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12:05 AM
 
12:16 AM
Hi everyone! In this page https://en.wikipedia.org/wiki/Skolem_normal_form there is a part with
"... by the axiom of choice ..." - Is AC even required? Can't we just say that, there is an element in the model which satisfies this formula, let's say b, then we create the constant b function
 
Bob
Hi
 
Hi guys! I have two points A and B. I also have a point C so the $\angle ACB=120$. If I insert the triangle $ABC$ in a circle. How should it look like?
 
@vesii, just choose a portion of the circle which is 240 degrees, choose A and B from the boundaries and C randomly from the smaller side
 
I'm trying to find the "geometric place" of all points of $C$ so the angle is $ACB=120$ using analytical geometry. I took three points on a circle. So If the angle $ACB$ is 120 degrees then $AOB$ is also 120 degrees. But now I'm stuck.
As I understand I need to set $C$ to be $(x_C,y_C)$ and somehow to get an equation using only those points.
I though of using the cos theorem, but I also have $A$ or $B$ in the final equation.
Any ideas?
I need to find the "Locus"
 
12:35 AM
you can select a 240 degrees portion by creating a 60 degrees and adding a 180 degrees and then the placement of C doesn't matter, it's 120 degrees in all cases (for the smaller side)
 
But how I find the Locus ?
 
In geometry, an inscribed angle is the angle formed in the interior of a circle when two secant lines (or, in a degenerate case, when one secant line and one tangent line of that circle) intersect on the circle. It can also be defined as the angle subtended at a point on the circle by two given points on the circle. Equivalently, an inscribed angle is defined by two chords of the circle sharing an endpoint. The inscribed angle theorem relates the measure of an inscribed angle to that of the central angle subtending the same arc. == Theorem == === Statement === The inscribed angle theorem states...
 
Yes I know that I need to use the inscribed angle. Thats why I told $AOB$ is $120$.
 
@Thorgott oh my god.
 
12:54 AM
:)
 
so easy lmao
What I don't understand is that I thought about this
it was one of the first things I thought but, for some unknown reason, $\frac{mn}{d^2} = l/d$ didn't look like a solution
hmm, what about the last implication?
ah, forget it
$\left((ab)^d\right)^{l/d} = (ab)^l$
 
 
1 hour later…
2:19 AM
@LucasHenrique Yeah, you might be right. We just need to make sure we have $m/r$, $n/s$ relatively prime and their product has to be bigger than $n$.
And, yes, to create $r$ and $s$ I found it easiest to use prime factorizations.
 
I usually want to avoid prime factorization, for no reason at all
Commodity maybe
 
@vesii You have the answer. $C$ can be anywhere on that arc. You do still need to prove there are no other solutions.
@Lucas: Well, if you figure out a proof (recipe) without it, please tell me.
I think you need it to make sure that $m/r$ and $n/s$ are relatively prime. But maybe you can be clever.
 
@Thorgott's proof, just above, is very clever and doesn't deal with primes
 
Oh, is Thorgott being smart? :)
Oh, that looks cleverer than I was. Darn.
I worked with $rs=d$, by the way, so it would come out exactly right.
 
2:42 AM
but doesn't @Thorgott proof imply that $4+2 = 6$ has order $6$ over $\mathbb Z_{12}$?
the last implication is the problem. because, in fact, it says that among $(ab)^d, (ab)^{2d},\ldots, (ab)^{d\frac ld}$, the only term that equals $e$ is $(ab)^l$
that doesn't guarantee that there aren't other divisors of $l$, namely $r$, not being multiples of $d$, with $(ab)^r = e$
e.g.: in my example, $m = 3$, $n = 6$ so $d = 3$, $l = 6 \implies l/d = 2$ and $ab = 6$. It's correct that $(ab)^3 = (ab)^d \neq e$ but $(ab)^{6} = (ab)^l = e$. However, $(ab)^2 = e$, so the order is not $6$
@Thorgott: I'm happy I wasn't too picky when thinking about the last implication :P
 
3:05 AM
looks like I was too clever to be true
 
Here is what I think is a short argument, no explicit numbers or painful fiddling
Let A be a finite abelian group. Let x be an element of order m. Let y be an element of maximal order n. If (m,n) = m, then we are done, as this means m divides n. If m > (m,n) > 1, then replacing x by (m,n)x, we see that there is an element z of order k with (k, n) = 1. I claim this is impossible.
For consider the element z+y. We have kn(z+y) = 0, so ord(z+y) | kn. If k' is a proper divisor of k, then k'n(z+y) = (k'n)z, which is not zero, as k'n is not zero in Z/k (here we use that (k,n) = 1 and that k' is a proper divisor). Similarly for any divisor n' of n. Thus ord(z+y) = kn > n; this contradicts the assumption that y had maximal order.
 
