Mathematics - 2017-04-29 (page 5 of 5)

9:00 PM

thanks

that makes things a lot easier I guess! XD

maybe I could use partial derivatives then

it's not hard, just expand $$f(x+h)g(x+h)=(f(x)+df(x)(h)+o(\|h\|)) (g(x)+dg(x)(h)+o(\|h\|)) = f(x)g(x)+f(x)dg(x)(h) + g(x)df(x)(h) + o(\|h\|)$$

Sha Vuklia

yea right!

now that $m=1$ that is indeed easy

thanks :)

LeGrandDODOM

so the Frechet derivative of $fg$ at $x$ is $h\mapsto f(x)dg(x)(h) + g(x)df(x)(h)$

Sha Vuklia

didn't know that was the name :P but cool I guess

LeGrandDODOM

@DanielFischer have you ever done Fortran ? I was shocked to see that modern scientific Python libraries like numpy and scipy rely heavily on Lapack which is Fortran

@ShaVuklia have you done functional analysis yet ?

Sha Vuklia

9:07 PM

@LeGrandDODOM nope I haven't

from what I see on wikipedia, nothing is familiar

Daniel Fischer

@LeGrandDODOM Never done it myself. But why do you find that shocking? Fortran was made for fast calculations.

LeGrandDODOM

@DanielFischer I thought it was a thing from the past that had been superseded by C

Juan Sebastian Lozano

It's very analogous to how much of modern stuff is based on C/C++

Fortran has some advantages over C for computation, I think.

Daniel Fischer

@LeGrandDODOM When you have a working implementation in one language, it's not overly attractive to port it to a different language. And C only became a nice language in 1999.

C++ will probably never become a nice language.

3

LeGrandDODOM

what do you mean with "nice" language ?

Daniel Fischer

9:14 PM

A readable language with decent syntax.

Sha Vuklia

9:28 PM

@LeGrandDODOM I get:
$$\begin{aligned}
f(\vec a+\vec h)g(\vec a+\vec h)-f(\vec a)g(\vec a)=g(\vec a)f'(\vec a)\vec h+f(\vec a)g'(\vec a)\vec h+f'(\vec a)\vec h(o(\Vert \vec h\Vert)+o(\Vert\vec h\Vert)g'(\vec a)\vec h+ o(\Vert\vec h\Vert)
\end{aligned}$$
How can I show that $f'(\vec a)\vec h(o(\Vert \vec h\Vert))=o(\Vert\vec h\Vert)?$ Should I expand it as follows?:
$$
f'(\vec a)\vec h(o(\Vert\vec h\Vert))=h_1D_1f(\vec a)o(\Vert\vec h\Vert)+\dots+h_nD_nf(\vec a)o(\Vert\vec h\Vert).
$$

Astyx

Hi chat

Sha Vuklia

hi @Astyx

Astyx

Ugh what are we doing ?

Sha Vuklia

lol :P

showing the product rule in more dimensions

Astyx

Multivariable function ?

Sha Vuklia

9:30 PM

yes

I'm guessing what I'm doing is right

Astyx

A very bold guess :p

Sha Vuklia

we get $h_iD_if(\vec a)o(\Vert\vec h\Vert)=o(\Vert \vec h\Vert)$

lol why? :P

Astyx

Just kidding

Sha Vuklia

oh XD

Astyx

I feel your approach might be a bit brutal though

Sha Vuklia

9:33 PM

as in I did it over the top?

like, I slaughtered the function? XD

it's kind of what I have to do tho. i need to get rid of all those terms

Astyx

It's quite straightforward that $f'(a)ho(||h||)= o(||h||)$ (divide the RHS by $||h||$ and see it go to 0)

Oh and use some topology perhaps

Sha Vuklia

you mean LHS right?

Astyx

I do

Sha Vuklia

oh yea I see what you mean by "brutal" now XD

right, I sometimes forget what the $o$-symbol really means

Astyx

Wait, why do you have products between $f'(a)$ and $h$ anyway ?

Sha Vuklia

9:39 PM

I was writing it out again anyways, give me 1 minute!

Simply Beautiful Art

I'm sorta reaching the concept of "inaccessibly larger" ordinals! Woo!

