The h Bar

0celo7

12:00 AM

@ACuriousMind But don't you have to smooth a thing

Abuse.

ACuriousMind

...what?

0celo7

29 secs ago, by 0celo7

@ACuriousMind But don't you have to smooth a thing

I think I had a stroke there.

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

Welcome back @ArtOfCode more flags?

0celo7

I mean to say "but don't you have to prove that it's smooth"?

Wonder who is flagging!

ArtOfCode

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ Yeap.

ACuriousMind

12:01 AM

Oh, come on, that was a silly joke!

ArtOfCode

I know :/

0celo7

@ACuriousMind Remember I once got suspended for saying I like my women with a little pain.

ArtOfCode

But someone complained, so I toned it down.

Who?

Confidential, that.

12:03 AM

The Confrontation Clause of the Sixth Amendment to the United States Constitution provides that "in all criminal prosecutions, the accused shall enjoy the right…to be confronted with the witnesses against him." Generally, the right is to have a face-to-face confrontation with witnesses who are offering testimonial evidence against the accused in the form of cross-examination during a trial. The Fourteenth Amendment makes the right to confrontation applicable to the states and not just the federal government. The right only applies to criminal prosecutions, not civil cases or other proceedings....

Bull.

ACuriousMind

@ArtOfCode Yeah, it's fine, I don't blame you.

0celo7

I have a constitutional right.

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

Only in the USA

ArtOfCode

@0celo7 Nopenopenopenope

0celo7

A free society cannot work without this.

ArtOfCode

12:03 AM

You ain't on trial.

ACuriousMind

@0celo7 This isn't a "criminal prosecution"

ArtOfCode

And this ain't a free society.

0celo7

@ACuriousMind Of course it is.

You're out to get me.

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

Relax guys

0celo7

Nope.

I want to know who it was

ArtOfCode

12:05 AM

This is a meritocracy mixed with some benevolent dictatorship.

JD isn't here

Let it go

No

please

No!

12:06 AM

Let it go.

See now it's from a dictator :)

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

-_-

0celo7

If people stood up to Stalin, I can stand up to an SE mod.

ACuriousMind

I'd complain about the star wall, but it was pretty bad already :P

ArtOfCode

I ain't a pse mod

0celo7

I forget there are other SEs

ArtOfCode

12:07 AM

Damn edits :P

ACuriousMind

@0celo7 You need to broaden your horizon.

^

please

@ACuriousMind huh?

dnt hrt me

12:09 AM

Eminem - No Love (Explicit Version) ft. Lil Wayne

►

skill patrol

Dictators don't say plz

0celo7

@ACuriousMind It's not like you go on other SEs

ACuriousMind

@0celo7 Oh, I go there. I just don't participate there.

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

Lurking

ArtOfCode

@skillpatrol What? No, of course not. Who said plz? I must dictate at them immediately.

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

12:11 AM

:D

0celo7

skill patrol

Thanks for saying plz

ArtOfCode

Who's this ArtyCoder fellow? He needs a good dictating-to.

Or she. This may be a dictatorship but I'll be damned if it's a sexist one.

0celo7

I think Arty is ugly

Oh, sexist

I read that as "sexy"

ArtOfCode

Sexy dictatorship. Sounds fun.

ACuriousMind

12:16 AM

I'd say something about sexy dictatorships, but I don't want to get flagged again.

0celo7

It's a sad coward who flags from the shadows

ArtOfCode

#ModAdvantages: can't get suspended for anything I say

0celo7

@ACuriousMind Hmm, can I construct the bump function so it always kills $f'(0)$?

ACuriousMind

What's that in your profile picture, anyway @ArtOfCode?

0celo7

A spiral, @ACuriousMind

ArtOfCode

12:17 AM

@ACuriousMind Good question. Prizes for the first correct answer.

0celo7

A SPIRAL

ArtOfCode

...apart from the obvious ones :P

0celo7

so it's not a spiral?

