1:20 AM
@leslietownes Or, realistically, you fire up Maple or Mathematica.

1:36 AM
in this economy?

@leslietownes Sage?
Sage is free.

2 hours later…
3:11 AM
okay, you'd probably just resort to (admittedly uglier) formulas that do not need an arc length parametrization, in sage

4 hours later…
7:05 AM
Brute force has become more available in this economy.

2 hours later…
8:41 AM
@ThomasFinley The intersection of the regions $(x-a)^2+y^2\le a^2$ and $z^2\le k^2\left(x^2+y^2\right)$ is bounded.
sorry I was gone for so long

9:35 AM
wb

10:33 AM
hi

2 hours later…
12:18 PM
Any topological manifold whose universal cover is $S^{2n}$ has either trivial or $\Bbb Z/2$ $\pi_1$. This is because the only group that acts freely on $S^{2n}$ by Homeo is $\Bbb Z/2$. Is the converse true? meaning if the only nontrivial group that acts freely on a manifold by Homeo is $\Bbb Z/2$ then that manifold should be $S^{2n}$?

12:58 PM
Currently working the following problem:
> Let $(X,Y,Z)$ be a point chosen uniformly within the three-dimensional unit sphere. Determine the marginal distributions of $(X,Y)$ and $X$.
I'm tempted to consider the joint density $f_{XYZ}(x,y,z)=\frac1{4\pi}$ when $x^2+y^2+z^2=1$ (as the surface area of the sphere is $4\pi$), however, this function will integrate to $0$ over the unit sphere. So this can't be the joint density. I'm a bit lost on how else I should proceed with the problem. Any ideas are appreciated.

Given a fixed monometric orthogonal reference in space, consider the lines $r$ and $s$ defined as follows:
\begin{align*}
r: & \begin{cases} x = 2 \\ z = 0 \end{cases} \\
s: & \begin{cases} x = y \\ y=z \end{cases}
\end{align*} and the point $A(3,0,1)$. Represent:
$1.$ the plane for $r$ parallel to $s$.
- $\vec{d}_{s} = \langle 1, 1, 1 \rangle$
- $A(3,0,1)$
$(ax+by+cz+d = 0) \Longrightarrow (x+y+z+d=0) \Longrightarrow (3+0+1+d=0) \Longrightarrow (d=-4)$
So: $x+y+z-4=0$.
it's right ?
$\begin{cases} x = t \\ y = t \\z=t \end{cases}$ the coefficients of $t$ are $1$ so the direction vector of $s$ is correct, right?

1:20 PM
@onepotatotwopotato isn't any rational homology sphere a counter-example?

1:31 PM
@psie yeah this rv doesn't have a density

@Thorgott I don't know rational homology sphere even definition

maybe the author actually meant ball. I feel sphere makes the problem a lot harder.

Maybe, maybe not

@onepotatotwopotato having the rational homology of a sphere

We're all feeling a lot spherical today

1:34 PM
why is that a counterexample?

@psie think of the projection of the sphere onto the xy plane

@Jakobian ok, how do you know it doesn't have a density at all? I see how, if the density is constant $c$, then $c\int_A d\mu=c\mu(A)=0$ since $A$ is two-dimensional and $\mu$ is 3-dimensional Lebesgue measure, but perhaps the density could depend on $x,y,z$ in some weird fashion so that it integrates to $1$?

Could someone please check what I wrote above pls

@psie Because $P(S^2) = 1$
but $S^2$ is of Lebesgue measure $0$
I don't really understand your justification about density but lets just forget that part

@onepotatotwopotato Lefschetz implies that any fixed-point free map has degree $(-1)^{n+1}$, where $n$ is the dimension. so if $n$ is even, you can run the same argument as for spheres to show only $\mathbb{Z}/2$ could act freely.
that said, this example is incomplete
cause I haven't argued whether such a $\mathbb{Z}/2$-action does exist
I'm not familiar with the construction of even-dimensional rational homology spheres, so this is above my paygrade
but this is the sort of place where I would look

1:49 PM
sorry I meant $P((X, Y, Z)\in S^2) = 1$

ah ok, and any function integrated over a lebesgue null set is $0$, I understand I think

If $W = (X, Y, Z)$ had a density, then $P(W\in S^2) = \mu_W(S^2) = \int_{S^2} f_W d\lambda = \int_{\mathbb{R}^3} f_W\cdot 1_{S^2} d\lambda = 0$ since functions equal a.e. have the same integral and $f_W1_{S^2} = 0$ a.e.
where $\mu_W$ is the law of $W$
@psie yeah
well now arguably it is a bit tricky to derive the density of $(X, Y)$ (because this one has density). Because given a subset of the unit disk, you have to project it onto the sphere and calculate the area of that

hmm, ok

2:04 PM
So to calculate $P((X, Y)\in A)$ you'd have to calculate the area of $\{(x, y, z) : (x, y)\in A, z = \sqrt{1-x^2-y^2}\}$
Or equivalently this should be, up to normalization, volume of $\{(x, y, z) : \frac{(x, y)}{\|(x, y, z)\|}\in A, 0 <\|(x, y, z)\|\leq 1\}$
the area of the sphere can be formalized as calculating area of those "cones" anyway

could you explain why $(X,Y)$ has a density?

intuitively this is true

It's new to me that a random vector can't have a joint density but the marginal densities can exist.