Astyx

Is $f, g : \Bbb R^n \to \Bbb R$ ?

Sha Vuklia

yes

Astyx

Then you can't have $h$ in the RHS

Cause it's a vector

Sha Vuklia

oh shit

Astyx

9:44 PM

It should, you probably messed up somewhere :p

Sha Vuklia

yea it's a lot of terms

i'm still writing it out! XD

Astyx

In any case you have $f(a+h) = f(a) + df_a(h) + o(||h||)$ where $df_a$ is the differential of $f$ at $a$

It's dangerous to write $f'$ in dimension $\gt 1$

Then let $B$ be a bilinear application from the domain to another space (as multiplication is, but that's to avoid confusion with multiplication by a scalar)

Then $B(f(a+h), g(a+h)) - B(f(a), g(a)) = B(f(a+h)-f(a), g(a+h)) + B(f(a), g(a+h)-g(a))$

Sha Vuklia

i'm not sure which term i got wrong first, but the only limit i'm still struggling with is the following:
$$
\lim_{\vec h\to \vec 0}\frac{f'(\vec a)\vec h\cdot g'(\vec a)\vec h}{\Vert\vec h\Vert}
$$

Astyx

cot ?

Oh you scared me :p

Sha Vuklia

right i have to read what you said carefully :P

haha! yea I scared myself too

sorry I first had to write it all out neatly

Astyx

9:50 PM

Then simplifying you get $B(df_a(h) + o(||h||), g(a+h)) + B(f(a), dg_a(h)+o(||h||))$

ALannister

Is there any general thing that can be said about kernels of evaluation maps in rings?

Astyx

Expand this by linearity

Make $h$ go to $0$ and argue continuity of things by finite dimension and stuff

Sha Vuklia

i'm not sure if I know what you mean by this $B$, but is it really relevant?

Astyx

And find that $dB(f,g)_a (h)= B(df_a(h), g(a)) + B(f(a), dg_a(h))$ which is the product rule

Sha Vuklia

is it to make it explicit that we are not multiplying scalars here?

Astyx

9:53 PM

Well it's not exactly clear what multiplication is in dimension $\gt 1$ really

Sha Vuklia

well, in this case, just scalars, vectors, and matrices, right?

Astyx

Huh ?

You can't multiply two vectors

At least not without saying what you mean by that

Sha Vuklia

no okay, but we do multiply a matrix with a vector

$f'(\vec a)\vec h$

Astyx

Oh I see ...

Sha Vuklia

so your $B$ isn't quite necessary (for me at least now) as long as I pay attention to what I'm doing, right?

Astyx

9:55 PM

But what is $f(a) g(a)$ then ?

Sha Vuklia

lol idk just the matrix multiplication between the two?

wait no

that's just two scalars

Astyx

Oh are we in dimension 1 ? Right then

Yeah it's mostly about keeping things clear then

Well each time I wrote $B(a,b)$ replace it with $ab$ if you want

I find it more confusing than anything really

Sha Vuklia

haha yea

I'm still not sure how to do this tho:
$$
\lim_{\vec h\to \vec 0}\frac{f'(\vec a)\vec h\cdot g'(\vec a)\vec h}{\Vert\vec h\Vert}
$$
You said I should use continuity of "things by finite dimension"?

can't you just argue that $\vec h/\Vert\vec h\Vert$ is a unit vector ?

Astyx

Yes you can

Parry

Is anyone in here familiar with non-monotonic logic?

Astyx

10:00 PM

So you get $f'(a)u \cdot g'(a)h$

Sha Vuklia

oh right, then I know what to do! thanks for your help!

Astyx

left thing is bounded

Sha Vuklia

yea there I can use continuity, I see

Astyx

Actually you could even take $||h|| f'(a)u\cdot g'(a)u$

(I still find the use of $f'$ very terrible when talking about a differential)

Sha Vuklia

ah I see!

yea well, $f'$ is what my teacher uses, and I don't really see the problem myself :P

I guess I'm just used to it

Astyx

10:03 PM

Well it seems like it's a scalar when it's in fact a linear application

Sha Vuklia

hahahah yea,

but that's why you have to get used to it not being a scalar:P

Astyx

And not the same thing as what we mean by $f'(a)$ in single variable function

Sha Vuklia

technically it is the same, I would say

Astyx

It should at least be $f'(a)(h)$

Sha Vuklia

no, because linear maps are often denoted like that

$Ax$ e.g.

