Mathematics - 2020-03-07

Krull Dimension

01:37

I'm wondering whether the quartic curve 3x^4+3x^2y^2+y^4-x^2-y^2=0 has a particular name or belongs to a known family of curves

I emailed someone who has a webpage about ovals and egg shapes about it, and he said it was a hippopede, but the x^4 and y^4 terms have different coefficients, so I'm not sure how it would be possible to get the equation in that form.

adesh mishra

03:32

@Thorgott But I completed my Calculus without them, are they needed for anything coming in future? Like linear algebra, Vector Calculus, Topology

Slereah

11:07

Hello the math

Does anyone know a reference to external tensor product bundles that doesn't involve category theory?

Mike Miller

12:05

@Slereah No, but they're not hard to write down

(E o F)_{x,y} = E_x o F_y

Check local trivializations exist on products of charts where trivs exists for E and F

Slereah

I am particularly interested on the covariant derivative on them

ie for the external product of two tensor bundles

Mike Miller

what about it

Slereah

4

Q: Defining the covariant derivative on bitensors

Bitensors (tensors defined on two different points) are an extension of tensors found in some applications of general relativity, where objects such as the world function, parallel transport operator, heat kernel or propagator are bitensors of some kind. It is unfortunately not commonly done in t...

Basically this

It's a bit hard to find something in between Synge's physicist description of bitensors and the extremely category theoretical description of exterior tensor products

Edward Evans

12:49

@Lukas lol nice

thanks :)

adesh mishra

15:24

@Semiclassical Sammy sir are you there ?

Rithaniel

So, I'm trying to find the class group of $\mathbb{Z}[\sqrt{-33}]$. I've used the Minkowski Bound and found a total of 10 classes, but now I need to try and identify which ones are unique. I know that, to show that two ideals $P,Q$ are in the same class, I should find $\alpha,\beta$ such that $\alpha P=\beta Q$, but how do you find such $\alpha,\beta$? How do you show that no such elements exist?

Edward Evans

15:39

@Rithaniel the Minkowski bound in this case is just over $7$ so you only need to check the classes of $2, 3, 5$ and $7$

topologicalorientablesurface

Let $(X,d)$ be a (non-empty) complete space. Suppose $X=\bigcup_n F_n$ where each $F_n$ is closed. Then there exists $n$ such that $F_n$ has non-empty interior.

Rithaniel

Well, it's a little bit more complicated than that. For example, $(2)$ splits into $\mathfrak{A}_1\mathfrak{A}_2$, and therefore $(4)$ splits into $\mathfrak{A}_1^2\mathfrak{A}_1^2$, but we don't know if there is any repetition in this.

If the class group is isomorphic to $\mathbb{Z}_8$, with $\mathfrak{A}_1$ corresponding to $\overline{1}$ and $\mathfrak{A}_2$ corresponding to $\overline{7}$, then $\mathfrak{A}_1^2$ corresponds to $\overline{2}$ and $\mathfrak{A}_2^2$ corresponds to $\overline{6}$

So I have to check $4$ and $6$ as well

Leaky Nun

@topologicalorientablesurface baire reacc only

topologicalorientablesurface

Suppose for the sake of a contradiction that for each $n$, $F_n$ has empty interior. In particular, that would imply there exists an element $x_1\in F_1^c$ (otherwise $X$ would be empty). So, we may obtain a closed neighborhood $G_1$ of $x_1$ such that $G_1\cap F_1=\varnothing$, where we can specify that $diamG_1<\frac{1}{2}$ (this exists because a singleton is such an example). Consider $F_1\cup F_2$.

Since they have empty interior, we may find $x_2\in X\backslash (F_1\cup F_2)$, with a closed neighborhood $G$ of $x_2$ satisfying $diamG<\frac{1}{2^2}$, such that $G\cap (F_1\cup F_2)=\varnothing$. Set $G_2=G\cap G_1$. This set in particular, is a closed set containing $x_2$......

Edward Evans

@Rithaniel No, the class group is generated by the classes of prime ideals, so you only need to check the classes of the primes lying over rational primes less than the Minkowski bound

topologicalorientablesurface

15:46

Why does $G_1$ contain $x_2$?