3:45 AM
okay, primitive element theorem is good to go for finite fields. :)
why is separability needed to prove the primitive element theorem for finite extensions (with infinitely many elements)?
the proof I know uses the polynomial $p(t) = \prod\limits_{i=1}^r (\alpha+t\beta - \sigma_i(\alpha)- t\sigma_i(\beta))$, where $\sigma_i$ is an embedding of the extension into is algebraic closure and argues that since this polynomial has finitely may roots, there's a $c \in E(\alpha, \beta)$ which is not fixed by any of the embeddings
and I honestrly can't remember why, but this implies that $E(\alpha, \beta) = E(c)$. (maybe something using polynomials? is it where separability is needed?)
 
4:27 AM
I think what's going on is that the image of an element $\alpha$ under an embedding $\sigma$ is a zero of the minimal polynomial of $\alpha$ still. So if no $\sigma$ fixes $\alpha$, the extension $K(\alpha)/K$ has degree at least the number of $\sigma$s. For $\alpha$ to be primitive, you want this to be the degree of your initial extension, say $L/K$. There are precisely enough $\sigma$s for this to work (namely $[L\colon K]$ many) if the extension is separable.
 
5:20 AM
I think I get it. Why wouldn’t we have enough $\sigma$s, otherwise?
 
@LucasHenrique it's one of the characterisations of separability
but you can prove it for simple extensions and then use two tower laws
 
Also, mainly @Ted and @Thorgott: I apologize for being so insistent and bragging about the same topics over and over. I’m really interested in the topics I’m studying but my book really does not help me, and I don’t know better resource.
Thanks to everyone helping me. :)
@LeakyNun ok, that makes sense.
 
@LucasHenrique that isn't what bragging means
o que queria dezir voce?
 
Blabbering*
 
ah
ah, dizer not dezir
spanish is decir
 
5:29 AM
@LeakyNun hahaha. That would be “o que você queria dizer?”
 
oh you put the "voce" there?
 
Mobile chat often sends duplicated messages
 
como voce pode ver, eu nao falo portugues bem
 
That’s perfect
I don’t speak English very well so we’re even :P
@LeakyNun there’s a weird thing about verb conjugation in Portuguese. “I, you, he” and “they” have each different verb conjugations
To drink, for example, in the order I specified above: bebo, bebes, bebe, bebem.
 
em espanhol tambem
 
5:35 AM
The weird thing in portuguese is that we don’t use the “you” conjugation in informal language, because you $\mapsto$ tu, and we use “você” in informal language, opposed to Spanish which uses “usted”
That usually confuses people, even other members of Latin America
@LeakyNun asking questions is like claiming sentences but with an “?” in the ending
You like Ted $\mapsto$ Você gosta do Ted
Do you like Ted? $\mapsto$ Você gosta do Ted?
 
I see
@LucasHenrique voce usa "tu"?
 
Rarely
I live in São Paulo, and that’s unusual. The other regions use “tu”, but each in its own way
In Rio they use “tu” a lot, but not “tu bebes”, but “tu bebe”
 
eu acho que o sotaque brasilero nao usa "tu" jamais
interesting
 
Well, “tu” is never used “correctly”, but it’s used a lot
The pronoun that is really awkward to hear anywhere is “vós”, the plural of “tu”
No one knows how to conjugate the verbs for “vós”
It’s only used in very respectful and formal reunions, like in churches or in very specific, technical/bureaucratic contexts
Eu imagino que vós, leitores, estais deveras entretidos por um linguajar tão sofisticado e desnecessariamente pomposo! :p
 
5:58 AM
@LucasHenrique @LeakyNun Hello !
 
Hey there, @Jacksoja
 
Did you solve that question ? haha
 
Yes, after a long time
 
@LucasHenrique I did not think of it much since yesterday,have other courses
but ill give it a try today or tomorrow
we can compare proofs if you want
 
Sure.
 
 
4 hours later…
10:01 AM
i think i've finally understood the diagonalization.
 
