Mathematics - 2021-09-19

Semiclassical

1:06 AM

@BalarkaSen neat. constraints are definitely a thing for physicists, and i know that dealing with them can get tricky

though i mostly know they become tricky when it comes to quantization, e.g., en.wikipedia.org/wiki/First_class_constraint

love_sodam

1:29 AM

is there some analogue of reduced l.e.s. for cohomology groups?

Ishwaran

1:42 AM

@robjohn @robjohn we can then take triangle POT and QOS and prove that the Line TS and PQ is parallel

And then say Angle RTS = RPQ and angle RST = RQP

Since it forms transversal

And prove the triangles are similar by AAA similarity criteria

@TedShifrin Your play with Angles gave a giant leap towards the solution for the problem :D

Ted Shifrin

2:27 AM

@Ishwaran definitely false. Look at robjohn’s picture!

Paramanand Singh

2:41 AM

@Ishwaran I had mentioned earlier that the line TS can't be parallel to PQ unless the base angles of original triangle are equal. It is best to draw a random triangle instead of a symmetric figure and then you can see why TS can't be parallel to PQ.

Also you should note that you are supposed to prove similarity of RPQ and RST. While proving similarity and congruence the order of letters is very important and that tell which vertex of one triangle corresponds to which vertex of another triangle. Here P corresponds to S and Q to T

So the angle at P matches angle at S. For parallel lines you should have angle at P equal to angle at T

robjohn

here is a new picture, with $O$ included.

Paramanand Singh

Hello @robjohn!

This time you have drawn straight lines instead of wiggly lines :)

Ted Shifrin

Selling out!

robjohn

That's right, I sold out.

Paramanand Singh

2:58 AM

Also liked your proof which uses similarity of two right triangles with a common acute angle at R @robjohn. It appears to be rather very simple and direct.

robjohn

Thanks. As Ted said earlier, the diagrams with isosceles triangles are just awful and lead to easy errors.

@ParamanandSingh are you sure?

Paramanand Singh

Cool I like this one. The ones with straight lines are, what to say, too textbook-ish

Ishwaran

4:19 AM

Oh yeah

Error in the similarity

@robjohn looks legit

love_sodam

4:37 AM

In hatcher, for manifold $M$, there is a orientable covering space $M_R\to M$. Is this space have a name?

leslie townes

i think it's just called the double cover, or something like that. orientation cover?

this is not my field

oh, for a topologist

Ted Shifrin

A geometer will do. The orientation double cover.

leslie townes

cool. i'm giving my answer partial credit.

but writing in red ink $\color{#C00}{\text{TRY HARDER}}$.

Ted Shifrin

4:54 AM

Extra credit for principal $\Bbb Z_2$ bundle :)

love_sodam

Is it some kind of bundle?

Ishwaran

5:06 AM

@ParamanandSingh Can you explain your proof little briefly ?

copper.hat

dinner at the kensington pub tonight.

leslie townes

i hope they boiled the food enough, to your liking

Ted Shifrin

Potatoes and boiled cabbage?

copper.hat

i had a burger, suitable cooked for my taste :-)

outdoors, beside the circle :-)

leslie townes

that must've been great. i love that area.

copper.hat

5:14 AM

nice to have dinner w friends. especially since i am alone on the house for the week

hope the break is recovering peacefully

leslie townes

she's very fussy. she wants to do everything, but can't. i think she's healing. we haven't given her any pain relief medication in a week, and no complaints.

copper.hat

that is progress. it is tough all around, esp when they are small.

leslie townes

i talked to my dad on the phone today and she wanted to get on. she spent about 10 minutes telling him about her toys, and her leg, and the adventures of her imaginary friends.

Ted Shifrin

I’m sure that was a fun conversation for grandpa.

leslie townes

he did enjoy it. they don't talk very often, so each time there is a lot of vocabulary development that has taken place since last time. as a narrative it was pretty hard to follow. she doesn't fully appreciate how people on the phone aren't in the room, so she changes subjects or emphasizes things by pointing at stuff.

copper.hat

5:22 AM

my granddad has little interest, my grandmother seemed to find every detail fascinating.

leslie townes

but she did respond to his questions. it wasn't a monologue. she doesn't ask questions yet.

copper.hat

it is good for both of them to interact.

the kensington air knocked me out :-), good night folks!

leslie townes

goodnight.

copper.hat

nothing to with the alcohol, of course

leslie townes

i was going to say.

copper.hat

5:25 AM

:-) gn

Ted Shifrin

Night!

