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00:00
I know what irreducible varieties are.
you can't get that if you're hausdorff, e.g.
Yes, of course.
@TedShifrin irreducible varieties can't be written is a union of 2 proper closed subsets, it's that idea but expressed in terms of open sets and generalized to spaces
I've only heard it in the context of schemes and varieties.
As I said ...
Lol, in AG we defined it for spaces and just said it's a property that a variety, as a space, may or may not have
00:07
Let $A \subseteq \Sigma^*$ be a set of strings. Let $B_k$ be the set of strings with $k$ prefixes in $A$, i.e. $B_k = \{s \in \Sigma^* : |\text{Pref}(s) \cap A| = k\}$. Can I obtain a generating function for $B_k$ from a generating function for $A$?
Analytic Combinatorics addresses the question of obtaining generating functions for substrings on page 211, section III.7, example III.26, but it's only for fixed strings.
01:09
what's the difference between pullback(precomposition) and composition?
On wikipedia it says, "Precomposition with a function probably provides the most elementary notion of pullback: in simple terms, a function f of a variable y, where y itself is a function of another variable x, may be written as a function of x. This is the pullback of f by the function y. $f(y(x) \equiv g(x)$"
Isn't this the same as the compostion of f and y?
Yes, but pullback is a thing that makes sense for more than just functions.
oh, okay
01:27
indeeeeeeed it does
You can get 3 mathematicians in a room that all disagree what pullback is.
do I have to disagree?
Hey everyone!
hi @Perturb
Hey @TedShifrin :)
I'm trying to show that $\gamma '(t)$ never vanishes, and my approach was to take smooth charts on $\mathbb{T}^2$ and then compute $\gamma'(t)$ in terms of local coordinates, but the author identified the torus as a subset of $\mathbb{C}^2$ so that complicates things a bit, am I using the correct approach? I can't just take usual derivative of $\gamma$ and show that it's non-zero for all $t$ since $\gamma'(t)$ doesn't really mean that here.
01:41
No, @Perturb, your instincts are always way too complicated.
Just compute the derivative as a map to $\Bbb C^2$.
If the derivative is nonzero, you're done.
@TedShifrin Really it's that simple? The author defined $\gamma'(t_0)$ to be $d\gamma_{t_0}\left(\frac{d}{dt}\bigg|_{t_0}\right)$ though and since it wasn't said anywhere that this was equivalent to evaluating the derivative of $\gamma$ at $t_0$ I assumed I had to do things the hard way
No, you need to understand how to simplify things, not make them more complicated. If $\gamma\colon M\to N\subset\Bbb R^n$, then computing the derivative as a map to $\Bbb R^n$ suffices, because you know it'll land in $T_qN$.
pullback is a limit
(do I count as a mathematician?)
No and no.
great
01:47
I can't enumerate the number of times I told a student in office hours "you're making this too hard".
That was one of my most common hints to students, Mike.
I think it generally comes from not understanding concepts/definitions. Tautologically, I suppose.
Another one: "we only know so much". There is frequently only a short list of things one could possibly do, so it's not unreasonable to enumerate the list and decide which one.
oh hell, it's @Eric
Ah shit.
i aint here
01:49
Oh, good.
no worries
I may have Portuguese cuisine questions for you soon.
@Perturbative I just saw the relevant question on main. I'm confused about this. What is your definition of immersion that this isn't tautological?
01:51
@TedShifrin By it landing on $T_qN$ do you mean you're identifying $T_qN$ with something under an isomorphism?
Zee
Zee
@Perturbative the author defined it as a subset of C2 couse it looks nicer
No, no. I'm saying that you know the image lands there, so if it's nonzero sitting in $\Bbb R^n$, then it's nonzero sitting in $T_qN$.
Zee
Zee
Cos and sin are so 2010
@MikeMiller The definition of an immersion I was using was that given a smooth map $F$ between manifold $M$ and $N$, $F$ is an immersion if its differential is injective at each point. There's also a theorem that says if $\gamma : \mathbb{R} \to M$ is a smooth curve on a manifold $M$ then $\gamma$ is a smooth immersion iff $\gamma'(t) \neq 0$ for all $t$, this theorem is the main result I'm trying to use
That is NOT a theorem, @Perturb.
A linear map from $\Bbb R$ is nonzero iff it is injective.
01:55
Okay yeah it's too easy to be a theorem, call it a result
Don't call it anything :P
What does $\gamma'(t)$ mean...? $d\gamma_t(1)$?
It's just a fact ...
As far as I can tell your question is just about this fact
that's what happens when you keep doing everything abstractly and forgot the ideas behind everything
Well, there is the point I was making, @MikeM. You don't need charts on $N\subset\Bbb R^n$ to make this decision.
Often that is not the case.
irony hits me in the face so hard I'm sent back to '99