11:01 AM
Linear operator diagonalization?
 
Where do people often use symbols $\imath, \jmath$?
 
$\imath$ is usual to denote immersion from a ring into another one
Embedding*
 
No result for this...
Some results about this
 
11:17 AM
@LucasHenrique yeah
 
11:37 AM
@Andrews $\iota$ looks very similar to $\imath$ and is probably more well-known, so I guess people just use that for embeddings
 
Afternoon @Lukas
 
Buon pomeriggio!
 
Ah yes
Góðan dag
 
that looks like Icelandic
 
I think it's probably the same in Icelandic but it's Faroese lol
 
11:45 AM
Loquamur Latine! Non difficile est.
Quid hodie fecisti?
 
Nie moge mowic po lacinski
lol
 
Pōrando-go desu-ka?
 
si
Sagt man da als Antwort eigentlich einfach "desu" oder so?
lol
 
sō desu
oder
hai
oder
un
 
hai kannte ich
ah nice
 
11:50 AM
tiel falis la babilona turo
 
Ich will den Vorlesungsangebot fürs Sommersemester sehen
I don't think the trick you told me to use works anymore lol
 
oh damn
I'm doing it wrong hte
then*
 
Ĉu vi rekonas ĉi tiun lingvon?
 
Looks romance but also not
Or maybe latvian or smth
lol
 
11:53 AM
je suis esperantiste
 
Ah that was Esperanto?
 
yeah
"tiel falis la babilona turo", too
 
Ah nice
Quickly: what things do I need to change in the URL on the Vorlesungsverzeichniss?
with 1 s
lol
 
delete "&veranstaltung.semester=20192"
but it only seems to work on searches
so you can search for a particular lecture or seminar
and then deleting the above will give you the results for all semesters, including the next one
 
yeah that's cool
thanks hahaha
Ah Algebra 2 is with Böckle
 
11:58 AM
Böckle is great :)
there's also a Böckle seminar on noncommutative algebra
they also do Brauer groups in there
 
Ah nice, might do that too
 
so it's relevant for NT
and since it's a seminar and not a proseminar you can get masters credit, most likely
 
woohoooo
it doesn't look like AZT2 is there :(
 
yeah I think that was mentioned
Vogel has other teaching duties or something
but there'll be a CFT seminar at least
 
yeah I spoke to Rustam in the tutorial and he said it probably won't be offered
Oh there's a Seminar called Algebraische Zahlentheorie
so I guess that'll be the CFT seminar lol
Und Analytische Zahlentheorie!
 
12:13 PM
who's offering analytic NT?
oh, I see, it's Kohnen
then it'll be good :)
Kohnen is an internationally recognized expert on modular forms
there's even a book by Shimura which cites one of his papers
 
wtf
hahaha
That's nice then
 
there's even a Kohnen space of modular forms
 
and the Dozent for the AZT seminar is Leohardt
also woah
Quantenfeldtheorie, Teilchenphysik und Statistische Physik
lol
ah das ist ein Oberseminar
Nice ich glaub ich weiß jetzt was ich nehme hehe
 
12:32 PM
was denn?
 
uhm
Alg 2, AZT Seminar, Analytische Zahlentheorie, Funktionalanalysis
I can get credit for Alg 2 and Functional analysis
so that's fine
and I might take one more thing
Like Russian
or the non-commutative algebra seminar
or both lawl
 
@EdwardEvans Se esperanto intrigas vin, lernu.net vera bonas.
Ciao @Leaky
 
@LukasHeger hej
 
@EdwardEvans what's special about Ithkuil?
 
12:38 PM
@Edward lol, how many speakers does Ithkuil have?
 
It's an engineered language so probably 1
hahaha
It's a cool experiment though
96 grammatical cases
 
I like that for Esperanto it's not impossible to actually find other people to talk to, despite the fact that it's engineered
 
yeah indeed
although aren't there some native speakers of it?
Like a couple of thousand
in the sense that they grew up speaking it as their first language
 
yes, there are some native speakers
 
1:37 PM
Hello
Suppose a function f is holomorphic in some disc, then if I write the power series of f about the center of this disc, will the radius of convergence of this power series be atleast as much as the radius of the disc?
 
@feynhat yes
 
pretty sure you can strengthen that as such: the radius of convergence will be the radius of the largest disk, centered at the point of interest, on which f is holomorphic
 
@LukasHeger How do I show this?
 