Paramanand Singh

6:09 AM

@Ishwaran: since $P, Q, S, T $ lie on a circle the angles subtended by $PT$ on both $Q, S$ are equal. Hence angle $PQT$ is equal to angle $PST$

Subtract these angles from 90 degree to get angle $TPQ$ equal to $RST$

Ishwaran

Thank you @ParamanandSingh

Euler 2

7:34 AM

I was making a math expression parser
it was not working for negative numbers so what I did before parsing was to replace each minus sign of the negative numbers in the string with ~ (advice from SO)

now I dont want to publish it to github

I think I should modify the algorithm, rather than doing this weird hack

porridgemathematics

9:20 AM

@robjohn I think that would work if $du + i (pdx + dy) = gdz$ held globally in $\Omega$, but this only holds locally in $\Omega$, for not necessarily the same holomorphic $g$ at different points, unless I misunderstand your answer

Adil Mohammed

whats the name of the rule we use here? I know it has got to do something with leibniz and that's all

porridgemathematics

9:33 AM

fix some $x_0 \in \Omega$, then $f(z) = \int_{\gamma_{z_0 \rightarrow z}} du + i(^{\ast} du)$ (mod $2 \pi i$), and in a small disk around $x_0$, we may write $f(z) = f(x_0) + \int_{\gamma_{x_0 \rightarrow z}} du + i(^{\ast} du)$ (mod $2 \pi i$), which is equal to $f(x_0) + g(z) - g(x_0)$ (mod $2 \pi i$), then exponentiating we get something of the form $C exp (g(z) - g(x_0))$ valid in a disk around $x_0$ @robjohn im guessing this is what you were saying?

porridgemathematics

9:46 AM

uh actually sorry, it should technically be equal to $f(x_0) + G(z) - G(x_0)$ (mod $2 \pi i$), where $G' = g$ in a disk around $x_0$

so after exponentiating we get $C exp(G(z) - G(x_0))$ valid in a disk around $x_0$

Reacreational Mather

10:20 AM

0

Q: How to solve the problem on allegation by the assumption?

Question: " The average marks of the students in four sections A,B,C,D together is 60 %. The average marks of the students of A,B,C,D individually are 45 %,50 %,72% and 80 % respectively. If the average marks of the students of the sections A and B together is 48 % and that of the students of B n...

Somebody please explain the answer to the question in a more theoritical context.

The explanation seems to tough to comprehend.

My question is absolutely same.

robjohn

11:23 AM

@porridgemathematics you have that $\mathrm{d}f=g\,\mathrm{d}z$ locally. Then use monodromy to get what I said.

@porridgemathematics right

Ishwaran

12:29 PM

need help on 2nd sub division

Adil Mohammed

12:52 PM

@Ishwaran are you in 11th or 10th?

Ishwaran

10th

Adil Mohammed

This is a really good question, that i never saw while I was in 10th

Ishwaran

@AdilMohammed Actually this question is not from Examination point of view

Adil Mohammed

Makes sense why i never saw it lol

Ishwaran

Its something like a Brain teaser from the chapter trig

but it teases my brain a lot :P

Adil Mohammed

12:56 PM

The trick here is to notice there are 2 different angles, and the 3 formulas you learnt in grade 10 (sin2x+cos2x=1 and the rest) can only be used if they have the same angles

Ishwaran

@AdilMohammed yeah thats where Iam stuck

Adil Mohammed

So that rules out multiplying numerator and denominator by some fancy functions and then using the 3 rules , so that leaves us only with one choice ahead- convert Tanx into sinx/cosx and proceeding

Ishwaran

why not trying it by moving the whole denominator to RHS

Adil Mohammed

Oh you cant do that, if you did you would be saying LHS is already equal to RHS.... we are supposed to prove they LHS=RHS but we cant make the assumption they are already equal (at least in these sort of questions)

Ishwaran

then.. any other way ?

Adil Mohammed

1:05 PM

if you continued along that path you would be getting 1=1, if i am right... which technically means the LHS=RHS (I think) math teacher don't like if you do like

Ishwaran

of course... no maths teacher would

Adil Mohammed

@Ishwaran try turning Tan theta into sin theta/cos theta and take the LCM while adding the fractions

Ishwaran

so we must find an expression to multiply it

@AdilMohammed Let me try it

Adil Mohammed

@Ishwaran Oh no, i forgot to mention but the idea behind finding an expression to multiply the fraction is so that you can use the 3 formulas

(basically if you have a linear thing, you need to make it into a square thing before you use any of the 3 formulas, this is where a^2-b^2 formulas and all come into play)

but as we have already concluded... we cant use the 3 formulas... so there's no point finding the expression to multiply it

Ishwaran

@AdilMohammed then I will go on with this

Adil Mohammed

1:17 PM

yup did you get it?

