Mathematics

12:34 AM

Let $M$ be a riemannian manifold. If $f:\mathfrak{X}(M)\rightarrow C^{\infty}(M)$ is $C^{\infty}(M)$ linear then does there exists a smooth covector field $\omega$, such that $\omega(X)=f(X)$ for all smooth vector fields $X$?

Leaky Nun

do it locally

orientablesurface

such a $\omega$ exists, right? @LeakyNun

Leaky Nun

I don't know

check the details

Thorgott

12:50 AM

You want to show that if $X_p=0$ for some smooth vector field $X$, then $f(X)(p)=0$

The map $\omega\mapsto(X\mapsto\omega(X))$ is clearly injective and $C^{\infty}(M)$-linear, so surjectivity implies that it gives an iso $\Gamma(T^{\ast}M)\cong\Gamma(TM)^{\ast}$, which is pretty

skullpatrol

3:35 AM

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Mathphile

How would we prove that $p_{k-1}^2 \gt p_k$ where $p_k$ is the kth prime?

I can't seem to prove it through induction

Sophie

@Mathphile use the fact that for a positive integer $n$ either it has a prime divisor less than $\sqrt{n}$ or it is prime.

what a deceptively simple looking problem

Mathphile

3:55 AM

@Sophie not sure how I can use that fact in the proof

Is it possible to prove this using induction?

Sophie

4:08 AM

@Mathphile wow apparently I misremembered it. Look here: math.stackexchange.com/questions/1594162/…

haha Robert Soupe commented "What a maddeningly simple problem!"... exactly how I feel about it

Mathphile

woah this proof is much more complicated than I expected it to be

copper.hat

5:10 AM

hello

Vikas Ukani

5:59 AM

how to remove rows in a pandas DataFrame which is not available in other DataFrame ?

Mathphile

6:15 AM

@VikasUkani you should ask this at stack overflow. This is the wrong chatroom.

copper.hat

any convex folks in here :-)

abenthy

8:18 AM

Does the usual de Rham isomorphism $H_\text{sing}(M;\mathbb{R}) \cong H_\text{dR}(M;\mathbb{R})$ still hold if I replace the real coefficients in singular cohomology by complex coefficients and consider $\mathbb{C}$-valued differential forms on the real manifold $M$?

Leaky Nun

aren't you just tensoring with $\Bbb C$ over $\Bbb R$

abenthy

Yeah, I thought so too, I guess I'll have to look up the universal coefficients theorem again.

Leaky Nun

they're fields

everything works well with fields

abenthy

So for fields I can squeeze the tensor product into the homology functor, right? Like $H_\text{dR}(M;\mathbb{R}) \otimes_\mathbb{R}\mathbb{C} = H(\Omega_{\text{dR}}(M;\mathbb{R}) \otimes_\mathbb{R} \mathbb{C}) = H(\Omega_\text{dR}(M;\mathbb{C}))$

Leaky Nun

yes

abenthy

8:51 AM

Alright, thank you

I just read about hodge structure from Milne's notes, and this makes much more sense now

Mary Star

11:35 AM

Hello!! Does someone of you have an idea about my question about differential equations?

0

Q: For which $f(x)$ does the solution exist for an arbitrary $c>0$?

We have the problem $$y''+4y=f(x), \ y(0)=0, y(c \pi )=0$$ with $c>0$ and $f(x)$ an arbitrary smooth function. For which values of the parameter $c$ has the problem alswaysan unique solution? For which $f(x)$ does the solution exist for an arbitrary $c>0$ ? For the first part we have to check w...

ordinary-differential-equations

Anonymous

2:30 PM

If the positive subseries of a series diverges to $\infty$ and negative subseries diverges to a finite value, is there any standard way of showing that any rearrangement of the series will also diverge to $\infty$? I think it is similar in spirit to the Riemann rearrangement theorem though I can't find it in a textbook

Thorgott

much easier than Riemann rearrangement

you can bound the partial sums below by the partial sums of the positive terms minus the finite value of all the negative terms and this goes to infinity

Anonymous

2:45 PM

@Thorgott Ummm, not sure I understand. Let's try to do the proof like this. Suppose there exists a rearrangement of the series $(a_n)$ that converges to a finite value $L$, but its positive subseries diverges to $\infty$ and its negative subseries converges to a negative finite value.

