Mathematics - 2023-05-27

user 85795

00:15

🥳👍

geocalc33

Grammar question which I'm sure is already posted somewhere on grammar stack exchange...but is it grammatically correct to answer a question with a single contraction?

For example: "Is it sunny out today?" Answer: "It's."

noballpointpen

I am not a native speaker, but it sounds not natural for me.

shintuku

@geocalc33 'it is' is okay, 'it's' is not

geocalc33

I think it's not grammatically correct as well but I'm curious to the reason

noballpointpen

Then I don't think you visited the right place to ask... we are not language experts here.

geocalc33

00:29

I'm going to check grammar stack exchange

noballpointpen

Ted, if you don't want to answer/discuss, just let me know and I will be ok with it.

Since I am somewhat awaiting for an answer. Not sure if you're just busy.

shintuku

00:51

what's $\mathbb F_p$, is it always the field of elements of a field $F$ modulo $p$?

Silly Goose

01:08

in Dummit and Foote group actions are introduced and then it is found that any group action induces a representation of said group in the symmetric group over the set acted upon. But other sources say that it is Cayley's theorem which tells us that any group can be represented as a subset of the symmetric group; which is correct?

Cayley's theorem seems to prove more than is wanted since it proves that every group is isomorphic to a subgroup of a symmetric group

noballpointpen

I didn't study algebra yet, but can't these be just different ways of proving the same? Just a general observation from me.

shintuku

01:35

@shintuku no it is an alternative notation for $\mathbb Z_p$ that emphasizes the fact it is a field

robjohn

02:14

It’s good that someone could answer your question

one potato two potato

How can I reply to my own message?

shintuku

02:28

@onepotatotwopotato you can get the permalink number of your comment by clicking on the arrow that appears on the left when you hover over your message. then begin your message with :(permalink number), e.g., :30976263

one potato two potato

@onepotatotwopotato go

shintuku

gj you did it

one potato two potato

oh neat

Silly Goose

what theorem(s) tell us that "Every finite-dimensional unitary representation on a Hilbert space V is the direct sum of irreducible representations."?

one potato two potato

02:59

Given a group presentation, I think it's quite a hard problem to ask what are the normal generators of that group?

Sourav Ghosh

03:29

Thank you :)

porridgemathematics

04:50

im missing a simple argument here, can anyone help fill it in? $u$ is subharmonic on the unit disk, and $v$ is defined as $u(z^2)$ on the upper half of the unit disk, and $0$ on the lower half. Then $M_v(r) = M_u(r^2)$, where $M_f(r) = \frac{1}{\pi r^2} \int_{\Delta(0,r)} f dA$, $\Delta(0,r)$ is the open disk center $0$ of radius $r$

it should follow from a simple change of variables but im not able to make it so

porridgemathematics

05:05

instead I get $M_v(r) = \frac{r^2}{4} M_u(r^2)$..

oh crap, sorry im being a moron

ignore my question

it turned out $M_u(r)$ really meant $\sup_{|z| = r} u$

in which case it makes total sense

nickbros123

06:43

is the motivation for defining matrix multiplication the idea of creating a system of linear equations / row vectors $C_{m \times p}$ which is a matrix with m equations p unknowns, whose equations/rows/row vectors are each a linear combination of the rows of another system $B_{n\times p}$ which is a matri with n equations, p unknowns

Utkarsh

07:14

Hii I want to know what is the direction ratio of the line $\frac{x-2}{2} = \frac{2y - 5}{-3} = z+1$. The component values along x and y axis are 2 and -3. What is for z? 0 or 1?

Mr. Feynman

I have a question about the connection 1-form. As I define it, it is as a *matrix& of one forms. As we said yesterday, when we have a $G$-structure we also require that the connection 1-form is $\mathfrak{g}$-valued. How does that make sense in general? Not all Lie groups are matrix Lie groups, so I have no reason to expect that the Lie algebra is made up of matrices...

porridgemathematics

07:39

@Mr.Feynman connection one forms are only locally defined anyway, in which case they are $\text{End}(TG)$-valued one forms

so you can think of it as $\mathfrak{g}$-valued

for finite dimensional lie groups, the matrix viewpoint works

Mr. Feynman

08:14

@porridgemathematics I'm not very grounded on the matter and I have never encountered a definition involving endomorphisms of the tangent bundle, nor I know how we get to the Lie algebra from that... I suppose this comes from some consideration about left translations (?)

