Mathematics - 2022-12-23

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leslie townes

00:21

now about that lawvere theory

1 hour later…

Ted Shifrin

01:37

Gesundheit!

leslie townes

02:06

munchkin didn't want to go on a walk to look at holiday lights around the neighborhood, until my wife floated the option of having us push her around in a stroller (she still fits in it, although we haven't used it in ages). munchkin agreed to that.

user4539917

02:18

Sounds a bit overly privileged, but that shouldn't be an option next year...

leslie townes

yeah, we won't be able to make a habit of this.

user4539917

Enjoy the moment. 🎄

👼

Ted Shifrin

03:20

@leslietownes rolls $7^\pi+e^6$ eyes

shintuku

03:31

reality is a strict subset of the rationals

i can not believe some people say it is the real numbers

leslie townes

ted: we drew the line at taking her in the stroller up a flight of hillside stairs.

Ted Shifrin

Shocking!

user4539917

$7^\pi+e^6$ +1 eye for drawing the line somewhere :-)

Ted Shifrin

Only 1 extra?

user4539917

03:48

Ok, $7^\pi+e^6$ ± 9.999...

Are we allowed to roll imaginary eyes?

Ted Shifrin

04:10

You are suggesting rolling $i$s …

leslie townes

04:23

boo

4 hours later…

Yai0Phah

08:45

@CroCo This exercise does not make sense to me — the convexity seems a relative concept (a subset of a real vector space being convex).

1 hour later…

Yai0Phah

10:00

It should have been a subset of an affine space, not necessarily a vector space. There is a concept of convex spaces, but it is out of reach.

one potato two potato

10:34

@Feynman_00 $\nabla_Xf = Xf$ and product rules on tensors and on the pairing between a covector field and a vector field, I mean. The forms are the same. It's weird to expect they are exactly the same.

Feynman_00

11:21

@onepotatotwopotato Oh, I also forgot to mention the covariant derivative requires an additional structure, i.e. connection. The pairing rules are born forcing some sort of Leibniz rule IIRC. Although I wonder what would be the best intuitive explanation to say what makes $\mathcal{L}_X Y$ and $\nabla_X Y$ different and in what sense each of them is the "derivative of a field along the direction of the other"

ILikeMathematics

12:12

When introducing a function, why does one use the co-domain instead of the range?

Given $f(x) = x^2$, one would say

$f: \mathbb{R} \rightarrow \mathbb{R}$

instead of $f: \mathbb{R} \rightarrow [0, \infty)$.

robjohn

12:51

It is much easier to state the co-domain where a function takes its values. The range often takes more work to determine.

Koro

0

Q: In a normed space $X$, convergence of $\|y_1\|+\|y_2\|+\cdots$ may not imply convergence of $y_1+y_2+\cdots$

Hint to show the veracity of this statement is to consider subspace $Y:=$ the subspace of $l^\infty$ consisting of sequences which have almost all elements equal to zero; and $y_j\in Y$, such that $y_j=(x_n), x_j=\frac 1{j^2}, x_n=0\, \forall n\ne j.$ So clearly $\sum_{i=1}^n \|y_i\|=\sum_{i=1}^n...

real-analysis functional-analysis

user123234

13:05

can someone help me with the following identity:
I have $X$ a Hilbert space and $T:X\rightarrow X$ a bounded linear operator. I want to show that $||T^*||=||T||$. What I know is that $$||T^*||=\sup_{||x||=1} ||T^*x||$$. But $||T^*x||^2=\langle T^*x, T^*x\rangle=\langle x,TT^*x\rangle$ but this somehow does not help me further.

ILikeMathematics

Thanks, so there are multiple valid co-domains, right?

Given $f(x) = x^2$, saying $f: \mathbb{R} \rightarrow \mathbb{R}$ and $f: \mathbb{R} \rightarrow \mathbb{C}$ would both be valid, right?

