Mathematics - 2020-01-20 (page 2 of 3)

5:17 PM

This is not the usual definition of $2$-transitivity as far as I can tell. We usually require $g$ and $g^{\prime}$ to be the same (then $2$-transitivity means acting transitively on pairs of points). A $2$-transitive action as you defined it is still transitive though; take $y=y^{\prime}$ and then $g^{\prime}=e_G$ will always work.

Ted Shifrin

@Sha, I agree with @Thorgott that your definition is fishy.

Subhasis Biswas

@Thorgott the answer is extremely well written.

something that even a layman like me can kind of get a hold of.

Thorgott

@Subhasis Actually, I was being kind of sloppy in the answer by just saying "Borel-measurable" without making clear which Borel-algebra I was referring to at times. I should fix that.

codingnight

Hello, I am doing a problem and I came upon this: if i have the inequality aR<rcos(x)<R where a,r,R>0, what can I do to isolate x? Is cos^-1(aR/r)<x<cos^-1(R/r) correct? I am unsure about it because wolfram alpha says the inequalities cos(x)<1 and x<cos^-1(x) are not the same.

Ted Shifrin

The other way around, @codingnight, because $\cos$ is a decreasing function on $[0,\pi]$.

codingnight

5:21 PM

ahh of course, then if I had sin insted of cos my reasoning would be correct? thanks @TedShifrin

Ted Shifrin

Yes, that's right.

Mike Miller

@TedShifrin here the OP wants to know why a domain with Jordan curve boundary has closure actually a manifold. I outlined a proof which works to show that if $U \subset M$ is an open subset of a manifold, whose boundary $\partial U$ is a locally flat manifold (there are charts taking it to $\Bbb R^{n-1} \subset \Bbb R^n$), then $\overline U$ is a manifold with boundary. Do you see a quicker argument? I had to fiddle a bit.

Ted Shifrin

Agh. This is the kind of stuff I'm terrible at. :P

Mike Miller

OK

it's just point-set crap

It surprised me that it took so much work to show this for Jordan domains

Ted Shifrin

This is why I stick to the smooth (or complex analytic) category. :P

Mike Miller

5:32 PM

You'd need to give more or less the same proof in the smooth category

The question is whether regularity of both the interior and the boundary gives regularity of both at the same time

Ted Shifrin

Hmm.

That sort of thing seems non-obvious to me.

Mike Miller

OK, that's basically what I wanted. Then the proof I outline is probably necessary.

(You need to make some assumption to rule out the case of having a closed hypersurface $H \subset M$ and letting your domain be $M \setminus H$, but this is an edge case.)

Sha Vuklia

@Thorgott oh oops, sorry, that was my bad

I should have only included one $g$ (and no second $g'$)

Ted Shifrin

Hi @Sha. So, yes, it's clear that $2$-transitive implies transitive.

Sha Vuklia

Elo Ted

oh hm, let me figure that out for a sec, because for some reason it wasn't clear to me

Ted Shifrin

5:41 PM

OK, yell if you don't see it.

Sha Vuklia

you are assuming "my" definition right, that $x\neq x'$ and $y\neq y'$, and $\sigma_g(x)=x'$ and $\sigma_g(y)=y'$ for some $g$? @TedShifrin

oh I see it I think

Ted Shifrin

Yup, I'm assuming that.

Sha Vuklia

also @TedShifrin, are you somewhat familiar with how representation theory is used in physics?

(it's a shot in the dark, I admit that:p)

Ted Shifrin

Not really.

One of my best friends at UGA teaches such a graduate course in the physics department, though.

Sha Vuklia

hm rights, well if you don't mind, I would like to say why I was asking

ah cools

Alessandro Codenotti

5:48 PM

Hi everyone

Sha Vuklia

the thing is, I am graduating this year, and I have to chose a topic that lies at the intersection of mathematics and physics. now I found a professor in mathematics who is into algebra, and I guess in particular in representation theory

and I also found another professor, who came with the idea to explore a certain peculiarity in quantum theory that has to do with entanglement

which sounds cool, and I've already been looking at the paper

however, judging from the paper, I couldn't really see what kind of interesting mathematics I could get out of it (they were just using basic linear algebra, for the most part it seemed)

and while I'll have my appointment with this physics professor soon, I am asking around a bit

Ted Shifrin

Personally, of course, I would lobby for more differential geometry. It's all over modern physics.