Leaky Nun

@Rithaniel kconrad.math.uconn.edu/blurbs/gradnumthy/classgpex.pdf

Edward Evans

@Rithaniel also $2$ is ramified

Leaky Nun

@topologicalorientablesurface draw a picture

Rithaniel

$3$ is ramified, $2$ is not, because $2\nmid -33$

Edward Evans

$-33$ isn't the discriminant of that number field

$-4\cdot 33$

is

Rithaniel

15:48

Ah, right, discriminant, my bad

Edward Evans

also, the blurb @Leaky linked is great for examples of class group computations for quadratic fields lol

Used it like a billion times

and by "like a billion" I mean probably $\leq 100$

but that's basically a billion

topologicalorientablesurface

@LeakyNun still don't quite see it

Rithaniel

What do the prime factors of $(2)$ look like, then? I know $(3)=(3,\sqrt{-33})^2$

Abhas Kumar Sinha

Hello, I need some urgent help.

Is there any PhD here?

Edward Evans

@Rithaniel My guess is $(2, 1 + \sqrt{-33}) = (2, 1- \sqrt{-33})$

Abhas Kumar Sinha

15:53

any subject/matter/wagera wagera

Rithaniel

Ah, right, that's true

Edward Evans

@Rithaniel Prime because $\Bbb Z[\sqrt{-33}]/(2, 1 + \sqrt{-33}) \cong \Bbb F_2[X]/(1+X) \cong \Bbb F_2$

uhh is that right

ye

Rithaniel

What about $(6)=(3)(2)=\mathfrak{A}_1^2\mathfrak{A}_2^2$. Should I list $\mathfrak{A}_1\mathfrak{A}_2$ as another potential class?

Edward Evans

It's not prime is it?

Leaky Nun

what on earth is wagera?

Abhas Kumar Sinha

16:02

wagera = blah (in english)

@LeakyNun Are you a PhD?

Leaky Nun

no

Abhas Kumar Sinha

You just met a right guy

okay then...

I'm in a big problem

Edward Evans

@Rithaniel as I say, you only really care about the classes of prime ideals lying above rational primes less than the Minkowski bound, so you'll want to look at the classes of primes above $2, 3, 5$ and $7$. We know $2, 3$ are ramified and we know their factorisation in $\Bbb Z[\sqrt{-33}]$. How do $5$ and $7$ split?

Abhas Kumar Sinha

very big one and needs a PhD for that

Rithaniel

5 is inert and 7 splits. Also, my professor advised that I check 4, and you're just waving about a theorem we haven't proved yet, so I'm going to go ahead and list $\mathfrak{A}_1\mathfrak{A}_2$ as another potential class group

Edward Evans

16:10

o.o okie dokie

Rithaniel

So, is your reasoning that, because it's generated by those prime ideals, then the prime factorization of 6 is just also generated by those ideals, and so we would check those prime ideals anyways during the verification?

adesh mishra

Can I use mathematical induction when my formula to be proved involves two variable?

Rithaniel

Like, we know $(7)=\mathfrak{A}_3\mathfrak{A}_4$, so we need to check which class $\mathfrak{A}_3\mathfrak{A}_1$ falls into, right?

So we'd check $\mathfrak{A}_1\mathfrak{A}_2$ anyways

Edward Evans

I don't know what you're doing, $(7)$ factorises as a product of two distinct prime ideals (because it's split) so you have primes $\mathfrak{A}_7, \mathfrak{A}_7^\prime$; you need to check the classes of both of these

adesh mishra

Edward when you get free please just ping me for that induction question that I have asked above.

Edward Evans

16:19

@adeshmishra math.stackexchange.com/questions/983051/…

Rithaniel

So, we're trying to find the structure of the class group. If we have $(2)=\mathfrak{A}_1^2$ and $(3)=\mathfrak{A}_2^2$. Then we should check the ideal class of $\mathfrak{A}_1\mathfrak{A}_2$ eventually. If $\mathfrak{A}_1\mathfrak{A}_2$ is principal, then we know that $\mathfrak{A}_1=\mathfrak{A}_2$. If it isn't, then $\mathfrak{A}_1\neq\mathfrak{A}_2$

Same with all combinations of ideals that we have found

Edward Evans

I see what you mean now lol, yeah that's a way of eliminating generators

Rithaniel

(Yeah, reading back on what I wrote before, I could have been more clear)

Edward Evans

haha it's fine

adesh mishra

@EdwardEvans Thank you so much.