Ishwaran

working on....

Ishwaran

1:30 PM

@AdilMohammed Still ending up in 1 = 1

Adil Mohammed

Yahhi did you take the denominator to the other side? Or anything to the other side, for that matter? the correct method is turning LHS into RHS and then telling "Tada, look I turned LHS into something that looks like the RHS, therefore the equation is true"

Ofcourse others also prefer "Hence proved" to conclude

Koro

Alternatively, $\tan x=\frac {\sin x}{\cos x}$ and $\cot x=\frac {\cos x}{\sin x}$ and then using $\cos (a+b)=\cos a \cos b-\sin a \sin b$ gives the result directly.

If U and V are finite dimensional vector spaces $T\in L(U,V), S\in L(V,W)$ then dim null ST$\le $ dim null S +dim null T. How do I prove this?

Adil Mohammed

@Koro i thought about using that formula, but in his grade its not yet introduced so i decided against using it, but I think you can skip it if I am right (i just did an eyeballing of the solution, i am not sure)

Koro

@AdilMohammed I think by class 10 (in India), the formulae for sin (a+b) and cos (a+b) are introduced. Right?

Adil Mohammed

Nope, not as far as i know

maybe in coaching foundation programs, but not in general

Ishwaran

1:40 PM

@Koro nope

Koro

Well ,by 2012 I had those in my syllabus.

Adil Mohammed

@Ishwaran wait ill write and show you the answer, but later... you shouldn't be getting 1=1 when doing proofs

@Koro whaaaaaaaaaaaaaat

Ishwaran

@AdilMohammed I proved phi = theta

Koro

ok, the syllabus has probably changed now.

Ishwaran

hence I guess I proved the sum

Koro

1:41 PM

anyways again that sum formula is not needed at all.

Adil Mohammed

@Ishwaran god lord, how did you get that

Koro

Did you simplify cot and tan in terms of sin and cos and do the cancellation?

Adil Mohammed

yes exactly, i may not have been great at communications but that's what you do

Ishwaran

@AdilMohammed Wait a min I share my work

hope my hand writing is ok enough to understand

Adil Mohammed

oh exactly what i suspected (i used to do this a lot too)

Ishwaran

1:55 PM

:D

Adil Mohammed

Do LHS separately, do not make it "interfered" with the RHS

Also dont outright write LHS=RHS like you have done in the beginning

Ishwaran

yup I need to rewrite it neatly

Koro

@Ishwaran If x=x then it does not imply that x=1 so your last line in image needs work.

Ishwaran

@Koro hmmm in trig functions too ?

coz for each value of theta the ratio is fixed ryt

Koro

Without shifting terms from LHS to RHS. Just write LHS in terms of sin and cos, and you'll see cancellation and eventually you'll get RHS.

Ishwaran

1:59 PM

@Koro Yes.....

Koro

@Ishwaran No. Note that $\sin \frac \pi 4=\sin \frac \pi 4$, but each side (being the same) equals $\frac 1{\sqrt 2}\ne 1$

$\cot \theta +\tan \phi=\frac {1}{\sin \theta \cos \phi} (\cos \theta \cos \phi+\sin \theta \sin \phi)$ so $$\frac{\cot\theta +\tan\phi}{\cot \phi+\tan \theta}=\frac{\sin \phi \cos \theta}{\sin \theta \cos \phi}\Big(\frac{\cos \theta \cos \phi+\sin \theta \sin \phi}{\cos \theta \cos \phi+\sin \theta \sin \phi}\Big)=\frac{\sin \phi} {\cos \phi}\frac{\cos \theta}{ \sin\theta}$$

porridgemathematics

@robjohn thanks :)

Ishwaran

@Koro I dont know why I am seeing evrything with syntax

Koro

@Ishwaran do you see mathematical expressions in chat?

If not, then you can install mathjax bookmark to see maths here in chat.

Ishwaran

how to install it ?

Koro

2:04 PM

If you're on laptop, see the top right section and get the bookmark from LaTeX in chat link.