Anonymous

In this specific rearrangement, how do we proceed to show that it's invalid?

Anonymous

Is there any theorem like every rearrangement of a convergent series converges to the same value?

Anonymous

Umm, no I guess that's unconditionally convergent

Anonymous

Oh, I guess the triangle inequality could be helpful in an $\epsilon-\delta$ proof

Thorgott

I don't see a good reason to do this by contradiction

Anonymous

2:56 PM

@Thorgott What's your approach?

Anonymous

I think we have to show that there exists no rearrangement of this form that converges to a finite value. This sounds like a contradiction proof to me

Thorgott

I outlined mine above

You can also just show that every rearrangement diverges to $+\infty$ directly

Anonymous

@Thorgott I noticed, but there you are only showing that specific rearrangement diverges to $\infty$

Anonymous

That is, one with all the negative terms clubbed together at the beginning

Thorgott

no, take any rearrangement, look at a partial sum (which is a finite sum), separate that into positive and negative terms, bound this below by adding all the other negative terms (here we use that any rearrangement of just the negative terms converges to the same value) and then observe that lower bound goes to $+\infty$ (here we use that any rearragement of just the positive terms diverges to $+\infty$)

Anonymous

3:04 PM

Which theorem gives us this "any rearrangement of just the negative terms converges to the same value"?

Anonymous

Also why "any rearrangement of just the positive terms diverges to +∞)"?

Anonymous

I mean, it makes intuitive sense but I can't pinpoint the theorem

Thorgott

first one is the standard fact that absolute convergence implies unconditional convergence

don't know a reference for the second one, but it's an easy fact

Anonymous

@Thorgott Oh, it's absolutely convergent as all the terms of the negative subseries have the same sign right?

Anonymous

@Thorgott Do you know a simple proof?

Thorgott

3:10 PM

yes (and since it's convergent by hypothesis)
yes, you should try proving it yourself

Anonymous

@Thorgott Hint please? :P

Anonymous

Suppose we just have that $(a_n)$ diverges to $\infty$

Anonymous

I do a rearrangement

Thorgott

partial sums of both the series and the rearrangement are increasing sequences, so you just have to show unboundedness of one implies unboundedness of the other

Anonymous

@Thorgott Ummm, but how do we know that the rearrangement is increasing?

Thorgott

3:24 PM

all summands are positive

Anonymous

@Thorgott We don't know all terms of $(a_n)$ are positive though

Anonymous

Some might be negative too

Anonymous

But still all add up to $\infty$

Thorgott

"any rearrangement of just the positive terms diverges to +∞"

this is what we're talking about, no?

Anonymous

@Thorgott Oh yes!

Anonymous

3:27 PM

Oops

Akiva Weinberger

4:56 PM

How can I topologically show the 600-cell has 600 sides?

The assumptions are: tetrahedral cells, 2 cells per face, 5 cells per edge, 20 cells per vertex

Euler's identity in 4D is $C-V+E-F=0$

Solving for $C$, knowing a tetrahedral cell has 4 faces, 6 edges, and 4 vertices,

I get $C-\dfrac42C+\dfrac65C-\dfrac4{20}C=0\\$

But the left-hand size simplifies to $0$

so this is true for all $C$

So what gives

Leaky Nun

5:12 PM

each cell has 4 faces so in total there are 2400 faces, if not for the fact that each face belongs to 2 cells so we double-counted by a factor of 2, so there are in fact only 1200 faces

similarly there are 600 x 6 / 5 = 720 edges?

yeah wiki confirms 720 edges

@AkivaWeinberger

Akiva Weinberger

But how do you find 600 in the first place

Leaky Nun

what does 600-cell mean

Archer

1

A: general solution using variation of parameters

In fact this just involves some elementary functions simplifications. By variation of parameters, you can get the particular solution is $y_p=\sin x\int_0^xf(x)\cos x~dx-\cos x\int_0^xf(x)\sin x~dx$ Note that the particular solution can also rewrite as $y_p=\sin x\int_0^xf(t)\cos t~dt-\cos x\i...

Could anyone look at my comment on this post?

> The variation of parameters methods has no limits in the integral but you have put them. Does that make a difference?