Edward Evans

16:25

@Rithaniel defo take a look at the thing Leaky posted too, Conrad uses a bunch of different techniques for working out whether or not an ideal class is trivial

Rithaniel

I'm reading it as I go, yeah

Though, the Minkowski Bound gives us that every prime ideal class has a member with norm equal to one of the integers less than the bound. Regardless of whether these classes generate the class group, finding an ideal with each of the respective norms would give an exhaustive list of the ideals in the class group.

Edward Evans

Sure but.. that seems like more work?

Rithaniel

16:41

Yeah, I suppose so. The "generated" part would indeed make it much easier when working with large MBs.

The Terrible Puddle

17:09

So $(x,y)$ is an ideal generated by $x$ and $y$ but what does $(x,y)^2$ mean exactly?

Rithaniel

$(x,y)(a,b)=(xa,xb,ya,yb)$ (I'm gonna hedge my bets and claim that this only holds for sure in the commutative case with identity. It might hold in general, but I'm not familiar enough with that stuff)

The Terrible Puddle

Just found this https://math.stackexchange.com/questions/1521552/how-do-you-square-an-ideal
I think that helps

Rithaniel

Yep, Bernards answer confirms that it does hold in general

Rithaniel

17:42

Alright, the answer I got for the class group of $\mathbb{Z}[\sqrt{-33}]$ is $\mathbb{Z}_2\oplus\mathbb{Z}_2$.

Edward Evans

Nise

Rithaniel

Next up is $\mathbb{Z}[\sqrt{-37}]$, but it seems like nothing splits and there is just a single ramified prime

Yeah, $-37$ is not a QR mod $3$, $5$, or $7$, so there are no split primes

Edward Evans

17:58

I assume you mean less than the Minkowski bound lol

Ted Shifrin

@adeshmishra Adesh: You can actually do a proof by induction on one variable only (although you may need to do induction on the other variable in the base case, if there's no other way). Suppose your statement is $F(m,n)$. Then let your statement be "$P(n)$: $F(m,n)$ holds for all $m\in\Bbb N$" and proceed

Rithaniel

@EdwardEvans Yeah. It actually didn't occur to me that there might still be split primes in the ring even if no primes less than the bound split

Edward Evans

@Rithaniel interestingly, one expects half the primes in that ring to be split

Rithaniel

Because it's a HFD?

Ted Shifrin

What's a HFD?

Edward Evans

18:12

@Rithaniel By Cheboratev's density theorem!

but that's just an aside lol

Cheboratev*

loool

Rithaniel

Ted: Half-factorization Domain. An atomic domain is an HFD if any two factorizations into irreducibles must be of the same length, if they are equal

Edward Evans

and really I mean one expects half the rational primes to split in that ring*

Ted Shifrin

Oh, never heard of that before.

Rithaniel

Well I have another theorem to google, Edward

Edward Evans

Nice :) It's a cool result

@Ted me neither

also hey @TedShifrin :P

Ted Shifrin

18:15

Chebotarev also tells you that "almost every" integer irreducible polynomial is reducible over $\Bbb Z/p\Bbb Z$ for all $p$. We worked that out years ago.

hey @Edward

Rithaniel

I'm actually taking a topics course in Factorization (It's what this work is for), so I suppose I shouldn't use the terminology from that course and assume everyone knows what it all means.

Edward Evans

oh nice

Rithaniel

One thing is that a ring of algebraic integers is a HFD if and only if the order of it's class group is less than or equal to 2

Edward Evans

Cool, didn't know that

anakhro

Hi all.

Ted Shifrin

18:24

Hi one @anakhro.

Astyx

Hello

anakhro

What have you all been up to?

Ted Shifrin

Salut @Astyx

Stan Shunpike

18:47

hello

@TedShifrin hey ted!

William Sun

For a real-valued function $f$ defined on a closed interval, is the following condition equivalent to the Riemann integrability of $f$:

Ted Shifrin

heya @Stan

William Sun

For any $\epsilon>0$ there is a number $\delta >0$ such that for any partition $P$ with norm $|P|<\delta$, $U_f(P)-L_f(P)<\epsilon$.

