00:00 - 14:0015:00 - 00:00

12:03 AM
@loch aha the answer comes one page after
$\Lambda_M$ is filtered iff $M$ is flat

i see

3 hours later…
2:52 AM
if $\mathfrak a$ is an ideal in $R$ then $\mathscr F(N) := \mathfrak aN$ is a functor R-Mod => R-Mod?

@LeakyNun yes

and there is a natural transformation $\mathfrak a \otimes_R - \implies \mathscr F(-)$

3:11 AM
yes

3:43 AM
heyo chat

3:55 AM
I am reading Howie's Complex Analysis. There I see this remark: The observation that $c$ and $d$ are inverse points is the key to showing that every circle can be represented as $\{z:|z-c|=k|z-d|\}$. Suppose that $\Sigma$ is a circle with centre $a$ and radius $R$. Let $c = a + t$, where $0 < t any online tool to do the symbolic calculation ? I have two nonlinear equations in two unknowns any suggestions? 2 hours later… 5:40 AM 3 Oh no!! 5:51 AM Devils rep @Secret quack 1 hour later… 7:09 AM Ugg, I'm struggling with understand the tricks used in holder duals, they basically substitute in a second variable and maximize over it. Then the dual becomes the auxiliary space the second variable lives in? 8:06 AM Question Say we have two concentric circles, one of radius$r$and one of radius$R$Choose a point from each at random. What's the expected distance between them? just integrate some expression "bUt YoU cAN dO iT eLEMentARiLy ThO" In other words, if we have two planets in circular orbits, what's their average distance @LeakyNun I mean I dunno what the answer is so any technique is fine I've just been too lazy for the integral so far$\displaystyle \int_{0}^{2\pi} \sqrt{(r-R\cos\theta)^2 + (R\sin\theta)^2} \ \mathrm d\theta$this looks familiar is this right? do I need to divide by 2pi or something$\displaystyle \int_{0}^{2\pi} \sqrt{(r-R\cos\theta)^2 + (R\sin\theta)^2} \ \mathrm d\theta = \int_{0}^{2\pi} \sqrt{R^2 + r^2 - 2rR\cos\theta} \ \mathrm d\theta$(hey I derived cosines theorem!) I'd need some way of evaluating$\int_0^{2\pi}\sqrt{a+b\cos\theta}d\theta$yeah no you can't evaluate this integral directly, you need to recognize that this is actually the area of the large circle so it evaluates to$\pi R^2$8:12 AM That sounds wrong I'm fixing the point on the inner circle and moving the point on the outer circle around the line connecting the two points trace out the large circle That would only work if sweeping around the outer point kept the angular velocity of it around the inner one constant angular velocity is always constant no matter around which point of reference That's clearly not true why not? 8:16 AM If the inner point is right next to the circle then the angular velocity would be really fast as the outer point goes right by it yeah? and it would be slow as it's far away but you need to multiply by... wait wtf @MarioCarneiro hi! lol hi @AkivaWeinberger you need to multiply by the distance so it's ok @MarioCarneiro I hope you recognized me lol 8:17 AM yes I did no need to stop on my account This is like when you do a change of variables and you need to multiply by something to make the differentials work out @LeakyNun Maybe you could multiply by the distance from the inner point to the tangent line of the outer point in order to? @AkivaWeinberger You're doing$\int d~d\theta$yeah and I said the answer is$\pi R^2$but the area of that triangle isn't$d~d\theta$it's$h~d\theta$where$h$is the distance to the tangent so$\int h~d\theta$will equal$\pi R^2$but$\int d~d\theta$won't Also 8:25 AM hmm... The answer has to be symmetric in$r$and$R$Also Let's do the special case where$r=R$We have$\int_0^{2\pi}\sqrt{(1-\cos\theta)^2+\sin^2\theta}d\theta$Well I guess this is$r=R=1$are these on the circle or in the circle? you're right, the answer doesn't match up on the circle${}=\int_0^{2\pi}\sqrt{2-2\cos\theta}d\theta$when r=R=1 the answer turns out to be 8 8:29 AM = 8 which doesn't add up yeah no should be divided by 2 pi Oh that's true @MarioCarneiro but we get$4/\pi$which is still not the area of the circle anyway I've confirmed (numerically) that