Mathematics - 2021-11-16 (page 1 of 3)

copper.hat

00:12

when i was growing up you could hear the concord booms in the distance on a regular basis.

eigenvalue

Is anyone here skilled with eigenvectors/values and eigenspaces (in semi-Riemmanian manifolds)? I am stuck with a particular calculation (an example form a paper)... can anyone help? See below:

Given information: We have the type-changing metric $g=v(du)^{2}+(1+u+v)dudv+u(dv)^{2}$ and the signature change (where the metric becomes degenerate) happens at $D=\{(u,v)\in\mathbb{R}^{2}:1+2(u+v)+(u-v)^{2}=0\}$.

I am trying to calculate the eigenvalues and eigenvectors at some $(\hat{u},\hat{v})\in D$ in order to find out what the span is that defines the eigenspace. The eigenspace for the eigen…

Ted Shifrin

00:31

WTH does $\lambda_3$ mean?

eigenvalue

$\lambda_3$: this is how I denoted the 3th eigenvalue

Ted Shifrin

How does a $2\times 2$ matrix have $3$ eigenvalues?

leslie townes

yeah, there should only be 2, for any given (u,v). it's a quadratic in lambda.

i'm also not 100% sure of some of the algebra on the preceding line but have not checked.

eigenvalue

@TedShifrin Yes, you are correct. just double checked

Ted Shifrin

The algebra has to be wrong if you don’t find an eigenspace.

eigenvalue

00:40

both eigenvalues just yield the eigenvector zero, which shouldn't't be the case

Ted Shifrin

Proof that you are wrong.

eigenvalue

the eigenspace is supposed to be: span of $R_{(\hat{u},\hat{v})}=(1+\hat{u}+\hat{v})\partial_{u}-2\hat{v}\partial_{v}$ and I don't know how to get there. Maybe there is a different approach...

Ted Shifrin

Your algebra is wrong. We get a quadratic with leading coeff 1.

leslie townes

the iff in the above chain of things with the det can't be right. the last expression is zero irrespective of lambda if either u or v is zero, but the thing before it doesn't have to be.

that's an even simpler way to see it.

eigenvalue

@leslietownes could you point to the exact part (which section) your are referring to with "the iff in the above..."?

monoidaltransform

00:44

How does one show that the second homology of orientable compact 2 manifolds is just the integers using only singular theory?

leslie townes

i don't easily know how to link to equations within a display. the part: "eigenvalues for some u, v in D" followed by a line that begins with a det evaluation and then has an iff with a rewritten expression. that iff.

i assume that means without the classification of those things? or simplicial homology?

maybe there's something short of a complete classification that gives you some kind of decomposition of those things into connected sums, even if you don't know what the summands are. that might simplify the problem as long as you know singular homology is a homology theory.

i've forgotten all of this.

monoidaltransform

You can classify them, theyre just 2 spheres with handles attached

But not simplicial homology

leslie townes

so can mayer vietoris handle that? and induction? it's not clear to me if the classification is 'allowed' as part of this.

monoidaltransform

Both , MV and induction are allowed

leslie townes

poincare duality? :D

monoidaltransform

00:51

No thats not allowed lol

But out of curioisty

How do you show it using poincare duality

Calculating cohomology of compact orientable 2 manifolds is easier than homology?

Thorgott

the top homology of an orientable compact connected manifold is always $\mathbb{Z}$, generated by a fundamental class

monoidaltransform

Top homology?

Thorgott

in degree = dim(M)

monoidaltransform

And how do you relate singular homology to top homology?

And can we use poincare duality

Thorgott

top homology is literally just another term for the singular homology in the highest degree, which is the dimension of the manifold

Poincaré duality makes this statement trivial, but that's dishonest, because any of the usual approaches to Poincaré duality will require you to construct the fundamental classes first

Lukas Heger

00:58

poincare duality says that $H_2$ is isomorphic to $H^0$ which is $\Bbb Z$ because the manifold is path-connected

leslie townes

yeah. it turns into counting path components.

and is a 'backwards' way of looking at it. i don't understand that stuff very well anymore outside of theorems as black boxes.

Thorgott

yes, it's incredibly backwards

leslie townes

at least i was making the joke i thought i was making when i put in the ":D"

monoidaltransform

Cool. Thanks

Best source for this stuff is hatcher?

Thorgott

I mean, you can try very hard and prove Poincaré duality without talking about fundamental classes, but I can't think of a way to do this where "without talking about fundamental classes" really just means "sweeping them under the rug"

yes, read the section on local homology in Hatcher

he explains fundamental classes there

monoidaltransform

01:00

Cool. Thanks

Ted Shifrin

01:16

One can do a not-too-hard argument using differential forms and deRham cohomology, but I guess that's not quite "singular."

anakhro

Sanity check: they must be using some sort of convention on the second equality like saying that it is $\max(0,w_i - v_i)$

Thorgott

yeah, de Rham is a bit different in this regard

anakhro

Oh I think I misread this...