ArtOfCode

Ts&Cs: prizes may not actually be available. Ever.

That's how most competitions work anyway...

ACuriousMind

Hmmm...you generated it yourself, that much a Google image search tells me :P

0celo7

12:20 AM

google image search?

ArtOfCode

I actually didn't... I've made it my own, but it was originally a cc0 image from a google image search.

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

Don't you know what a Google image search is @0celo7

0celo7

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ I can't even find Wiki articles

ArtOfCode

Hey @HDE

HDE 226868

@ArtOfCode Howdy.

0celo7

12:22 AM

@ACuriousMind I have to solve a quadratic equation for physics...

@ACuriousMind Did you say that if I write $\tilde f=\phi_\eta *f$, then if $\lim_{\eta\to 0}\phi_\eta"="\delta$, then $\tilde f\to f$?

because that's what I'm finding

ACuriousMind

Yes, where the latter limit is w.r.t. any $L^p$ norm.

@ArtOfCode Okay, my google-fu fails me. It looks nice, though ;)

0celo7

@ACuriousMind Yes, I'm finding $|\tilde f-f|\le K\eta$, where $\eta$ is the "width" of the support of $\phi$.

ArtOfCode

@ACuriousMind Mine does too, I have no idea what it is or where it's from.

0celo7

Wait, that's not an $L^p$ norm, is it

under the assumption of $\int \phi=1$.

ACuriousMind

@0celo7 The $L^p$ norm of $f$ is $(\int \lvert f\rvert ^p)^{1/p}$.

0celo7

12:31 AM

@ACuriousMind I know what an $L^p$ norm is >:[

ACuriousMind

@ArtOfCode Haha, very well, then it will stay a mystery

0celo7

but I don't need an $L^p$ norm for some reason!

It's probably because $f$ is Lipschitz.

ACuriousMind

Well, convergence in the $L^p$ norm implies pointwise convergence almost everywhere, I think.

Bah, it's only after passing to a subsequence, but good enough.

0celo7

@ACuriousMind Ok, here's the full derivation: Note that $$\int \phi(x-y)(f(y)-f(x))\,\mathrm{d}y=\tilde f(x)-f(x)$$

because $\phi$ is normalized to one

then take the absolute value, move it inside of the integral

then move $\phi$ outside of the absolute value because it's positive

apply the Lipschitz condition

Then you get $$|\tilde f-f|(x)\le K\int \phi(y)|y|\,\mathrm{d}y$$

Do you agree?

And that integral can be made as small as we want by making the support of $\phi$ small.

ACuriousMind

Looks alright

0celo7

12:38 AM

Ok, so I can make $\tilde f$ "close" to $f$.

But how do I make $\tilde f'$ "close" to $f'$, in some sense?

I'll figure out the GR later, I want to understand the calculus first.

I need to find the derivative of $\tilde f'$ on $(-\epsilon,\epsilon)-\{0\}$.

Then by continuity of the derivative I can extend it to $0$.

Well, $$\lim_{\eta\to 0}[\phi_\eta \ast f]$$ might work.

But is that smooth

And now I have dumb limit stuff, and he said it shouldn't require Lebesgue integration

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

12:58 AM

When are you going to change your avatar @0celo7?

0celo7

When I find something to change it to

I think he had something in mind with this that just doesn't work

ACuriousMind

Change it to something that represents something you enjoy

0celo7

No matter what I do, I get nonsense or something that isn't well defined!

i.e. nonsesne!

@ACuriousMind Hmm, I haven't had my dog as my avatar yet

ACuriousMind

Must...not...make...lewd...joke

0celo7

What?

@ACuriousMind Is there in general no way to use mollifiers and control the derivative of the smoothed thing?

imgur.com/a/mUuU1

ACuriousMind

1:06 AM

Well, as I said, I think the derivative of the mollified functions also converges to the derivative of the original function, if that exists.