Ted Shifrin

Yes, @William.

William Sun

The norm of a partition is the maximal length of the subintervals in the partition. $U_f(P)$ is the upper sum.

I can only prove the trivial case that "there exists a partition $P$ such that for each refinement $P'$ of $P$ we have the variance sum $<\epsilon$."

Rithaniel

18:53

Heya Anakhro, Astyx, and Stan

Ted Shifrin

So you know that for every $\epsilon$ there is a partition $P'$ with $U_f(P')-L_f(P')<\epsilon/2$. From that you can deduce what you're saying.

Oh, I guess this is the criterion you were calling your trivial case.

William Sun

We have for a partition $P$ and a refinement $P'$ of $P$ the following inequality $$L_f(P)\leq L_f(P')\leq U_f(P')\leq U_f(P)$$

And the "trivial case" follows immediately from this inequality.

Ted Shifrin

Yes, but I want you to use my partition $P'$ to construct a $\delta$.

Stan Shunpike

@TedShifrin I want to learn more about causality and how it relates to probability. know any good resources for that?

Ted Shifrin

As a hint: You'll need to know how many partition points there are in $P'$.

Nope @Stan.

William Sun

19:00

Let me write the whole proof down here. First for any $r>0$ there is a partition $P_0$ such that $U(P_0)-L(P_0)<r/2$. It is sufficient to find a number $\delta>0$ such that for each partition $P$ with $|P|<\delta$, $Var(P)-Var(P\cup P_0)<r/2$, where $Var(P)=U(P)-L(P)$ and $P\cup P_0$ is the refinement of the two partitions.

I guess I should start with the number of partition points of $P_0$.

Since $f$ is bounded we can find $\delta$ small enough...$\delta<(\max |f|)/(2n) should work.

$n$ is the points in the partition $P_0$. Thank you for the hint @TedShifrin

Ted Shifrin

You're welcome.

You need an $\epsilon$ in there somewhere, don't you?

William Sun

Oh yes there should be an epsilon on the numerator.

Ted Shifrin

And probably a $4$ in the denominator?

William Sun

Yes I should write it down instead of doing it in head...

Ted Shifrin

Probably :P

FuzzyPixelz

19:23

Hey @TedShifrin

Ted Shifrin

hi @Fuzzy

FuzzyPixelz

Interested in some good old funky questions?

Ted Shifrin

That's a bit vague.

Rithaniel

So, if a principal ideal factors in an algebraic number field, under what conditions do the prime ideals it factors into end up also being principal?

FuzzyPixelz

Of course it is, I wrote a question earlier but no one answered, math.stackexchange.com/questions/3572368/…

Ted Shifrin

19:27

There are various ways of proving that you can differentiate under the integral sign.

If you don't know any Lebesgue theory, the easiest trick is to use Fubini's Theorem to change an order of integration.

FuzzyPixelz

I was trying to follow the hint I was given

Using the chain rule

Yeah... I'm not supposed to use Fubini's theorem either

Ted Shifrin

The chain rule is not relevant to this issue.

So you have to do some compactness uniform convergence argument with difference quotients, then.

Oh, and the Mean Value Theorem.

FuzzyPixelz

That's for when the integral bounds are constant I think?

Ted Shifrin

Sure. Which they are in the proper proof.

FuzzyPixelz

Here, en.wikipedia.org/wiki/Leibniz_integral_rule#Proof_of_basic_form

Ted Shifrin

19:32

Huh?

FuzzyPixelz

But @TedShifrin, the whole point of this assignment for me is to apply the chain rule, and I already know the Leibniz rule for constant bounds.

Ted Shifrin

Oh, then the rest is easy.

Yes, you define $F(x,y,z) = \int_y^z f(x,t)\,dt$ and use the chain rule. What is your problem?

FuzzyPixelz

Showing that it's differentiable

Ted Shifrin

It has continuous partial derivatives.

FuzzyPixelz

The partial derivative w.r.t $x$ has to be continuous.

Oh yes, I'm not sure of my proof for that one

Ted Shifrin

19:35

That's because $\partial f/\partial x$ is continuous on a compact set, hence uniformly continuous.