Thorgott

or maybe if you're thinking of an explicit argument like in Spivak, proving a volume element generates the top cohomology is somewhat similar to constructing a fundamental class

anakhro

YEAH I MISREAD IT.

Ted Shifrin

01:23

Yes, I’ve presented that proof in classes a lot of times.

Need the cohomology of spheres, however.

anakhro

@Thorgott spivak's diff geo books, or calculus on manifolds?

Ted Shifrin

Vol 1 diff geo

He does the Thom class and PD, too.

anakhro

I didn't know he did the Thom class.

I guess I mostly just looked up the stuff I needed when I skimmed through those.

leslie townes

01:54

i'm still proud of my topology joke. i learned most of what i have forgotten about that from spivak and bott/tu.

i was in a class sequence that made extensive use of hatcher but i always found it more useful as a kind of, i dunno, formula book or reference for statements of key theorems. i found it very hard to learn from. not a critique, just not my style.

some of those exercises are absolutely brutal for non topologists.

anakhro

Agreed.

I don't like Hatcher. Also his font is horrible.

leslie townes

i'd forgotten about the font. it looks like lucida. i think we've ranted about that before.

anakhro

You can never rant too much about fonts.

leslie townes

i do like that it is available for free in e-form and in print for cheap. having something text searchable is great. indexing is kind of a lost art.

anakhro

02:09

Indeed, that much I appreciate.

leslie townes

the paperback binding was also relatively good. not a given these days.

anakhro

*death stare @TedShifrin *

leslie townes

ted and i agree on almost everything relating to book formatting and printing.

oh, you meant that one of his books isn't free online. haha.

he needs to pay for his yacht somehow.

copper.hat

wonder what sort of yacht Ted has?

still no traction on my stupid penumbra problem.

dc3rd

I hope it's one that has him in a good mood because I need to ask about skew lines and systems of eqns

Ted Shifrin

02:16

I should add up my royalties from three books. I’m guessing less than 25K in over 25 years.

leslie townes

solid gold yacht money.

dc3rd

that's not lucrative at all.....especially for the efforts expended

anakhro

Not to mention all the flak you receive online for it.

dc3rd

might cover the maintenance for a day on Bezo's mega yacht

Ted Shifrin

Uh huh. And I did the typesetting pretty much.

Gotta leave in 10 minutes.

leslie townes

02:17

most math books amount to waaay less than minimum wage, if you thought of it that way. unless you somehow sell the next big calculus book.

dc3rd

flak?....who gives flak for the books?

copper.hat

It definitely does not reward the effort, but you are making a lot more than many authors in the same boat.

anakhro

@leslietownes Stewart is making millions

dc3rd

10 mins? well then......I'm doing the question asking to show through each point of the third skew line $_3$ there is a single line that intersects $l_1$ and $l_2$.

Ted Shifrin

Stewart is dead.

Intersect plane and line.

Probably not a necessary exercise for you. More algebra.

Chapter 6 far more important.

dc3rd

02:21

every exercise is necessary in terms of practice . So I should think of the plane created by $l_1$ and $l_2$ then....I think I kinda did that but not sure....I'll try this out.

they're also fun because of making sure to apply what I've been learning and not blindly.

anakhro

@TedShifrin Have you seen his house?

Ted Shifrin

Yes.

No, skew lines do not form a plane, dc3rd.

copper.hat

who is stewart?

dc3rd

I was just thinking that too Ted

Ted Shifrin

You need a point and a line.

leslie townes

02:25

copper, james stewart. author of a very popular calculus text, probably slightly after your time.

berkeley used it in 1A/1B/53.

copper.hat

thx

anakhro

@leslietownes Correction: a very popular doorstop

Ted Shifrin

Much copying from Edwards & Penney, but not proved in court.

copper.hat

i suspect skew lines are in my evening entertainment

leslie townes

haha. i hate long books as much as the next person but for what it is, it's pretty good. i've heard the rumors about edwards and penney.

dc3rd

02:29

So let me explain my idea I attempted: I drew a sketch and see I'll have to take a point on $l_3$ and to have a line I need a direction vector so that is unknown.

...Oh....planes are going to be created by my intersecting my line $l_s$ with each of $l_1$ and $l_2$

anakhro

Leslie, no :((

copper.hat

this is why we need vr headsets to do 3d sketchs

dc3rd

your point is valid now copper....whiteboards may not suffice

leslie townes

anak, i'm sorry. if it makes you feel better, i no longer have my copy of stewart.

anakhro

The future of high school math classrooms is filled with vr headsets

copper.hat

02:31

:-). i use string and foam core sometimes when stuck...

last night in fact, with my penumbra problem.

anakhro

Invigilators will have to confiscate vr headsets from first year calculus students who neglect to read the accepted devices lists for exams.

copper.hat

vr would make electronic communications much more interesting

trying to be well behaved here.