the answer does depend on r also so it's def not the area of the circle 8:30 AM Oh, it couldn't be the area of the circle anyway 'cause the largest distance is$2$the pi r^2 expression doesn't have the right units heh? it's definitely true when r=0 right Oh well I think he would've updated his conjecture to$(\pi R^2)/(2\pi)$still not the right units this is very interesting 8:31 AM it's a distance, if you double r and R the answer should double Hm when$r=0$you get$R$'cause it's constant @LeakyNun but what happened to my area interpretation but whatever, your conjecture was that it doesn't depend on$r$, only$R$which means our example with$r=R=1$still means it's wrong ($4/\pi\ne1$) anyway wolframalpha refused to compute the integral so let's call it a day it's an elliptic integral 8:33 AM (You use$1-\cos\theta=2\sin\frac\theta2$to finish that example) @MarioCarneiro Yeah that's what I got which I guess means it depends on the perimeter of some ellipse? 21 mins ago, by Leaky Nun$\displaystyle \int_{0}^{2\pi} \sqrt{(r-R\cos\theta)^2 + (R\sin\theta)^2} \ \mathrm d\theta = \int_{0}^{2\pi} \sqrt{R^2 + r^2 - 2rR\cos\theta} \ \mathrm d\theta$(hey I derived cosines theorem!) there must be some 3B1B way of somehow turning this into an actual ellipse or something something like that, although the modern study of elliptic integrals has diverged somewhat from actual ellipses Right so extra credit would be to find the average distance from Earth to Mercury and to Venus Hm. Fixing$R$, varying$r$, what does the graph of this look like I hope it's strictly increasing Wolfram Alpha lets you plug in "distance from Earth to sun" and "distance from Mercury to sun", doesn't it @AkivaWeinberger it's... actually linear?? r = 0 to 1, R = 1 8:37 AM that looks like matlab also didn't divide by 2pi also it actually wobbles afterwards $$\frac{r+1}{\pi } E\left(\frac{4 r}{(r+1)^2}\right)-\frac{r-1}{\pi}E\left(-\frac{4 r}{(r-1)^2}\right)$$ You want sin(t) not sin(x) oops, changed, it no longer wobbles 8:40 AM Really close to$\sqrt{x^2+1}$looks about right. It's too flat to be$\sqrt{x^2+1}$though the second derivative of the function is discontinuous at 1 So the reason I was thinking about this was 'cause I saw this I didn't read the whole thing but it sounded wrong Oh and now it won't load so I can't look at it again If it loads for one of you can you read it and tell me what it says Venus is not Earthâ€™s closest neighbor Calculations and simulations confirm that on average, Mercury is the nearest planet to Earthâ€”and to every other planet in the solar system.$\sqrt{(r-R\cos\theta)^2 + (R\sin\theta)^2} = \sqrt{r^2+R^2 - 2rR\cos\theta} = \sqrt{r^2+R^2-2rR(2\cos^2(0.5\theta)-1)} \\ = \sqrt{(r+R)^2 - 4rR\cos^2(0.5\theta)} = \sqrt{(R-r)^2 \cos^2(0.5\theta) + (R+r)^2\sin^2(0.5\theta)}$8:50 AM Wait a moment Wait did your graph actually confirm that 'cause it's increasing in$r$(Units in AU so$R=1$) and$r_{\rm Mercury}<r_{\rm Venus}$@LeakyNun @LeakyNun The half-angle thing only works for$r=R$I think what? I said you could evaluate it using the half-angle identity but unless$r=R$you still end up with an integral with no closed form yeah but now$\displaystyle \frac1{2\pi} \int_{0}^{2\pi} \sqrt{(r-R\cos\theta)^2 + (R\sin\theta)^2} \ \mathrm d\theta = \frac1\pi \int_0^{\pi} \sqrt{(R-r)^2 \cos^2\alpha + (R+r)^2 \sin^2\alpha} \ \mathrm d\alpha$so it's half the perimeter of an ellipse with major axis R+r and minor axis R-r Oh is that the formula for the perimeter of an ellipse which we all know is (approximately)$2\pi\sqrt{\dfrac{(R+r)^2+(R-r)^2}2}$so the whole integral becomes$\approx \sqrt{r^2+R^2}$confirming your approximation of$\sqrt{x^2+1}$(it actually shocked me to find out that there is no closed form for the perimeter of ellipse: I only found out just now) 8:57 AM For$(r,R)=(1,1)$we had$4/\pi$Oh Semimajor axis$R+r$and semiminor axis$R-r$oops so for$(1,1)$we have a line of length 4 (so "circumference" 8) and then 8/2Ï€ =4/Ï€ okay, so now @LeakyNun come up with a 3B1B style explanation that relates this value to the perimeter of an ellipse lol no I'm still annoyed that the article won't load for me anymore Punishment for my skepticism Did they mention it being related to the perimeter of an ellipse? 