Mike Miller

most things people call physics are questionably labeled

of course, i am not arguing against labeling GR that way

Ted Shifrin

There's no question that representation theory shows up in QM and elsewhere in physics, but I'm not advanced enough.

Sha Vuklia

@TedShifrin yea this was also on my mind for a while

and I think the math professor I have, has also led a project last year that revolved around diffgeo

Ted Shifrin

5:52 PM

@MikeM: You're opposed to thinking of E&M as a $U(1)$-bundle? :D

Sha Vuklia

but yea, I guess I'm quite clueless when I start seriously thinking about topics, on how to implement an interesting math component

but alright, I'll just await the appointment in any case

Ted Shifrin

My favorite elementary example (which I put in my notes) is explaining the Foucault pendulum in terms of parallel translation (which is diff geo), @Sha.

But that's super elementary.

Thorgott

reminds me of this image (which I'm sure has been shared here before)

Mike Miller

@TedShifrin you know what i mean

Ted Shifrin

"commutative field theory"?

@MikeM: I never got far enough to understand how the physicists arrived at all the enumerative geometry phenomena that they did.

Mike Miller

5:54 PM

neither did they

Ted Shifrin

LOL ... I mean, I don't know even the physical framework in which they unearthed it.

Mike Miller

good

Ted Shifrin

LOL, bi***y mood today? :D

Subhasis Biswas

@Thorgott now, is $[a,b]$ is perfect w.r.t the Lebesgue measure, where the sigma algebra is formed by the Lebesgue measurable sets?

rather than Borel sets?

Thorgott

Yes, the Lebesgue-algebra only differs from the Borel-algebra by adjoining measure zero sets

anakhro

6:07 PM

@TedShifrin I am sure you'd be able to pick up a quantum physics book and recognize some of the stuff they are doing in terms of the formalization in mathematics.

Like the Clebsch-Gordan coefficients, for example.

Ted Shifrin

Yeah, Pauli matrices, etc.

@MikeM: With regard to a post, it seems that Hirsch doesn't prove (easy) Whitney embedding in the non-compact case. Sigh.

Mike Miller

Bredon does

Ted Shifrin

I don't know (nor own) that book. But I've heard good things.

Mike Miller

First half is quite good

Exposition of cohomology is too formal

Products mainly

Ted Shifrin

I guess the whole trick is to find an exhaustion function in the non-compact case.

Mike Miller

6:12 PM

Something like this.

Ted Shifrin

Odd that Hirsch doesn't do it. He would have needed to put partitions of unity into chapter 1, I guess.

Mike Miller

It surprises me a little yeah

Akiva Weinberger

I draw a 60 degree angle on the ground. What is the set of points from which, if viewed from that point, it will appear to have 90 degrees in perspective?

We can look at the intersection of that set of points with a sphere

I'm guessing we end up with a circle with diameter equal to the distance between the endpoints of the angle

(on the sphere)

Ted Shifrin

That's an interesting question.

Akiva Weinberger

The conjecture can be thought of as looking at the aperture angle of a cone, from another point on the cone

Hm… now I'm less sure about it

OK let's do some vectors

I have $\angle p0q$

Ted Shifrin

6:27 PM

So you're basically saying it's just the phenomenon that the locus of points at which a $90º$ angle is subtended is the arc of a semicircle.

Akiva Weinberger

viewed from $v$ with $\|v\|=1$

@TedShifrin Yeah but a spherical version

$p$ projected onto a plane perpendicular to $v$ is $p-(p\cdot v)v$, right?

Ted Shifrin

If $v$ is a unit vector.

Akiva Weinberger

Yeah

So I want $90^\circ=\angle(p-(p\cdot v)v),0,(q-(q\cdot v)v)$

which means those vectors are perpendicular

so $\big(p-(p\cdot v)v\big)\cdot\big(q-(q\cdot v)v\big)=0$

Ted Shifrin

with a \cdot in there

Akiva Weinberger

Yeah whoops

So $p\cdot q=(p\cdot v)(q\cdot v)$ I think?