@porridgemathematics For the time being I want to focus on this piece. Why does this work? There are non matrix Lie groups that are finite dim

porridgemathematics

08:58

@Mr.Feynman for the first question, you could think about the christoffel symbols , so if I write (for an affine connection), $\nabla_{\partial_i} \partial_j = \Gamma_{ij}^k \partial_k$, then I could equally write in local coordinates, $\nabla = (\Gamma_{ij}^k) (\partial_k \otimes dx^j) \otimes dx^i$, which is a covector field taking values in $TG \otimes T^{\ast}G = End(TG)$

Ajay

Are Darboux and Riemann sums the same thing?

porridgemathematics

@Mr.Feynman currently what im saying just applies on any smooth manifold,

and for any affine connection

im rusty on connections of principal $G$-bundles, but the logic should be very similar

the lie algebra should come out of $T_{e}G$ and left translations yes

the details someone else will have to help you out with or you will have to fill in yourself, since im pretty far removed from DG right now

I just said finite dimenisonal lie group because then $End(TG)$ is a finite dimensional vector bundle, so matrices correspond to endomorphisms, i cant remember if something like ados theorem matters here

oh yeah, dont you have that the connection one form should be invariant under left or right translations (which one I forget)

leslie townes

ajay: i am not sure that they are consistently defined across all sources, but in the cases i've seen, they are not the same thing, although they give rise to the same notion of integrability and the same values of the integral

porridgemathematics

so that should explain why $End(TG)$ coincides with $End(\mathfrak{g})$ @Mr.Feynman

darboux sums are just extremal riemann sums

which is why they give the same integral

but anyway im pretty sure the connection one form for you will be an End(g) valued one form, not a g-valued one form...

you could say its a matrix of g-valued one forms though

one potato two potato

09:38

I need to choose one of two classes: TDS (topological data analysis) and Probability theory.

robjohn

@Utkarsh exactly what do you mean by "component values"?

Mr. Feynman

@porridgemathematics Ooh, this is basically the definition of connections as functions taking sections of a bundle into 1-forms valued in such bundle [...] and then you just played with tensor products

@porridgemathematics That's something about which I got confused yesterday, damn it

porridgemathematics

09:53

yeah

Mr. Feynman

In that case I'm getting confused on why in the case of vector bundles the connection form is a matrix of real valued one forms and for a principal bundle we have a matrix of $\mathfrak{g}$-valued one forms

What is the reason why for a principal bundle we wouldn't be content with a connection form made up of real valued forms? I mean, even without technicalities...

porridgemathematics

i think there is confusion in using terms here, to me a connection is locally as I wrote

terminology aside, it looks like $\nabla = (\Gamma_{ij}^k) (\partial_k \otimes dx^j) \otimes dx^i$

the matrices in those 'blah is a matrix of blah valued whatevers' is really always referring to $\Gamma_i$ with components $(\Gamma_{i})_{jk} = \Gamma_{ij}^k$

Mr. Feynman

What I call connection form (at least for vector bundles) is the matrix whose entries are $\omega^j{}_{k}:=\Gamma^j_{ik}dx^i$

porridgemathematics

right, but there is some abuse going on there

because we are presupposing a fixed coordinate frame, and the effect of applying the connection to a vector field, in those coordinates, is the same as writing the vector field in those coordinates, and matrix multiplying

the $\partial_k \otimes dx^j$ is thus being hidden

so sure, its a matrix of scalars if you want

but to say so you are smuggling away those tensorial components

in both the principal G bundle case or the usual case

so thats why you can always say a matrix of $TM$ valued one forms

because the components of the matrix really are $\Gamma_{ij}^k \partial_k \otimes dx^j$

so this way there is no special cases involved

the principal G bundle is only different because you can replace $TM$-valued with $\mathfrak{g}$-valued

actually I think matrices are going to just cause confusion when thinking about why you can go between some description and another description