9:01 AM it loaded for me, eventually... it looks like they got the right answer, with the elliptic integral and everything @AkivaWeinberger net neutrality starter pack try this Fun fact, the minimum-energy trajectory from Earth to Mars starts when they're furthest apart from each other Actually I think that's wrong Never mind oh yeah what the actual, it's$\sqrt{r^2+R^2}$so it increases with$r$so yeah actually mercury should be the closest to all other planets lol but that's not very useful is it I'll just go like oh mercury is r=0.3 and venus is r=0.7 and earth is r=1 so like "morally" venus is closer to earth like that's not what people actually mean when they say venus is closest to earth right they mean the orbits are the closest "orbits are closest" = what? @AkivaWeinberger does my link work for you? 9:06 AM I guess they have small Hausdorff distance right the distance being exactly R-r @LeakyNun It does (though the images aren't there) Thanks PCM = point-circle method 9:08 AM @AkivaWeinberger Earth at launch time is furthest way from Mars at arrival time would probably be the more accurate way of saying it @LeakyNun Doesn't show for me @MarioCarneiro What's the Hausdorff distance Thanks That works @MarioCarneiro That's not$R+r$(for concentric circles), is it? it's how you can measure "distance between sets" in R^2 or other metric spaces @AkivaWeinberger it's R-r it's$\max_{a\in A}\min_{b\in B}d(a,b)$9:10 AM How's the Hausdorff distance defined @MarioCarneiro Ahh Wait why is that symmetric in$A$and$B$in other words "you are always somewhat close to a point on the orbit of venus" oh, it's actually the max of that and its symmetric counterpart You need to restrict to compact subsets if you want to get a real metric though @AkivaWeinberger Sure @MarioCarneiro anyway I'm glad to see you here lol 9:12 AM There we go Thay actually does look like$R+r$no it's R-r the inf is R-r at any point, so it's taking the sup of a constant Inf isn't constant in x It's like a two player game. You want to maximize$d(a,b)$, I want to minimize it. I go first and choose$a$, you go second and choose$b$it is, it's always R-r you always choose the co-radial point @LeakyNun$d(x,y)$, where$y$is fixed and$x$varies, depends on$x$@LeakyNun The player who wants to maximize it goes second because sup is on the outside 9:14 AM$\forall x \in X, \inf_{y \in Y} d(x,y) = R-r$player 1 (maximizer) chooses x, player 2 chooses y player 2 chooses the co-radial point, achieving$R-r$@LeakyNun Ohh Oh OK you're right I get it now Sorry So the game goes outside-in then Oh crap I'm back to 98 tabs I thought I was recovering how to count the number of tabs On the bottom I have about 3000 tabs open in FF 9:19 AM Also Guess what I'm doing on Friday I have like 80 tabs Guess what I'm incredibly scared of doing on Friday marathon in israel? good luck! 9:21 AM (Not the best translation but I guess they don't want to pay a translator) Gonna tell you right now I'm not gonna win any prizes it isn't about the prizes Will probably take me like 90min or something It is about finishing without collapsing @TobiasKildetoft I just learnt yesterday that a surjective endomorphism on a f.g. module is in fact bijective and I'm like wat I still haven't learned modules What are they for 9:24 AM @LeakyNun With no other assumptions at all? they're like vector spaces but your scalars don't form a field, just a ring they come up like everywhere @TobiasKildetoft yes @AkivaWeinberger if k is a field and G is a group then k[G]-modules are k-representations of G (so study of modules subsume (modular) representation theory) Not sure I ever came across that before. Z-modules are abelian groups (so study of modules subsume study of abelian groups) Anyway, I will