Mathematics - 2023-12-20 (page 1 of 2)

Xander Henderson

03:44

@Jakobian No idea. I don't observe that holiday.

leslie townes

Merry Tuesday!

Soumik Mukherjee

Merry Wednesday

Ted Shifrin

04:12

Happy Grinch!

4

robjohn

05:30

@TedShifrin It's better that you are a retired bum than still an active bum.

Ted Shifrin

@robjohn I don't know about that; just so long as I'm not a homeless bum.

robjohn

At least we live in an area where it does not get freezing cold outside very often.

Ted Shifrin

Indeed. That's why we're so popular. :(

leslie townes

05:46

whole lotta neerdowells retire here from places like georgia :(((((

psie

I have a super basic question. When someone writes something like $L=\lim\limits_{n\to\infty}\frac1{a_n}$, then $a_n$ is allowed to be $0$ for sufficiently small $n$, right? I'm asking because there is a ratio test for power series, and I'm reading two different statements of this test, one where they claim that $a_n\neq0$ for all $n$ and in the other where they claim $a_n\neq0$ for sufficiently large $n$.

robjohn

@leslietownes peachy

leslie townes

psie the details of what is 'allowed' might depend on the author of a particular text or classroom setting, but for purposes of defining/considering the limit, one wouldn't need to require a_n nonzero for all n to make sense of things, as long as a_n is nonzero for all n sufficiently large

specifically in the context of power series, or even finite sums, you might wonder what is leading to people allowing a_n to be 0 and what sense they hope to make of things in that case

but i'd think of that as separate from a limit existence/calculation

psie

ok 👍

leslie townes

it might be as simple as an author just wanting to generate quick examples (in the spirit of "just put a random polynomial as a denominator" or "just put ln(ln(ln(whatever))" and they don't want to bother checking if some arbitrarily chosen example maybe has some issues that affect only finitely many terms

psie

06:04

makes sense

Thomas Finley

06:34

0

Q: Let $x$ be a nilpotent element and $y$ be a unit in a commutative ring $R$. If the elements $x,y$ do not commute, then is, $x+y$ is a unit in $R?$

Let $x$ be a nilpotent element and $y$ be a unit in a commutative ring $R$ with identity. Show that $x+y$ is also a unit in $R.$ If $y=1,$ is commutativity of $R$ necessary? Justify: If the elements $x,y$ do not commute, show that the above property does not hold. My solution goes like this: Let...

I need some help with this problem a bit.

leslie townes

06:49

uh, where you have written "does not hold" in the statement of the original problem, surely only "may not hold" is intended. it is possible to give examples of rings containing elements x and y that do not commute for which x + y is nevertheless a unit. the exercise is presumably to show, by example, that when x and y do not commute, the sum x + y is not necessarily a unit.

which is done e.g. in the answer someone just provided there :)

lots of static in the problem phrasing there. a commenter pointed out another issue with it.

Koro

07:23

I watched a movie where a dog gets depressed and stops eating so the dog is taken to a vet whom the owner tells that '"we gave our parrot with its cage to our neighbour and ever since it's behaving like this. And the parrot cage was always at a height, the dog never even saw the cage." The vet advises to bring the parrot back and install cctv in that room and they find out that the parrot speaks ''your meal is ready, your meal is ready" and the dog starts eating.

the parrot picked these lines from the owner and used to speak those lines in the absence of the owner and the dog would think it's the owner telling him to eat. :-)

I never saw such a movie before.

leslie townes

was there other stuff in the movie too or was it just that?

Koro

07:48

there was. But I liked this bit most.

:)

leslie townes

oh. if it was just that, i was thinking, that truly does sound one of a kind :)

Koro

it was a Russian movie (show?) called 'Soul healers' in english.

Koro

09:24

why is it taking so much time to render?

is there any fix to this?

leslie townes

i have no idea, but what kind of psychopath would put "det" outside of math mode like that

Koro

Ohh I didn't realise there was \det as well.

leslie townes

why \cdots but literal "..." instead of \dots in the inline "v_1, ..., v_n"

Koro

I put \det inside now.

leslie townes

three backslashes in a row also marks you as a madman, just because latex ignores certain whitespace, that's no reason not to put it there

haha but in actual answer to your question i dunno

Koro

09:31

leslie townes

"p\in S" instead of "p \in S" also looks demented to me

Koro

Leslie, actually I put \\\ there because somehow \\ wasn't working.