(e.g., if it's holomorphic for |z|<1 and its power series about z=0 has radius of convergence 2, then it's also holomorphic for |z|<2)
 
Is there any reference for this? I have checked Conway and Papa Rudin, and couldn't find anything.
I tried using identity theorem
 
1:47 PM
the identity theorem won't help
you have to use the Cauchy integral formula
supppose wlog that the center of the disk is $0$ to make notation less cumbersome. Let $C$ be circle around $0$ traversed once and of radius $<r$, where $f$ is holomorphic on the disk with radius $r$. Then we have the Cauchy integral formula $f(z)=\frac{1}{2\pi i}\int_C\frac{f(w)}{z-w}\mathrm{d}w$
Now the idea is the following: use that $\frac{1}{w-z}=\frac{1}{w} \frac{1}{1-\frac{z}{w}}=\frac{1}{w}(1+\frac{z}{w}+\frac{z^2}{w^2} + \dots+(\frac{z}{w})^{n-1}+\frac{(\frac{z}{w})^n}{1-\frac{z}{w}})$ (this is just a partial geometric series)
 
@LukasHeger denominator should be $w-z$
 
oops
thanks
too late to edit
 
happens
 
$|z| > r$, right?
 
no, $|z|<r$
so we obtain $\frac{1}{w-z}=\frac{1}{w}+\frac{z}{w^2}+\dots+\frac{z^{n-1}}{w^n}+\frac{z^n}{w^n} \frac{1}{w-z}$
now we plug that into the integral:
$f(z)=\frac{1}{2\pi i} \int_C \frac{f(w)}{w-z} \mathrm{d}w=\frac{1}{2\pi i} \int_C (\frac{f(w)}{w}+\dots+\frac{f(w)z^{n-1}}{w^n})\mathrm{d}w+\frac{1}{2\pi i} \int_C \frac{z^n}{w^n}\frac{f(w)}{w-z}\mathrm{d}w$
Now recall the generalized Cauchy integral formula: $f^{(n)}(z)=\frac{n!}{2\pi i}\int_C\frac{f(w)}{(w-z)^{n+1}}\mathrm{d}w$
(which is obtained from just differentiating the usual Cauchy integral formula)
apply that with $z=0$ and compare with the terms we have in our integral, then we see that $f(z)=\sum_{k=0}^n \frac{f^{(k)}(0)z^k}{k!}+R$, where $R$ is the remainder term $\frac{1}{2\pi i} \int_C \frac{z^n}{w^n} \frac{f(w)}{w-z} \mathrm{d}w$
so all that is left to show is that $R \to 0$ as $n \to \infty$
I'll leave that to you :P
it's just some standard inequalities you have to deal with a lot in complex analysis
use that continuous functions attain a maximum on a compactum etc.
but that's the idea, at least
hope this helps @feynhat
 
2:08 PM
We can bound the integral by the perimeter of C times the supremum of the integrand. But the integrand vanishes as n tends to infinity because |z| < |w| = r... right?
 
yeah exactly
 
Right so this shows that f equal to the power series for any |z| < r.
So, the radius of convergence is atleast r.
 
2:56 PM
Can someone please help with the following: Is [max(0,x^2 sin(1/x))]^2 differentiable at all points?
 
3:08 PM
are you familiar with the function $x^2\sin{1/x}$ as its own? @gustaffIR
 
@sevdaicmis Its differentiable everywhere but derivative is discontinuous?
 
3:32 PM
it's not differentiable in 0
wait what am i saying :)
 
3:59 PM
let's call it $f$, $f(x)=x^2\sin{1/x}$ for $x\neq 0$ @gustaffIR
the domain of this function can be extended while keeping it continuous. do you know about that?
 
4:36 PM
you can ping me when available @gustaffIR
 
4:47 PM
Anyone here mind tackling a conceptual issue in algebraic geometry I'm having?
 
5:00 PM
@JamalS depends on the issue
 
@LukasHeger I'm trying to understand the concept of a generic point corresponding to the zero ideal, as it pops up in the definition of a principal divisor.
@LukasHeger In Hartshorne, a principal divisor is defined as $(f) = \sum \nu_Y(f) Y$ where $\nu_Y$ is a discrete valuation for the discrete valuation ring $\mathcal O_{\eta,X}$ where $\eta$ is the generic point on $Y \subset X$.
I understand it in the case that the prime divisors are points, then the discrete valuation is just the order of the poles or zeroes. But in the general case, in practice, what do you take as the discrete valuation?
 