FuzzyPixelz

What about $y$ and $z$ ? Shouldn't we argue about continuity in $\mathbb R^3$ ?

Ted Shifrin

Yes. So where do you get stuck? That just uses the fact that the integrand is integrable, hence bounded.

FuzzyPixelz

I didn't get stuck, I'm just looking to verify my proof

Ted Shifrin

Your proof forgot the $x$ continuity, didn't it?

The $yz$ part is fine.

FuzzyPixelz

I used the max norm?

Ted Shifrin

19:41

Huh?

No, you have a mistake.

FuzzyPixelz

Wait, are we talking about the first or the second post?

I wrote an answer down..

Ted Shifrin

I'm talking about your answer. I ignored everything else.

FuzzyPixelz

I'm sorry, what is my mistake exactly

Ted Shifrin

Where is $x_0$ on the right hand side?

FuzzyPixelz

Oh, I should've replaced $x$ with $x_0$ in the $\frac{\partial f}{\partial x}(x,t)$ inside the second integral, but I only used the fact that $\frac{\partial f}{\partial x}$ is bounded afterwards so it should be fine?

Ted Shifrin

19:50

Well, are you claiming that the function is independent of $x$?

You better pay a little more attention to the algebra you did.

FuzzyPixelz

I see.. it's definitely wrong

CaptainAmerica16

Hi

Ted Shifrin

It's not hard to fix, @Fuzzy, but it needs to be fixed.

Hi, @CaptainAmerica.

FuzzyPixelz

I would have to use the uniform continuity of $\frac{\partial f}{\partial x}$ like you said @TedShifrin

Ted Shifrin

Yup.

FuzzyPixelz

19:54

That would add $M|x-x_0|$ on the RHS

Ted Shifrin

Well, you need to do the trick of adding $0$ twice.

No, no $M|x-x_0|$, I think.

Do it carefully.

FuzzyPixelz

I'll do it now

CaptainAmerica16

Ted, you dont do research anymore, do you?

Ted Shifrin

Nope.

CaptainAmerica16

Ah, ok. I figured

I'm going to apply to the honors program for my degree. I'll have to write an undergraduate thesis. Have you advised an undergrad?

Idk how any of this works really lol

Ted Shifrin

20:03

I advised a few Honors theses, but in math I generally discouraged people from doing them (and masters theses). The US is different from Europe — we don't need a MA to apply for a PhD. But it's good experience writing stuff. For an honors thesis, it's generally not much more than a reading report, not much "new" expected.

anakhro

Why did you "discourage" them?

CaptainAmerica16

I guess good for networking more than anything

Ted Shifrin

Because generally for those going on to a PhD it was more significant to take graduate courses than to write a thesis. For people quitting with the undergrad or masters, sure, no problem.

CaptainAmerica16

Oof, I was hoping it would help me prepare for the PhD

Ted Shifrin

At the end of my career, one of my star students wrote up her summer REU research (which was significant) as an Honors thesis. She did the work elsewhere, but I advised that Honors thesis, and it was a lot of time/work getting it publishable.

anakhro

20:08

It was because of my undergrad thesis that I actually continued with a M.Sc., having had the experience with math closer to research math. I suppose it really depends on your supervisor for the undergrad thesis, and what topic it is, though.

Ted Shifrin

@CaptainAmerica: If you're in a position to take graduate courses with actual doctoral students in the classes, that's the best to impress people when you apply to grad school.

Letters from professors from such courses. If not, then a thesis is good.

anakhro

I was taking grad courses at the time of doing my undergrad thesis.

Ted Shifrin

Of course, people who go to a small liberal arts college with no graduate program can't do as I suggest.

CaptainAmerica16

I'm going to a research school, thankfully

Ted Shifrin

But a letter (such as I have written) saying that an undergrad outperformed grad students in first-year graduate courses (with specifics, of course) goes a long way.

And if one doesn't do well in such courses, then maybe one should contemplate one's decision more.

CaptainAmerica16

20:11

Do you think I could get undergrad research positions in math as a CS major?

Ted Shifrin

I don't know what undergrad research positions are.

But if you're a CS major and contemplating grad school in CS, you should be doing what I'm saying in CS.