I don't know why mse isn't rendering the mathjax as I type it. It's taking time to render.

too much time

leslie townes

i do know that the previewer code seems to be a different beast from what the renderer actually uses. or at least was for a long time (certain things which were maybe not orthodox mathjax would render one way in the previewer and another in posted answers)

\\ somehow not working feels like a sign that maybe something else is up. does mathjax know pmatrix? i suppose it must

Koro

I did pmatrix but then it placed v_i's in a single row

I disabled my adblocker and it started working fine again.

leslie townes

odd, but cool that you found a fix

Jakobian

10:04

@XanderHenderson I love the spirit of Christmas :D

For me its not religious or anything. More like tradition

robjohn

10:34

@Koro instead of "..." you might use \dots, or \cdots, or better \vdots since it is a vertical array.

Koro

@robjohn sure, thanks.

robjohn

As for the slowdown in rendering, I noticed that happening a long time ago, but I haven't seen it recently.

Koro

me too, it happens sometimes.

I guess it's due to adblockers.

disabling it fixed it for now.

Koro

11:00

0

Q: Orient the space $B$ such that the map $f: A\to B$ is orientation preserving.

Setup: Suppose that $f: S\to \tilde S$ is a smooth map, where $S, \tilde S$ are oriented $n$- surfaces in $\mathbb R^{n+1}$ with smooth unit normals $N_1$ and $N_2$. Suppose that $Df_p$ is non singular for every $p\in S$. The derivative $f$ is said to be an orientation preserving map if for all...

calculus multivariable-calculus differential-geometry

@Thorgott could you please take a look at it? I saw your message but decided to respond with some calculations and definitions :)

Sine of the Time

Can someone check if my computation is correct (if you want)? I have to compute $f\star g$ where $f(x)=x\chi_{[0,+\infty[}(x)$ and $g(x)=\sin x\chi_{[0,+\infty[}(x)$ and I found that $f\star g=(x-\sin x)\chi_{[0,+\infty[}$

Koro

11:16

why do you have $\chi_{[0,+\infty[}$ in the end for f\star g?

Sine of the Time

when $x\le 0$, I found that the convolution is $0$

Koro

how?

$f\star g(u)= \int_{-\infty}^\infty f(t) g(u-t) dt= \int_0^\infty t g(u-t) dt= \int_0^u t \sin(u-t) dt$, no?

Sine of the Time

$f\star g(x)=\int_0^{+\infty} (x-y)\chi_{[0,+\infty[}(x-y)\sin y dy$ if $x\le 0$ then $x-y\le 0$ since $y\ge 0$ and $\chi_{[0,+\infty[}(x-y)=0$ for $y\ge 0$

Koro

For convolution definition, don't you take integral from -\infty to \infty?

Sine of the Time

Yes, but since there's $\sin y \chi_{[0,+\infty[}(y)$ I restricted the integral

Koro

11:29

hmm, right!!

@Koro the last equality is not true for u<0. For u<0, g(u-t)=0 as $t\ge 0$.

@SineoftheTime it looks correct to me.

Sine of the Time

thank you for your time :)

Thorgott

12:03

@leslietownes I agreed with your previous points, but this is something I do too...

@Koro $df_p\colon S_p\rightarrow\tilde{S}_p$ is an isomorphism of vector space, so it is always true that if you fix an orientation of $\tilde{S}_p$, then there is some choice orientation of $S_p$ that makes $df_p$ orientation-preserving. this has nothing to do with manifolds yet.

the crux with manifolds is that you want the orientation of $S_p$ to "continuously depend on $p$" in some sense

this is ensured by taking orientations induced from a normal vector field, which continuously (or even smoothly) depends on $p$

however, consider $df_p$ as before. for each $p$, it is either orientation-preserving or orientation-reversing.

using continuity of the determinant, we see that if $df_p$ is orientation-preserving or -reversing, the same holds for all $df_q$ for $q$ in some neighborhood of $p$

then, in the connected case, it follows that $df_p$ either preserves the orientation for all $p$ or reverses it for all $p$

in the former case, $f$ is orientation-preserving, in the latter case $f$ is orientation-reversing and becomes orientation-preserving if you switch the orientation on one of the two spaces (e.g. by inverting the normal vector field)

mick

12:27

0

Q: Is this 3D algebra $T$ power-associative?

Consider a commutative 3D algebra $T$ where the nonreal units $x,y$ satisfy $$x^2 = A_1 + A_2 x + A_3 y $$ $$xy = B_1 + B_2 x + B_3 y$$ $$y^2 = C_1 + C_2 x + C_3 y$$ where all the parameters $A_1,A_2,..$ are real ofcourse. So for instance we get $$(a + b x + cy)^2 = a^2 + b^2 A_1 + 2bc B_1 + c^2 ...

abstract-algebra 3d associativity perfect-powers

some 3D algebra LHF i think

Xander Henderson

12:53

@Jakobian That's great for you. But it kind of indicates that you grew up in a Christian-dominant culture, and are not specifically non-Christian. I'm Jewish. Christmas doesn't mean much to me. Just a lot of commercialism and bad music.