Adil Mohammed

A "shortcut" to circumvent that is simply copying the equation and pasting it on the "Ask a question" section, or give answer section to get the latex "translated"

Joe Shmo

@TedShifrin Let $(\Sigma, \mathcal{A}, \mu)$ be a probability space. Let $(A_n)_n$ be a sequence in $\mathcal{A}$. Let $f: \Sigma \rightarrow \mathbb{B}$ be measurable such that $\int_{\Sigma} |\mathbb{1}_{A_n} - f| d\mu \rightarrow 0$ as $n \rightarrow \infty$. Prove there exists $A \in \mathcal{A}$ for which $f = \mathbb{1}_A$ a.s.

The claim is easy if $A_n \searrow A$, or $A_n \nearrow A$. But if, say, $A_{2n} = [-2, 1]$, and $A_{2n+1} = [-1, 2]$, then I don't even see a set $A$ that satisfies the requirement $\mathbb{P}(A_n \triangle A) < \epsilon$ for any $n$.

sorry, $f: \Sigma \rightarrow \mathbb{R}$.

in fact, the claim follows if $(A_n)_n$ converges, not even necessarily monotonically.

SMC

2:23 PM

I hope the question is not too much naive but there is any way to get this paper w/o paying? I need to consult it for my thesis but I'm not found on how pubblication works londmathsoc.onlinelibrary.wiley.com/doi/10.1112/plms/…

Thorgott

2:34 PM

academic.oup.com/plms/article-pdf/s3-11/1/469/4693043/…

isn't this it?

@JoeShmo it suffices to prove the set $f^{-1}(\mathbb{R}\setminus\{0,1\})$ has measure zero

SMC

It is but my uni is not listed as Institution

i guess no hope

Thorgott

oh, I'm stupid, I didn't realize I was logged in there...

have you tried scihub?

Joe Shmo

3:14 PM

@Thorgott I was planning on doing that, actually, but I still can't see an explicit set in the example I gave

Thorgott

if the set where $f$ takes values different from $0$ or $1$ has measure zero, there's an obvious candidate for an indicator function $1_A$ that is a.e. equal to $f$

Joe Shmo

suppose $f = \mathbb{1}_A$ for some $A$ in the example I gave. What is $A$?

$A_{2n} = [-2, 1], A_{2n+1} = [-1, 2]$

I'm not even trying to prove the claim at this point, just verify it.

porridgemathematics

3:47 PM

@JoeShmo $\int_{\Sigma} |\mathbf{1}_{A_n} - f| d \mu \rightarrow 0$ implies $\mathbf{1}_{A_n}$ converges to $f$ in measure, since you are working in a probability space (a finite measure space), this implies that there is a subsequence of the $\mathbf{1}_{A_{n_k}}$ converging to $f$ almost everywhere, clearly then, $f$ is equal to $\mathbf{1}_{\liminf_{k} A_{n_k}}$ almost everywhere

Joe Shmo

@porridgemathematics, so in the example, $\lim \inf A_{n_k} = [-1,1]$?

porridgemathematics

i dont know, because you haven't given me an $f$ in your example

Joe Shmo

if you take a subsequence involving both the evens and odds

porridgemathematics

the subsequence I took depends on $f$

Joe Shmo

ok

well the claim is that $f = \mathbb{1}_A$ a.e. for some $A$, no?

the choice is pretty much on the set $A$

porridgemathematics

3:53 PM

yes, and I claim that for some subsequence $A_{n_k}$ of your sets $A_n$, the candidate $A = \liminf_k A_{n_k}$ works.

Joe Shmo

ok, well we don't have much choice luckily

porridgemathematics

that subsequence itself is extracted depending on how the $\mathbf{1}_{A_n}$ converge in probability to $f$

Joe Shmo

either $\lim \inf = [-1,1]$, or $\lim \inf = [-2,1]$, or $\lim \inf = [-1,2]$

porridgemathematics

sure

Joe Shmo

none of these work

Thorgott

3:56 PM

cause there's no $f$ in your example

porridgemathematics

exactly

Joe Shmo

cool, thanks to both of you!

porridgemathematics

your question started with 'there is an $f$ to which the $\mathbf{1}_{A_n}$ converge in $L^1$, but with the sets you have, there is no candidate for $f$

np

Joe Shmo

gotcha. meaning the condition would never hold on these sets to begin with

porridgemathematics

thats right

love_sodam

4:02 PM

How can I show : If $M\setminus\{p\}$ is orientable then $M$ is orientable ?

