the probability distribution of Y (independent of X) $P(Y)$ and the optimal positions of X and Y which their distance (defined by the number of squares between them) is minimised is then given by picking the highest value of $P(X)$ and $P(Y)$
such that the entry of their correlation matrix is also maximised under the above constraints.
even "defined by the number of squares between them" needs some more definitions
imagine a 3x3 board. Person 1 is on (1,1) and person 2 is on (3,3).
What's the number of squares between them?
yeah, I don't think it get into that detail, or at least not explicit.
However, dreams like these are repository of ideas, and I often use them to create new questions after getting rid of all the funny use of terminologies by the dreams themselves
That's a super interesting approach. I like it.
heya Eric
One possible sensible version of the above question I can think of that is based on the dream's but without all the vagueness is to simply convert this discrete problem into a continuous problem, so instead of considering positions as a integer lattice, we can consider the plane $\Bbb{R}^2$ and ask about the same question where the notion of distance is then given my the euclidean metric.
However, I am not sure if there exists a continuous version of a correlation matrix for this problem to be well defined, though

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