Jakobian

I did. We've used to go to Church when I was little. Well I'm an atheist but I still associate Christmas with good memories like decorating the Christmas tree

Xander Henderson

There ya go.

user587860

13:28

Does hyperbolic 3-space H_3 admit an isometric embedding into C^4?

Thomas Finley

14:02

Does $\Bbb Z/2\Bbb Z\cong 5\Bbb Z/10\Bbb Z$ as rings?

I feel that this is not true

Jakobian

Did you have a look in this survey article? I don't think such an embedding is possible. A Riemannian $n$-manifold of constant negative sectional curvature cannot be isometrically immersed in $E^{2n−2}$. — Dietrich Burde Feb 24, 2018 at 16:54

@Supersymmetry

Thomas Finley

I think they are only isomorphic as groups

This question was given to us by our professor, and I think that it's incorrect.

Jakobian

@ThomasFinley Why wouldn't they be?

Xander Henderson

@ThomasFinley Rather than asking us if something is true, why not just write it out.

Neither set has all that many elements. It shouldn't be hard.

Xander Henderson

14:20

No response?

Thomas Finley

@XanderHenderson $$\Bbb Z/2\Bbb Z=\{2\Bbb Z,2\Bbb Z+1\}$$ and $$5\Bbb Z/10\Bbb Z=\{10\Bbb Z,10\Bbb Z+5\}$$

Did I miss something?

I am feeling strange

Jakobian

did you hydrate properly?

Thomas Finley

@Jakobian maybe no

Xander Henderson

I didn't ask you to write out representations of the two rings. I asked you to write "it" out. Write out the problem.

Check the addition and multiplication.

Do something for yourself. Don't just rely on others to spoonfeed you the answers.

Thomas Finley

@XanderHenderson wait, the representation I wrote is not correct, no? Reading jakobian's comment makes me believe I wrote it incorrect.

Xander Henderson

14:27

@ThomasFinley I don't care if it is "correct" or "incorrect". I made no comment on that.

Thomas Finley

@XanderHenderson But before that, I need if I am making no mistake with the basics.

@XanderHenderson That makes it more confusing, you know!

Xander Henderson

Moreover, I don't even know what "correct" or "incorrect" would mean. You have two sets, each of which has two elements. I don't really care how you label those elements. You are asking a question about rings---there is more structure there.

What happens when you add two elements?

Or when you multiply them?

Jakobian

Never mind making mistakes, the more important is your thought process about getting from point A to point B, that is, you're claiming the two rings are not isomorphic, why is it so?

Thomas Finley

@XanderHenderson I know the bijection $2\Bbb Z+1\mapsto 10\Bbb Z+5$ and $2\Bbb \mapsto 10\Bbb Z$ and I verified this is an isomorphism.

Xander Henderson

I didn't ask you about bijectiions or isomorphisms. I asked you to describe the ring structure on each of the two sets you described. But if you have constructed an explicit isomorphism, what else is there to do?

Thomas Finley

14:31

@XanderHenderson Ha ha, nothing. Ok, I got it. I was correct after all.

Maybe I got that strange feeling just like that. I must ignore them (henceforth).

@XanderHenderson and @Jakobian But thanks for correcting me, when I was too hasty.

Xander Henderson

I wouldn't ignore such feelings, but part of developing mathematical maturity is understanding when what you have done is correct.

You have to think.

Thomas Finley

@XanderHenderson I agree.

Xander Henderson

In any event, I really hate your notation. It think it obfuscates things. $\mathbb{Z}/2\mathbb{Z}$ consists of two sets: $[0]$ and $[1]$ (the equivalence classes of $0$ and $1$, respectively). If you write things out in this way, it is much easier to do the arithmetic (thinking of addition and multiplication modulo $2$).

You can do something similar for $5\mathbb{Z} /10\mathbb{Z} = \{[0],[5]\}$. it is then much easier to verify that sending $[1]$ to $[5]$ does what you want it to do. Or, alternatively, you can use the uniqueness of rings with a prime number of elements.

(which is something you were working on yesterday)

Jakobian

Yeah. That the only (non-commutative, non-unital) rings with $p$ elements are the ring with trivial multiplication and $\mathbb{Z}/p\mathbb{Z}$

So you'd have to verify multiplication in $5\mathbb{Z}/10\mathbb{Z}$ is non-trivial, say by computing $[5]^2$

Thomas Finley

@Jakobian yes, never thought it like that.

Koro

14:43

@Thorgott I don't understand why df_p is orientation preserving? Can you please explain that?

Xander Henderson

@ThomasFinley I mean, I wouldn't have generally taken that approach, but you were developing that tool yesterday. Most books are written in such a way that the material presented in one section is likely to be relevant in the next section or two. [Good] authors choose to present ideas so that they flow.

When you learn about a new tool or definition or result, you should expect to use that tool or definition or result in the future. It should become part of your "kit".