Semiclassical

6:30 PM

that seems wrong

take $v=e_z$

Mike Miller

what is the degree in perspective

Semiclassical

then that's $p\cdot q=p_xq_x+p_yq_y +p_zq_z$ whereas $(p\cdot v)(q\cdot v) = p_z q_z$

Ted Shifrin

Yeah, I'm puzzling over what the question means. Ordinarily I need a place for the eye, but I still need a viewing plane.

Semiclassical

oh, never mind, I misread you

Ted Shifrin

You forgot cross-terms, @Semiclassic.

Semiclassical

6:32 PM

what?

Akiva Weinberger

Draw the plane through your eye and the line, for each line. What is the dihedral angle where the planes intersect?

That's my definition

Mike Miller

eh

Akiva Weinberger

I think it's the same as drawing the plane through the vertex of the angle whose normal points towards your eye, and projecting

Does $p\cdot q=(p\cdot v)(q\cdot v)$ have a direct interpretation?

Ted Shifrin

I don't understand that sentence.

Akiva Weinberger

$p$ and $q$ are fixed

@TedShifrin I accidentally a phrase

Semiclassical

6:33 PM

switching over to p,q as column vectors:

Ted Shifrin

There are cross-terms in your dot product.

Semiclassical

the cross-terms cancel

Ted Shifrin

No, they add.

Unless I'm being unusually stoooopid.

Semiclassical

$$\big(p-(p\cdot v)v\big)\cdot\big(q-(q\cdot v)v\big)=p\cdot q-(p\cdot v)(v\cdot q)-(q\cdot v)(p\cdot v)+(p\cdot v)(q\cdot v)(v\cdot v)$$

and $v\cdot v=1$

Akiva Weinberger

Thus setting to $0$ gives $p\cdot q=(p\cdot v)(q\cdot v)$

Ted Shifrin

6:35 PM

Oh, I see. The three terms are all the same up to sign. Duh.

Akiva Weinberger

and I see the geometric interpretation of $p\cdot q=(p\cdot v)(q\cdot v)$

Ted Shifrin

OK. Ted is stooopid.

Thanks.

Semiclassical

But there's a simpler way to write this: Let $P_\perp = I_3 - vv^\top$

Akiva Weinberger

Complete $v$ to an orthonormal basis $\{v,v_1,v_2\}$

Semiclassical

(so projection onto the orthogonal complement of $v$)

Akiva Weinberger

6:36 PM

Then this basically says $p_0q_0+p_1q_1+p_2q_2=p_0q_0$

Ted Shifrin

Not $I_2$, Semiclassic.

Akiva Weinberger

so $p_1q_1+p_2q_2=0$, which is what I want

Semiclassical

yeah

Ted Shifrin

OK, so our viewing plane is orthogonal to $v$.

Akiva Weinberger

Yes

In any case. Given fixed $p$ and $q$, where does $p\cdot q=(p\cdot v)(q\cdot v)$ for unit $v$?

Semiclassical

6:37 PM

then $p-(p\cdot v)v=P_\perp p$ and $q-(q\cdot v)v = P_\perp q$, so we want $(P_\perp p)^\top (P_\perp q) = p^\top P_\perp q=0$

Ted Shifrin

So that's a quadric surface on which $v$ lives, intersected with the sphere.

@Semiclassic: So that gives what DogAteMy wanted: orthogonal vectors when projected into that plane.

Semiclassical

yep

i just like writing things in terms of projections, heh

Akiva Weinberger

$p^\top q=p^\top vv^\top q$

Is that useful?

${}=v^\top pq^\top v$

Ted Shifrin

You have Semiclassic's projection matrix in there.

Semiclassical

yeah

Ted Shifrin

6:39 PM

Except this is projection on $v$.

Semiclassical

I can't say anything terribly useful about quadric surfaces in 4D space, though

Akiva Weinberger

Since $v$ is unit, I'm basically solving $v^\top(pq^\top-(p\cdot q)I)v=0$ for $v$ I think

so $v^\top Av=0$ for $A=pq^\top-(p\cdot q)I$

Semiclassical

oh, so you want it as: Given $p,q$, find $v$ such that blah blah blah

Akiva Weinberger

Yes

Ted Shifrin

DogAteMy: So the algebra of it tells us that you're going to get the intersection of two quadrics (the sphere and the other one). So we'll have a curve, which may have two components. Have to decide the signature of the quadratic form.