How about, $X$ has density and $W = (X, 0)$
its a similar situation

ok

2:15 PM
you know, its challenging for me to formalize this
it was even when I was learning it
I'm trying my best to justify everything formally, but its hard
so even if I adhere to intuition, I don't see why you'd want to criticize that
we'll get to it eventually, hopefully

ok, right now I don't understand why $(X,Y)$ has a density, but it feels ok not to understand it right now

I mean, imagine the projection on the unit disk and this should help
now the neat thing about a set of the form $\{(x, y, z) : \frac{(x, y)}{\|(x, y, z)\|}\in A, 0 < \|(x, y, z)\}\leq 1\}$ is that we can use polar coordinates on it
In mathematics, a spherical coordinate system is a coordinate system for three-dimensional space where the position of a given point in space is specified by three numbers, (r, θ, φ): the radial distance of the radial line r connecting the point to the fixed point of origin (which is located on a fixed polar axis, or zenith direction axis, or z-axis); the polar angle θ of the radial line r; and the azimuthal angle φ of the radial line r. The polar angle θ is measured between the z-axis and the radial line r. The azimuthal angle φ is measured between the orthogonal projection of the radial...
we need to see how this set transforms under those
so of course $r = \|(x, y, z)\|$ and the second inequality just says $r\in (0, 1]$
and $\frac{(x, y)}{\|(x, y, z)\|}\in A$ means that $(\sin\theta \cos\varphi, \sin\theta\sin\varphi)\in A$

indeed

the Jacobian of this is $r^2\sin\theta$
we can integrate over $r$ first and this will always lead us to a constant value of $1/3$, but up to a constant we don't care about the value of this integral anyway
So what interests us is $\iint_{(\sin\theta\cos\varphi, \sin\theta\sin\varphi)\in A} \sin\theta d\theta d\varphi$
I think we can get rid of this annoying $\sin\theta$ from here

2:31 PM
how can we ditch it?

one could substitute cosine I guess, it'd lead to something like $\iint_{(\rho \cos\varphi, \rho \sin\varphi)\in A} d\rho d\varphi$
and now I think we should substitute back from polar coordinates to cartesian ones
maybe the proper choice was to use cylindrical coordinates from the start
I'm not sure

hmm

so we want $x = \rho\cos\varphi, y = \rho\sin\varphi$ now
and now we need to calculate the Jacobian because I guess I don't know what it would be
so for that we'd have to differentiate functions $\rho, \varphi$ in terms of $x, y$
$$J = \det\begin{bmatrix} \frac{\partial r}{\partial x} & \frac{\partial r}{\partial y} \\ \frac{\partial \rho}{\partial x} & \frac{\partial \rho}{\partial y} \end{bmatrix} = \det\begin{bmatrix} \frac{x}{\sqrt{x^2+y^2}} & \frac{y}{\sqrt{x^2+y^2}} \\ -\frac{y}{x^2+y^2} & \frac{x}{x^2+y^2} \end{bmatrix} = \frac{1}{\sqrt{x^2+y^2}}$$

couldn't we use the inverse function theorem in some way?

probably since what I got is pretty much $1/r$
But our desired integral is now $\iint_A \frac{1}{\sqrt{x^2+y^2}}dxdy$
and that means, we have found our density!
up to a multiplicative factor
$\int_{x^2+y^2\leq 1} \frac{1}{\sqrt{x^2+y^2}} dxdy = \int_0^{2\pi} \int_0^1 drd\varphi = 2\pi$
okay so we got $P((X, Y)\in A) = \iint_A \frac{1}{2\pi \sqrt{x^2+y^2}} dxdy$

2:49 PM
awesome sauce!

well... if we want to be proper, we need to justify what we did
I kind of really don't want to
but its probably not as scary as it seems

Well, how do you obtain the density of $X$? By integrating out $y$?