Thomas Finley

@XanderHenderson Actually, I am comfortable with either of those notations but I just use them interchangeably. But yes, I never thought about the second approach to the problem as you mentioned.

@XanderHenderson One thing, I thought it not so important, because of my lack of experience with working on problems, maybe. But yes, I will add that tool now, to my "kit".

Soumik Mukherjee

I don't think using notations interchangeably is a good practice

Koro

The latest question that I asked on mse will remain unanswered.

Xander Henderson

@ThomasFinley The point of notation is to more clearly communicate ideas. The advantage of writing equivalence classes as $[n]$ (rather than thinking of them as sets with elements like $m\mathbb{Z} + n$) is that you can actually do arithmetic with $[n]$ much more concisely, which means that the ideas are communicated more clearly.

A ton of problems in mathematics are solved simply by using the right notation.

Thomas Finley

14:48

@XanderHenderson Oh! I will surely keep it in mind.

one potato two potato

15:15

by using the right definition

Akiva Weinberger

15:35

More progress on this question!

14

Q: How many ways to arrange $n$ points in $(\Bbb F_q)^2$ with no three collinear?

How many ways are there to arrange $n$ points in the finite field plane $(\Bbb F_q)^2$ with no three of the points collinear? An easy upper bound is $(q^2)^n=q^{2n}$, but of course it's less than that. (Of course, if I asked the same question over $\Bbb R^2$, it would be infinite.) The collineari...

combinatorics finite-fields inclusion-exclusion mobius-inversion finite-geometry

Soumik Mukherjee

15:55

I just noticed that the mean square has become the green square:)

Xander Henderson

@SoumikMukherjee Well, if you were average, you might become envious.

Soumik Mukherjee

16:06

:"[

Ted Shifrin

@SoumikMukherjee But the snarl indicates green grinch.

Xander Henderson

@TedShifrin Which sounds like "Gingrich". OH NO!

Ted Shifrin

16:23

You had to go there~

Jakobian

@SoumikMukherjee I think Thomas meant that he's comfortable with both of them

Soumik Mukherjee

Well he wrote about using them interchangeably also

Thorgott

16:39

@Koro I didn't say that

keep in mind that "being orientation-preserving" is not a property of linear maps between vector spaces, it is a property of linear maps between vector spaces with chosen orientations

Xander Henderson

@Thorgott Yet another proponent of critical manifold theory!

Rithaniel

Okay, so, given a ring $R$ and a (finite) list of pairwise-zero-multiplying idempotents $e_i$ (so $e_i^2=e_i$ and $e_ie_j=0$ when $i\neq j$), then $\sum_{i=0}^ne_i$ acts as the identity on each $e_i$ used in the sum. Is there an additional condition that would be sufficient for this sum to be the identity?

Xander Henderson

I don't think it is appropriate to be talking about orientations in this room! This is a public forum, and is meant to be largely apolitical. Get your "chosen orientations" out of here!

Thorgott

17:09

lmao

@Rithaniel they should generate the ring

Rithaniel

17:21

Yeah, that one's certainly sufficient. Then you can easily show that the sum acts as the identity on any element of the ring

Thorgott

oh, there's an issue, you didn't specify non-trivial, so your list contains the identity

I think it should work if you assume you have a maximal list of non-trivial idempotents as long as it is non-empty

Rithaniel

It can probably be refined into a necessary condition, too. When I read "they should generate the ring," I assume that's in the context of "generate it as a $X-$module," where $X$ is some ring such that $R$ is expressible as an $X-$module. Perhaps some reasonable restriction on $X$ would make it necessary

robjohn

@TedShifrin did someone ring?

Koro

@Thorgott then what did you mean?

anyways, I looked at one solution and they have also done it for some basis that works.

robjohn

@XanderHenderson Should the US map be put on a Mobius strip to denote non-orientability?

Koro

17:33

Either the definition of orientation preserving in the book is wrong

or the solution is wrong.

Xander Henderson

@robjohn NOT WITH MY TAX DOLLARS!!!!!11!!!!!!!!!!!11!!111!!!!

Koro

I believe the former

@robjohn on a Mobius strip and then inside Klein bottle.

Xander Henderson

@Rithaniel Grr.... $X$-module. The hyphen is a hyphen, not a subtraction nor a negative sign. $X$-module, not $X-$module. X(

TYPSETTING CARRIES SEMANTICS! X(

Rithaniel

$-$-$-$-$-$

Koro

I asked many people and all of them do it with some basis.

not for all bases.

robjohn

17:36

@Koro I got one of those from a friend a few years ago. I have trouble filling it up.

Koro

suggesting that the book's definition is wrong.

Xander Henderson

@Koro And the plural of "basis" is "bases". Why is everyone so terrible today?!

:'(

robjohn

@XanderHenderson and I'm terribler!