Mr. Feynman

At this point I think it is really about terminology as you say, because for a vector bundle with a local frame $\{e_i\}$ (so we get rid of any confusion with $\partial_i$), a connection is such that $\nabla e_k= \Gamma^j_{ik}dx^i\otimes e_j$, so the point is my definition only takes into account the first factor while in the other case we are considering the whole thing

porridgemathematics

10:04

I would always use the pure tensor notation to see why you can do that

I believe it really is down to terminology yes

theres nothing special going on in the principal bundle case

besides the additional constraints

Mr. Feynman

I'm using multiple references and they might not agree on these conventions so I'll take my time to check things a little bit more and I'll be back on this either here or directly asking a question on the site. For the time being, thank you for your help @porridgemathematics

porridgemathematics

no worries, and yeah I should have used $e_i$, the $\partial_i$ only refer to coordinate vector fields

Utkarsh

11:05

@robjohn Direction ratios along axis.

robjohn

11:30

@Utkarsh then $2:-\frac32:1$

note that $4:-3:2$ works, too

Thomas Finley

12:36

Hey everyone! I have a question regarding the definition of a field and the operations within it. In the definition, it states that the operations of addition and multiplication should be well-defined. I'm a bit confused about whether these operations refer to the usual sense of addition and multiplication or if they can be defined arbitrarily.

Specifically, I'm wondering if it's possible to define addition as simply a name given to an operation that has no connection to the conventional notion of addition. Could addition, in this case, represent any operation unrelated to the traditional …

shintuku

14:28

any comprehensible nonhausdorff spaces other than the line with two origins?

on an unrelated note: jstor.org/stable/25099190

Utkarsh

16:01

@robjohn Oh yeah. Thanks.

Mr. Feynman

16:32

In the definition of the Maurer-Cartan form, Frankel's book does the following:

Can someone help me understand why do we need $dg$ at all in the way he arrives to it? It maps a vector to itself, so why would I compose $(L_{g^{-1}})_*$ with it instead of just having $\Omega=(L_{g^{-1}})_*$?

> Thus $dg$ takes $Y$ at $g$ into $Y$ and $g^{-1}$ left translates [...]

In fact, seeing the intrinsic construction on Wikipedia, I wouldn't even think of putting $dg$ there

user 858770

17:02

> Only the clearest of minds are the first to think of something nobody would think of.

Ted Shifrin

17:13

@Mr.Feynman You need a $\mathfrak g$-valued $1$-form. On $T_eG$, it will be the identity map (viewing it as a $(1,1)$-tensor and then translating into a vector-valued $1$-form). Think of $dg$ as being like $dx$ on $\Bbb R^n$. The first structure equation on $\Bbb R^n$ writes $dx = \sum\omega^i\otimes e_i$. This is also writing the identity map as a $(1,1)$-tensor.

Wolgwang

> $f:[0,1]\to\Bbb R$ is a continuous function such that $f(1-x) = \dfrac1{f(x)}$ for all $x\in[0,1]$. The possible values of $\displaystyle\int_0^1 f(x)\mathrm dx$ are :
a) $\pi$
b) $\dfrac7{22}$
c) $e$
d) $\dfrac 1e$

$$I=\displaystyle\int_0^1 f(x)\mathrm dx\\I=\displaystyle\int_0^1 \dfrac1{f(x)}\mathrm dx\\2I=\displaystyle\int_0^1 f(x)+ \dfrac1{f(x)}\mathrm dx\ge \displaystyle\int_0^12=2\\I\ge1$$

Is this right?

I assumed it to be positive and applied AM/GM 🤡

Ted Shifrin

Yeah, that's right. I don't like the question, though. "The possible values" suggests this is an exhaustive list, which it obviously is not. $1$ is missing. "Which of the following are possible values of ..." would be valid.