5:18 PM
@JamalS if you have a regular point on a curve, then the stalk at that point is automatically a discrete valuation ring with quotient field $\mathcal{O}_{\eta,X}$ (which is the function field of the $Y$)
 
@LukasHeger So I can choose any regular point on a prime divisor, in order to define a discrete valuation?
 
no wait, I was just thinking of curves
I think in general the answer to your question is just an algebraic coincidence: if $X$ is normal, then $\mathcal O{\eta,X}$ happens to be a discrete valuation ring, because its Krull dimension is $1$ and it is a normal domain and these happen to be discrete valuation rings
there is not necessarily a simple description of the discrete valuation in geometric terms
 
So actually computing the discrete valuation on some function isn't an algorithmic step, it requires more of a case by case treatment?
 
well, if you know the maximal ideal of your discrete valuation ring, then the discrete valuation is easy to describe: if $f \in \mathcal O_{\eta,X}$, then the normalized discrete valuation is the biggest integer $n$, such that $f$ is contained in the $n$-th power of the maximal ideal
all I'm saying is that there is no particular "geometric" description for the discrete valuation
 
@LukasHeger Oh okay, I'm a physicist so maybe my commutative algebra is lacking - I didn't realise you could approach it that way.
 
5:26 PM
the geometric intuition comes from the special case where you actually have poles and zeroes
 
I see, so the special case is more the exception, in general there isn't a nice geometric picture.
 
yes, the intuition comes from the special case
@JamalS maybe it's helpful if you look at a somewhat non-geometric example, say $X=\mathrm{Spec}(\Bbb Z)$, pick a prime $p$, then we have a closed point $(p)$ in $X$ which is a prime divisor if we consider it as the closed subscheme $\mathrm{Spec}(\Bbb F_p)$
the stalk is the localization $\Bbb Z_(p)$ and the discrete valuation is the usual $p$-adic valuation
Now if you take a rational number, say $\frac{a}{b}$, then you can reasonably think of the $p$-adic valuation as counting poles and zeroes of that rational number as a "meromorphic function" on $X$
for example, $\frac{5}{6}$ has a zero of order $1$ at $(5)$ and poles at $(2)$ and $(3)$
at least we can think of it like that through analogy
we can think of $\frac{5}{6}$ as a "meromorphic function" on $X$, because if we have a point $(p) \in X$, then if $p$ doesn't divide $6$ (where this function has a pole), then $\frac{5}{6}$ makes sense in $\Bbb Z/p\Bbb Z$
so in this sense, the notions of poles and zeroes do kind of make sense in general
I don't know if this is more confusing or helpful
 
@LukasHeger That makes it much more clear, I can see the connection now between the general case, and poles/zeroes.
 
glad to help :)
 
Hartshorne can sometimes be too brief about details like this.
 
5:33 PM
yeah, it's not an easy read
as a number theorist, I like to look at examples such as $\mathrm{Spec}(\Bbb Z)$ or $\mathrm{Spec}(\Bbb Z[i])$, it helps me to see how thinks work out if you no longer have a base field
 
@LukasHeger Funny you say that, after Hartshorne I'm planning to work through Milne's Etale Cohomology, to understand things like the Weil conjectures.
 
ah, funny you mention that, I use étale cohomology in my bachelor thesis
 
@LukasHeger That's considered advanced for a BSc, no?
 
yes
quite advanced
 
I tried to find the least category theory intensive book on etale cohomology, and it seems like Milne is probably the best choice in that regard
 
5:37 PM
Milne is quite good, I used it as my main source for étale
 
 
2 hours later…
7:24 PM
@MikeMiller TIL an instanton is just flowline of a gradient
I was reading Raoul Bott's article, "Morse theory indomitable".
He explains what he calls the Thom-Smale-Witten complex, where given a Morse-Smale function $f : M \to \Bbb R$, you let $C_k^f(M)$ be the free abelian group generated by the index $k$ critical points and $\partial : C_k^f(M) \to C_{k-1}^f(M)$ is given by $\partial a = \sum_b \varepsilon_{ab} b$ where $b$ varies over index $k-1$ critical points and $\varepsilon_{ab}$ is the signed count of the number of instantons (flowlines of $\nabla f$) exiting $a$ and entering $b$.
 