Jakobian

13:39

@user123234 this holds in any Banach space and follows from Hahn-Banach theorem

robjohn

14:13

@ILikeMathematics While there is a subspace of $\mathbb{C}$ that is homeomorphic to $\mathbb{R}$, not everyone considers $\mathbb{R}\subset\mathbb{C}$, so it is not clear that both could be considered co-domains of the same $f$.

Goku カカロット

@ILikeMathematics Something tells me you like mathematics.... But I just can't quite put my finger on it

XZYZ

what is the name of chatroom where I can go to help regarding my question to reopen?

Goku カカロット

CURED

ILikeMathematics

14:41

@robjohn If we defined $\mathbb{R}\subset\mathbb{C}$, then both $\mathbb{R}$ and $\mathbb{C}$ would be co-domains of $f$, right?

DIRAC1930

15:01

How does Cartan know that his parameterization of $X_i$ by $\xi_i$ here google.co.uk/books/edition/The_Theory_of_Spinors/… can obtain all isotropic vectors of complex Euclidean space just by specifying $\xi_0$ and $\xi_1$?

1 hour later…

Jakobian

16:09

@robjohn I really don't know who doesn't consider $\mathbb{R}$ as a subfield of $\mathbb{C}$

or for what reason

@ILikeMathematics I'd say it's 100% valid

ILikeMathematics

16:48

@Jakobian Thank you

@robjohn Thank you too

1 hour later…

shintuku

18:04

is there any way to decompose a linear map so that it is obvious, geometrically speaking, to see what it's effects are on the basis vectors of another matrix?

the geometrical interpretation is obvious when you just consider a single matrix, where each column can be thought as specifying a modification of the unit vectors of a space where the unit vectors are $(1,0,..,0)$, $(0,..,1,..,0)$, etc., and become whatever the columns of the matrix are

what I'm not having as easy of a time finding a geometric interpretation for, in terms of the effects on the basis vectors, is the composition of matrices

say the matrix A =$\begin{bmatrix} 2 & 8 \\ -5 & 2 \end{bmatrix}$

it's nice, for a geometric interpretation, to think that it defines a vector space where the two basis vectors are $(2, -5)$ and $(8,2)$

which we can very easily visualize, making a quick sketch

what's not obvious to visualize is what then happens when you take another matrix and do $BA$

leslie townes

if you apply another matrix B on that to the left to get BA, its columns are B (2,-5)^T and B(8,2)^T, so to 'visualize' what those look like in general (assuming (2,-5) etc have been chosen more or less at random) you basically need to visualize what happens to any pair of vectors

tiny nit also but you keep saying 'basis vectors' of a matrix where i think you might just mean 'columns' (i.e., no reason or need for the columns to form a basis in the linear algebra sense)

if i understand it correctly, you want to basically be able to 'see' the matrix multiplication without performing something akin to matrix multiplication. maybe possible for matrices of specific types, but not in general

shintuku

oops sorry for the vague language

leslie townes

ie, matrix entries list out what a matrix does to the standard basis. that's what you see when you stare at B and A individually. to 'see' it for BA, you really do need to do that matrix multiplication, because seeing what B does to e_1 doesn't tell you what B does to A e_1 (which may in general be a mix of all of the basis vectors, that you figure out and 'see' by computing the matrix product)

slight abuse of notation there because in the case of non square matrices those e_1's may also have different lengths and not be the same 'e_1'

shintuku

yeah by basis vectors of a matrix i meant what happens when you do $A\hat \imath$ and $A\hat \jmath$, where $\hat \imath, \hat \jmath$ are unit vectors $(1,0), (0,1)$

@leslietownes this is what I'm looking for, yes

@leslietownes right, I'm trying to find if there is any work done to 'see' the effects, when composing matrices

without doing the multiplication itself

i thought, maybe there is some way to decompose matrices where this becomes more obvious

oh well, gonna have to do that multiplication

thanks leslie

leslie townes

18:20

i don't know of one in general (again, if you have a specific B or A in mind, particularly with some structure, there are probably things you can come up with for that product, but i don't know of anything that generally works for products)

if you have an application in mind you also have to be sensitive to computational concerns, where it is hard to beat 'matrix multiplication' for a fairly well optimized thing you can find off the shelf

shintuku

it is sort of quick, for the things it makes possible

Koro

Does the set of reals have a Hamel basis?

leslie townes

koro: all vector spaces have bases (under the usual axioms anyway) and some of them have no-effort bases that you can just write down. how are you considering R as a vector space?

