Mathematics - 2021-03-17 (page 1 of 2)

12:12 AM

K

geocalc33

12:29 AM

What did Minkowski put his 2 year old daughter in to drive to the store? A "boost"-er seat. Where do all the Galois cohomologists go to find holes? The étale site. An algebraic topologist was explaining how to open a bottle of soda. He said "first grab the cap, Dehn twist."

$\Huge\text{Happy St. Patrick}^{\!\!\color{#090}{\raise{1pt}\unicode{x2618}}}\!\!\!\!\text{s Day}$

3

12:50 AM

Aren't you early, @robjohn? Cute froggy shamrock.

hello all

Howdy.

@TedShifrin the chordal vs. great-circle distance on the sphere: it's geometrically obvious to me that, say $\text{chord}(z,w)/\text{circle}(z,w)$ approaches 1 as $z\to w$. But I don't see an obvious relationship between the two in order to see this algebraically as easily.

I guess wikipedia might have a formula up its sleeve somewhere for this.

Use trig.

Forget the sphere and look at a circle.

Just think. Don't wiki.

Oh, I guess I can connect the points with radii to the centre of the circle and then trig would be more applicable, and I can actually find the length of the arc part then

Good point!

12:58 AM

Arc is the easy part.

Chord is harder?

Needs trig.

Can't you avoid trig by using the complex plane?

Then the arc needs inverse trig?

Oh, so it's one or the other, I guess.

1:04 AM

If you find another way, educate me.

Will see what I can do. All hypothetical for now.

This is all related to the $(\sin\theta)/\theta$ limit in basic calc.

I see. So it's easy to get the great-circle distance from the chordal one using cosine law, but then you have to take a trig limit like sin x/x like you said, but with arccos.

1:19 AM

I think using the angle is less arcane.

arcane sounds fun

Well you have to do a 1-sided limit.

Nah.

I am doing it differently then you, I think, then.

You mean arccos as $x$ going to $1$?

I get that circle(z,w) is arccos(stuff with a chordal(z,w) term), then I let x = chordal(z,w) and take the limit as x to zero from the + side.

1:25 AM

Yeah, OK.

In particular, $\lim_{x\to 0^+}\frac{\arccos(1 - x^2/2)}{x} = 1$ is what I want to show.

Way harder than my suggested way. This is a degree 2 T.P.

T.P.?

Taylor

Oh. I guess that is one way to show the limit.

1:27 AM

It's the usual manipulation that shows up in the derivative of $\cos$, but still one step harder.

There is nothing I take more pleasure in than doing everyone one step harder than Ted's way.

@TedShifrin I'm going by SE (UTC) time.

in the true Irish spirit!

So after ten mins of walking in the Toronto cold, this dawned on me @TedShifrin, since we stated that $\mathbf{x_{0}}$ is orthogonal to $\mathbf{a}$ at the beginning that would mean $\mathbf{a} \cdot \mathbf{x_{0}} = 0$ and another mistake I made was the denominator. so the scalar is actually: $ t = \frac{b}{||\mathbf{a}||^{2}}$

Both correct.

@robjohn UTC is LaJolla, so our time zone :)

1:42 AM

@TedShifrin LaJolla? How so?

"land of holes" or "jewel city"

UTC is Greenwich (London)

Doesn't work here

UTC = University Towne Centre

@TedShifrin Ah.

I thought you might have been misled.

1:51 AM

This means my $\mathbf{x}$'s will be of the form $\mathbf{x_{0}} + t\mathbf{a}$ where $t$ is just as stated above. So I got a couple of thoughts/questions reflections on this:

1) Where did we use the area of the parallelogram in this? The only place the cross product shows up is in letting us know that $\mathbf{a}$ and $\mathbf{x}$ are going to be orthogonal to the normal vector $\mathbf{c}$

2) Why are we allowed to assume the existence of a perpendicular vector $\mathbf{x_{0}}$ to $\mathbf{a}$?

Coordinated Universal Time

Coordinated Universal Time or UTC is the primary time standard by which the world regulates clocks and time. It is within about 1 second of mean solar time at 0° longitude, and is not adjusted for daylight saving time. It is effectively a successor to Greenwich Mean Time (GMT). The coordination of time and frequency transmissions around the world began on 1 January 1960. UTC was first officially adopted as CCIR Recommendation 374, Standard-Frequency and Time-Signal Emissions, in 1963, but the official abbreviation of UTC and the official English name of Coordinated Universal Time (along with the...