Rithaniel

$X\sim$module or $X$~module? (X-$module$)

Koro

@XanderHenderson I'm badder.

robjohn

17:37

I'm worsest

Thorgott

@Koro I meant what I said. If you fix the orientation on one space, then there is one (and only one) orientation on the other space s.t. the isomorphism is orientation-preserving.

Koro

what do you mean by orientation preserving? What is your definition?

I ask because it seems different than mine.

Suppose that $f: S\to \tilde S$ is a smooth map, where $S, \tilde S$ are oriented $n$- surfaces in $\mathbb R^{n+1}$ with smooth unit normals $N_1$ and $N_2$. Suppose that $Df_p$ is non singular for every $p\in S$.

The derivative $f$ is said to be an orientation preserving map if for all $p \in S$ and for all positively oriented bases vectors $v_1,...,v_n$ of the tangent space $S_p$ (i.e., with the property that $\det\begin{pmatrix}v_1\\ v_2\\ \vdots\\ v_n\\ N_1(p)\end{pmatrix}>0$), we have $\det\begin{pmatrix}Df_p(v_1)\\ Df_p(v_2)\\ \vdots\\ Df(v_n)\\ N_2(f(p))\end{pmatrix}>0$.

I use this definition.

Here is what AI had to say about it. It doesn't seem correct.

@Thorgott when you say the isomorphism is orientation-preserving, you mean $Df_p$ is an isomorphism that preserves orientations.

But that's a problem as per my definition because then one also has to consider $N_2(Df_p (u))$ at some u

Jakobian

17:56

@Koro Asking AI about math is like rolling a dice

Thorgott

@Koro if $V,W$ are oriented vector spaces, an isomorphism $f\colon V\rightarrow W$ is orientation-preserving if it takes a (equivalently, any) positively oriented basis of $V$ to a positively oriented basis of $W$

Koro

@Jakobian what to do when there's no one to answer!!

Thorgott

you have to understand orientations of vector spaces before talking about orientations of manifolds

Koro

@Thorgott OK. You see I didn't know this definition before.

Jakobian

Won't it be something like, an orientation on a manifold is an orientation on all of its tangent spaces (+ compatible)?

Koro

18:00

I suppose f in your message should be defined as orient. p iff determinant(f)>0.

Jakobian

And then orientation preserving map will be something that preserves orientation on tangent spaces ?

Thorgott

determinant does not make sense without choosing bases

Koro

but you said oriented so...

Thorgott

if you choose positively oriented basis for $V$ and $W$, $f$ will be orientation-preserving if and only if the determinant of its matrix representation in that basis is positive

Koro

V,W are oriented so we can take 'ordered' basis.

Jakobian

18:01

@Thorgott I thought determinant was invariant of change of basis?

Koro

with positive determinant.

Thorgott

@Jakobian yes

@Jakobian yes, but we're looking at two different vector spaces

the determinant is only intrinsic for automorphisms $V\rightarrow V$

Jakobian

oh, so they have to be the same

Koro

@Thorgott yes, I meant that.

Thorgott

and then you want to take the same basis on domain and codomain

and the resulting matrix has determinant invariant under change of basis

Jakobian

18:02

yeah that makes sense

Thorgott

@Koro ok, that's fair then. I'm just being precise because "the determinant" is not well-defined in this context

Koro

now the point is: suppose v_1, v_2,...,v_n is an ordered basis of V. hmm, how do you say it's positively oriented?

what is positively oriented here?

For R^3 one can say e_1,e_2,e_3 is positively oriented because determinant (e_1, e_2, e_3)=1>0

but what can we say in general vector space?

Thorgott

I said "oriented vector space"

that means a vector space together with a chosen orientation

Koro

what is orientation in case of vector space?

Say, on the space of 2 by 3 real matrices.

Thorgott

an equivalence class of bases, where two bases are equivalent if their change of basis matrix has positive determinant

so there's exactly two equivalence classes

the members of the chosen equivalence class are called positively oriented bases

the members of the other equivalence classes are the negatively oriented basis

Koro

18:08

@Thorgott ah, I see so that's the definition.

Thorgott

yes, you just define "orientation" to mean precisely what it has to mean so that "orientation-preserving" means "positive determinant" (in the sense discussed earlier)

Koro

I see, I understand. Thanks for clarifying it up.

Thorgott

yeah, sorry for not explaining this first

Koro

@Thorgott ok, understood.

Thorgott

I hope you can see now how the unit normal field on a surface induces an orientation on every of its tangent spaces

Koro

18:15

You see, I know this as a definition: basis vectors v_1,...,v_n in S_p are said to be positively oriented if det(v_1,...,v_n, N(p))>0.

S_p is n dimension tangent space at p.

N is unit smooth normal field.