@TedShifrin right
01:57
@Leaky: You sound like me lecturing you.
@TedShifrin I learn from you :P
I'm not sure I'm flattered.
Zee
Zee
You become a catagory theorist
@MikeMiller hit me send me back this future sucks
@MikeMiller So $\gamma'(t_0)$ is defined to be $\gamma'(t_0) = d\gamma_{t_0}\left(\frac{d}{dt}\bigg|_{t_0}\right)$
01:58
god
Which I guess is what you said
$d/dt = 1\in\Bbb R = T_q\Bbb R$.
could we stop making everything so complicated
We're going to do role reversal. Some day I will try to be categorical and as complicated as possible.
Maybe after I die.
Zee
Zee
If you do catagorical things , you are* already dead inside
02:00
$\gamma'(t_0)$ is simply the "speed" of the curve at $t_0$
damn, Zee, you need to learn basic spelling.
no, @Leaky, it's the velocity.
Big difference.
we invented $D$ only to generalize it to functions with multivariable domain
sorry, velocity
don't "define" $\gamma'(t_0)$ using $(D\gamma)(t_0)$
You set up the machine so that it does the work for you...
@LeakyNun That's how the author has defined it, not me
02:03
This is from Lee's Smooth Manifolds book
When $f\colon \Bbb R\to\Bbb R^n$, most authors will write $Df(t_0)(1) = f'(t_0)$.
Zee
Zee
This reminds me of lee smooth
Yup
Much as I love him, sometimes Lee overindulges in pedantry.
But I taught him when he was a grad student, so maybe it's all my fault.
@TedShifrin :0
Zee
Zee
I guess you couse me pain even in class then
02:04
Never would've known you taught Lee
@Perturbative go read spivak
He taught out of my diff geo notes one year and complained to me numerous times about the lack of rigor ... we have different styles.
Zee
Zee
It’s a good reference but a terrible book to read cover to cover
Spivak is so wordy and spends so much time tangling and untangling notation. Lots of books have issues.
aha, I know what to do
02:06
Then Kobayashi-Nomizu is so terse it's impossible to read unless you understand it already.
@Perturbative go watch Ted's lectures
I do not like his book so much but understand why it's popular.
I read Lee because his books seem more rigorous than most
I don't like any intro manifolds book.
No, @Leaky. Wrong course.
02:06
@Ted so what should one do?
His book is equally rigorous as all of the other references.
doesnt Tu have a manifolds book ppl like
Sometimes being ultra-rigorous is counter-productive.
He just writes more.
idk if it's smooth
02:07
Yes, Eric, it's smooth. I haven't looked personally.
the book?
Every manifolds book is a smooth manifolds book unless otherwise stated.
his diff geo book is (y)
Zee
Zee
Aubin has a nice short book
I'm rather fond of Boothby's Intro to Manifolds and Geometry book.
I think it has the right level of pedantry.
02:08
@MikeMiller i didnt know if it was like the first Lee book
If it's not you're either reading Lee 1 which is really a point set topology text, or you've gone off the deep end and are reading Kirby-Siebenmann.
There is nothing you can say about topological manifolds at an elementary level that isn't in Hatcher
I guess you could classify 0- and 1-manifolds. Great.
u could classify surfaces if u want
02:09
classify exotic R^4
does hatcher do that? ider
No you couldn't. You'd need to triangulate them first, so you'd need Schoenflies.
Tee hee
i thought lee 1 did that
where are all the algebraists gone :c
02:10
I would be shocked if he doesn't assume existence of a triangulation.
I'll look it up out of spite.
iunno i opened it once and never read it
Zee
Zee
@LeakyNun I been getting a little ring-y lately...
@Ted might you be familiar with representations of S_n by any chance?
that's the symmetry group on n letters
I know what $S_n$ is, dammit.
Zee
Zee
Can’t you only triangulate compact manifolds ?
02:12
sorry
I'm no representation theorist.
But Weyl's Classical Groups will answer your question.
Zee
Zee
Only important one is SL2R
lol wtf no
@TedShifrin Ohh I think I get what you're saying now. so $\gamma'(t_0)$ non-zero in the sense of normal calculus means that $d\gamma_{t_0}(1)$ is non-zero in $T_{\gamma(t_0)}(\mathbb{T}^2)$
right
it lands tautologically in $T_{\gamma(t_0)}(\Bbb T^2)$.
02:14
tautological?
i.e. the tautological vector field on $TM$?
shaddup.
does it actually exist though
I'm not very sure what $TTM$ and $T^\ast T^\ast M$ look like
@TedShifrin Should I worry about giving a formal proof of that (non-zero derivative in terms of usual calculus passes to non-zero differential) at the moment or just accept it and move on ?
02:18
You already have the formal proof by Lee's definition.
Zee
Zee
You have one variable here , so the differential is the entries differentiated
Okay so because we're working on subset of $\mathbb{R}^n$ and $\mathbb{C}^n$ essentially we can just work with usual multivariable calculus instead of relying on the abstract manifold stuff (since there's a correspondence in any case)
Thanks for your help @TedShifrin :)
02:33
Hey @Daminark. How's the blood pressure?
 