0celo7

But it doesn't!

I think this is far less trivial than he had in mind...

Or maybe he expects me to know functional analysis...

Haha!

JD is a master!

ACuriousMind

What's definitely true is that the convergence holds also in the Sobolev norms for $L^2$. Since the mollifer converges in the $L^2$ norms, that means the $L^2$ norms of the derivatives all converge separately

0celo7

I really respect the guy

@ACuriousMind Sadly that all means nothing to me.

I'll stop by his office tomorrow

ACuriousMind

Smart move, guys :P

0celo7

(removed)

"Buseman function" heh

JM97

1:20 AM

Please answer this physics.stackexchange.com/questions/227121/…

0celo7

no

ACuriousMind

@JM97 It's not really clear what you're asking there.

JM97

If electric field inside the pn junction oppose the diffusion current then if I apply Kirchhoff's law along the circuit I would gain some positive potential in forward bias

@ACuriousMind I have edited the question hope you understand the question now

0celo7

1:38 AM

@Slereah My conjecture on the causal relations thing was wrong. $M=\mathbb{R}\times S^2$ is simply connected...but something

Dammit I forgot what Freire said about it :(

Oh, it's got a compact factor...or something

GR is the worst

@ChrisWhite I need to visualize $\mathbb{R}\times S^2$

user54412

That's $\mathbb{R}^3$ with a point removed...

0celo7

really?

proof?

user54412

eh or something similar

0celo7

@ACuriousMind ?

user54412

actually take that back

user54412

1:44 AM

R, not R^+

JM97

@ChrisWhite could you please answer this physics.stackexchange.com/questions/227121/…

0celo7

huh?

@JM97 no

He doesn't know any physics, he's here for the oatmeal

JM97

Why no

0celo7

> He doesn't know any physics

JM97

Okay do your Know?

0celo7

1:45 AM

I'm just a kid

JM97

But It doesn't mean that you don't know physics

0celo7

I don't

I know a minimal amount of math

JM97

I have been waiting for two months to get a satisfactory answer

0celo7

I know enough math to be frustrated at not knowing enough math to learn more math.

JM97

But your reputation at physics.se is higher than mine so it implies that you know better physics than me . Please have a look at it and answer, if you feel easy

0celo7

1:50 AM

Why did you invite me to a room

user54412

actually I take back taking it back - $\mathbb{R} \times S^2$ is topologically just $\mathbb{R}^3 \setminus \{0\}$

0celo7

Proof?

Alfred Centauri

@0celo7 I know enough math to be frustrated at not knowing enough math to know how much more frustrated I could be if I knew enough math to be that frustrated.

JM97

By mistake

user54412

the homomorphism is just bi-continuously mapping $\mathbb{R} \to \mathbb{R}^+$, with e.g. the exponential, right?

0celo7

1:51 AM

@AlfredCentauri Lol

@ChrisWhite erm, sure

the two are homomorphic anyway

your point?

user54412

well what more do you want?

0celo7

Oh

Very good...

Written in my notebook for future considerations, thanks.

I'm trying to figure out the causal separation thing for 2-dim manifolds homeo to $\mathbb{R}^2$

this is very, very hard

I have no clue why it should even be true for Minkowski.

Theorem: if $p\ll q$, no Cauchy surface contains $p,q$. Proof. Can't be bothered, it's true.

Theorem: In Minkowski space, $p,q,$ are contained in a Cauchy surface $\Leftrightarrow$ $p,q$ have spacelike separation. Proof. No clue.

So if $p\ll q$, they are not in a Cauchy surface and don't have spacelike separation.

user54412

@JohnRennie The problem is, an infinite, homogeneous, nonempty universe is incompatible with Newtonian physics. The force on a test particle is undefined, and I'm pretty sure Newton acknowledged this himself at some point but I don't have the reference.

user54412

That "derivation" of Friedmann assumes spherical symmetry about an arbitrary point, and uses that to order the spherical shells of matter in such a way that they always cancel. But the sequence of partial sums of forces is not absolutely convergent, so I can reorder them to get and real number I want, or no limit at all.