My definition of orientability is defined using homology (local orientations)

dimension of $M$ is greater than 1

Thorgott

4:19 PM

pick a ball around $p$, every point in the ball except $p$ has a local orientation. argue these all come from the same orientation on the ball. then, if you equip $p$ with the local orientation induced by the orientation of that ball, they all glue together to an orientation of $M$.

the dimension assumption comes into play when you use the deleted ball is connected in dimensions >1. the claim is still true in dim 1, but for different reasons

Rithaniel

I've got a question: Given a big (uncountable) algebraically independent set $A\subseteq\mathbb{R}$, can you say anything about the outer measure of $A$? My instinct tells me that the two are independent topics, and the measure could be trivial or infinite or anywhere in between, but I don't legitimately know

Thorgott

4:45 PM

it can be 0, no clue if it can be positive

love_sodam

5:09 PM

@Thorgott You mean if $p\in B$ is an open ball then there is a unique generator $\mu_B\in H_n(M,M-B)$ such that $i^x_*(\mu_B) = \mu_x$ for each $x\in B\setminus\{p\}$? Here, $\mu_x$ is a local orientation and $i^x:(M,M-B)\to (M,M-x)$ is an inclusion map.

Thorgott

yes

sufficiently small open ball, we should say

love_sodam

5:25 PM

@Thorgott Is that can be proved easily? I saw some similar statement for section in Hatcher (Lemma 3.27) which is not very easy to prove at first sight.

Thorgott

totally different statements, this one is easy

the point is that we know a priori that $H_n(M,M-B)$ is cyclic and $i^x$ induces an iso on $H_n$ for each $x\in B$, so the only potential issue is consistency, but local consistency holds by definition and passing to global consistency is a connectedness argument

love_sodam

5:53 PM

@Thorgott the global consistency you mean if $B'$ is another open ball that contains $x$ then $i^x_*(\mu_{B'}) = i^x_*(\mu_B)$ ?

Thorgott

no

I mean that you can find $\mu_B$ such that $i_{\ast}^x(\mu_B)=\mu_x$ for all $x$

love_sodam

Oh I knew that as a local consistency

love_sodam

8:28 PM

@Thorgott seems there is a useful fact in Hatcher prop 3.25

Koro

8:46 PM

W is finite dimensional vector space. $T_1,T_2\in L(V, W)$. It is to be proven that null $T_1\subset $null $T_2$ iff there exists $S\in L(W,W)$ such that $T_2=ST_1$.

Ted Shifrin

Hi, @robjohn. You doing better?

Koro

Hi professor Ted and professor Robjohn

Ted Shifrin

So, "it is to be proven." Do so.

Koro

I proceeded like this:

For converse: Suppose there exists $S$ such that $T_2=ST_1$. Let $x\in $null $T_1$ then $T_1 x=0$ hence $ST_1(x)=S(0)=0$ (because every linear map maps 0 to 0) whence $T_2x=0$ so $x\in null T_2$. This shows null $T_1\subset $null $T_2$

Ted Shifrin

Yes, that wasn't the interesting direction.

Koro

8:54 PM

Now for the other direction: Since W is FDVS, let $w_1,...,w_k$ be a basis for W. I define $S:W\to W$ by $Sw_i=w_i$ (identity) and then I claim that $T_2=ST_1$. For any $x\in V$ we have $ST_1(x)= T_1(x)$, I am stuck here.

Thorgott

yeah, you can argue using the double cover as well

the point in both arguments is roughly the same

namely that you can extend the orientation (a.k.a section of the oriented double cover) through the puncture if it has enough codimension

Ted Shifrin

You're stuck because you're wrong.

leslie townes

taking S = I is unlikely to work unless T_2 = T_1.

Thorgott

^

Koro

yeah, $T_1$ does not have to be $T_2$. I figured.

Ted Shifrin

8:57 PM

For example, $T_2$ could be the zero map.

Koro

hmm, I was wrong in taking S=I

Ted Shifrin

You're too experienced to do something like that.

leslie townes

i dunno how to think about this without giving it away. S is gonna have to take stuff that T_1 acted on and move it around so it lines up with what T_2 does. maybe a starting point is to pick a basis for the range of T_1 (if it is nonzero; if it's zero that is an easy special case). and fiddle around with first defining S on that. bear in mind you will need to use the hypothesis about nullspaces somewhere.

Ted Shifrin

Shhhh, @leslie. Hush.

Koro

Finite-dimensional nature of W has to do something with it and the fact that a linear map is determined completely by its value on basis of domain.

leslie townes

8:59 PM

i'll hold my tongue.

Ted Shifrin

Go play with the munchkin, @leslie.