2 hours later…
04:20
as in, still have yet to find out anything about grad school? or still waiting anxiously.
 
2 hours later…
06:35
Is there a recurrence relation for oeis.org/A199571?
06:51
Or is it regular?
centralBinomialCoefficient[x^2] integers[(x catalan[x^2])^N] is the generating function I found
 
1 hour later…
08:21
@Dair sadly nothing
09:05
@user76284 under section "Formula"
 
1 hour later…
10:32
Hey @Rudi! Long time no see
10:59
Wel well wel, it seems that a counter example has been found to Donkin's tilting conjecture. Good thing I am leaving academia or this would have required some serious overhaul to my research statement.
hello!
does anyone know how to use Coefficient in wolfram alpha
$Coefficient[(\frac{(1-x^9)*(1-x^10)^4)}{((1-x)^(5)}), x, 9]$
raw code: Coefficient[((1-x^9)*(1-x^10)^4)/((1-x)^(5)), x, 9]
its not working and is giving 0
what should it give?
The coefficient of $x^9$ in the expression
non zero obviously (as i think
11:03
why would that be obvious?
@TobiasKildetoft I opened and wrote some terms of individual series
It is not even obvious to me that you had a polynomial there
$\frac{(1-x^9)\times(1-x^{10})^4)}{((1-x)^{5}}$
@TobiasKildetoft using binomial theorem I used expanded form
ahh, actually sure, it is not hard to see it is a polynomial
So what did you get the coefficient to be?
I have not calculated it, as actually I needed to calculate coefficient of $x^{40}$ which would have many cases, but WA gives 0 for any power of $x$
except 0
11:09
Ahh, then it does seem something it wrong (but calculatinga few terms would not have told you that it was non-zero as there will potentially be a lot of cancellation).
Anyway, the fastest way is probably to differentiate a suitable number of times
@TobiasKildetoft!
@LeakyNun ?
might you know about irreps of S_n and polytabloids?
not sure about the last word, but sure for reps
(asking for a friend)
11:12
yeah, I noticed that this morning
well apparently you can make irreps of S_n using young tableaux
yes, that I know
just not sure what poly or oid are doing there
might you have any good references for that?
Sagan's book is fairly thorough for the classical approach using Young symmetrizers to define Specht modules
There is also another approach using induced modules, but I am not aware of a good reference for that except for more general references which really deal with Hecke algebras and Soergel bimodules in general
@Tobias I did it wih Series command on WM :D link
11:17
@jeea Why would you use power series expansion to just expand into a polynomial? Surely there are faster methods?
(especially since I am fairly certain that method will calculate what you need along the way anyway)
I just needed the answer, It was a combination question
@LeakyNun So the best reference will depend on how advanced it is allowed to be.
I think he likes it advanced
unfortunately you two seldom come here so you can't talk directly
Then maybe reading something like the exposition in Geck-Pfeiffer on characters of Coxeter groups is a good approach
Actually, there is also a third approach using that the group algebra is positively based in the KL basis and that all the left cell modules are simple.
@LeakyNun What time is he usually on?
no idea
@TobiasKildetoft what is Donkin's tilting conjecture?
11:34
It states that certain indecomposable modules for reductive algebraic groups in positive characteristic remain indecomposable when restricted to a certain subgroup
(probably the vaguest statement of it I have ever written)
@Astyx!
Hi
@TobiasKildetoft!
@Astyx Ahh, you are here
Well that's lucky !
Did you see my question ?
11:44
Yeah, I wrote some suggestions a bit further up, for Leaky to give to you
The best reference really depends on how advanced you want it, and how generalizable it should be.
Well I was wondering whether this process was applicable for any group G given a subgroup H, and viewing the Specht space as $l^2(G/H)$
In order to understand the machinery behind all this better
my mission here is fulfilled, bye
Also I wanted to know whether there was an "easy" proof that the specht space of the dual is almost the original specht space. if that makes sense
@Astyx Not sure what you mean. The dual is always the same as a vector space. We only change the action
Anyway, you cannot in general expect to have nice enough properties to get everything via subgroups and some extra process (at least not in any uniform way). At least outside Coxeter groups
Also not sure what you mean by $l^2$
I mean, given a partition $\lambda$, $S^\lambda \sim S^{\lambda^*}$ where $\lambda^*$ is the dual of $\lambda$
Where, in this article, it means you flip the diagrams so that collumns becomes rows and vice-versa
11:52
right, this is something very particular to the way we have indexed the reps
and this indexing is not nearly as canonical as it may seem
(though it does seem to be the one that appears in any sort of uniform treatment, even those that seem completely unrelated)
(By $l^2(E)$ I mean function $E\to\Bbb C$ with some kind of scalar product, which appears in the paper I linked)
Hey, I got a quick and maybe stupid question
So you're saying that my concept of duality isn't as obvious as I think it is ?
@Astyx No, I am saying that the fact that dualizing the rep corresponds to dualizing the partition is something very special
@Astyx Anyway, $l^2(G/H)$ will almost never be the correct thing, as it will rarely be irreducible. You need some way to single out an irreducible constituent in it.
Right
I meant viewing the Specht space as part of $l^2(G/H)$, but I do not have a generalized view of what that part ought to be
11:59
There are some different ways to do this, depending on the group, but there will rarely be any really good one (outside Coxeter groups)
That space is really just the induced rep, and we induce from a Young subgroup. The Specht module is the unique one that does not occur in any such induced modules for larger partitions (in the dominance order)
and it even occurs with multiplicity $1$ the first time it appears
I have a collection of processes that will each end at some random real-valued point in time, and at a particular time I am going to examine all the data I have collected for these processes.