Koro

R as a vector space over R.

Xander Henderson

Then $\{1\}$ seems like a reasonable basis...

leslie townes

18:30

{1} is a basis of k as a vector space over k

Xander Henderson

Too slow, @leslietownes! :P

leslie townes

i needed the extra time to answer for all fields

Koro

Oh of course. I know that. I was thinking something else.

Ted Shifrin

Seeing what a matrix does is the point of diagonalization/Jordan form, but with a product you’re up a creek unless they!re simultaneously diagonalizable.

Xander Henderson

@leslietownes What? Are you a category theorist? Who needs all that generality.

Koro

18:32

Every vector space ( not fd ) has a basis.

leslie townes

xander: what would lawvere theorize about all of this?

Koro

I'm yet to learn its proof.

I'm sure it would require Zorn's lemma.

leslie townes

koro: the standard example of a goofy basis would be one of R as a vector space over Q

Xander Henderson

Anyway, there are really only a couple of fields: $\mathbb{R}$ and $\mathbb{C}$ (and $\mathbb{Q}_p$ for the real perverts). Other fields are dumb, and probably don't exist.

Koro

yeah. I know that R over Q is not finite dimensional.

I'll wait till I learn the proof of 'all not finite dimensional vector spaces' also have a basis to see how Hamel basis of R/Q looks like.

Xander Henderson

18:35

Feels axiom of choice-y.

Ted Shifrin

= Zorn

smth

https://en.m.wikipedia.org/wiki/Horocycle
Why is horocycle circle in Poincare disc model? I somehow feel that it could be but I am not sure why perfect circle, why not ellipse or something like that?

Xander Henderson

@TedShifrin Sure. But I'm thinking of an axiom of choice-y construction.

$[x] = \{ y : x-y \in \mathbb{Q}\}$ is an equivalence class; choose one representative from each class; there's your basis.

Or something like that.

Ted Shifrin

Feels more Zorny to me … maximal linearly indep set and all.

Xander Henderson

@TedShifrin Sure, that would work, too.

Ted Shifrin

18:39

Your quotient $\Bbb R/\Bbb Q$ as abelian group …. Are elements linearly independent? I doubt it.

Xander Henderson

See, that's why I don't do algebra.

leslie townes

rolls $\pi^e + 11/\gamma + \mathbb{Q}$ eyes

Koro

I tried to prove that- in a normed space X if every 'absolutely cgt. series implies cgt. series' then X is complete.

I tried to prove it like this:

Let $(x_n)$ be a Cauchy sequence in $X$. It is known that if a subsequence of $(x_n)$ converges to some $l\in X$, then $(x_n)$ also converges to $l$. Moreover, one can find a subsequence $(x_{n_k})$ of $(x_n)$ such that $\|x_{n_{k+1}}-x_{n_k}\|\lt \frac 1{k^2}$ for every $k\ge 1$. Therefore, WLOG we can assume that for all $n$, the following holds:

$\|x_{n+1}-x_n\|\lt \frac 1{n^2}.$

It follows that the series $\|x_1\|+\sum_{i=1}^\infty \|x_{n+1}-x_n\|\le \|x_1\|+\frac{\pi^2}6$ converges by comparison test. It follows therefore by the hypothesis that the series $x_1+\sum_{n=1}^\infty (x_{n…

I think this is correct. :)

leslie townes

19:07

i skimmed that iwthout rendering chatjax but the idea looks correct. for simplicity, maybe just use 'any convergent series of positive numbers', or one that sums easily right from the definitions (e.g. 2^{-k} or 1/(k(k+1)) instead of one that has a fancy sum..

maybe a question of aesthetics as to whether bringing pi^2/6 into it improves, or detracts from, the argument :D

Koro

@leslietownes :D

thanks for the review :-).