from density of $(X, Y)$
@Jakobian sorry I mean the density is $\frac{1}{2\pi \sqrt{x^2+y^2}}$ for $x^2+y^2\leq 1$ and $0$ otherwise
its not supported on all of the euclidean plane

makes sense

oh okay yeah we integrate the $y$ out

2:55 PM
yeah

$\int \frac{1}{2\pi \sqrt{x^2+y^2}} 1_{B^2}(x, y) dy$
If $x\notin [-1, 1]$ then this is $0$ and if $x\in [-1, 1]$ then $1_{B^2}(x, y) = 1_{[-\sqrt{1-x^2}, \sqrt{1-x^2}]}(y)$
so we get $\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \frac{1}{2\pi \sqrt{x^2+y^2}}dy$
which is some arcsin type of integral

yummy

ah no its not arcsin
that one has a minus
for this one you just have to substitute $u = \sqrt{x^2+y^2}$ I think
$udu = ydy$
or maybe $u = x^2+y^2$
$\text{arcsinh}(y/x)$
so it'll be $\frac{1}{\pi}\text{arcsinh}\frac{\sqrt{1-x^2}}{x}$

so that is the density of $X$?

yeah
for $x\in [-1, 1]$
though I think I've done something wrong since the density is negative for $x\in [-1, 0]$ from this
ah okay I know, the integral is $\text{arcsinh}(y/x)$ for $x > 0$ and $-\text{arcsinh}(y/x)$ for $x < 0$, or $\text{arcsinh}(y/|x|)$
so it'll be $\frac{1}{\pi}\text{arcsinh} \frac{\sqrt{1-x^2}}{|x|}$ for $x\in [-1, 1]$
$\text{arcsinh}(z) = \ln(\sqrt{z^2+1}+z)$ so maybe it simplifies a little when you plug it in

3:16 PM
ok 👍

$\frac{1}{\pi}( \ln(|x|+\sqrt{1-x^2}) - \ln(x^2))$
I mean this first term it does seem like it could be written as some inverse hyperbolic function
$\frac{1}{\pi}(\text{arccosh} |x| - \ln(x^2))$
okay here you go

silly question maybe, but the density of $(X,Y)$, can we also write it as $$\frac1{2\pi}\frac{1}{\sqrt{1-x^2-y^2}},$$ for $x^2+y^2\leq 1$. I notice in WolframAlpha it also integrates to $1$. I don't understand why though, seems to be a different function.

no we can't
why would we
$\int_0^1 3x^2dx = 1$

yeah ok, silly me

$\int_0^1 2x dx = 1$
both would be densities but they represent different random variables

3:26 PM
so, as a final note maybe, do you think it would have been easier to start with cylindrical coordinates and ditch spherical?

I don't know

4:22 PM
When they say, " From these subspaces, we select ordered bases whose union is an ordered basis $\beta$ for $V$ such that..." ------ How are they so sure that such a choice is possible ?
Is there some kind of a theorem that are supposedly using?

4:44 PM
@Pizza you computed the plane to $A$, and if it's parallel to $s$ it's wrong that $d_s=(a,b,c)$

wait
i fixed before i think
Given a fixed orthogonal monometric reference in space, consider the following lines $r$ and $s$:
\begin{align*}
r: & \begin{cases} x = 2 \\ z = 0 \end{cases} \\
s: & \begin{cases} x = y \\ y=z \end{cases}
\end{align*}
and the point $A(3,0,1)$. Represent:

$1.$ The plane for $r$ parallel to $s$.
$(1)$ To find the normal vector to the plane for $r$ parallel to $s$, we first need to find the direction vector of $s$, which will also be parallel to the plane we are looking for.
The plane we are seeking will be parallel to $s$, so it will have the same normal vector as $s \to \langle 1, 1, 1 \ra it was for$r$i did for$A$is now correct? what are you computing now?$1.$The plane for$r$parallel to$s$. i did that (if its correct how i did) the plane will not have the same normal vector as$d_s$it has to be orthogonal it says plane for r parralel to s wait the direction vector is correct though, right? 4:55 PM of$s$or of the plane? it's correct but in this case do I need to know the normal vector? of the plane? yes 4:59 PM how do you find the plane without knowing its normal vector? I got confused with the direction vector, yes Take the space of smooth suspensions of$S(S^1)$(for endpoints$p,q$) over the set indexed by radii$r>0$s.t. the collection of suspensions partitions the unit 3-ball minus the line$(-1,1)$. Take all suspensions$\Omega$s.t. the endpoints define a unit sphere bounding the 3-ball. On first pass, a 3-ball minus the lines pivoting about the origin only leaves the bounding sphere. However, the suspensions$\Omega$have infinite cardinality. What would$\Omega-$the set of intervals passing through the origin be diffeomorphic to? so i need to find the normal vector to the plane that is orthogonal to the line s @ThomasFinley They are explaining what they're going to do in that section so the answer is most likely "keep reading" 5:16 PM orthogonal so the dot product must be zero, so$(1,1,1) \cdot (-1,1,0) = -1+1+0 = 0$The normal vector of the plane will therefore be$(-1,1 ,0)$? @SineoftheTime Two vectors$\text{u,v} \in V$are said to be orthogonal if:$$\text{s(u,v)}=0$$ should be correct right? 5:32 PM @Thorgott Thanks! I will follow your advice. 5:52 PM @Pizza what you found is one vector, but you don't know if it the right one. You have to use the condition that$r$is in the plane 1 hour later… 7:14 PM I'm pretty sure there is no such theory 8:07 PM scratching everything in its way, including my arm. 8:25 PM koro: that's great. last night our cat (7 years old, raised almost all of that time in civilization) clubbed my son in the face, because he suddenly raised his hand toward her head in a way that probably looked like a paw strike her claws weren't out, but i heard the hit, which is not something i usually say about cat paws I can imagine the situation here :). The mother of the cat in the image knows her strengths. She never raises her paws. She never shows any discomfort. @leslietownes I hope he is not injured These kittens are very different from the other cats I have seen. These are ok with being lifted up. @SoumikMukherjee the claws weren't out, luckily. Leslie: I think that your 7 yo cat also knows her strengths. That's why no claws. soumik: he looked as though he might cry for about 15 seconds, then forgot it There is one question though that I've not yet found an answer to. 8:39 PM Oh that's good, he is strong:) Does a cat/dog know that humans exist? Or they look at humans made up from several entities -hands, legs, face etc? Like say if you touch a cat with a stick in your hand, does the cat know that it's you moving the stick, or they think that the stick is moving on its own? If I take my foot close to two cats here, they either attack the foot or they stare at the foot or they just run avoiding trouble. They don't look at my face while they are at it. So they perhaps look at feet as different entity from the human? @SineoftheTime 👍I'll see better tomorrow, thanks a lot anyway @Koro Maybe not, we see each other in the face because we know that's where the communication(if any) will happen. But why would a cat look at your face in a fight? For them, your foot is the place where the probable attack will come from Cat fights another cat with paws. But during that fight, cat isn't looking at the paws of the other cat. Perhaps they just look at only what moves... @Koro I think it depends , the cat could know your movements. 8:49 PM in the situation above, foot movement catches their attention naturally. @AlessandroCodenotti @Thorgott By the way, I'm pretty sure this is correct and so its an example where quotients and products don't commute, up to homeomorphism. When you feed the cat, the cat knows that you are the one moving the food I've read it, checked the details But about the existence part, I guess there is no answer and perhaps we'll never know because cats don't talk. @Pizza or "my hands are moving the food"? like does a cat know that my hands are mine? What if they can talk, but not interested to talk to inferior beings (humans ) 8:52 PM or to avoid voting and taxes then they would have a specific part of brain responsible for communication like dolphins @Koro In math term, Is koro simply connected? @Koro From the smell I think so? or do you mean that the cat thinks that your hands are perhaps objects? However, I am a pizza, not a cat. free pizza everyone @user85795 thanks. Om nom nom 9:05 PM :-D @user85795 can't eat myself:( 3 then interpret it as a call for your freedom, pal @user85795 hope I can escape from your stomach @user85795 you know what happens to pizzas when they are eaten, I hope I don't end up like that too so yes, free pizzas I'm going anyway, bye! 9:21 PM cya pal 9:47 PM @Koro i had a cat where, if you were using a toy or your hand to play with him and he got too amped up, there would be a definite moment where you could feel his attention shift from the toy/hand to you usually followed by him backing up, squaring up, and jumping at your shoulder or some other part of your body with both paws out we need ted here with his screech stories 10:17 PM I am barely familiar with differential forms, but should I think of$\sin\theta d\varphi\land d\theta$as an "area vector element" of the unit sphere? Is there any efficient way in transforming this "area vector element" into cartesian form? I don't understand the difference in simply working with$\sin\theta d\varphi d\theta$, i.e. the scalar area element. 10:53 PM (removed) If I have a submanifold$M\subset\mathbb R^N$and a smooth function$f:\mathbb R^N\to\mathbb R$, I can obtain the derivative of the restriction$f|_M$by projecting$df_p=\sum_h \frac{\partial f}{\partial x_h}dx_h$to$T_pM$. Is there a similar way of obtaining the second derivative of$f$(using the connection induced by the Riemannian metric inherited from$\mathbb R^N$) using the partial derivatives$\frac{\partial^2 f}{\partial x_i\partial x_j}\$?
Preferably not using charts

11:37 PM
But charts are the BESTEST!

Hard disagree

charts for some, miniature american flags for others

11:56 PM
@leslietownes Woo-hoo!
We're going forward, not backward!
Upward, not forward!
And always spinning! Round and round!

@Jakobian Impressive, Moishe is always reliable