Math

02:54

Can one find a bijection between the set of all functions from N->{0,1,2,3,4,5,6,7,8,9} and the set of all functions from N->{0,1} without using Cantor Schroeder Bernstein ?

dc3rd

Instead of being vague this is what I was working with.

the question asks to show that through each point of $l_3$ there is a single line that intersects both $_1$ and $l_2$

$$l_1: \mathbf{x} = s(1,0,0)\\ l_2: \mathbf{x} = (0,1,0) + t(1,0,1) \\ l_3: \mathbf{x} = (1,2,2) + u(1,0,2)$$

I defined $$l_s: \mathbf{x} = l_3 + w\mathbf{v}$$

where $w$ is just a scalar and $\mathbf{v}$ is my direction vector and $l_3$ will be fixed given any scalar $u$ I have chosen.

So the line $l_s$ would intersect $l_1$ and $l_2$ if:

leslie townes

some of that notation (e.g. the 'definition' of l_s and equating what appear to be names of lines) is a little opaque to me.

dc3rd

which part needs clarification?

oh...I see....So I was using $l_3$ in $l_s$ as my "point" from the line $l_3$ from which my line $l_s$ begins............I see how this could come as vague

leslie townes

03:10

l3 is described as "x = (1,2,2) + u(1,0,2)" but is elsewhere described as something that "will be fixed given any scalar u i have chosen." i can't tell if l1, l2, l3 are point sets (lines) or unknown points on known lines, with the points to be found later.

dc3rd

I interpret it as the second because for any arbitrary scalar I will have a point on the line.

leslie townes

i do like the idea of fixing a point on l3 for starters, that's just fixing a value of s, then algebraically characterizing what it would mean for a line through that point to also intersect l1 and l2. this is where your unknown direction vector comes in and i guess a new scalar. i'm already noticing that scaling the direction vector and the scalar in opposite directions will produce many symbolic solutions that correspond to the same line, if it exists.

i think it's just a question of clarity in notation. you seem to have a sense of what the goal is, but i cringe just seeing x = three different things. there's no need to call any of them x and that's one less letter then to stare at. :)

dc3rd

understandable

leslie townes

i'm not sure that a more geometric approach might not clarify this more than just setting up systems of equations and solving. knowing ted, it wouldn't surprise me.

dc3rd

Actually I got this whole idea from drawing a sketch.

leslie townes

03:15

but the program you have loosely outlined is an OK one. just maybe don't write the l's and x's and just work in terms of your unknown variables that you might actually want to be solving for. you won't be solving for l2, but for some point on l2. etc.

and yes if l2 is just my name for a parametrization in terms of another variable, in some sense it's all the same thing. i just think symbolically it is clearer to reduce the letters involved.

math, certainly, although that is certainly a quick way. i don't know what you'd want to "do" with the bijection.

Math

@leslietownes I just want a constructive bijection

dc3rd

I see what you're saying about solving for a point on each of these lines leslie. That means that the scalars for each of the lines is important because they are dependent on each other since my line $l_s$ will be intersecting them at some particular point

leslie townes

04:16

dc3rd WA can do it although obviously it sheds little light on 'why' this is happening. tinyurl.com/fs8vvaab

i've noticed that WA is pretty bad with vectors and matrices. every time i input them, even using what i think is correct mathematica, i seem to get it wrong the first time.

it seems to struggle with stuff of different types, mixing scalar multiplication with vectors, or transposing a matrix. probably because i'm neglecting to tell it what my variables are.

dc3rd

you right about the little light being shed on things. I put that question away for a bit to work on another. I know I got the idea right, but it is framing it in the right way to use the "tools" that I have at my disposal

leslie townes

it didn't like t and u being outside of the vectors instead of inside them, but it was fine with k. go figure.

dc3rd

because I thought through the question more and I'm seeing it is completely dependent on my scalars. but I'm concerned about the direction vector of the line.

leslie townes

oh. in the way i posed it to WA, i just took an arbitrary pair of points on both lines, parametrized the line connecting them, and asked it to solve for what parameters would make that thing lie on the third line. the formulas turn out particularly simple although that might be because the numbers are 'choice.'

i actually may have scrambled the roles of l1, l2, l3 there. i hope the idea is clear.

dc3rd

Yes when trying to simplify my question to try and solve it more I was able to see what you were talking about, but I was just wondering how the "direction vector" is captured in this?..............actually it just kinda hit me.........if I have points on each of those lines I can get my direction vectors.........

leslie townes

04:26

tinyurl.com/4a5bh4a5 tracks the logic of the problem a little better.

when you get old you randomly permute lines.

dc3rd

is the reason you used $(1-k)$ because it gave you a way to capture the intersecting point on the two lines?