Astyx

10:05 PM

They shouldn't :p

Sha Vuklia

hahahah, well, I hope one day you can decide on all the symbols used in maths XD

and conventions*

Astyx

IMHO you should chose between dealing with matrices or dealing with abstract vector space and stick with it

And I believe the transition with infinite dimension vector space is "weirder" if you're used to matrices etc

Sha Vuklia

but matrices are a subset of abstract vector spaces, no?

Astyx

Anyhow, I'll stop whinning and go to bed now :p

Sha Vuklia

at least, the space of matrices

yea good idea

me too

i'm really thankful for all the help you've provided over the months tho!

it's really cool

good night in any case!

Astyx

10:08 PM

Yes there is an isomorphism between any space of endomorphism of a vector space of dimension $n$ and matrices of size n

(and more genrally linear maps from dim n to m and $m\times n$ matrices)

Glad to help, really :)

Cya

Sha Vuklia

bye!

Leullame

10:49 PM

If I have a homomorphism $f$ from the additive group of a field $F$ into the complex numbers, and $f(ax)=f(bx)$ for all x in F, does that mean a=b?

Eric Stucky

Not without some more conditions

If f is the zero homomorphism, for instance, that's no good

Leullame

Yes, I actually have a nonzero homomorphism.

(I think nothing more is assumed)

("nonzero" = not identically 1)

Eric Stucky

Is it just a group homomorphism, or a field homomorphism as well?

Eric

If it were a field homomorphism there would be nothing to show right; all field homomorphisms are injective or zero

unless that's what Leullame is showing

Eric Stucky

oh, I guess that is a thing :P

Eric

10:59 PM

but that would be trivial to show because the kernel is an ideal, and fields only have two ideals

mhm

Leullame

It is a homomorphism of groups.

(The target group is multiplicative, complex numbers without zero)

When the field is finite, am I right that this follows since all homomorphisms are of the form $f(x)=e^{2 pi i x / |F|}$?

Ted Shifrin

Oh, now have both @Eric s :P

:P

lol

Hi, both.

11:11 PM

heya Ted

Eric Stucky

Leu: I think that it's true for finite fields, but the argument you gave needs a little finesse

what you've written is true for Z_p, in particular

But not for F_4, for instance.

Ted Shifrin

But what if $a$ is the characteristic of the field and $b=0$?

Eric Stucky

o.O

those are the same, as field elements

Ted Shifrin

Oh, where do $a$ and $b$ live?

Eric Stucky

oh I see what you mean

Ted Shifrin

11:14 PM

Totally ambiguous question.

Eric

@Ted I assumed $a, b$ lived in the additive group of the field since that made the most sense to me

Ted Shifrin

I assumed they were integers :P

Leullame

$a,b$ live in the field

Ted Shifrin

You're multiplying elements of the additive group? Doesn't sound good.

Leullame

This is a multiplicative group of a field

so it is defined

Eric Stucky

11:15 PM

(additive)

Leullame

(Yes)

Eric

welp I gotta go to dinner so farewell chat

Bubye.

eat well, mate

Bye

11:16 PM

Hey @Ted I finally wrote out the bit about the quadric

Leullame

We believed we proved it for fields of prime order, and want it say for all finite fields

Ted Shifrin

I haven't thought about it intelligently, @Leullame.

@Danu: What bit?

Danu

The identifications $G_2/U(2)_-\cong \widetilde{\operatorname{Gr}}_2(\Bbb R^7)\cong Q$

Ted Shifrin

Oh.

Eric Stucky

Leu: my finite field theory is quite weak, but I know that the multiplicative group is cyclic. Can you choose a basis with nice properties on the powers? I want the argument to start "Let $x^{a_1}, x^{a_2},\dots, x^{a_n}$ be a basis; f is completely determined by which $p^\text{th} root of unity it maps these elements onto."