CaptainAmerica16

I dont really know either, the school just says that undergrads can assist willing professors with research. Nvrmnd, that was a silly question.

Ted Shifrin

Too early to be contemplating this.

CaptainAmerica16

I honestly want to do pure math for grad school. CS is just to have something to fall back on

You're right, I've just been thinking a lot lately. Lol

Ted Shifrin

As I said, too early to contemplate. Then you should double-major.

FuzzyPixelz

20:14

$$\left | \frac{\partial \phi}{\partial x}(x,y,z)-\frac{\partial \phi}{\partial x}(x_0,y_0,z_0) \right |=\left |\int_{z_0}^z \frac{\partial f}{\partial x}(x,t)dt - \int_{y_0}^{y} \frac{\partial f}{\partial x}(x,t)dt +\int_{y_0}^{z_0} \left (\frac{\partial f}{\partial x}(x,t)-\frac{\partial f}{\partial x}(x_0,t) \right )dt \right |$$ Since $\partial f/ \partial x$ is uniformly continuous we can write $| \frac{\partial f}{\partial x}(x,t)-\frac{\partial f}{\partial x}(x_0,t)| \le k |x-x_0|$

So the LHS becomes less than or equal to $$\max(M, k|y_0-z_0|) \| (x,y,z)- (x_0,y_0,z_0)\|$$

Ted Shifrin

See how challenging math classes go for a few years and then see.

CaptainAmerica16

Ok

FuzzyPixelz

Not really Lipschitz continuous this time, but it's continuous

(Hopefully)

Ted Shifrin

I still don't see quite where that right hand side came from.

Oh, maybe you added a different 0.

Yeah, you did. I get it.

Looks ok.

FuzzyPixelz

$$\int_y^z=\int_y^{y_0} + \int_{y_0}^{z_0}+\int_{z_0}^{z}$$

Is what I did

Thanks @TedShifrin you're awesome.

Ted Shifrin

20:23

Yeah, but how the $x$ and $x_0$ work in too ...

FuzzyPixelz

Another error?

Ted Shifrin

I'm just saying that what you just wrote down doesn't suffice.

Because you have to keep track of $x$ values, too.

But I think we're done. Anyhow, I am.

FuzzyPixelz

You mean the zeros I added? I only did it for the integral with $x$ not $x_0$

Ted Shifrin

OK, and then you group the middle term with the $x_0$ integral. Got it.

user76284

Anyone have an idea for how to sum the series $\sum_{n \geq 1} \frac{\mu(n)}{n} \log (n+1)$ where $\mu$ is the Möbius function?

user76284

20:56

I know that $\sum_{n \geq 1} \frac{\mu(n)}{n} \log n = -1$.

Jack Ohara

@LeakyNun Hello

@LeakyNun Do you know how to construct elliptic curves of char k finite, of any order?

Rithaniel

Having some difficulty. I am looking at $\mathbb{Z}[\frac{1+\sqrt{-59}}{2}]$, in which I know that $(2)$ is a prime ideal that splits, but I can't figure out the ideals it splits into. First two I tried were $(2,1+\sqrt{-59})$ and $(2,1-\sqrt{-59})$, but their product is $(4,2+2\sqrt{-59},2-2\sqrt{-59},60)$, and I don't see how $2$ can be written as a linear combination of those elements.

In fact, I don't believe the norm of any element in $\mathbb{Z}[\frac{1+\sqrt{-59}}{2}]$ can attain a value congruent to $2$ mod $4$, so nothing of the form $(2,a)(2,\overline{a})$ is going to work

Rithaniel

21:40

Wait, perhaps 2 is inert

The minimal polynomial for which $\frac{1+\sqrt{-59}}{2}$ is a root of is $x^2-x+15$, which is $x^2+x+1$ mod 2

I forget what this information is suppose to give me, but I do know that $x^2+x+1$ is irreducible over the field of 2 elements

user193319

21:59

2

Q: Group Convolution is Associative

Let $G$ be some locally compact group and $\mu$ its associated Haar measure. I am trying to adapt this proof that convolution on the locally compact group $(\Bbb{R},+)$ is associative. Here's what I have so far: $$((f \ast g) \ast h)(u) = \int_{G} (f \ast g)(x)h(x^{-1}u) ~d \mu (x)$$ $$= \int_{...

integration functional-analysis harmonic-analysis locally-compact-groups haar-measure

Rithaniel

What exactly is $\mathcal{O}_K$ anyways? I keep seeing people use that terminology, but I've yet to find a definition for it.