leslie townes

yeah, no real need to consider the world of all direction vectors or have symbols for that, just the ones of lines that are gonna intersect l1 and l2. saves a variable. :)

oh that's just the formula that one way of parametrizing the line between p and q is kp + (1-k)q, k ranging over the reals.

then that hits the line, if at all, if the system is solvable for u. which it turns out to be, uniquely, no matter what u is.

dc3rd

So I see you didn't solve for $u$, is that because we are already making our other scalars dependent on what our value of $u$ will be?

leslie townes

put another way, a direction vector from q to p is p-q, and the line through p and q goes through, for example, q. so {k(p-q) + q: k in R} will be the line.

yeah, i'm hoping that no matter what u is (no matter what point on the third line i've got), i can solve for the other variables. WA is pretty good at this, if things don't get too nonlinear.

dc3rd

I'm going to have to play around with things more just to get more comfortable with ideas more deeply...............man every time progress is made I'm reminded of how far I still have to go...😭😭...gonna be 50 still figuring things out....

leslie townes

04:34

so we definitely hope that u will be 'free' variable in solving that system, and not something determined by the relationship in some way. which would mean that maybe for some u's we have a solution and for other u's we don't.

Math

@leslietownes can you give me a bijection without using CSB?

dc3rd

always an element of uncertainty in math....always,,,,

leslie townes

math: i don't see a simple formula for one. this is why i was asking if you had a particular goal in mind. i think you'd have to be clever to get one without an inductive construction.

inductive constructions are within the realm of what i regard as constructions, even CSB maybe even falls under that rubric. i'm guessing you want something really 'clean' and i don't know of anything.

easier if you replace {0,...9} with {0,...,7} of course. :)

Math

Hmm one of my ideas was to get a explicit bijection between N->{0,1} and N->{0,1,2} then to N->{0,1,2,3} and so on but I can’t still see one

leslie townes

yeah, that's a thought. 'just add one' and induct. i don't see a clean way. combinatorics people are often really good at stuff like this, but i am not among such people very often.

Math

04:49

Why do you say it’s easier if 9 is replaced with 7?

leslie townes

then you concatenate bit strings. represent 0-7 as three-bit sequences and send f(n) to the 0-1 sequence given by concatenating the bit strings f(1), f(2), f(3), ... and thinking of that as a function on N (value at n = the nth bit in that concatenated string). this is a bijection, and maybe even a not too bad formula for the nth entry as a function of n.

you can do this kind of idea with {0,...9} too except you need four bits per number, and you only get an injection and not a surjection because you're not 'using' the bit strings corresponding to 10-15.

there might be some clever way to 'compress down' so you fill in the gaps but i couldn't think of one.

so the issue why this idea won't work simply is that 10 is not a power of 2, while 8 is.

this is not that different from your idea of 'just add one.' we can do 8 elements in the codomain pretty easily and just want to add two more.

Math

So the sequence 2,2,2,2,…. Will map to 1,0,1,0,1,0,1,0,…. Is that what your saying ? Then where will 1,0,1,0,1,0….. map to ?

leslie townes

oh in the 8 case? i meant thinking of all of the 0-7 as three bit strings. all of our new symbols for 0-7 have the same length. so 2 is 010.

i should have mentioned that.

without that it's not clear and probably not true that you even get an injection. if the first value is 1 and the second is 11, might concatenate the same as the function whose first value is 11 and the second is 1. can't have that.

Math

05:20

Nice. I still feel that it’s possible to get a bijection between the set of all functions N->n and N->n+1

leslie townes

yeah, definitely. i was imposing the additional constraint that the bijection be 'nice' which is subjective. no funny business that would amount to proving CSB in a particular case. vague goal.

copper.hat

my penumbra problem was actually a straightforward computation after all.

i am thinking of adding and answering a question nonetheless.

leslie townes

05:37

straightforward or not, cool that you got it.

copper.hat

just getting slower, unfortunately.

robjohn

06:28

@Math Isn't that essentially the same as mapping reals in base $n$ to base $n+1$?

copper.hat

there, i asked a convex psq.

robjohn

@copper.hat were you going to post the answer you got?

leslie townes

give or take a little with the reals. handling that distinct sequences can be the same. i was thinking it might be easier to work directly with "digit" strings.

copper.hat

@robjohn yup :-). i got distracted while fixing an answer to another question.

is that kosher?

robjohn

@copper.hat Yes. There is even a checkbox in the question asking form that will allow you to post both at the same time.

copper.hat

06:36

i am happy if someone else answers, i'm not in a rush.

thanks!

robjohn

@copper.hat If you post your own answer, it reduces the PSQ-ness, if you're worried about that.