You have the additional equation $\|v\|=1$.

Akiva Weinberger

6:44 PM

I'm given an angle, and I want to view it in such a way that it looks like a right angle @Semiclassical

Semiclassical

an interesting case would seem to be $p\cdot q=0$, in which case $A=pq^\top $ is rank 1.

so which viewpoints preserve the right angle

and the answer is seemingly "any viewpoint perpendicular to $p$ or $q$?"

Ted Shifrin

@Semiclassic: I don't understand that remark. It always has rank 1.

Semiclassical

$A=pq^\top -(p\cdot q)I$?

Akiva Weinberger

"In which case $A=pq^\top$, which has rank 1"

Ted Shifrin

Oh.

Semiclassical

6:46 PM

$A$ isn't symmetric, which is weird

Akiva Weinberger

Let's do a specific example

Johan Liebert

Would a question related to this(👇) one be answerable on this site?

4

Q: How can the angular velocity vector be obtained from angular displacements that are not vectors?

My physics book (The Fundamentals of Physics) while explaining vector-ness of angular quantity (formally "Are Angular Quantities Vectors?") states that angular velocity and angular acceleration are vectors. But the turning point comes when it talks about angular displacement and states that this ...

Ted Shifrin

If you're doing orthogonal projection, it better be symmetric.

I'm thinking about my two quadrics.

Akiva Weinberger

$p=(2,-1,-1)$ and $q=(-1,2,-1)$, which has a 60 degree angle

Ted Shifrin

WLOG, you can make $p$ and $q$ unit vectors, of course.

The curve on the sphere, as DogAteMy commented an hour ago, passes through the normalized vectors $p$ and $q$ (and their negatives).

Aha, so we have to have a hyperboloid intersecting the sphere.

Akiva Weinberger

6:50 PM

I think $A=\begin{bmatrix}1&4&-2\\1&1&1\\1&-2&4\end{bmatrix}$ in that case

Semiclassical

would be nicer if the first two rows were swapped

Akiva Weinberger

grapher.mathpix.com/…

Tell me if this link works

Semiclassical

yeah

Ted Shifrin

Oh, you drew my picture.

So, you're getting a quartic curve with two components.

Akiva Weinberger

The intersection looks lopsided

@TedShifrin And a cusp at the origin

Ted Shifrin

6:54 PM

Lopsided?

Akiva Weinberger

Oh sorry curve

Ted Shifrin

Oh, wait, this shouldn't be homogeneous. It shouldn't go through the origin.

Akiva Weinberger

@TedShifrin It's not symmetric along the xy plane - but then again neither are p and q

Ted Shifrin

Your equation $(p\cdot v)(q\cdot v) = p\cdot q$ is a hyperboloid, not a cone.

You just graphed a cone.

It's not a cusp. It's a cone point at the origin.

Akiva Weinberger

Yes - I used the assumption $v\cdot v=1$

Ted Shifrin

6:56 PM

Well, the intersection with the sphere is the same.

Akiva Weinberger

OK so how about this question

For what $k$ is $p+q+kp\times q$ a solution

Basically asking, if I stand at their sum, and then move perpendicularly away from the plane generated by them, how far up do I have to go

And again the equation is $p\cdot q=(p\cdot v)(q\cdot v)$

Oh wait

No that's not the equation unless we know that $v$ is a unit vector

We have $p\cdot q=(2-\|v\|^2)(p\cdot v)(q\cdot v)$ in general

Maybe I should've just said $p+q+kr$ for $r$ unit perpendicular to $p$ and $q$, to make it easier

Then the magnitude of that is what, $\|p\|^2+\|q\|^2+k+p\cdot q$

Ted Shifrin

Wait. How did $(p\cdot v)(q\cdot v)=(p\cdot q)\|v\|^2$ turn into that equation?