hello. If $R$ is a ring, and $M$ is an $R$-module with generators $e_1,\dots,e_n$, and $M$ is free (and we ask in that case that the generators are linearly independent), then the map $f:R^n\to M$ given by $(r_1,\dots,r_n)\mapsto r_1e_1+\dots+r_ne_n$ is an isomorphism. If $M$ isn't free though, then that map is only surjective, but has some kernel, giving us $R^n/\text{ker}(f)\cong M$. Is this what is meant by giving generators and relations for $M$, i.e. that...cont
$M=<e_1,\dots,e_n>/<q_1,\dots,q_m>$ where $q_i = r_{1,i}e_1+\dots + r_{n,i}e_n$ for generators $(r_{1,i},\dots,r_{n,i})\in\text{ker}(f)$ of the kernel of $f$.

(I know how one can formulate this more normally, but I was specifically wanting to relate it to the quotient of a free module we get for finitely generated modules)
 
I dunno how to formally do the signed count though. I think it's like, for any critical point $c$ let $S_c$ and $U_c$ be the stable and unstable manifolds of the flow of $\nabla f$ at $c$. Since $f$ is Morse-Smale, $S_a$ and $U_b$ are always transverse for any pair $a, b$ of critical points. So $\dim (S_a \cap U_b) = \lambda_a - \lambda_b$, difference of the indices. If this difference is $1$ then we have finitely many flowlines from $a$ to $b$.
Now if $\gamma$ is a flowline from $a$ to $b$, then for any $x \in \gamma$, there is a short exact sequence $0 \to T_x \gamma \to T_x S_a \oplus T_x U_b \to T_x M \to 0$
 
he calls those instantons? funny
 
The orientation on $T_x \gamma$ comes from the direction of the flow of $\nabla f$, and the same is true of $T_x S_a$ and $T_x U_b$. I guess I count $+1$ if the exact sequence preserves orientation, $-1$ otherwise
It seems a little fidgety but I get the picture
@MikeMiller Ya
 
@BalarkaSen In addition to the ambient orientation, you need to specify in advance an orientation on each stable manifold --- this ends up being irrelevant up to iso
There is a trick to get around this but it's nice to think in terms of specifying the orientations
 
7:39 PM
Isn't the orientation on the stable manifold naturally coming from the direction of the flow?
 
Not sure I follow
The direction of the flow on the stable manifold is "inwards to 0"
 
Because $S_a = \{x \in M : \lim_{t \to -\infty} \phi^t(x) = a\}$, so it's a union of a bunch of flowlines which "converge" to $a$, so doesn't those tangent directions to the flowlines give some orientation?
I suppose not
 
Naw, a nonvanishing vector field doesn't give an orientation
 
Ya figured
 
The Klein bottle has a nonvanishing vector field
 
7:41 PM
I'm dumb
 
you're good homie
 
Each term in $\partial^2 a$ counts the total number of 2-broken flowlines from an index $k$ critical point to an index $k-2$ critical point. These should be zero somehow, because $M(a, b) \cup M(b, c)$ for a triple of critical points $a, b, c$ each having index less than the previous one is a manifold which can be naturally "compactified" to a manifold with boundary consisting of the 2-broken flowlines from $a$ to $b$ to $c$.
Once I quotient by the $\Bbb R$-action by flow, it's a 1-manifold with boundary.
That has an even number of boundary components, cancelling in pairs. Something of this sort.
Little hard to imagine this
 
The tilted torus example is always nice to think about
 
I was fiddling with that picture a little
 
For orientations, if $b$ and $a$ have index difference 1 and $\gamma$ is a flowline between them, we have as vector spaces $T\gamma = TU_b \cap TS_a$, right? So the sign should be comparing the orientation of these.
 
7:50 PM
True
 
The first is oriented by direction. For the second we need a consistent rule to orient the intersection of transverse subspaces. There is some standard way, but I don't know what properties you want it to have; probably associativity, and I am too lazy to check if this way does that.
 
I can definitely see that in the tilted torus all the boundary maps are zero, for example, so I intuitively get it. The complex is $0 \to \Bbb Z \to \Bbb Z^2 \to \Bbb Z \to 0$, all maps zero.
Essentially because pairs of flowlines come from opposite directions cancelling each other
 
But let $V,W \subset T$ be transverse subspaces of an oriented vector space. Write $C$ for their intersection. Then we may write $V = C + V'$ and $W = C + W'$ for subspaces $V',W'$ which intersect $C$ trivially. We have $T = C + V' + W'$. There is a unique way to orient $C$, and thus $V',W'$, so that the above isomorphism with $T$ is oriented.
There's probably a better way to say this.
 