need to step out of the room shortly, as I will be having a phone meeting negotiating my starting salary and stuff like that. if M is a manifold then (locally free)$C^\infty(M)$-modules are vector bundles @AkivaWeinberger structure theorem for finitely generated abelian groups say that they are all$\Bbb Z^r \oplus \bigoplus_{i=1}^t \Bbb Z/p_i^{n_i} \Bbb Z$the generalization is structure theorem for finitely generated modules over principal ideal domains (Z being a PID) they're all$R^r \oplus \bigoplus_{i=1}^t R/p_i^{n_i} R$and this provides a "straightforward" proof of Jordan Normal Form theorem because a$k$-vector space with a linear transformation to itself is just a$k[X]$-module, and$k[X]$is a PID 9:38 AM > I just learnt yesterday that a surjective endomorphism on a f.g. module is in fact bijective That sounds like a variation on the pigeonhole principle really If you restrict to vector spaces, then it's saying that the basis is finite, so any surjection from the basis to itself is a bijection sure, but what is shocking is that it holds for modules (that's handwaving but should give the intuition) because one expects modules to be very ill-behaved submodule of f.g. module isn't f.g. etc 9:42 AM I think of modules as vector spaces... quotients don't have sections etc no division is like, okay i.e. submodules don't have complements I usually think of that stuff as a miracle / quirk of AC that this happens at all in vector spaces aha constructive vector spaces :P how about C being a C[X]-module i.e. a "smaller" thing somehow being a module over the "bigger" ring 9:45 AM are you interpreting X as some random number? yeah sure or is the multiplication some crazy nonconstructive thing no,$p \cdot m$is just "evaluate$p$at 7 and then multiply by$m$" yeah okay, so this is true in general... if you have a ring hom$R \to S$then$S$-modules are$R$-modules, and the hom need not be injective or surjective, so "bigger" doesn't matter right, so how do you translate this to vector spaces :P 9:49 AM why wouldn't this still hold? I mean, you can't have a non-trivial vector space that's "smaller" than the field oh, this is just the thing about field homs being injective yeah actually I don't know how I should visualize modules as vector spaces... like year 1 vector spaces just ignore the fact that there is no division and do linear algebra anyway lol 9:58 AM all the structure is there, but some of the theorems require more assumptions every module is flat and projective and injective and torsion-free and free those are all words for "modules aren't vector-space-like enough, I need this property of them for my argument to go through" I guess category theory does the same thing... when an argument needs an assumption, give it a name I see Say we have a great circle that passes very close to the poles but doesn't touch them I guess a manifestation is in the Künneth formula @MarioCarneiro 10:01 AM and we take the inverse image of that in the exponential map What does the result look like where if your base ring is a field then you get to omit a term from the short exact sequence so Tor is kinda like the "measure for how much your ring is away from being a field" also if every module over your ring is flat and your ring is local then your ring is a field @AkivaWeinberger type mismatch are you interpreting your sphere as$\Bbb CP^1$The exponential map from Riemannian geometry which is defined for all manifolds (well, with a basepoint, but I was assuming the basepoint was the north pole) really For$v\in\Bbb R^n$,$\exp_p(v)=\gamma(1)$where$\gamma$is the geodesic with$\gamma(0)=p$and$\gamma'(0)=v$aha is it a flow along some vector field? 