There is a probability density function that governs how long these processes each last, p(X|\mu), and I want to estimate \mu. The values that have ended correspond to points in the density function, but those that have not ended correspond to points in the CDF, ergo actual probabilities. I'm not sure how to reconcile these into a single likelihood function. Is anyone knowledgable?
Sorry if I'm kinda dumping that in the middle there, I feel like it's a stupid question with an easy answer, but I'm a very stupid person, and therefore outmatched by it
A Young subgroup is a subgroup that leaves the rows of a tableau untouched right ?
Yes, that is one way to see it
It is just the natural way to associate a subgroup to a partition (to $(n_1,\dots, n_m)$ associate $S_{n_1}\times\cdots\times S_{n_m}$)
This phenomenon is really a special case of what happens for Soergel bimodules (or, since this is a Weyl type, for projective functors on Category $\mathcal{O}$).
So you said that this does not generalize nicely to any group, and the reason it works for the symmetric group is because it's Coxeter right ?
basically yes
You need some nice collection of subgroups to work with, and you won't have that in general
But for Coxeter groups you have parabolic subgroups, which do the trick
12:07
Do you know where I can find a proof for Coxeter groups in general then ?
The Young subgroups are then the parabolics for $S_n$ (up to conjugation)
Geck-Pfeiffer is a good source for Coxeter groups in general, via Lusztig's $j$-induction
(which basically specializes to this whole thing for $S_n$)
Do we also have that the Specht representations are all the irreducible representations of any Coxeter group or is that specific to the symmetric group ?
Well, they are not called Specht for other types
But we do get all of them via a similar process via induction from parabolics and picking out suitable factors
But the parabolics will be indexed by some slightly more complicated sets (pair of partitions in type $B$ for example).
And as far as I recall pairs of partitions with the same parity length in type $D$ (or something like that).
(I mean lengths both odd or both even)
Interresting
I have to go right now, otherwise I won't be able to eat anything this lunch. Thansk a lot for your time, I'd love to continue this conversation later on ! And I'll look Geck-Pfeiffer up.
Bye !
12:36
@Astyx Bye (sorry for the delay, my son wanted to play a game).
13:31
0
Q: Linear algebra, OpenCV distance point plane.

user8469759I'm reading through an opencv example that performs camera alignment. More specifically the homography is being estimated in such code, but there are these few lines that aren't quite clear to me: //! [compute-plane-normal-at-camera-pose-1] Mat normal = (Mat_<double>(3, 1) << 0, 0, 1); Mat norma...

can anyone give me an insight?
13:52
@ÍgjøgnumMeg: Yeah I was quite busy the last half year. I'm fine. Hope you too!
14:28
@Rudi :D Hur är läget med dig då?
allt bra
jette bra
:D skönt att höra det
@Rudi jag fick avslag från universitetet i Heidelberg så måste jag studera på ett annat universitet :(
lol
 
2 hours later…
16:23
@ChthonicOne
You said
"However, they are mixing up infinity as a set, and infinity as an element of the set of infinity."

and immediately after you said:

"X + infinity = infinity only because on the left we have one element from the set of infinity "

and then you say:

"I did not say the set of infinity"
"Thus infinity + 1 = infinity only makes sense if we understand that each infinity in the equation is a different element in the set, where their difference is exactly 1."