Akiva Weinberger

Oh wait you're right

I took it from an earlier calculation, not realizing that it also had assumed $v$ was unit

$(p\cdot v)(q\cdot v)=(p\cdot q)\|v\|^2$ looks like the right equation

Ted Shifrin

7:11 PM

Yeah, that's the equation that arose from my stoooopidity earlier.

Akiva Weinberger

$(\|p\|^2+p\cdot q)(p\cdot q+\|q\|^2)=(p\cdot q)(\|p\|^2+\|q\|^2+k+p\cdot q)$

$\|p\|^2\|q\|^2=kp\cdot q$

$k=\dfrac{\|p\|^2\|q\|^2}{p\cdot q}$

If $p$ and $q$ are unit, then that's $1/\cos\theta$

So I can stand on their sum and move vertically an amount equal to their secant

Ted Shifrin

The quantity you're considering is far from invariant under scaling $p$ and $q$.

Akiva Weinberger

This is true, but neither is the setup — I'm looking at the solution $p+q+kr$ for unit $r$

Ted Shifrin

Why is $r$ unit? If $p$ and $q$ are, $r$ won't be unless it's a 90º angle.

Akiva Weinberger

That's how I defined it

I was doing that instead of $p\times q$ because that seemed too annoying

Ted Shifrin

7:17 PM

Oh, ok.

I'm not really paying detention.

Akiva Weinberger

This makes no sense actually

Argh

Semiclassical

I still have my (general) spherical problem which I now have a citation for (but not a genuine proof)

Akiva Weinberger

Hm - I actually meant $k^2$ there

Still feels off

Oh duh I'm an idiot

$(\|p\|^2+p\cdot q)(p\cdot q+\|q\|^2)=(p\cdot q)(\|p\|^2+\|q\|^2+k^2+2p\cdot q)$

$\|p\|^2\|q\|^2-(p\cdot q)^2=k^2p\cdot q$

Semiclassical

Let $A,B,C,D$ be points on the sphere and $AB,BC,CD,AD$ be their geodesic distances on the sphere. Then the set of possible such distances is characterized by the following inequalities:

Akiva Weinberger

$\|p\|^2\|q\|^2(1-\cos^2\theta)=k^2\|p\|\|q\|\cos\theta$

$\|p\|^2\|q\|\sin^2\theta=k^2\cos\theta$

$k=\|p\|\|q\|\tan\theta$

and that, finally, feels like it makes sense!

Semiclassical

7:30 PM

$$0\leq AB+BC+CD+AD-2\theta \leq \pi, $$ for $\theta=AB,BC,CD,AD$

(i'm writing it in that way to avoid having 8 different inequalities. )

Akiva Weinberger

Deja vu… I feel like you've asked something like this before

Maybe not

Semiclassical

yeah. this is the most general version

I'm pretty sure the other cases I've talked about correspond to restrictions of this, where one or more of the distances are equal

now to find a proof that isn't just "It's true because of Theorem X in book Y"

Akiva Weinberger

No fuck

$k=\sqrt{\|p\|\|q\|}\tan\theta$

OK better

Semiclassical

I do have at least a partial proof of this, though

Ted Shifrin

DogAteMy: You still have an error, unless there's more typo.

Is it $\cos^2\theta$?

Semiclassical

7:34 PM

First, one should be able to make a triangle inequality argument that $AB+BC+CD\geq AD$

and similarly for the other three cases.

Ted Shifrin

I don't think so.

Akiva Weinberger

…Yes but it corrects itself

What is it with me and typos today

Semiclassical

should be $\leq 2\pi$, woops

Akiva Weinberger

I even had trouble typing English earlier

Semiclassical

to get the other bound, replace B and D with their antipodes. then all angles get replaced by their supplements

so $3\pi -AB-BC -CD\geq \pi -AD\implies AB+BC+CD-AD\leq 2\pi$

Akiva Weinberger

7:37 PM

@Semiclassical So basically, you're saying that, if you want to get from $A$ to $D$, but make a detour to two stops along the way, at most you walked $\pi$ more than you needed to

user76284

Is there any connection between en.wikipedia.org/wiki/Derived_set_(mathematics) / proofwiki.org/wiki/Definition:Set_Derivative and the ordinary derivative from calculus? i.e. can the latter be rephrased as an instance of the former?