@TedShifrin I was. Wikipedia misled me

@TedShifrin okay, we run into the problem that I don't get any degree 1 terms in the power expansion because of x^2 in the argument. Is that why you mention degree Taylor polynomial? I don't see how going to a higher degree helps.

Why not the four leaf clover @robjohn? Mathematica wouldn't allow you to do it?

No. All vectors is no problem. I've answered the area question multiple times, even within an hour. How do we know there's a vector orthogonal? Is every vector a scalar multiple of $a$? If not, go back to projections.

@dc3rd Too busy, and not very common. Plus, harder to make a face from.

1:56 AM

Westfield UTC

Westfield UTC (formerly known as University Towne Centre) is an open-air shopping mall located in the University City community of San Diego, California built in 1977. It lies just east of La Jolla, near the University of California, San Diego campus. It is owned – except for the ex-Sears parcel – by the Unibail-Rodamco-Westfield. Its anchors include Macy's, Nordstrom and a 14-screen Arclight Cinema. == History == Ernest W. Hahn first proposed building UTC in 1972. Upon opening in 1977, the anchor stores were Robinson's (later Robinsons-May), The Broadway (now Macy's), and Sears. In 1984...

@anakhro You can do the $(\cos u - 1)/u^2$ limit just with the usual.

@user85795 I remember University Towne Center. My uncle and aunt used to live in University City. I see that Westfield bought that mall up as with many malls all over.

One near me, too.

@TedShifrin I don't quite follow what you mean by that. I am dealing with arccos not cos---how do I make the change?

Change of variable, of course.

2:01 AM

I figured that much, but I don't see what the change of variable is to force a swap between cos and arccos

Aw come on.

i feel so blind

$\sin^2(\theta)+\cos^2(\theta)=1$

This looks like it would be a good hint.

Try substituting $\theta=\arcsin(u)$ and see where that leads you.

Try $\theta=\arccos(u)$ as well. see where each goes

2:16 AM

That always seems to throw in an arcsin and arccos together.

@anakhro one of those should get you somewhere. I'll let you work out which one.

I appreciate it.

bloody hell, i'm i'm confused.....smh.

@dc3rd what are you confused about?

don't worry @ankharo, I need to just piece it out cause this is going on too long for my satisfaction...

2:21 AM

worries

some long roads are more satisfying to be walked alone on

withers up.

I get cute identities for sqrt(1-x^2)

Semiclassical

2:47 AM

o/

@anakhro should get something $\arccos(x)=\arcsin\left(\sqrt{1-x^2}\right)$

It's St. Patrick's Day in Ireland...

@copper.hat I posted the banner soon after that.

:-).

3:09 AM

i'm having green tea in honor of the occasion

we used to just wear a sprig of shamrock and go to mass.

attend a parade if there were younger kids around.

we will have corned beef and cabbage tomorrow.

we didn't really associate a particular meal with the day. main thing was that it was a day off school with the mass downside.
bacon (back bacon) & cabbage would be a more typical meal if there were such a thing.

corned beef was up (down) there with spam for us. partly because my dad worked in the meat industry.

now spiced beef, that was an entirely different kettle of fish.

Spiced beef was fishy?

3:31 AM

@TedShifrin Not at all. It was a salted beef (involving a week in the fridge in a mix with spices, salt & saltpeter). Then you boil & simmer for a couple of hours to serve.

Irish food was not particularly heart healthy.

Sorry, took me a moment to understand why you mentioned fish :-). My expression on the previous line did not ring a bell.

You are not responsible for anything you utter!

Q: Finding torsion subgroups of elliptic curves over finite fields

Have I suffered enough yet @TedShifrin? All I got are these fragments but I can't bring the picture together. Let alone even draw the picture of what is happening correctly.

Martin Ødegaard

4:07 AM

Is there a mistake in this question?

1

Finding torsion subgroups of elliptic curves over finite fields. Given $y^2=x^3+x+1$ over $F_3$ I need torsion subgroup of $E[3]$ $E[3]$ is either trivial or isomorphic to $\mathbb Z_3$ The points $(1,0),(-1,0),(0,0)$ are each of order $2$, so useless, but the point $(3,1)$ has order $4...

elliptic-curves

I'm confused because the points $(-1, 0)$ and $(0,0)$ are not on the curve surely, as $0^2 \bmod 3 \not \equiv 0^3+0+1 \equiv 1 \bmod 3$ and $0^2 \bmod 3 \not \equiv (-1)^3-1+1 \equiv -1 \bmod 3$?