Koro

these two facts will show me the light.

Ted Shifrin

Finite-dimensionality isn't actually relevant.

leslie townes

sadly, she's napping. tired out after a visit to the duck pond.

Ted Shifrin

Oh, y'all went to the duck pond, cast and all? Cool.

leslie townes

9:00 PM

we put her in a stroller. hadn't done that in a while.

Ted Shifrin

Ah.

So, follow up on what you said, @Koro. What sort of basis for the domain should you take?

Koro

I'm trying to complete the solution professor Ted. I'll share that shortly.

Ted Shifrin

Oh, well.

Did the ducks remember her, @leslie, or did the cast cast a pall on the matter?

leslie townes

they kept their distance, i think because she couldn't just go and hang out with them without us pushing her there in a stroller. they did not seem hostile.

Ted Shifrin

Right, tolerating stroller + parents is too much for the ducks to contemplate.

I'd love to see a picture of her holding court with them :P

leslie townes

9:09 PM

i might be able to screencap a video.

koro's exercise does go through in the linear algebra sense for infinite dimensional spaces but can fail in normed linear spaces if you want "S" to be bounded.

Ted Shifrin

Well, I wasn't thinking in such detail. But if $T_i$ are bounded operators, won't $S$ then be bounded?

If one of them isn't, of course ...

@leslie Re video, for munchkin's privacy, best not to make this public.

Thorgott

hmm, I believe this should go through in the bounded context, but you need to quote Hahn-Banach

Koro

@TedShifrin It seems complicated. My experience in linear algebra so far has been in matrices/determinants mostly. It's probably the first time I'm reading linear maps (not necessarily from R^m to R^n, wherein I could write linear map in form of matrix times a column vector).

Thorgott

my favorite approach to this question would be through quotient spaces, but that's probably a bit too fancy

Ted Shifrin

Well, even if you use matrices, you still need to think about bases, @Koro. Go back and answer the question I asked. What's a smart way to choose a basis for the domain?

Too fancy by far, @Thor, but yes.

Why are you doing problems at random if they're not consistent with what you've learned?

Koro

9:18 PM

I read linear algebra for the first time few years back when I was at college. Now, I'm not at college. I started reading linear algebra and after going through vector spaces, now I'm at linear maps (chapter 3 in linear algebra done right).

Ted Shifrin

Oh, my least favorite book. Well, then he should have taught the relevant tools in the chapter, eh?

Focus on the question I've asked twice.

Koro

basis for range $T_1$ can be considered first that is $T_1v_1,...,T_1v_k$ for some k and then this can be extended to a basis of W. Is that correct?

Ted Shifrin

I asked for a basis for the domain, did I not?

An intelligent choice thereof.

Koro

domain of S, right?

Ted Shifrin

No.

Domain of $T_i$.

Koro

9:31 PM

I am confused there because $T_i$ can be infinite dimensional also :(

leslie townes

i do think there can be a problem in the bounded case. T_1 can be injective but also crunch norms of nonzero inputs arbitrarily close to zero (e.g. "multiplication by x" on C[0,1] with inputs supported in tiny neighborhoods of zero). taking T_2 = I, it seems that your S would potentially have to blow the norms of some vectors up by arbitrarily large factors.

this might go away if you require T_1 to have closed range.

Ted Shifrin

Ah, right @leslie.

Did you ever comment on that Laplacian question? I think your first objection was right, since the OP wrote down $\Delta u$ to start.

Thorgott

the real question is what the injective objects in the category of banach spaces are

Ted Shifrin

@Koro I believe Axler's book is all finite-dimensional, so don't worry about this.

And I assume he's talked about bases, dimensions, probably nullity-rank (although you do not need that here).

If not, the question is inappropriate.

leslie townes

on review of that comment, i think he was just being sloppy with notation. but i'm not sure that his operator is closed, because he didn't impose anything resembling boundary conditions on the domain and you usually need to do that. i couldn't figure out if he was in R^1, where it might not make a difference, or R^n.

Ted Shifrin

9:37 PM

I thought $\Bbb R^n$.

Thorgott

ok, I think I see what goes wrong

Ted Shifrin

Now I'm gonna go back and look.

Thorgott

we're fine if there is a bounded projection onto the range

leslie townes

generally if you say "let the domain be the set of things for which both sides of this formula make sense," where the formula involves a differential operator, you don't get even a closable operator without boundary conditions or regularity conditions. even for a one dimensional candidate for d/dx.

but that's at the level of vibes and not real math.