It’s OK to Ask and Answer Your Own Questions

Jeff Atwood on July 01, 2011

The FAQ has contained one key bit of advice from the very beginning: It’s also perfectly fine to ask and answer your own question, as long as you pretend you’re on Jeopardy! — phrase it in the form of a question. So … if you have a question that you already know the answer to…

copper.hat

@robjohn much appreciated!

leslie townes

e.g. given a binary sequence, mentally insert a carriage return and line feed after each 1, and then after any string of b consecutive zeros. the sequences of lengths of those are sequences in {1,....,b} and might be all of them. the inverse mapping would be backspacing. no identification of distinct sequences. i haven't thought that through.

copper now i'm flashing back to some problem in convexity i had on my homework in grad school. might dig it up and imitate you.

Math

@robjohn Yes but we are looking bijection between them not using CSB theorem

robjohn

06:53

was I invoking CSB?

copper.hat

i don't like that you can no longer see when the last time someone visited the site. i know it has been discussed on meta, i'm just whining about it. maybe it is some gdpr thing.

robjohn

@copper.hat they said it had to do with making the profiles more responsive. I think it also has to do with some privacy issue.

Math

@robjohn your idea is too give a bijection between [0,1) and the base n sequences and hence show that they are also bijective right?

N\ge 2

leslie townes

to the extent it is motivated by privacy concerns it's the kind of thing that strikes me as an 'abundance of caution.' i don't know gdpr or eu variants but there is a lot of received wisdom about what people couldn't or shouldn't do under US law that has no basis in fact and often does not even originate in the legal department.

copper.hat

@robjohn I imagine it is the European GDPR regulations?

leslie townes

06:57

if it's just 'uh, on balance, consensus was no reason to keep this,' i sort of understand it more. i certainly don't mind that people can't see my info and if it were togglable i would toggle it off.

i do get how it can interfere with response that might be scaled in proportion to whether someone is likely to read it soon.

robjohn

It used to simply bring up a "calendar" that one could scroll through to see when people had been here. I don't think it was really a response sink.

leslie townes

people who work in academia seem to be doing fairly well with not being judged for activity under their own name. when i was in grad school / teaching this was a concern of mine. i thought it was a really bad look that people i knew to be fairly mediocre in the classroom were publicly spending a lot of time with their grad school friends on other stuff.

but that seems to be less of a concern than it used to be, maybe in some places it is even encouraged.

Math

@robjohn is that what your saying right?

the fact that some numbers have two representations can easily be avoided

leslie townes

a lot of people in media are basically expected to spend time on twitter. this too is an inversion of what i would have thought a good idea, but it seems to be fine. outside of when people get cyberbullied or harassed or whatever.

PM 2Ring

It wasn't really motivated by GDPR, but they had received a few complaints over the years that it was a bit creepy, and enabled stalker behaviour. They recently restored "Last seen", but it now has a granularity of a week.

leslie townes

07:02

stalking is sadly a genuine concern and often entirely outside the scope of a legal regime.

copper.hat

apparently twitter (or rather its algos) gives its biggest boost to right wing legislators.

my offspring would not tell me a bf's name unless i promised not to stalk.

leslie townes

and chatterboxes, yeah. i don't know if it's deliberate. i understand why if you measure 'engagement' all of that stuff is going to outperform, like, some mild video that is not going to rile people up. or, god forbid, a written article.

PM 2Ring

Indeed! And there have been incidents of stalking on the network, and members being pestered or harassed off the network by other members.

copper.hat

i harass myself from time to time.

i had a serial downvoter once.

i had my suspicions and the last time visited helped with the correlations.

PM 2Ring

One of our regulars in the Python room (who teaches coding at a college) actually received death threats over stuff that happened in connection to the Monica debacle. He ended up deleting his account. And he's not the type that gets scared easily.

copper.hat

07:09

wow.

PM 2Ring

So I'm sympathetic to the idea that Last Seen was revealing a bit too much info. I was hoping they'd quantize it to a day, but a week's better than nothing, I guess.

leslie townes

notices AMS had what i believe to be at least a second article touting mathoverflow as a place to talk about research. it helpfully did not directly encourage the use of real names. i vaguely recall some of the earliest promotion around similar sites was less measured about this. i was once 'called out' for using a pseudonym although i was not trolling under it. it was just 'OK, creep anonymous internet person who isn't using a real name.'

struck me as very tonedeaf. the idea that everyone is on better behavior if we're all real all the time has been, i think conclusively disproved by how many use facebook.

copper.hat

yup, the all or nothing approach to personal information can't end well

leslie townes

to say nothing of the zillions of studies on how inferences about a name on a piece of paper can make a difference in hiring decisions or how people are otherwise treated.

PM 2Ring

@copper.hat He's a Marine who fought in Iraq. He got some permanent injuries when the APC he was in got hit by a bomb, but at least he survived, unlike a few of his buddies.

copper.hat

07:13

that is disturbing.

anonymous threats are hard to deal with.