Paramanand Singh

19:57

@Koro I am not sure of the normed space thing, but the proposition is True for ordered fields and has a very non-trivial proof which originated on mathse

57

Q: In which ordered fields does absolute convergence imply convergence?

In the process of touching up some notes on infinite series, I came across the following "result": Theorem: For an ordered field $(F,<)$, the following are equivalent: (i) Every Cauchy sequence in $F$ is convergent. (ii) Absolutely convergent series converge: $\sum_n |a_n|$ converges in $F$ $\i...

real-analysis sequences-and-series ordered-fields

The proof above was later put in a paper jointly by asker and answerer

Koro

@ParamanandSingh thanks for sharing that. Right now, it's beyond my access. I don't know Archimedean fields yet (I know R is one). Besides, how would one define 'convergence' on arbitrary fields?

By normed space X above, I mean that X is a vector space and has a norm defined on it which induces a metric topology on X.

@ParamanandSingh nice :).

leslie townes

not on arbitrary fields, ordered fields :) totally ordered sets always have at least an order topology.

Koro

sadly, neither the asker nor the answerer seem to be active on mse anymore.

leslie townes

you can even come up with a basis for the topology using, you guessed it, intervals

it all comes back to intervals

Koro

20:18

@leslietownes arghh. I meant that only. Please replace 'arbitrary field' by 'arbitrary ordered field' in my earlier comment. :)

@leslietownes oh yes. We can even come up with subbasis -the open rays :).

hence we can define the notion of convergence

leslie townes

another victory for intervals

AMDG

Is there a name for factorizing a particular power with exponent from an integer?

Like if I take all factors of 2 out of an integer $x$, can I say that I have computed the power of two factorization of $x$ or something?

leslie townes

sometimes called the height or the p-adic valuation or order of the integer, at least for primes p.

the exponent itself, rather, is called that. and not e.g. the 'factorization' or the process by which you do that.

AMDG

21:20

Ah ok

Yeah that's what I've been looking at alot as you can imagine. A204983 still. Yes, I'm also aware it's called the multiplicative order of p modulo q.

There's a relation between it and oeis.org/A002326

Well, so I conjecture, but empirically it is correct for something like 78 terms, and logically it's kind of obvious to me but I just... find it difficult to put into formal words (a proof). oeis.org/A204983

I'm trying to take an approach to prove it based on the intuition that $1 = 0.(1)_2$ where $0.(a)_b$ denotes an infinite repeating numerical expansion in base $b$ and $a$ is an algebraic expression.

Now I am getting to the part where I prove that we have some $\frac1x = 2^k\cdot 0.(p)_2$ and that $\frac1x$ has a factor $\frac{1}{2^q - 1}$ as a result.

Believe it or not, yes, I'm still doing more research on an optimal division algorithm for binary computers (two years later) and this is probably the best lead I've had to date. If it's true, then I think it's kind of obvious from there that there's infinitely many Mersenne primes as well.

Well, regarding my question of words, I suppose I'll use the wording I presented and then follow with an algebraic representation as well.

Also, I got the idea that maybe describing things algebraically after describing them with words is a better way of doing it for the reader to get a logical understanding of the expression. Any objections to that?