Wolgwang

Ok Thanks :)

Ted Shifrin

17:32

@noballpointpen I find your argument very hard to read and difficult to follow. Your second sentence has a typo "or has a fixed point" — "or if $f$ has a fixed point".

You say "but this is impossible." Why is it impossible? Isn't that the argument I wrote out? Why do you say it is so obvious? Is what's following then showing that $f(\sup L)\in R$ leads to a contradiction? You say "we have $r$ and $e$ with ... Why?

Leslie already said it was very hard to read. I concur.

Ted Shifrin

17:54

@Mr.Feynman Yes, for $v\in T_gG$, $\Omega(v) = L_{g^{-1}*}(v)$ is correct. But you want a differential form so that you can differentiate and wedge it :)

冥王 Hades

18:05

If my mother saw me on the phone at 5 AM like this she'd have been pissed

Mr. Feynman

Dear @TedShifrin I'll haven't answered yet because I've had a nap and I'm quite fuzzy but I'll make sure to reply as soon as this feeling stops :)

Ted Shifrin

@Mr.Feynman No need to respond ;)

Feel better!

3

@Hades Your phone should be taken away from you.

monoidaltransform

If $G$ is lie group then the identity component can be written as $\{ e^{tA}: t \in \mathbb{R}, A\in \mathfrak{g}$ \}$?

Ted Shifrin

@monoidal What does the group action have to do with anything?

You're asking if $\exp\colon\mathfrak g\to G$ maps onto the identity component?

monoidaltransform

sorry by accident included action

yes

Ted Shifrin

18:10

It's true when $G$ is compact, of course, but it needn't be true.

There are invertible real matrices and positive determinant with no logarithm.

冥王 Hades

@TedShifrin that's precisely what she'd say. Deja vu

monoidaltransform

Sorry, why is this obvious for when G is compact?

Mr. Feynman

If I recall correctly the Lie group also has to be connected other than compact. It is not obvious, it's a theorem in Lie theory.

Ted Shifrin

@monoidal I don't know how "obvious" it is. It is relatively obvious if you use a little bit of Riemannian geometry.

We're talking about the component of the identity, so that is connected, @Mr.Feynman.

@monoidal For me, it follows from the fact that one-parameter subgroups are geodesics when you have a bi-invariant metric on $G$.

Mr. Feynman

Oh, my bad :)

noballpointpen

18:25

@TedShifrin Hello, Ted! I don't know why you don't see why it is impossible, or why I am wrong. If I am wrong, I am ready to admit it, but only after I understand exactly what was wrong. Read my comment on this point after "Indeed" here; tell me which inference is wrong or makes you wonder why I made it, I will try to justify&argue it from my point of view.

By definition of supremum, for any element $x \in S$, if $\sup L < x$, then $x \notin L$.

Ted Shifrin

I grant that it is correct. I'm just saying it's very hard to read what you write. You don't lay it out clearly and wander in and out of contradictions. Anyhow, I'm tired of this problem.

noballpointpen

Good. I will not bother you with it anymore then.

shintuku

why don't you post a question on main

Ted Shifrin

18:42

It's all done with, shin.

shintuku

ok who won

Ted Shifrin

noballpoint won ... the mystery problem has at least one solution.

Mr. Feynman

@TedShifrin Reading this message and the previous one I'm still failing to appreciate why the differential ${L_{g^{-1}}}_*:T_g G\longrightarrow T_eG=\mathfrak{g}$ alone isn't already a $\mathfrak{g}$-valued form. It is linear and $\mathfrak{g}$-valued. There is clearly something I'm missing, I don't see what though

Ted Shifrin

You missed my final point. How will you take the exterior derivative of that form? Or wedge it with another $\mathfrak g$-valued form? If I write $\Omega = g^{-1}dg$, then I can easily apply the rules of differential calculus/differential forms.

Have you done any computations with differential forms and moving frames in geometry? Look at the short section in my diff geo text on moving frames and surfaces. Section 3 of chapter 3.

noballpointpen

Huh, it's not that I am stubborn. I already said that I adore Ted's solution. It is just me who wanted to know if my logical thinking is flawed and I am making shit mistakes in the process of inferring. I don't even think it's my muddled writing that bothered me much; years of experience and I think I will learn how to write concisely and without moving in-out of contradictions; just wanted to know if I proved it. Also, Ted, not about that problem, but would you give me a bad grade?