PM 2Ring

Back in the ancient days of FidoNet, you had to use your real name (in most places), although of course it was just an honour system, nobody did any checking. But you just couldn't use an obvious pseudonym.

leslie townes

a friend of mine was being stalked and didn't know it. the cops just showed up at her house one day because they found all of this stuff involving her while arresting a guy for some other creepy stuff he was doing. didn't even know the person, he just lived nearby.

i used my real name on local bbs'es and things connected to them. it didn't hurt that i personally knew most of the people i was interacting with and the world was just a slightly smaller place.

copper.hat

searchpeoplefree.com

usa only

leslie townes

haha. those kinds of sites are usually a few years behind me, but still surprisingly good. i don't submit change of address forms with USPS and get almost no junk snail mail. i kept a lot of bills i was paying out of my name for a long time. they still scrape it all up eventually.

copper.hat

i used to pay cash to reduce the trail, but the amount of cash i had to carry got silly

more an inconvenience than a security thing

leslie townes

07:21

i last used a site like that for applying to the bar. i couldn't remember the addresses of many of the places i had lived, or landlords names, or anything, and the form required something like a 10 year lookback.

two clicks later, oh yeah, i remember living there.

copper.hat

i had forgotten i lived in albany in 85, such sites reminded me

PM 2Ring

The online world was certainly a lot smaller back then. One day, I bumped into an acquaintance (who I only knew by first name) at a restaurant. Somehow, the conversation turned to computers, and it turned out that we both had Amigas. And that we both used the same BBS. And had been chatting online with each other for several months without realising that we'd known each other IRL for several years. :)

copper.hat

that's funny

i think i have only met an online acquaintance once, and that was from mse!

in caffe med on telegraph

PM 2Ring

In the early 2000s I participated in a science forum hosted by the ABC (our BBC). It was associated with a science popularizer who had regular radio & TV appearances. The core population was around 100, and we often organised group meet-ups. It was interesting to meet people in the flesh when you already feel like you know them fairly well. And then you learn all sorts of interesting tidbits that they would never mention online. ;)

leslie townes

i had something kind of like that. my pediatrician for most of my early childhood was also a computer hobbyist. i didn't know this. later "met" him on a local bbs in the context of swapping software and talking about nerd stuff. we found out who we really were at a pc users group meeting.

i remember how funny and awkward it was. "wait, uh, were you a pediatrician?" "yes! do you still like dinosaurs?"

PM 2Ring

07:31

Nice. :)

leslie townes

i don't still like dinosaurs, and don't know what i really saw in them. my daughter likes them a lot. i like birds, which are basically dinosaurs.

copper.hat

it bugs me a little when someone answers a question an hour after you with essentially the same answer.

PM 2Ring

They are! They've evolved a bit in the last 65 million years, but they still know they're dinosaurs. ;) So we shouldn't be surprised when they do stuff that seems vicious to human standards.

copper.hat

i think humans need no assistance on that front...

PM 2Ring

Some 1 rep newbies will even post an almost exact duplicate of an existing answer on the same page. I guess they're hoping to score upvotes from their classmates.

copper.hat

07:40

are turtles considered dinosaurs?

PM 2Ring

Sure. Humans are far and away the most dangerous & vicious creature on the planet. But some of us try to behave in a civilised fashion...

copper.hat

i admit i add ice to my white wine from time to time.

utterly barbaric

leslie townes

the most likely interpretation is people get so excited to independently come up with an answer that they don't read the ones that are already there. there isn't time for that.

PM 2Ring

I think turtles predate dinosaurs. Or maybe they're related to pleiosaurs, or something... I'd have to check.

leslie townes

copper i once asked a houseguest not to do that in front of me. apparently completely normal in some parts of the world.

copper.hat

07:42

i only do it when no one is looking.

but only once did i knowingly consume reconstituted mashed potato which has got to be the vilest sin around.

leslie townes

the california condor is definitely a dinosaur.

PM 2Ring

My mum's mum would put sugar in white wine.

copper.hat

apparently andean condors can reproduce asexually

eeewwww

i saw some california condors at pinnacles

leslie townes

a california condor recently did that. must be a dinosaur thing.

copper.hat

seems pointless

all the pain with none of the fun

when is mother's day?

PM 2Ring

07:44

Some lizards have reverted to asexual reproduction. Eg, some whiptail species.

copper.hat

9 months after father's day

leslie townes

sigh.

copper.hat

i'm working on my social sensitivity

i only took the training once

apparently a guy reported me for lack of sensitivity

PM 2Ring

New Mexico whiptail

The New Mexico whiptail (Aspidoscelis neomexicanus) is a female-only species of lizard found in the southwestern United States in New Mexico and Arizona, and in northern Mexico in Chihuahua. It is the official state reptile of New Mexico. It is one of many lizard species known to be parthenogenetic. Individuals of the species can be created either through the hybridization of the little striped whiptail (A. inornatus) and the western whiptail (A. tigris), or through the parthenogenetic reproduction of an adult New Mexico whiptail. The hybridization of these species prevents healthy males from forming...

copper.hat

they are all called mary

PM 2Ring

07:49

I had several primary school teachers called Mary...