For an example, consider what I've written around the expression $$2^{\lfloor\log_2(2^m - 1)\rfloor + 1}\cdot 0.[2^m - 1]_2\cdot \sum_{n=1}^{\infty} 2^{-nm} = 0.(2^m-1)_2.$$ (apologies for the mess; I'm going to structure it after I've actually figured out a more or less succinct proof): incongruous-yew-7c4.notion.site/…

AMDG

22:00

You know what... I think I can get a simpler and shorter proof doing what I should have from the start: we can compute the integer $y$ that multiplies $x$ as given in my formula there in A204983 directly via the 2-adic valuation of 2 mod 2n + 1 (we'll call this $a$), and the existing means to iterate over all residues for exponents [0, a) and represent it as a finite sum. From there, it isn't hard to prove the factor of $2^k$ as mentioned, and the factor of $\frac{1}{2^x - 1}$.

I'll be honest though... I'm not sure how little I have to prove or just provide as lemmas with proofs, but for now, those things that are proven, I'm referring to by "it is known that".

D Left Adjoint to U

22:29

Yes, use Hensel's

:D

AMDG

22:46

What is Hensel's?

Me not know words. Me know nombres... or hombres if you prefer...

I think he's referring to Hensel's lemma which I just looked up. I don't know why I went to Wikipedia. MathWorld is better. Infinitely better.

My head is too foggy right now to even read this. It's easier for me to just intuit what I already know and work through it logically. Besides (as you can clearly see), I don't know theorems pertinent to what I'm working on. (Is there an online encyclopedia of theorems organized by topic?)

Bruh, see Dr. Weisstein here is literally doing what I suggested but in reverse giving a formal followed by informal definition of Hensel's lemma.

Xander Henderson

23:27

@AMDG You can google it.

The term you want to google is "Hensel's lemma".

AMDG

Cool, so I found the right thing. Thanks.

Honestly I wish I could just defer the formal side of things and pointing out relevant theorems to a mathematician. I kinda just want to get this over with and move on.

robjohn

@Jakobian People often equate the subfield of $\mathbb{C}$ that is isomorphic to $\mathbb{R}$ with $\mathbb{R}$, but saying they are the same can lead to problems.

For example, a complex number is $x+iy$ where $x$ and $y$ are real numbers.

This becomes circular if $\mathbb{R}$ is truly a subset of $\mathbb{C}$

Xander Henderson

23:43

@robjohn You could make it non-circular by defining $\mathbb{C}$ differently. For example, take $\mathbb{A}$ to the set of roots of of polynomials in $\mathbb{Q}[x]$, and define a valuation $\nu$ on $\mathbb{Q}$ in the "obvious" way (e.g. $\nu(x) = |x|$ on $\mathbb{Q}$).

robjohn

@XanderHenderson What is this $\mathbb{Q}$ of which you speak?

Xander Henderson

I'm pretty sure that you can show that the valuation extends to $\mathbb{A}$ in a unique way (I'm not 100% certain of this claim, and I certainly don't know how to prove it, but I'm willing to bet its true). Then complete $\mathbb{A}$ with respect to the metric you get from the valuation.

@robjohn An element of $\mathbb{Q}$ is an equivalence class of ordered pairs of integers, where $(a,b) \sim (c,d)$ if $ad = bc$.

It is the field of fractions of $\mathbb{Z}$, if I remember the terminology correctly.

robjohn

At some point, one will want to say that $\mathbb{Q}\subset\mathbb{R}\subset\mathbb{C}$. There may be a way to define these in a non-circular way, but it is not the usual way of doing it.

Xander Henderson

@robjohn Sure.

No argument there.

I think that the usual "non-circular" way is via embeddings, n'est-ce pas?

E.g. if $x \in \mathbb{R}$, then the map $\iota : \mathbb{R} \to \mathbb{C} : x \mapsto x + i0$ preserves all the right structures, so that $\iota(\mathbb{R})$ is isomorphic to $\mathbb{R}$ in all the ways that matter.

So we get lazy and say that they are really the same thing, and then we stop caring. :D

Anywho, I'm off.

robjohn

Yes, if one is willing to consider isomorphic things equivalent but perhaps not "equal"

but loosely speaking, it is so

@XanderHenderson yes

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