Akiva Weinberger

6:06 PM

1 hour ago, by Akiva Weinberger

The assumptions are: tetrahedral cells, 2 cells per face, 5 cells per edge, 20 cells per vertex

It means that^

From that information, how can I determine the number of cells

Leaky Nun

6:22 PM

@AkivaWeinberger I thought 600-cell means it has 600 cells

Akiva Weinberger

7:11 PM

@LeakyNun Right yes but pretend I didn't know it's called that

Leaky Nun

7:27 PM

yeah I see your point, I don't know

@AkivaWeinberger what if it isn't a topological property

S^3 looks like R^3 with all infinities identified right

so imagine you draw a tetrahedron inside, and take its complement

I think the outside is also a tetrahedron

so I think you can "triangulate" (in a very loose sense) S^3 with 2 tetrahedra

Akiva Weinberger

Sure but that's two cells per edge

Leaky Nun

oh, I forgot those restrictions

@AkivaWeinberger why don't we start with 3D solids

why does the cube (3 faces per vertex, 2 faces per edge) have 6 faces

Akiva Weinberger

The biggest difference

is the Euler characteristic of S^3 is 0

and the Euler characteristic of S^2 is 2

Let's do the cube

$V-E+F=2$

$\dfrac43F-\dfrac 42F+F=2\\$

$\left(\dfrac13\right)F=2\\$

$F=6$

Leaky Nun

ok, fair enough

Akiva Weinberger

Compare

3 hours ago, by Akiva Weinberger

I get $C-\dfrac42C+\dfrac65C-\dfrac4{20}C=0\\$

That gives $0C=0$

which is unhelpful

Leaky Nun

7:40 PM

right

so the closest analogue is S^1

when in fact you do have infinitely many regular hollow polygons

Akiva Weinberger

Right

Leaky Nun

so is the 600-cell the only triangulation of S^3 that satisfies those properties?

can I subdivide each tetrahedron?

Akiva Weinberger

If you delete a tetrahedron,

all its edges suddenly have only four cells

so if you substitute back in a subdivided tetrahedron

it has to only bring that back up to five

Leaky Nun

but you subdivide all of the tetrahedra

Akiva Weinberger

I feel like that'll increase the cells per edge

Leaky Nun

7:43 PM

the S^1 analogue is breaking each line into two, so an n-gon becomes a 2n-gon

Akiva Weinberger

Right

Mike Miller

@AkivaWeinberger does passing to the convex hull help you?

Akiva Weinberger

Of what?

Mike Miller

the 600-cell in $\Bbb R^4$

Akiva Weinberger

The 600-cell is already convex, no?

I mean I guess you add in the inside

Mike Miller

7:44 PM

that's what the convex hull is, yeah

Leaky Nun

I'm not convinced that 600 is unique, but again I don't know anything

> The convex regular 4-polytopes were first described by the Swiss mathematician Ludwig Schläfli in the mid-19th century. He discovered that there are precisely six such figures.

ok then it should be unique lmao

Balarka Sen

The proof of uniqueness of platonic solids in 3D uses Euler chi, so that's Akiva's point

It doesn't seem to detect the number 600

in 4D

Leaky Nun

right, that's the problem

Balarka Sen

I didn't think though

Leaky Nun

(what if it isn't a topological problem but a trigonometrical one)

Akiva Weinberger

7:48 PM

So if they're geometric tetrahedra then I think you can do geometry

If they're bendy topological tetrahedra then I dunno

Mike Miller

Meh the convex hull contains exactly the information that the 600-cell itself does

Akiva Weinberger

I feel like all topological ones should be isomorphic to the geometric one. If that's true, that could be a proof

I kinda want a pure topology proof though

Balarka Sen

Why is classification of platonic solids the same problem as classifying finite subgroups of $\mathrm{SO}(3)$ (except degenerate cases like cyclic and dihedral) but different one dimension up?

Akiva Weinberger

especially 'cause in 3D they're so nice

Leaky Nun

we're stepping into triangulation theory

Balarka Sen

7:50 PM

$\text{SO}(4)$ has quite a number of finite subgroups... they are all compact $3$-manifold groups.

Mike Miller

Is that true? I didn't know that

Balarka Sen

Elliptization, right?

Mike Miller