Ted Shifrin

Oh, right, closed on its domain.

He didn't specify where he's working, either.

It seems like standard Sobolev fiddling, but I can't remember enough to read the sloppy post.

Plus his "affirmation" and "actualization" must come from some other language. I have no idea what he's talking about.

Koro

9:44 PM

@TedShifrin Not worrying about V. range T1 is FDVS so let $Tv_1, Tv_2,..., Tv_k$ be a basis for range $T_1$. Given any $c_i$'s (scalars), if $\sum c_i v_i=0$ then $\sum c_i T(v_i)=0$ whence $c_i$'s are $0$ that is $v_i$'s are LI in V. Now, for any $v\in V$ there exist $d_i$'s such that $v=\sum d_i v_i$ which gives $T(v- \sum d_i v_i)=0$ that is $v-\sum d_i v_i \in null T_1$ so the smart way to write a v in V is to write it as sum of linear combination of $v_i$'s and an element from null T1.

no, there is still one mistake.

Ted Shifrin

Not worrying about $V$?

Then you're doing the problem wrong.

Where is the nullspace of $T_1$?

Koro

@TedShifrin Not worrying about V. range T1 is FDVS so let $Tv_1, Tv_2,..., Tv_k$ be a basis for range $T_1$. Given any $c_i$'s (scalars), if $\sum c_i v_i=0$ then $\sum c_i T(v_i)=0$ whence $c_i$'s are $0$ that is $v_i$'s are LI in V. Now, for any $v\in V$ there exist $d_i$'s such that $Tv=\sum d_i Tv_i$ which gives $T(v- \sum d_i v_i)=0$ that is $v-\sum d_i v_i \in null T_1$ so the smart way to write a v in V is to write it as sum of linear combination of $v_i$'s and an element from null T1.

Ted Shifrin

I'm giving up on giving hints. Start with a basis for the nullspace of $T_1$. Do NOT start with random bases.

Koro

@TedShifrin I mean its dimension can be infinite also.

Ted Shifrin

OK, since you keep ignoring me, I'm just going to quit. Someone else can work with you.

robjohn

10:11 PM

@TedShifrin Still feeling funny in the tummy, but food is not so offensive.

leslie townes

10:39 PM

koro, any thoughts on how to use your view of elements of V to define S?

Balarka Sen

doing homework in my head: let $E_{ij}$ denote the matrix whose $i, j$-th entry is $1$ and everything else is $0$. then $E_{ij}$ has all rows zero except the $i$-th row which is $e_j^t$, so $E_{ij} e_j = e_i$, and $E_{ij} e_k = 0$ otherwise.

$E_{ij} e_k = \delta_{kj} e_i$

leslie townes

is it really only in your head if you type it out? :)

Balarka Sen

i guess not

leslie townes

i absolutely have to write stuff like that out, even when i know it and the composition law is simple and 'easy to do' in one's head.

Balarka Sen

i can never matrix multiply in my head

leslie townes

10:45 PM

i do my part for the pen and pencil manufacturing industries.

Joe

Sorry to interrupt, but does anyone know how to pronounce this symbol? It appears after the word "collection".

Balarka Sen

$e_i^t \otimes e_j = e_i^t \otimes E_{kj} e_j = e_i^t E_{kj} \otimes e_j = (E_{jk} e_i)^t \otimes e_j = 0^t \otimes e_j = 0 \otimes 0$

Success!

leslie townes

joe, i'd just pronounce it C. it looks to be a script C.

Balarka Sen

What happened to you @robjohn? I hope you're doing better

Joe

@leslietownes: Thanks. What was Paul Halmos thinking using that symbol??

leslie townes

10:50 PM

in many books script letters are often used to denote things that are larger than sets and cannot be sets. that might be what is going on here. what is it a collection of?

if it's in his set theory book, that's probably what he's doing. otherwise, who knows. every author seems to like at least one form of alternate typesetting, whether script or fraktur or who knows what.

Joe

Yeah it's from his set theory book. I think he uses lower-case letters to denote sets, and then upper-case letters to denote sets which contains sets, and the weird letters to denote sets which contain sets which contain sets

leslie townes

sometimes called 'proper classes' although it would surprise me if he got into that in his book. en.wikipedia.org/wiki/Class_(set_theory)

copper.hat

Halmos was just a wild guy.

(Halmos ended up at Santa Clara University, I used to travel with a prof. from the math. department.)