Found it: ring of integers for $K$ (they could have used Z, but nope)

geocalc33

22:23

0

Q: Branched manifolds in the context of general relativity

Does anyone study Branched manifolds in the context of General Relativity? I would like to define the union of two transversal spacetimes $G:=M \cup S.$ And I would also like to define a convolution operator $R:=M \star S.$ I want the convolution operator to operate on lines of constant...

reference-request mathematical-physics general-relativity manifolds-with-boundary semi-riemannian-geometry

@user193319 nice post I'm looking at it now

user450595

22:51

hi i'm getting a different result in wolfram than by hand, can someone have a look?
https://math.stackexchange.com/questions/3573024/solve-y-prime-frac-y2x2-frac-yx-cant-find-my-mistake

Astyx

$(v/x)^2/x^2 = v^2/x^4$

not $v^2/x^3$

user450595

okay i changed it see my edit

user450595

but what i am getting is

user450595

y=2x+c/x

user450595

which seems to be incorrect

Astyx

22:58

${1\over{-{1\over 2x^2} + c} }\ne2x^2 + c'$

it's $2x^2\over (c x^2 - 1)$

Rithaniel

Shouldn't that be $2c$ in the denominator?

Astyx

It's an arbitrary constant so it doesn't matter

Rithaniel

Yeah, this is true

Astyx

But yeah, you should probably give it another name

user450595

it seems that the relation which was missing is the v's and x's

user450595

23:01

i was able to find the last mistake

user450595

a question i have

user450595

when i got a riccati equation

user450595

which solution should i try to guess?

user450595

solutions*

Astyx

That seems like a wild question

user450595

23:04

what it can be anything?

Astyx

Wikipedia says to substitue $v = y q_2$ when $q_2$ is the coefficient of the term in $y^2$

Then $v = ({1\over u})' = {-u'\over u}$ gives you an ODE for $u$

If you manage to solve it you can work out the solutions to your original equation

I haven't done the maths though, you probably should

topologicalorientablesurface

Question

Astyx

Answer

topologicalorientablesurface

F is the countable union of sets if and only if the index set is $\{$ $1,2,3...N$ $\}$ or $\mathbb{N}}$?

Victor

0

Q: An inverse matrix problem on a statistics book

How to derived the inverse matrix I highlined?

linear-algebra statistics statistical-inference

topologicalorientablesurface

23:12

or does it mean that the index set has a one to one correspondence with the naturals or a finite subset?

Astyx

What's F ?

topologicalorientablesurface

a set

how does one define countable union, I suppose?

Astyx

oh

I think it's just that the set of indices has to be countable

topologicalorientablesurface

@Astyx yeah thats what I thought, but one of the arguments in Munkres book said that the set of indices is either $\{$ 1,2...,N $\}$ or $\mathbb{N}$. You can assume that without loss of generality, am I correct?

Astyx

Yes

topologicalorientablesurface

23:16

ok thanks. Just wanted to confirm my suspicions. Thanks @Astyx

Astyx

If you're countable, you have a mapping $\phi: I\to\Bbb N$ or $\phi :I\to [1, n]$

So you can change the indices $a_i = a_{\phi(i)}$ for $i\in I$ and get integer indices

Rithaniel

Figured out the cause of my headache before: I was relying on a statement which was only in certain circumstances equivalent to the real criteria of a splitting prime

Now I've gotten myself sorted, I believe

Lucas Henrique

23:40

hey chat

is it false that $I^2 = I$ for all two-sided ideals $I$?

I'm not sure, but it looks like $(2\mathbb Z)^2 = 4\mathbb Z$

(which is not $2\mathbb Z$, clearly)

I'm trying to prove that $(a+I)(b+I) = ab + I$, that would be $ab + aI + bI + I^2 = ab + I$

But I'm not sure how

EnjoysMath

23:56

Hey @Jack

@JackOhara

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