4:51 AM

@Astyx youtube.com/playlist?list=PLgAugiET8rrLB7WbPKgYJhfGOCyD9KQ0v

might be interesting for you

user93353

7:51 AM

If a is a quadratic residue mod p, then is -a always a Quadratic non-residue?

@user93353 $\left(\dfrac{-1}{p}\right) = (-1)^{(p-1)/2}$

user93353

@LeakyNun -thank you

8:09 AM

There seems to a problem here according to me, I cannot seem to pinpoint it.

fix your latex

In the class that I am TA'ing they were asked to prove that if $\{a_1,a_2,\cdots ,a_m\}$ is a basis of F^n, then $m = n$.
One of the solutions was, since $\{a_1,a_2,\cdots ,a_m\}$ is a basis, every element $y \in F^n$ can be written as $x_1a_1 + x_2a_2 + \cdots + x_ma_m = y$. Now they created a matrix whose ith-column was $a_i$, and this basis property reduces to the statement $Ax = y$ has a unique solution where $x$ is the column vector with entries as the coefficients. Then since every $y$ has a unique such representation in terms of the basis, this matrix has to be invertible and hence $…

I don't see any problem

$Ax=0$ has unique solution means that $A$ is injective

$Ax=y$ has solution for all $y$ means that $A$ is surjective

or are you suspecting that it's a circular proof?

Yeah I am suspecting it is circular somehow

but what theorems are supposed to be available then?

there's a problem with these "x without y" questions

do they have access to rank-nullity?

8:14 AM

Yeah they do

RREF?

that's reduced row echelon form

The thing is whenever this dimension is well defined argument being presented, I have seen it and done it as a change of basis matrix which will be $m \times n$ and since this guy is invertible $m=n$.
Yeah they know RREF

then they could argue that in the RREF of A, every column must have a pivot (by nullspace = 0) and every row must also have a pivot (by surjective)

so m = number of columns = number of pivots = number of rows = n

Yeah that's the standard argument

so it isn't circular

but you could as well argue that it is not clear from his presentation that this was his intention

do you have the big theorem about equivalent definitions of invertible?

8:18 AM

Hmm yeah true.

@LeakyNun Nope they do not

iirc I think I saw something like 18 equivalent definitions in one single theorem

details vary by authors

they could as well have been stuck and just wrote "this matrix has to be invertible and hence m=n"

it is not clear that they know the non-circular proof

Ah yeah that is a very good point

Maybe I will add that comment

i am not responsible for any misjudgement

Lol that's fine

Oh one more way to say this is, that this is precisely the change of basis argument but the change of basis matrix is from the standard coordinate basis on $F^n$ to this new m-basis

au contraire this is from the new basis to the standard basis

because A ei = ai

ei is the representation of ai in the new basis

A [ai]_newbasis = [ai]_standardbasis

8:33 AM

Yeah right I meant that

9:10 AM

hello, if $v_n(x)= \begin{cases} u_n(x), x\in \Omega\\ 0,n\notin\Omega\end{cases}$ can we say that $v_n(x)\to 0$ almost every where ? where $\Omega$ is bounded?

9:21 AM

someone here?

Benny

9:31 AM

I'm having some troubles when moving a problem involving definite integrals from $\mathbb R^2$ to $\mathbb R^3$, I would be glad to get some help...somehow the community seems to miss my question, although it's a good written question. Check it our over here: math.stackexchange.com/questions/4064768/…

9:50 AM

hi @Astyx

@SayanChattopadhyay The question is how do you define "basis"? When I first learned about those, I was told a basis is a generating free family of vectors. It's not too hard to see that two basis have the same cardinal (otherwise, you can always find a linear dependance relation between the vectors of the one with higher cardinal). Once you know that, you only need to check that there is a basis of cardinal n in $F^n$, which is the canonical basis. Then you're done

I think arguing about matrix invertibility is way too complicated for the problem at hand

@EdwardEvans cheers. Youtube tells me I already watched some of those apprently (although I don't remember)

hi Vrouvrou

have you an idea about my question ?

no

10:42 AM

@Astyx lol fair, I just got my talk for a seminar on Brauer groups

Your talk?

yeah right, as in, the talk I will give for the seminar

Oh, in the future

I thought you just gave a talk on Brauer groups

haha nah that'd be fast

I'll be talking about that exact sequence actually

when's the seminar?