Semiclassical

@AkivaWeinberger $2\pi$, but yes (fixing my typo from earlier)

Akiva Weinberger

I'm not sure - maybe it's "set of points you can take the derivative at" @user76284

i.e.

Ted Shifrin

@user76284: No, not really.

Akiva Weinberger

if you have a function $f:S\to\mathbb R$

Ted Shifrin

7:38 PM

Not really, DogAteMy.

Akiva Weinberger

then you can't define the derivative of $f$ at an isolated point

but you could define the derivative of $f$ at a limit point because you can take the limit of $h\to0$

Ted Shifrin

You can't do it in higher dimensions with only one sequence, either.

Semiclassical

where I have no idea is if you go from angles $AB,BC,CD,AD$ to all six mutual angles

Akiva Weinberger

Oh, I was thinking of subsets of $\mathbb R$. You're right

@user76284 So then no, I don't think so

Semiclassical

I know (from the theorem I alluded to) that things become more restrictive in that case than you'd expect otherwise

but I don't know what inequalities are needed to characterize the set of allowed angles $AB,AC,AD,BC,BD,CD$

Akiva Weinberger

7:40 PM

Weird - shaking my mouse on my Mac makes the cursor bigger (briefly)

user76284

I'm wondering if there's a topological definition of the derivative like there is for limits (en.wikipedia.org/wiki/Net_(mathematics)#Limits_of_nets).

Akiva Weinberger

There's a question like that on the main I think

[main]

Mathematics

Semiclassical

for my upper bound, the better statement is probably proof by contradiction: assume one can pick A,B,C,D which would violate the upper bound, and show that A,-B,C,-D would violate the lower bound

user76284

I think you mean this one math.stackexchange.com/questions/272278/…? I didn't find the answers very satisfying.

Akiva Weinberger

Found it:

165

Q: Why can't differentiability be generalized as nicely as continuity?

The question: Can we define differentiable functions between (some class of) sets, "without $\Bbb R$"* so that it Reduces to the traditional definition when desired? Has the same use in at least some of the higher contexts where we would use the present differentiable manifolds? Motivation/...

general-topology derivatives differential-geometry differential-topology

Different one

user76284

7:44 PM

Thanks, I'll take a look.

Regarding the first question I linked to, my understanding is that there's a purely type-theoretic notion of the derivative, for instance, that doesn't seem to be taken into account by the answers claiming that smoothness (for ordinary notions of "smoothness") is necessary.

Semiclassical

@AkivaWeinberger as a special case of this, suppose A=D. Then the longest path ABCD possible (following geodesics each time) is the circumference of the sphere.

Akiva Weinberger

I see

Semiclassical

(that is, it collapses to $AB+BC+CA \leq 2\pi$)

user76284

or arithmetic derivatives.

sevdaicmis

hey how does $\langle v,x\rangle\cdot w$ read? it's from a linear algebra book. they're both scalar product. so why they're both used? confused

user76284

7:52 PM

I'm also wondering why $\mathbb{R}$ seems to play such a central role in these definitions when one can instead use algebras like $\mathbb{Q}_p$. To me this seems to hint at a more general structure.

Semiclassical

My guess is that they're using $\cdot$ as multiplication between scalars

e.g. $\langle v,x\rangle =1$ and so $\langle v,x\rangle \cdot w=1\cdot w = w$

i don't really like that but it seems plausible

user76284

There's also arxiv.org/pdf/1804.00746.pdf which I found interesting.

sevdaicmis

ah, right. after all, using them both scalar is not even type theoretically correct :)

Stan Shunpike

@TedShifrin Hey ted!

Akiva Weinberger

@user76284 I'm not gonna read through that, but I guess it's getting computers to differentiate functions algebraically rather than numerically?

This feels kind of like differential Galois theory

(a field I know nothing about)

user76284

7:58 PM

It's automatic differentiation for a category of derivative-augmented functions.

Automatic differentiation being different from both symbolic differentiation and numerical differentiation.

Akiva Weinberger

What's the difference between automatic differentiation and symbolic differentiation?

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$Mathematics$ Mathematics