Danu

11:21 PM

I gave you explicit credit for the proof too

(and also referenced Chern's proof)

Ted Shifrin

Why'd you do that? I just explained to you why the oriented Grassmannian is a quadric.

Danu

Yeah, that bit

I made that into a separate proposition

Ted Shifrin

Oh, well ...

I suspect the $G_2$ stuff is harder :P

Danu

well that was explicitly in some other paper :P

Ted Shifrin

Oh, so you hijacked the hard stuff.

Danu

11:22 PM

I just happened to come across it...

Ted Shifrin

nods knowingly

Danu

It's this $S(U(1)\times U(2))$ thing I was talking about earlier :P

The proof is honestly not hard

I can tell you right now

Ted Shifrin

Nah, my brain is dead now.

Danu

it's like 3 lines!

Leullame

Are all homomorphisms from a finite field $F$ into $\Bbb C^*$ are of the form $f(x)=e^{2 \pi i (x_1a_1+...+x_na_n)}$ for some $a \in F$, where the components are from the base field? Does this prove the original claim?

Eric Stucky

11:25 PM

1) Definitely, yes. 2) Maybe :P

LegionMammal978

For any given real number, is its sequence of convergents always Cauchy?

Intuitively, this would appear to be the case

Eric Stucky

Legion: I'm not sure what you mean by that, but it is true that all convergent sequences (in any metric space) are Cauchy

arctic tern

@Danu by 2 equivalent definitions of G_2, do you mean being automorphism group of cross product and octonions as the two defs?

Ted Shifrin

Continued fractions, @EricS.

Danu

So you define $G_2$ as the automorphism group of the Lie algebra $(\operatorname{Im}\Bbb O,\times )$ ($\times$ being the cross product on imaginary octonions, i.e. what you get when you multiply imaginary octonions and subtract the real part). $G_2$ acts transitively on oriented planes, even on triples that generate $\operatorname{Im}\Bbb O$ (bring all of them to $i,j,\ell$ where $\Bbb O=\Bbb H\oplus \Bbb H\ell$).

Ted Shifrin

11:27 PM

oh, and hi, tern.

arctic tern

also, $(\Bbb R^7,\times)$ is not a lie algebra

Danu

Now assume $g\in G_2$ fixes the oriented plane spanned by $\{i,j\}$. Then its restriction to that plane is an $SO(2)$ rotation, i.e. $g(i)=\cos\theta i-\sin\theta j$ and similarly for $j$. Because $g(i\times j)=g(k)=g(i)\times g(j)$, we compute explicitly that $g(k)=k$. So $g$ fixes $k$.

Now the orthogonal complement of $k$ inside $\operatorname{Im}\Bbb O$ is six-dimensional and you can equip it with a complex structure given by left-multiplication by $k$. Then the $\{i,j\}$ plane is spanned as a complex line by $i$, and preserved by $g$. $g$ also preserves standard Hermitian metric and the orthogonal complement of the line, so it is an element of $S(U(1)\times U(2))\subset SU(3)$

@arctictern it isn't?

Why isn't it?

arctic tern

nope. cross product satisfying jacobi identity is equivalent to multiplication being associative

Danu

Oh yeah the Jacobi

derpaderp

@arctictern Is there a name for the kind of algebra that it is?

arctic tern

not really standardized. there is some different operation you can convert it to and from that does have a name that I don't remember but I didn't like it.

Danu

11:34 PM

What's funny though is

that lots of things still work just as if it's a Lie algebra

The inner product is proportional to the would-be Killing form, etc

The analogy is still insightful for me

Maybe just because I don't have any intuition for algebra at all :D

arctic tern

also, to me the equivalence of defs of G_2 comes straight from $xy=-\langle x,y\rangle+x\times y$, since the inner product is already around when talking about octonions or cross products individually

Danu

Once you show that $\langle x,y\rangle$ is expressible in terms of $\times$

(which is through that would-be Killing form)

I guess this is obvious to you?

arctic tern

to me, a cross product is a thing that satisfies identities on an inner product space, so the inner product already exists if the cross product does

(there are other cross products that are not binary operations as well)

Danu

So you mean that your definition of the automorphism group would include invariance of the inner product?

arctic tern

ah, I see what you're getting at

I had thought about this before but don't remember what my thinking was