This complex $C_k^f$ itself is isomorphic to the cellular chain complex that comes from the cell decomposition induced by $f$ on $M$ in general, yeah?
@MikeMiller Makes sense
 
I think this isn't associative
Whatever
Surely the orientation on a transverse intersection is in G&P
Ah here's better. We have C + T = V + W. Orient C such that this is oriented.
 
8:05 PM
I don't know exactly what you want to do. If V, W are oriented subspaces of T then there is a short exact sequence $0 \to V \cap W \to V \oplus W \to T \to 0$. I oriented $V \cap W$ according to this rule.
That's the standard technique
I thought the problem here is we want a canonical orientation on the stable manifolds
(So $V$ and $W$ don't have canonical orientations given)
I wonder how one would build such a complex for a stratified space. I suppose a Morse function $f : W \to \Bbb R$ on a Whitney stratified space $W$ would be a stratumwise Morse function such that $df$ doesn't vanish on the limiting tangent planes on the lower strata.
I can have flowlines entering lower strata. Even worse, I might have critical points on different strata with equal index
I suppose $U_a$ and $S_a$ still make sense, they'll just be substratified spaces instead.
No, I cannot have flowlines entering lower strata, because if $A < B$ is a pair of strata, $d(f|_B)$ would not vanish on the limiting tangent planes to $A$. I can have flowlines converging to broken flowlines which enters lower strata, I suppose
Too complicated
I expect $\overline{M(a, b)}$'s to be stratified.
 
8:43 PM
@BalarkaSen I just wanted an associative rule for orienting transverse intersections.
 
a transformation not having two linearly independent eigenvectors cannot be diagonalized, right?
 
anyone here know about Wilk's theorem for $-2\ln(\Lambda)$? I'm confused how the convergence in distribution works.
 
9:01 PM
@sevdaicmis unless it's on a one-dimensional space, of course
 
oh yeah, i was thinking strictly for $\mathbb{R}^2$ for whatever reason.
but got your point
 
also, I don't know Wilk's theorem, but as far as convergence in distribution is concerned, it may (or may not) be more instructive to think of it in terms of weak convergence of measures
 
9:30 PM
Hello
What does the notation $min_{I\in \mathbb{N} (a_I\neq b_I)$ mean?
 
presumably the smallest $I\in\mathbb{N}$ such that $a_I\neq b_I$
 
yeah that's what I thought, but usually the subscript denotes the condition
right?
@Thorgott
 
10:06 PM
ah, I see what you mean
hmm yeah, it's weird
 
@topologicalmagician: It's garbage.
You're minimizing $a_i\ne b_i$ over all $i\in\Bbb N$.
Context?
 
@TedShifrin I'm trying to show that $d(x,y)=0$ if x=y and $2^{-k}$ when $k=min_{I}(a_i\neqb_i)$ is a metric on the cantor space, where $x=0.a_0a_1....$ and $y=0.b_0b_1....$
 
Oh, they wrote it wrong. They mean $\min\{i\in\Bbb N: a_i\ne b_i\}$.
The first place in which the two expansions differ.
 
Oh
thanks Ted
 
Sure :P
 
10:15 PM
I'm really annoyed I'm taking the most advanced analysis class atm and nowhere do they talk about differentiation in $\mathbb{R}^n$... nor did I take it any class prior to that one
 
Yeah, that's ordinarily the second semester of analysis (if taught at all). Guess you need to read my book or watch my videos :P
 
@TedShifrin yeah, I have your book and I try to watch your videos whenever I have time.
 
Aww ... such a patriot :P
 
The multivariable analysis course at UGA used to run very rarely. Once I started my course, all the potential undergraduate audience for that course had already had my course, and the one or two grad students who might have taken it weren't enough, obviously.
 
10:25 PM
I have a large framed photo of Ted above my fire place.
 
Ha ha ha ha ha.
Good for keeping the bats and thieves at bay.
 
I thought bats were fond of vampires???
 
Well, that distracts them from you.
 
Yes, it's a handy tool.
Okay, I love thine Ted. I must go.
BYE
 
Bye.
 
 
1 hour later…
11:54 PM
Hi chat.
 

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