(Since I am being serious for once, I should clarify that Halmos was not a wild guy in my sense of the word.)

leslie townes

11:14 PM

i never met him. he was about but in declining health during my time.

there's a nice photo of him with a cat out there. no mathematician who likes cats can be all bad.

Koro

@leslietownes yes, taking inspiration from what I want to prove ($S_2=ST_1$). Defining S:W$\to W$ as $S(T_1v_i)=T_2v_i$ (for all $1\le i \le k$) determines S completely and uniquely (from a theorem that says that a linear map F:X->Y, where X is FDVS is completely determined by its values on basis of X.).

Now for any v in V, there exist $c_i$’s (scalars) and v’ in null T_1 such that $v=c_1v_1+…+c_kv_k+v’$, which gives $T_1v= c_1 T_1v_1+…+ c_k T_1 v_k$ so $ST_1v= T_2(c_1v_1+…+c_kv_k)$. Since v’ is in null T1 which is a subset of null T2, it follows that v’ is in null T2. So we get $ST_1v = T_2 (c_1v_1+…+c_kv_k)+T_2 (v’)= T_2( c_1v_1+…+c_kv_k+ v’)= T_2 v$

1 hour ago, by Koro

@TedShifrin Not worrying about V. range T1 is FDVS so let $Tv_1, Tv_2,..., Tv_k$ be a basis for range $T_1$. Given any $c_i$'s (scalars), if $\sum c_i v_i=0$ then $\sum c_i T(v_i)=0$ whence $c_i$'s are $0$ that is $v_i$'s are LI in V. Now, for any $v\in V$ there exist $d_i$'s such that $Tv=\sum d_i Tv_i$ which gives $T(v- \sum d_i v_i)=0$ that is $v-\sum d_i v_i \in null T_1$ so the smart way to write a v in V is to write it as sum of linear combination of $v_i$'s and an element from null T1.

leslie townes

koro, note that your recipe seems to define S only on the range of T (note that it is only well defined at all because the T v_i are linearly independent). but you are definitely on the right track.

Koro

@leslietownes this is the central idea.

leslie townes

yes, indeed.

Koro

@leslietownes ah yes, you’re absolutely right. The working above can be fixed if I just remove the word S is uniquely determined. We know S exists (by the existence theorem for linear maps) and that’s it.

Also the representation of v in V as the sum $v=(c_1v_1+…+c_kv_k)+v’$ , where v’ is in null T1, is unique as sub space spanned by v_i’s and the subspace null T_1 have only 0 in common.

leslie townes

11:25 PM

is that clear from your construction, or is it something you proved?

Koro

I proved it. Let x be in span ( $v_1, v_2,…, v_k)\cap $ null T_1. Then $x=\sum c_iv_i$ which gives Tx=0 (as x is in null T_1 also) that is $\sum c_i Tv_i=0$ whence all ci are 0 as Tv_i ‘s are LI. So x=0.

Balarka Sen

Algebra people: Divisible abelian groups are projective + injective. Is the other direction true? Just for my own personal edification, not homework.

I don't care enough to think either just T/F would do lol

Koro

Leslie, I hope my answer to my question is complete now?

leslie townes

11:47 PM

i think so. if i were writing this up for future personal reference i might re-order the argument and make the various steps more distinguished from one another. things tend to spill out in fits and starts in a chat window.

news flash, my daughter has announced that after she turns into a cat she is going to turn into an elephant.

copper.hat

@Koro Here is another, more direct way (I think): One direction is immediate. For the other, let $v_1,..,v_p$ be a basis for $\operatorname{im} T_1$ and add $v_k$ as necessary to make a basis for $W$. Suppose $x_1,...x_p$ are such that $v_k = T_1 x_k$ (note that the $x_k$ are linearly independent), and define the linear $S v_k = T_2 x_k$ (and zero on the rest).

Now pick some $x$, we need to show that $T_2x = ST_1 x$. Note that for some $\alpha_k$ we have $T_1 x = \sum_k \alpha_k v_k = T_1(\sum_k \alpha_k x_k)$ and hence $x-\sum_k \alpha_k x_k \in \ker T_1 \subset \ker T_2$. Hence $T_2 x = T_2(\sum_k \alpha_k x_k) = \sum_k \alpha_k S v_k = \sum_k \alpha_k ST_1 x_k = ST_1 x$.

(I'm procrastinating from work :-).

leslie townes

copper, interesting piece in the chronicle this morning. an editor was cleaning out some old papers and found letters from the unabomber asking about places to go in south america (the editor had written a book on traveling there). before he began unabombing.

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$Mathematics$ Mathematics