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

How goes the job hunt?

user54412

2:05 AM

So I'm less surprised when the "right" answer comes out, given that we could have gotten any answer at all.

user54412

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ whose?

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

Yours.

user54412

Feb 13 at 0:42, by Chris White

@0celo7 joint offer by Berkeley and Santa Barbara

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

nice

Congratulations

user54412

:)

user54412

2:12 AM

@JohnRennie I've ranted about this a bit here -- note all the links in the first paragraph

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

:)

user54412

@0celo7 If $p \ll q$ then the causal line connecting them intersects the "Cauchy" surface more than once, so it isn't Cauchy.

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

Oops, meant to say thanks for the update.

0celo7

2:32 AM

@ChrisWhite I know.

I mean I couldn't be bothered to type the proof.

@ChrisWhite Wait

I think you're responding to the wrong thing there

0celo7

2:49 AM

52 mins ago, by 0celo7

Theorem: In Minkowski space, $p,q,$ are contained in a Cauchy surface $\Leftrightarrow$ $p,q$ have spacelike separation. Proof. No clue.

$\Rightarrow$ Trivial...?

@Slereah @ChrisWhite Is a Cauchy surface necessarily spacelike or can it have null parts?

Hmm.

What if we look at the causal diamond $J^+(p)\cap J^-(q)$

It must be compact if not empty.

We know that $p$ is contained in some Cauchy surface, call it $\Sigma_p$.

Then $J^+(\Sigma_p)\cap J^-(q)$ is compact.

Hmm, but that only works if $q\in D^+(\Sigma_p)$. Ok, assume that, we'll go for a contradiction.

We also know that $C(p,q)$ is compact.

Suppose that $\Sigma_p$ is properly embedded.

Hmm, let $[\Sigma_p]$ be the set of Cauchy surfaces containing $p$.

Ach, this is really hard.

Hmm, what if I do it in coordinates?

Set up global Minkowski coordinates with $p=0$.

Theorem: Then $p\vartriangle q$ iff $\exists v\in T_0M$, $\eta(v,v)>0$ s.t. $\exp_0 v=q$.

Maybe I like the notation $p\asymp q$ better.

Yes.

So the geodesics in Minkowski are just $x^\mu(\lambda)=v^\mu\lambda +a^\mu$.

And we're picking $x^\mu(0)=0$, so $x^\mu(\lambda)=v^\mu\lambda$.

Any two curves (with the same endpoints) in Minkowski space are homotopic.

There is a smooth homotopy taking a smooth spacelike curve into the above geodesic.

Proof:

(N/A)

Convex normal neighborhoods in Minkowski are big

So if $p\ll q$, then there exists a timelike $X\in T_pM$ s.t. $\exp_p X=q$.

Aha!

If I can show that the above thingie, then I'm done because $\exp$ should be a global diff on Minkowski space

Well then, let's see.

I need to find some way of determining $q$ in terms of $p$ and a spacelike curve...

Well, I know I'll have $q^i=p^i+x^i(1)$.

And $x^i(1)=\int_0^1\frac{\mathrm{d}x^i}{\mathrm{d}t}\,\mathrm{d}t$.

Yessssss

Yes, the proof is complete.

Awesome.

John Rennie

6:07 AM

@ChrisWhite thanks Chris. So the derivation has been deliberately chosen to reproduce the Friedmann result and it's really a retrodiction not a derivation.

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

The old zero divided by zero gives you anything you want trick ;)

John Rennie

6:48 AM

Oh dear, oh dear, oh dear:

bbc.co.uk/news/magazine-35861334

2

Really?

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

7:09 AM

:-/

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

7:39 AM

Hi @Secret long time no see :)

Slereah

8:52 AM

@0celo7 Why tho?