I'm not sure where you are getting that from.
@ÍgjøgnumMeg det låter ingen bra. Sorry to hear that.
16:51
$2557^{341} \mod{5681} $, can any clever thing be done about this?
@Lozansky 5681 has three distinct prime factors, Chinese remainder theorem it perhaps
2557 is a prime
17:39
@ÍgjøgnumMeg just watched a movie about the chinese remainder theorem (av.tib.eu/media/19882). Is there a relation between solving ODEs and solving modulo equations?
I mean in both cases you can search for a general solution by adding the solution "homogeneous" problem to a particular solution of the inhomogeneous problem.
18:24
@Rudi_Birnbaum yes and the connection is jordan normal form
@LeakyNun: cool, thank you!
19:28
@Rudi_Birnbaum: The phenomenon you're observing holds whenever you have a linear problem and you compare solutions of the homogeneous and inhomogeneous problems. So it applies to linear ODEs and to linear systems of equations in general.
hi @MikeM
@LeakyNun Your Jordan normal form comment seems totally out of place here.
19:44
@Rudi_Birnbaum Yeah I have the formula, I was just wondering if one can get the generating grammar from it.
I also have a simpler (I think) generating function than the one that's listed on OEIS.
Zee
Zee
ODEs are the hardest branch of math, even schemes are trivial in comparison
Hate ODEs
Hi @Ted @Mike @Zee et al
ODE an die Freude
3
Zee
Zee
+1 Idk what that means , but it sounds violent
lol, that's the German title of the piece "Ode to joy"
In German, the piece is used to model various physical phenomena
Zee
Zee
Huh?
19:57
idk just making a crappy joke
Zee
Zee
I don’t think you can use music like math.
Have you heard of the piece of music called "Ode to joy"?
Zee
Zee
Ya
I hate that piece . So overplayed
yeah I'm not a fan
20:22
@Zee lol
@ÍgjøgnumMeg :-)
user image
2
(c) T. Tao
hi @Daminark @Rudi_Birnbaum
@Rudi nice !
21:09
0
Q: Grammar of words with exactly $k$ prefixes in another grammar

user76284Given a context-free grammar $G$, how can one systematically construct a grammar $G_k$ such that $$ L(G_k) = \{w \in \Sigma^* : |\text{Pref}(w) \cap L(G)| = k\} $$ where $\text{Pref}(w)$ is the set of $|w|+1$ prefixes of $w$? Assume $G$ is unambiguous, if needed.

Hello @LeakyNun (sorry I forgot my manners)
 
1 hour later…
22:13
Does anyone know what this notation represents? gyazo.com/65f2764c3a1b1aa14ca42a2d4506950c
22:25
@WilliamOliver it's a block matrix
the big X means "we don't care about what it is here"
the big 0 means "everything here is 0"
@LeakyNun Hm okay thanks
It's a tic-tac-toe.
@LeakyNun And does the bottom right means "Identity matrix" ?
Thanks for the help!
22:46
Let's say I have some $R$-module $M$ which has generating set $E \subseteq M$ and another $R$-module $N$, can I define a $R$-linear map $f : M \to N$ by defining $f(e_i) = \text{something in $N$}$ for all $e_i \in E$ and then extend by linearity to get a well-defined $R$-linear map?
no, have you tried any examples where $M$ isn't free?
M can be free and it still isn't guaranteed to work.
Good catch, but at least in that case there are easier conditions one can add to make it true :)
They are the same in either case.
Oh, I didn't think it through.
23:08
You need to pay attention to the relations if it's not a basis, right. But at least one can say "basis" in that case, and it's in general hard to determine the relations.
I was just thinking of having the given $F_j\to N$ factor through the implied $F_j\to M$ :P
jk
@MikeMiller I haven't tried any examples where $M$ isn't free, but I will now
@KarlKronenfeld I just meant things get irritating if you chose too many generators because you just threw generators at a dartboard to see what sticks.
23:26
Then, you have those self-annihilating torsion darts
I think that's when you throw a second dart and hit the other one straight in the middle, but the other one was an exploding dart and suddenly they're both gone
This is a useful metaphor
It's like come on, I just wanted to divide by 2

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