One of my cousins became a nun, but I haven't heard from her in decades.

copper.hat

surprisingly there are no marys in either branch of my family

leslie townes

she holy ghosted you

i was thinking, i don't think we have any either.

copper.hat

spirit, did vatican 2 mean anything to you

PM 2Ring

She was the white sheep of the family.

copper.hat

one uncle was a parish priest in livinston just outside edinburgh

his catchment was the area where the panam flight 103 blew up

leslie townes

07:53

oof.

copper.hat

awful event

leslie townes

i just heard all of this noise from downstairs, crashing around. i thought the cat had found a rodent. no, she'd found some of my daughter's crayons and was chasing them around.

copper.hat

whoa, someone voted my penumbra answer up and IT WAS NOT ME

i am not sure i could survive in the same house as a cat

for long

dc3rd

So apparently these two matrices are symmetric.......so what does symmetric mean in the complex sense? isn't it $A = A*$?...........

copper.hat

it is not hermitian though

it needs to spend more time alone

dc3rd

07:56

Lol.....that is stated in the next line

copper.hat

hermitian means $A = A^*$

dc3rd

I see what you mean.......but I guess since I've never worked with complex matrices I guess I never thought about what it means to be symmetric in the complex space

copper.hat

symmetric means the same irirregardless of what field you are standing in

generally the important property is being hermitian which happens to be transpose with real matrices.

dc3rd

then I need to go look at the definition of symmetric again. Because the only one I have in the book up to this point was $A = A^t$

PM 2Ring

It should be spelled "Hermitean".

copper.hat

07:59

that is what symmetric means

i prefer the idea of a matrix up in some cave somewhere making humming sounds

irregardless

wowser, i have 2 votes for my answer and 1 for the question. must be someone here?

Leslie?

dc3rd

but if that is what symmetric means then what is being said in my picture? those aren't equal....

yes "conjugate transpose"

leslie townes

copper i hit the answer but not the question.

copper.hat

complex and symmetric is not complex and hermitEAN

leslie townes

it's not saying that A and A* are equal, only that A is its transpose (and A* too)

you can operator-ify some of this stuff if you hate matrices but i will bow out of that.

copper.hat

thanks for the upvote. i'll be getting mse swag soon

$x^* A y = (A^* x)^* y$

i'm inexplicably happy with my little question. i must have spent 4 hours on the stupid thing.

PM 2Ring

08:07

You were talking earlier about a 3D whiteboard. You can do simple interactive 3D sketches in Sage fairly easily. Sure, it's not VR, but it can be better than a plain 2D diagram. I did one yesterday for this Astronomy answer: astronomy.stackexchange.com/a/47503/16685

copper.hat

that answer is the tops

PM 2Ring

:D

copper.hat

love this story en.wikipedia.org/wiki/Gimbal_lock#On_Apollo_11

that is why i visited broom bridge last summer.

PM 2Ring

I spent about 4 hours today on a related problem: finding where the Moon's orbital plane crosses the equatorial plane. Then I finally realised that it's just a simple cross product.

copper.hat

it bugs me that such problems often end up in a trigonometric mess.

never understand the point of the bonus 100 points when you join another exchange

dc3rd

08:12

@copper.hat So a take on Ted's favorite formula...

copper.hat

i believe so. i am not sure why its his favourite...

it is the essence of duality

PM 2Ring

I get dizzy trying to visualise this stuff, after the 3rd or 4th rotation. :) Making those anims helps a bit, though. At least, it gives me a way to test stuff to make sure I haven't screwed something up.

copper.hat

@PM2Ring i love the xkcd comment

ok folks, my bedtime. i need my ugly sleep.

dc3rd

more groking is going to be needed on my part...........

copper.hat

don't read too much into it.

dc3rd

08:16

i'm off too...peace.

copper.hat

gn & gl

dc3rd

not that specifically but the whole body of work

copper.hat

you need to find a little project of interest that requires you to write some matlab code or similar.

or octave, the open source version

Math

@robjohn I invited you to another room to discuss the comment[just making sure you receive my invite]

PM 2Ring

The Moon's orbit is really tricky, mostly because it's so big relative to Earth, and it's rotating in a plane close to the ecliptic rather than the equator. From en.wikipedia.org/wiki/Lunar_theory "The number of terms needed to express the Moon's position with the accuracy sought at the beginning of the twentieth century was over 1400"

In a way, that had a beneficial effect on the development of science because the astronomers 2000 years ago could fairly easily tell that their model of lunar motion was rubbish. :)

@copper.hat Sweet dreams.