$J^+(p) \cap J^-(q)$...

Let's see all the points in the Cauchy surface

$\bigcup_{r \in \Sigma p} J^+(r) \cap J^-(q)$

Is the union of two compact sets also compact?

Finitely many is, apparently

But this is not finite

Hm

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

There are infinitely many paradoxes about the infinite :P

Slereah

yeah infinite things are kind of shit

Always full of counterexamples

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

Indeed.

Slereah

9:07 AM

Guys let's agree to only use finite sets of integers for everything

Danu

@Slereah there is a guy on hsm who agrees with you!

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

But then we couldn't count "everything." ;)

Hi @yuggib

yuggib

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ they are not paradoxes, they're fun!

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ \o

@Slereah noooooo....finitists are even worse than intuitionists :-P

Slereah

FROM NOW ON ALL MATH WILL BE INTUITIONIST, CONSTRUCTIVIST AND FINITIST

None of your fancy math

yuggib

I am about to puke

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

9:15 AM

Some people don't enjoy that kind of "fun." o/

Slereah

oh and also 3 dimensions max allowed

none of those fanciful higher dimensions!

yuggib

@Slereah I remark that intuitionism means no law of excluded middle

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

@yuggib What's wrong?

Slereah

Yes

yuggib

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ intuitionist, constructivist and finitist math is a pain in the ass to do

a three line proof in ZFC becomes a three pages proof

Slereah

9:27 AM

Could be worse

I've seen a guy do physics from Hilbert's geometry axioms

It works but goddamn it takes forever

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

I meant, are you really "about to puke?" @yuggib

Slereah

Maybe he has a tummyache

did you eat the kebab, @yuggib

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

I wonder if Hilbert ever even tried to do physics with his axioms?

yuggib

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ no, it was caused by @Slereah bold statement

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

Ok :-D

Slereah

9:32 AM

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ Nah

That book is written by a crazy philosopher man

Who is opposed to the concept of using NUMBERS and FUNCTIONS in physics

Because those are ARTIFICIAL CONSTRUCTS

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

Is he the one who called Cantor a corrupter of youth?

Slereah

Nah

That sounds more like Kronecker

Kronecker was kind of a prick

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

Right.

Slereah

And now Kronecker functions are often right next to Dirac distributions

That will teach him

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

He's the hotel guy :)

Slereah

9:38 AM

Man I wouldn't want to go to Cantor's hotel really

Having to switch room every time a new guy arrives

Plus imagine the commute when an infinite number of people leave

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

27 mins ago, by yuggib

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ they are not paradoxes, they're fun!

10 hours ago, by ACuriousMind

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ Fun is a primitive, you can't define emotions.

hubot

Hello!

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

Bonjour :)

hubot

What's properly tensor calculus?

What's difference between tensors and vectors?

Slereah

A vector is a type of tensor

hubot

9:51 AM

en.wikipedia.org/wiki/Tensor definition is too difficult for me

DeNiSkA

@hubot tensor are having lots of direction e.g pressure where as force is vector only 1 direction (am i correct @Slereah h)

Slereah

Well, what is a vector, to you

DeNiSkA

geometric object with ine direction on euclidean plane

hubot

@DeNiSkA "geometric object with ine direction on euclidean plane" is vector

Slereah

Direction and magnitude

DeNiSkA

9:53 AM

@Slereah ya!! i forgot magnitude

@hubot i didn't get you , sorry

Slereah

Do you know what a dual vector is

DeNiSkA

nah

hubot

Are Tensors expressed by matrices and are vectors expressed by single column matrix?

Slereah

Some tensors can be expressed by matrices, yes

hubot

Tensors probably we express by square matrices

Slereah

9:56 AM

To employ the terminology of matrices

A rank (0,0) tensor is a scalar

A rank (1,0) is a column vector

A rank (0,1) is a row vector

And a rank (1,1) tensor is a square matrix

But the rank can be arbitrarily high

hubot

If you can treat vectors as matrices then can I treat complex numbers as vectors (complex numbers can be expressed by single column matrices)?