@Thorgott I'm starting to understand it now. But I want to know what your definition of orientation preserving map is.

I have stated mine above.

And yours' still seems to me a different one.

Thorgott

ok, so now we have establish that an oriented surface has a natural orientation on each of its tangent spaces

so we call a diffeomorphism orientation-preserving if its differential at every tangent space is orientation-preserving in the previous sense

Koro

yeah

@Thorgott so just to clarify: positively oriented basis vectors in V and +ly oriented basis vectors in W. f was a linear map so its determinant w.r.t. to these bases is positive.

and you are are defining orientation preserving for a diffeomorphism only.

in my case it is for smooth maps between n-surfaces which by inverse function theorem are also local diffeomorphisms so ok.

hmm, so suppose that $v_1,v_2,...,v_n\in S_p$ are positively oriented in the determinant sense that I mentioned. Suppose that $w_1,..., w_n$ are also positively oriented in the same sense. I want to show that the change of basis matrix has positive determinant.

I think I understand it now. Let A be a matrix that does that, i.e., $w_i= Av_i$ for all i and also $N(p)= AN(p)$. So $\det(w_1,..., w_n, N(p))=\det (Av_1,..., Av_n, AN(p))=\det(A)\det(v_1,...,v_n, N(p))>0$

so $\det(A)>0$.

so my definition is consistent with your definition.

Thorgott

18:37

@Koro yeah, local diffeo is fine too

but if it were not an isomorphism on the tangent spaces, it would not make sense to ask whether it preserves or reverses orientation cause it doesn't map bases to bases

@Koro what kind of matrix is $A$ supposed to be? this does not quite make sense as written, I think

Koro

So $S_p\subset R^{n+1}$.

A is an (n+1) by (n+1) matrix.

A is like change of basis matrix.

Thorgott

ah ok, you should probably add indices to the normal fields

but that my definition is consistent with yours does not require an argument. it is tautology

Koro

ohh

it's tautology but so is 'differentiability implies continuity'.

🤔

@Thorgott equivalently any? f takes a positively oriented basis v_1,...,v_n to say w_1,...,w_n respectively. Now, let $u_1,...,u_n$ be positively o.b.. Let $P:V\to V$ be the linear map $P(v_i)=u_i$ for all i. I want to show that $f(u_i)$'s are positively ob.

Ted Shifrin

18:54

No, differentiability implies continuity is not tautology. Just ask Jakobian with the stuff he's reading in Banach spaces.

Thorgott

"tautology" in the sense that it is literally the same thing, just some words are substituted with synonymous words

@Koro here, you are just saying a positively oriented basis is taking to a positively oriented basis, which is the same as my definition of orientation-preserving

Koro

@Thorgott yes, I realised that.

Now, I'm trying to understand how to go to 'for all positively ob' from 'a positively ob'.

$f(u_i)= f\circ P(v_i)$ should be positively ob.

These are basis vectors, yes.

To show these to be positive oriented, I should show that there exists a linear map T:$W\to W$ such that $T(f(v_i))= f(u_i)$ with det(T)>0.

Thorgott

in the basis v_1,...,v_n of V, the u_1,...,u_n are represented by a matrix A with positive determinant (this is the definition of both bases defining the same orientation). now fix a positively oriented basis w_1,...,w_n of W. with respect to the bases v_1,...,v_n and w_1,...,w_n, f is represented by a matrix T.
the images f(u_1),...,f(u_n) are represented in the basis w_1,...,w_n of W by the matrix TA. now, det(TA)=det(T)det(A) is positive iff det(T) is positive as we already know det(A) is positive.

Ted Shifrin

I'm so glad Thor is here to take care of the differential topology basics, so I don't have to :D

Koro

I feel like I jumped straight into sea without learning swimming.

Ted Shifrin

19:02

By the way, @Thor, I meant to ask. Have you proceeded into the Ph.D. program by now? When we first met you, I was under the impression you were not yet a graduate student.

@Koro I think you always over-panic.

Koro

yeah, I should fix this.

But determinant of T is 1

@Koro det T=1 because matrix of T w.r.t. the bases {f(v_i)} and {f(u_i)} is the identity matrix.

this looks wrong. This will imply every change of bases transformation is of + determinant to some extent.

Thorgott

@TedShifrin I'll start working on a Master's thesis soon. The plan is to start a PhD next fall. My advisor is in the process talking to some of his colleagues.

Ted Shifrin

So you're staying at the same uni?

Thorgott

no, I'll be moving

Ted Shifrin

Ah. Are you already accepted?

Thorgott

19:13

Nope, I've not yet applied anywhere. The current stage is figuring out where to go.

Koro

Thor, open your own uni.

:)

Ted Shifrin

Wow. In our world, applications should already be submitted.

Maybe by next fall you meant fall, 2025.

Koro

@Koro is something wrong with this?