Ted Shifrin

18:55

Huh? Where did grades come from? You kept asking for critique, so I finally went and tried to read it more carefully. I think it takes practice to write mathematics clearly. You have to put it aside and read it again a day or two later and say, "If I were just seeing this for the first time, would I follow it all clearly and understand?"

Having a valid proof in mind and explaining it to someone clearly are two different things. And I don't claim that every time I answer a question in here I make things as clear as I should. I certainly do not.

noballpointpen

Grades... just imagined if I gave you that proof as my homework answer. I don't know how the teachers grade proofs. So I asked.

Ted Shifrin

I would probably give 6/10. Valid ideas but exposition not crystal clear.

Plus, that was jumping into it after plenty of discussion here.

Mr. Feynman

@TedShifrin No, I haven't used moving frames. I will look at that section, thanks

noballpointpen

Hm... is it considered a bad/low grade for a proof class?

Ted Shifrin

I grade homework tough so that students will be motivated to improve. But I never graded such courses on the standard American system that you need at least 9/10 to get an A.

4

noballpointpen

19:05

I like your approach.

冥王 Hades

Isn't A usually 9/10?

Ted Shifrin

Did you read my sentence?

冥王 Hades

Oh. My bad.

Please understand that I'm barely awake and on the edge right now so my reading comprehension is near 0

Ted Shifrin

You get 1/10 for having tried to read.

noballpointpen

What was a common/average grade in grad classes you teached?

Ted Shifrin

19:15

Grad classes in the US don't get graded seriously. First-year grad courses tend to have real grades, but not more advanced ones. However, I was very unpopular in that I wouldn't give automatic As in the advanced ones, whereas most faculty did.

noballpointpen

I mean, on your 1-10 scale. How you viewed their solutions?

Ted Shifrin

They varied. Certainly far from all 10s.

How did we get to graduate level from beginning undergraduate material?

noballpointpen

Because you said recently that your undergrads didn't work hard enough, ha-ha :)

And what about undergrads then?

Ted Shifrin

In what context did I say that?!

noballpointpen

You said they spent way less hours on the classes, so I interpreted it that they didn't work super-hard.

Ted Shifrin

19:19

I had some classes where I gave no A's or very few, but in my last year of teaching the Honors multivariable math class (which is where the videos came from), essentially have the students got A's or A-. The last few years of that class were some extraordinary students.

Huh? You're taking everything out of context.

I didn't say "way less hours" ... you asked for a number and I gave you one, and I said that my students generally worked harder in my classes than in most all the other classes.

noballpointpen

Sorry, it is probably my not so concise and communicative writing again.

Ted Shifrin

Weird.

noballpointpen

Why weird?

冥王 Hades

@TedShifrin did they also work harder than the previous ones?

noballpointpen

People sometimes take the words "you said X" too seriously... it's not that I wanted to muddle your words, just a non-formal and not-so-precise talk. Uh...

shintuku

19:30

0/10, no proof, no hypothesis, no conclusion

冥王 Hades

I'll die if I don't get A's

noballpointpen

In the classes where no A's were present, from-what-to-what grades on your scale usually varied?

shintuku

pass, distinguished pass, or fail

heard they use this system somewhere

to keep teenagers from killing themselves

leslie townes

20:10

i read on facebook that there's a school where 2+2 = 2 because the government decided 2 is the only number anymore

shintuku

the government is always right, so it is okay

after all, it comes from the word to govern. otherwise we'd have a chaorment

user 726941

wtf kind of fb pages are you reading, sir

shintuku

not only is the government always right, but so is facebook

after all, it's called facebook. if it didn't expect you to have your face it in it, why would it be called that?

user 726941

lest we forget that the storming of the capitol was organized through social media

shintuku

it was supported by the government so it's not that bad

hell even the president supported it, what more legitimacy could you want

user 726941

20:18

yeah just like 9/11

2 + 2 = 2

2

leslie townes

20:56

i heard that ted shifrin graded so harshly that he was put on triple double probation and had to wear an ankle monitor

Ted Shifrin

Nah, never that harsh. Interestingly, the person who's taken over my multivariable math class barely finished with 5 people, I think, and I finished with 15.