And quaternions as tensors?

Slereah

There is a relation between them, yes

You can treat complex numbers as vectors in $R^2$

And quaternions as vectors in $R^4$

hubot

What do you mean by $R^2$ and $R^4$?

Real numbers as R?

Slereah

Yes

$R^2$ is the plane

$R$ is the real number line

hubot

And $R^4$ is the surface?

Slereah

10:02 AM

$R^4$ is four dimensional

hubot

Ok

Is it true that fractional dimensions e.g. 3/4 D, 1/2 D, 2,5 D etc. are fractals?

Slereah

Depends on how you define dimensions.

Those are Hausdorff dimensions

Which can be fractional

hubot

How are Newton laws to 4D, 5D and higher dimensions space? I don't mean about kinematics in 4D or 5D because it's too easy to define. I mean about for example how decomposes friction on higher dimensional spaces or how is gravity in e.g. 4D space (not Minkowsky space)?

Is it possible that use Minkowsy space in classical mechanics (my 2 question)?

Slereah

The force is $\propto \frac{1}{r^{n-1}}$

Where $n$ is the number of dimensions

hubot

What's r?

Slereah

10:18 AM

Distance from the mass

Jester Tran

11:16 AM

Hi everyone

hubot

@JesterTran Hi!

Jester Tran

My roommate and I are having a house party. We invite 15 couples. At the party, everyone must shake hands with everyone whom they do not already know. At the end of the party, I can't remember the number of people I have met, so I go ask all other people at the party how many hands they have shaken, and they each tell me a different number. You do not shake hands with yourself or your 'partner' (the person with whom you came). How many hands did my roommate shake?

Are you familiar with that brain teaser?

hubot

No, I'm not familiar with this.

Probably 1 couple didn't know 14 couples so 24 hands are shaked?

*28

Jester Tran

How many hands did my roommate shake?

hubot

28?

Jester Tran

11:26 AM

Sorry, that is incorrect

hubot

But 1 couple didn't know other couples?

Jester Tran

So how do you know my roommate shook 28 hands?

0celo7

12:03 PM

Yes.

0celo7

12:44 PM

Any TeX people around?

Slereah

Sure

0celo7

Why is TeX doing this?

Slereah

What part is bad

0celo7

Why isn't it going to Prop. 3

Slereah

Is it the first proposition?

0celo7

12:47 PM

Well, yes, but that's not how numbering works

Slereah

From context nothing seems wrong

You can force number them

0celo7

It should be like this

How do I force number them?

I shouldn't have to do that though

user54412

28

Q: Theorem/Definition/Lemma problem --- Numbering

I am trying to work on the numbering of the theorems/definitions/lemmas etc., and I have some problems with the numbering. I would like the theorems, propositions, corollarys, definitions, conjectures, examples to follow the same numbering, and to reduce the numbering. For example, in my code b...

Secret

@Slereah I still found it amazing why the fact that complex numbers have this $i^2=-1$ property that essentially allow the imaginary and real components to mix with each other (which has no analogue in $R^2$ as there is no notion of vector multiplication (unless you define one to work just like the complex numbers)) never seemed to arise in physical problems, thus allowing us to use compelx numbers and $R^2$ vector interhangeably

0celo7

@Slereah what does that even mean

Secret

12:54 PM

-same thing for the quternions, as $R^4$ vector components cannot normally multiplied with each other without first defining vector multiplication

Slereah

$C_H^d(S):=\inf\Bigl\{\sum_i r_i^d:\text{ there is a cover of } S\text{ by balls with radii }r_i>0\Bigr\}$

$\dim_{\operatorname{H}}(X):=\inf\{d\ge 0: C_H^d(X)=0\}$

For $S$ a subset of a metric space

Transcript for