Thorgott

applications are around march/april, I think

Koro

it looks wrong to me because I didn't use the fact v_i's are positively oriented.

Ted Shifrin

19:19

For fall, 2024, Thor? Crazy.

Thorgott

@TedShifrin that is my understanding, yes

Xander Henderson

@robjohn Banned!

robjohn

I remember that November was when I needed to have applications (undergraduate) for the next fall.

Xander Henderson

@TedShifrin I've heard that other countries have later and/or "rolling" application deadlines.

Because, like, they treat a phd program like a job, rather than training.

Thorgott

@Koro this is what I warned you about earlier. you have to be careful about which bases you choose. any linear isomorphism is represented by the identity matrix in a silly choice of basis.

Ted Shifrin

19:24

Well, @robjohn, but applications to Ph.D. are typically due December or January, at the latest. We have a national mandated deadline of April 15 for all decisions to be submitted.

robjohn

Ah, I don’t remember when my graduate applications were due.

Ted Shifrin

Having spent 35 years writing recommendations, I'll tell you they were all due in December and early January.

For undergraduate, there's now all this early action/early decision stuff. Those admissions are made December 15.

robjohn

I believe you. Just no personal recollection.

Ted Shifrin

Yeah, for you and me (particularly) it was a long time ago. And, of course, in those days, places like Princeton wouldn't decide until NSF decisions had been made (also early in January, I think).

Sine of the Time

Ted is it hard to get into Princeton to do a PhD?

Ted Shifrin

19:30

Very.

Things may have changed, but it's a small department, small graduate classes, and in my day they didn't even teach first-year graduate courses. They assumed that students had learned it all already.

Sine of the Time

I see

robjohn

Yeah, there were no first year courses in 1981

Sine of the Time

I've read that you must have no less that 3 recomendation letters

Xander Henderson

@SineoftheTime That is pretty typical, yes.

Ted Shifrin

Yes, minimum 3, some places 4.

Sine of the Time

19:34

from an economical point of view, can you live only by the "salary" that they give you? I don't know how much costs the life in the US

Koro

@Thorgott how did you get the last line? Why by the matrix TA?

Xander Henderson

@SineoftheTime It depends, but usually, no.

Koro

@Thorgott yes, noted.

Ted Shifrin

It's not easy. When I went to Berkeley, apartment rents (even sharing) used up more than half my money each month. I imagine it's worse now.

Sine of the Time

what's the average price for a rent?

Xander Henderson

19:35

@SineoftheTime Graduate student stipends in the US tend to be around $30k per year.

Sine of the Time

In my homecountry, around 1200€ per month

Ted Shifrin

I don't even know what rents are now, other than the obscene amount I'm paying in San Diego.

Xander Henderson

@SineoftheTime That varies a lot by state (and locale).

Sine of the Time

@XanderHenderson I guess so

Ted Shifrin

Places like UGA aren't quite as bad, but even there rents are not as cheap as they should be because there are so many students (undergraduate and graduate) ...

Xander Henderson

19:37

Places with big universities tend to be more expensive, but not universally so. California and New York are obscene.

Sine of the Time

@XanderHenderson do you have an idea in Princeton?

Xander Henderson

Expensive. It is practically New York.

Google can get you better stats.

Ted Shifrin

You can find out by googling, @Sine. That will be more informative than asking us.

Xander Henderson

rentcafe.com/average-rent-market-trends/us/nj/princeton

dejavu

A quick question: Given a graded algebra $R$ with a graded derivation $d \colon R \rightarrow R$ and a graded $R$-module $M$, is there a standard name for maps $\mu \colon M \rightarrow M$ which satisfy a Leibniz rule with respect to $d$ (i.e, $\mu( r \cdot m) = dr \cdot m \pm r \cdot \mu(m)$)?

When $d^2 = 0$ and $\mu^2 = 0$ then this compatibility is what needed to turn $\left( M, \mu \right)$ into a DG-module but I'm interested in the more general case. It seems tempting to call $\mu$ a "module derivation" but googling it gives mostly derivations of the form $R \rightarrow M$ when $M$ is a bimodule and not of the form $M \rightarrow M$.

Thorgott

19:43

@Koro cause the composition of linear maps corresponds to the multiplication of representing matrices (if you make the effort to choose the bases consistently)

the matrix $A$ represents the isomorphism $V\rightarrow V$ sending $v_i\mapsto u_i$ in the basis $v_i$ on both domain and codomain

user587860

If we have a group action of a group G on a vector space V, does this extend to a group action of G/N on V? Where N is a normal subgroup of G.

Ted Shifrin

@dejavu This is too technical for us here, I think. You should probably post that on the main site.

Koro

@Thorgott I know that. I mean if $f(u_1)= c_1 w_1+...+c_n w_n$, then why is $(c_1, c_2,...,c_n)^t$ the first column of TA?