I think Munchkin would grade with an iron fist.

user 726941

5? Wow! Things are getting tougher "after" the pandemic.

Was the class run online.

leslie townes

ted has been retired since shortly after they invented the telegraph

Ted Shifrin

@Mr.Feynman If you're actually looking, for fun you should compute that the $\omega_{ij}$ really are the entries of the Maurer-Cartan matrix for the group $SO(3)$ (pulled back by a section to the surface). In my case, I'm working with the right-invariant Maurer-Cartan form $dg\cdot g^{-1} = dA\cdot A^\top$, which gives you the $de_i\cdot e_j$ if you treat the $e_i$ as rows, rather than columns.

No, not online. UGA never took the pandemic that seriously in the first place and now things are well past it.

shintuku

back then the students sent in their answers through morse and you had to hire someone to decode it

ted is talking about UGA not taking the spanish flu seriously

user 726941

21:05

Right, republican turf.

leslie townes

he finished with 5 people because 10 of them died of "ague"

that's a tiny class. it is upper division? or just not required for the science/engineering folks?

Ted Shifrin

It is a sophomore-junior level course, but mostly freshmen in it. But it's the hardest possible option for linear alg + multivariable. Not quite on the level of Harvard's 55, but analogous.

Mr. Feynman

@TedShifrin I'm not looking right now but I have a reminder on my whiteboard to do so. I would like to dig more into diff geo but as of now it's just self-learning and I have actual courses and exams to take care of... Why did I write all of this? I don't know, I just felt like it!

Ted Shifrin

No problem.

shintuku

21:38

any examples of nonhausdorff spaces other than the line with two origins

leslie townes

no, just the one

they should really call it the anti-line-with-two-origins axiom

less sarcastically, if X has more than two elements, {emptyset, X} is a popular example of a non hausdorff topology on X

en.wikipedia.org/wiki/Sierpi%C5%84ski_space is the simplest nontrivial example of a non-hausdorff topology

lots of examples on finite sets

sometimes run into 'real world' examples when you quotient something hausdorff by something

shintuku

i need the intuitions for the zariski topology

and i have none of them

leslie townes

oh thats definitely its own little world

noballpointpen

shintuku, are you in school or self-studying?

leslie townes

that's arguably a good short summary of what an intro course in comm alg / alg geom is. it's developing intuitions for that topology

Ted Shifrin

21:49

Sheaves of germs of continuous functions give a very natural non-Hausdorff space. Try separating the (germ of the) function $f(x) = \begin{cases} 0, & x\le 0 \\ x, & x>0\end{cases}$ at $0$ from the (germ of the) $0$ function.

Actually, that leads to a wonderful paradox. The topology of a sheaf (projection being a local homeomorphism) tells you that the set of $x$ such that a section is $0$ at $x$ is an open set. But I always thought the zero set of a continuous function was a closed set. (We're talking about real-valued functions here.)

shintuku

22:19

i am not in the sheaf world yet

Cotton Headed Ninnymuggins

22:30

I’m rolling $5$, $6$-sided dice and I want to know the chances of getting at least $1$ repeated number. I thought about this in $2$ ways.

...the first is by thinking about the chances of not getting any repeats subtracted from 1: $1-5/6*4/6*3/6*2/6 = 90.7$%.

...the second is by counting the number of ways to get no repeats divided by number of rolls. There's $6$ ways to get no repeats (we have to roll five different numbers with one of the six missing). There's ${5+6-1\choose 6-1}$ possible rolls. $1-\frac{6}{{5+6-1\choose 6-1}} = 97.6$%

What went wrong?

Ted Shifrin

22:53

No clue how you’re counting there.

It looks like you’re doing bars and stripes, but I can’t imagine how/why.

Transcript for

$Mathematics$ Mathematics