10:09 AM You can also think of it as$\gamma(\|v\|)$where$\gamma$is the normalized (unit-speed) geodesic with$\gamma'(0)=v/\|v\|$oh, so it's just the polar angle for each point on the circle Basically that map projection Notice that everything of distance$\pi$away from the origin maps to the south pole Also it's defined for the whole plane, not just the disc of radius$\pi$, but beyond that it just starts repeating itself I can get an exact figure, but you get something like a rounded half-circle Probably The reason it's called the "exponential map" comes from Lie theory (Fun fact: if you take a radius and circle centered on the origin and apply$\exp_p$to them, the images are still perpendicular) Tissot ellipses They're all squished radially This is true for all manifolds @MarioCarneiro do you come here often? 10:14 AM (Well, if it has constant negative curvature, they'd all by stretched radially) @AkivaWeinberger have you heard of Morse theory? Heard of it, don't know much about it Inverse exponential map of the hyperbolic plane Another pic of the exponential map Man, negotiating salary is hard. Especially when my experience is in something completely different. I'm merely an undergraduate so I don't know how to respond I can't life I am now having to learn how to live in the real world, rather than the nice and sheltered world of academia. It is scary out there. I mean, they actually expect you to do work once in a while. 5 10:21 AM that's when you go to grad school @MarioCarneiro I did that already Also the vid it refers to postdoc? Then I stayed in academia for another 4 years 10:21 AM keep pumping that well @AkivaWeinberger cool @TobiasKildetoft can I think of a local ring as a "field with some noise"? @LeakyNun I suppose can we do galois theory with local rings? no idea. We can try @LeakyNun So a quick check, what goes wrong here. I take the polynomial ring in infinitely many variables, I kill the first variable and identify the result with the original ring by shifting all indices by one. Why is this not a surjective map that is not a bijection? so$f: k[X_1, \cdots] \to k[X_1, \cdots]$with$f(X_1) = 0$and$f(X_{n+1}) = X_n$etc 10:32 AM right I think this isn't$k[X_1, \cdots]$-linear might be the issue$f(X_2 X_2) = X_1^2$but$X_2 f(X_2) = X_1 X_2$possibly we are killing more than intuition tells us when we kill off a variable out of infinitely many A$k[X_1, \cdots]$-module is a$k$-vector space with countably infinitely many endomorphisms and a linear map is a linear map that commutes with each endomorphism 10:36 AM Ahh, so it is the shifting of the variables that is the issue here right which provides an isomorphism of rings, but not of modules, since we would have to twist the action on the target, which would make it no longer an endomorphism. maybe @TobiasKildetoft if I set$\mathcal O_\Bbb Q = \Bbb Z_{(p)}$then what would happen to my number fields? @LeakyNun I have no idea what that means like for$L/K$fin. gal. you first set what$\mathcal O_K$is and then define$\mathcal O_L$to be the integral closure of$\mathcal O_K$inside$L$11:01 AM Is a subgroup of a Hopfian group itself Hopfian? Obviously there are no finite counterexamples (I initially foolishly tried searching through finite groups for a counterexample). @user193319 no Shoot. Do you have a counterexample in mind? it's in the link$F_2$contains a copy of$F_\infty$, the former is Hopfian, the latter is not Shoot...I am trying to prove that all finitely generated free groups are Hopfian and I was hoping to reduce it to the$F_2$case. 2 hours later… 1:07 PM 20k rep! 1:48 PM Guvnor' I have an interesting integral equation problem, I've been sent here from Phys.SE So I'm looking at the following integral equation $$\rho(\lambda) + \int_{-\infty}^\infty d\lambda' \rho(\lambda') \mathcal{C}(\lambda-\lambda') = (2\pi)^{-1}$$ Numerical integration does not exactly work with "infinity". One may choose a finite symmetric interval of integration and take the limit afterwards (which is the traditional way of doing these integrals), but Mathematica is not equipped to solve this integral completely symbolically Although I conjecture that it is theoretically possible Regardless, I wish to find a constant$\lambda_F\$ defined implicitly by
$$N = \int_{-\lambda_F}^{\lambda_F} \rho(\lambda) d\lambda$$