Ted Shifrin

@Supersymmetry That is not extending. That is inducing or reducing to. $N$ has to act trivially for this to make sense.

@Koro Go back to beginning linear algebra. How do you write the matrix of a linear transformation, given bases?

dejavu

@TedShifrin Okay, thanks! Since this is a terminology question, I thought I might get a quick answer here instead of littering the main site but I'll try.

user587860

19:47

@TedShifrin Thank you. By "N has to act trivially", do you mean that the action of N on X look like nx = x for n\in N and x\in X?

Ted Shifrin

@dejavu The main site is littered with "do my homework" questions with zero effort.

This is not so bad.

@Supersymmetry Yes. Think about what an action of $G/N$ would entail.

user587860

Thanks.

Koro

@Thorgott crisp!! thanks a lot.

I still have one confusion: why is det T>0?

I think that det T=1

$f: V\to W$ and in your notations: matrix of f=T with respect to bases {v_i} of V and {w_i} of W.

fv_i= w_i so matrix of f is identity.

Thorgott

@dejavu I would probably still just call it a derivation (and explain what I mean)

Koro

@Thorgott is this correct?

Thorgott

19:59

I guess the real question is what to call the pair $(R,d)$, cause then $(M,\mu)$ is clearly the right notion of what a module over $(R,d)$ should be

@Koro no?

at no point did we assume that f maps v_i to w_i

Koro

@Thorgott I used this.

So $f(v_i)=w_i$, no?

dejavu

@Thorgott Yeah, that's what I did for now but I find it hard to believe that nobody talked about such objects before and invented some more or less accepted terminology

Thorgott

no? the v's and w's were arbitrary positively oriented bases of V and W respectively

Koro

yeah, and I assumed that $f$ maps v's to w's.

Thorgott

@dejavu yeah, I've had similar issues before. I would encourage you to post a question

Koro

20:05

using the linked definition

Thorgott

@Koro then $f$ would be orientation-preserving to begin with

you're doing a circular reasoning

Koro

@Thorgott no, I'm trying to prove 'equivalently, any' part in your message.

So I assume that f is orientation pres. for positively oriented bases v_i's in V and w_i's in W.

then I'm trying to show that: for positively oriented bases u_i's in V and $f(u_i)$'s in W, f is positively oriented.

@Thorgott why not? this is the definition?

is there any book where it is given?

Thorgott

ah, now this makes sense

Koro

Surely, Thorpe's differential geometry book has none of it

Thorgott

yeah, you can assume $T$ is the identity, that's fair

Koro

20:11

$\ddot\smile$

thanks a lot, I get it now.

Thorgott

my explanation was suboptimal in that regard, sorry

Koro

nvm thanks, my 3 months old confusion is clear now because of you.

Ted Shifrin

Thorpe's book is not my favorite at all, although I never tried teaching out of it. For standard texts, look at Boothby, volume 1 of Spivak, among others.

Thorgott

this should be discussed in every book

Tu certainly has it in great detail, Lee has it without of a doubt as well

Ted Shifrin

20:28

I don't know their books, although I know the authors. Lee took a graduate course in complex geometry from me, but turned out to be far more prolific as a math author than I ever did. :)

Jakobian

If $f:A\subseteq E_1\times E_2\to F$ is differentiable at $x_0$, then $D_1f(x_0)$ and $D_2f(x_0)$ exist and $Df(x_0) = D_1f(x_0)\circ i_1 + D_2f(x_0)\circ i_2$ where $i_k:E_k\to E_1\times E_2$ are canonical injections. If $Df$ exists and is continuous at $x_0$, then $D_1f, D_2f$ are continuous at $x_0$. If $D_1f, D_2f$ exist and are continuous at $x_0$, then $Df(x_0)$ exists.

Ted Shifrin

That doesn't sound very interesting.

Jakobian

I just don't like the assumptions of continuity on the whole domain

Ted Shifrin

Still doesn't sound very interesting.

Jakobian

Yet?

Are there applications of the pseudo-derivative $PDf(x_0) =\sum_{k=1}^n D_kf(x)\circ i_k$ assuming that the partial derivatives exist?

say, for sufficently nice function $f$, but not nice enough for this to coincide with the derivative

Ted Shifrin

20:46

Not that I am aware of. I also do not know examples of functions that have partial derivatives on an open set that are nowhere differentiable on that open set.

I suspect there must be a Baire category type thing that it must be differentiable most places. But there's a good question for you to wrestle with.

Jakobian

If the function is continuous and all partial derivatives exists, then their points of continuity are co-meager sets. The intersection of all of those sets is non-empty

So by theorem I just mentioned, since partial derivatives are continuous at any point of that set, the derivative must exist somewhere

well, that doesn't say anything about the ones that can be discontinuous

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