Xander Henderson

13:24

The president of my institution sent out an email yesterday. In a nutshell: "Vaccine mandates are coming. You are either going to have to get vaccinated or agree to weekly testing (probably at your own expense). If this is something over which you would resign, please let your supervisor know by the first of December so that we can plan for the spring semester."

Unspoken: Don't let the door hit you on the way out.

Honestly, I am rather shocked by this, given how rural we are, and how popular Trump remains. Yay for the new president, sticking his neck out!

ypercubeᵀᴹ

13:38

@copper.hat would you prefer if you had just 1 point when joining a new SE site?

with all that comes with that (eg not being able to comment, ...)

Xander Henderson

@copper.hat I think that the idea is that the first 100 XP is awarded for simply learning how the SE platform works. Once you have figured that out, you can be trusted to operate on other sites in the network (i.e. you have proved that you are not a spambot).

So that 100 XP bonus is there to give you basic rights everywhere (upvote, comment, etc).

Neel Rayal

14:09

On a set of integers, a*b = a. Identify all the properties of this operation.

This satisfies closure, but I am confused if this satisfies associative or not.

Xander Henderson

What makes you think that your work is incorrect?

Neel Rayal

don't know, but had a gut feeling that this is wrong. Sorry I know this sounds non-sensical

Xander Henderson

Well, sometimes you have to trust your gut. Other times, you should ignore it.

Neel Rayal

@XanderHenderson Can you give a hint towards finding its identity element

Xander Henderson

In this case, I don't like your presentation, but the basic idea seems to be there: if $a, b, c \in \mathbb{Z}$, then $(a\star b)\star c = a\star c = a$ and $a\star (b\star c) = a\star b = a$. Thus associativity holds.

What makes you think that there is an identity element?

Prithu biswas

14:19

((a+b)+c) = (a+c) = a = (a+b) = (a+(b+c))

Neel Rayal

It's a practice problem. I have to find all the properties that this structure satisfies. I don't know to prove/ disprove that identity exists or not

Xander Henderson

Well, what about if there is an identity, what property must it satisfy?

Neel Rayal

a * e = a

Xander Henderson

@NeelRayal Is that it?

Neel Rayal

I guess..

Oh wait

a * e = e * a = a

Xander Henderson

14:21

@NeelRayal Exactly. So, can that happen?

Neel Rayal

No because a * e = a and e * a = e

Is this right approch?

Xander Henderson

And if $a \ne e$, then you are in trouble, n'est-ce pas?

Prithu biswas

∃e ∈ G ∀x ∈ G (e.x = x.e = x)

@XanderHenderson Is this the correct definition for identity?

Neel Rayal

@XanderHenderson Is that a doesn't equal to e?

Xander Henderson

@NeelRayal Yes.

tinyurl.com/cfqcvpc

@Prithubiswas Sure.

Neel Rayal

14:29

@XanderHenderson okay so if $a≠e$ identity element doesn't exist. So this is a monoid, rt?

Xander Henderson

@NeelRayal I can never keep track of what is a monoid and what is a semigroup and whatnot. I'm an analyst... I don't really care about these algebraic structures which don't play a role in my work.

Why don't you write down the definition of a monoid, and check which boxes are ticked by this structure?

That being said, I feel like monoids have identities?

Neel Rayal

Argh! closure and associative is satisfied, but identity isn't. So it's a semi-group.

I was so eager to reply fast that I jumbled the definitions.

Xander Henderson

@NeelRayal I mean, I have a phd and can't be arsed to remember the difference between a semigroup and a monoid.

Neel Rayal

I got that, the fault was on my side. Thank you for clearing the doubts.

Xander Henderson

@NeelRayal Nonono... what I'm saying is that the fine grained distinctions between these structures is not something that most folk really care about, unless they go into a field where they do matter. You should know the definitions for whatever class you are in, and should know that the definitions exist (so that you can look them up later), but it isn't a big deal in the long term if you can't pin down the definitions.

Prithu biswas

14:43

@XanderHenderson Do we have to prove that there ¬∃e ∈ G ∀x ∈ G (e.x = x.e = x) ?

Xander Henderson

@Prithubiswas Who is "we"? And in what context?

Prithu biswas

@XanderHenderson "we" is "me". And it is in the context of your discussion with Neel.

Xander Henderson

Well, Neel seemed to be trying to determine which properties were satisfied by the operation $a\star b = a$. So it is necessary to either prove that there is an identity, or prove that there is not an identity.

So... if your goal is to determine which properties are satisfied by this operation, and you claim that there is no identity, then yes, you should prove that.

Prithu biswas

14:59

If e is an identity of G , then for all x ∈ G,
x = x.e = e.x = e
x = e
So if G has an identity , then G is a trivial group?

@XanderHenderson Is this correct?

Xander Henderson

Looks fine to me.

Transcript for

$Mathematics$ Mathematics