10:44 AM

determining Br of a number field

the coming semester

With Q/Z and stuff?

right

it's on Albert-Brauer-Hasse-Noether

cool

You'll have to tell me too then :p

and then determining Br of a number field

yeah sure lol

in theory I should know something about Br by then

lmao

man I just sent a big email in swedish to a prof who is responsible for the swedish language courses at the university and she replied "Sorry I don't speak Swedish"

lmao

11:13 AM

@EdwardEvans oof

I put lots of effort into it too rofl

11:28 AM

@robjohn Yeah, I got that, but I figure it was the wrong track because I want to convert the arccos into a cos, not an arcsin.

@anakhro $\frac{\arccos(1 - x^2/2)}{x} = \frac{\arcsin\left(\sqrt{1-\left(1-x^2/2\right)^2}\right)}{x}=\frac{\arcsin\left(\sqrt{x^2-x^4/4}\right)}{x}=\frac{\arcsin\left(\sqrt{x^2-x^4/4}\right)}{\sqrt{x^2-x^4/4}}\frac{\sqrt{x^2-x^4/4}}{x}$

so...

${\sqrt{(\cos\theta - 1)^2 + \sin^2 \theta}\over|\theta|}\to_{\theta\to 0} 1$

11:45 AM

@robjohn I am really sorry. I don't see how this gets it any simpler.

11:55 AM

Is it possible to convey the intuition of what you are doing? You got it to arcsin (but that seems no better than arccos) so I am not sure where you are headed.

Why are you using inverse trig functions?

Well I suppose I could do it "Ted's way" which doesn't get an inverse trig function, but I thought why not try it this way, maybe I would learn something.

But currently the problem is to find the limit of what robjohn mentioned above, as x goes to zero from the right.

But in the Taylor expansion of arccos(1-x^2/2) you do not have a degree 1 term and so you can't do it so easily.\

So Ted proposed I could turn it into an expression depending only on cos, rather than arccos, using a change of variables.

And robjohn gave me a hint to help with this because I was stuck before. And I still don't quite see how I can go about such a subsitution.

Do you understand his equations?

I do understand his manipulations.

I don't see why he's making them or how it makes it any simpler.

Then you can just look at arcsin x/x and $\sqrt{x^2-x^4/4}/x$

12:03 PM

@anakhro do you know what $\lim\limits_{x\to0}\frac{\arcsin(x)}x$ is?

Oh, so this isn't related to Ted's change of variables?

hint, it's $\lim \arcsin(\sin\theta)/\sin\theta$

@robjohn no but let me try!

Okay, I see now. So the object was not to do Ted's proposed change of variables, but was rather to break down the limit into something you could evaluate more easily.

Thanks @robjohn and @Astyx!

henceproved

if you allow me to be politically incorrect, is it fair to say that integration happens on measure space and differentiation happens on norm spaces and these two concept coincide/related only when the underlying spaces is both norm and measurable space.

12:32 PM

Normed spaces are a lot more rigid than is required for differentiation. e.g. most (smooth) manifolds are not normed spaces. Do you mean to play off of the Frechet derivative when you bring up normed spaces?

@anakhro in the end it's the same computation

It's $\sin t\over t$

@Astyx yeah, it's just that I was looking at it completely wrong as a result.

12:49 PM

hi

if we say define by extension $v_n(x)=\begin{cases} u_n(x), x\In \Omega\\ 0, x\notin \Omega\end{cases}$

what is the sens of extension here please

1:41 PM

@Vrouvrou le sens est prolonger (extend) le domaine de la fonction

@EdwardEvans sit terra tibi levis

if $X_n \in L^1$, $Y_n \in L^2$ are both sequences of iid rv (i.e. the $X$'s are iid, and the $Y$'s are iid but they need not have the same distribution), then does $\sqrt{n} \frac{\sum_{j=1}^n X_j}{\sum_{j=1}^n Y_j^2}$ always converge in distribution to a normal random variable?

and of course the $Y_j$ are not zero almost surely

oh and both sets of rvs have mean zero, forgot to say

Thorgott

hmm, that's interesting

are the $X_n$ and $Y_n$ also independent?

they don't have to be, only within each group

if I didn't add the mean zero condition even setting $Y_j = 1$ it is false

you could even go for $X_n \sim N(1,\sigma^2)$ and that normalized sums characteristic function won't converge everywhere

Thorgott

the $Y_j=1$ case is just CLT

but thats only if the $X_j$ have mean zero right?

Thorgott

1:55 PM

yes, but that's what you said, no?

"both sets of rvs have mean zero"

oh sorry, I thought you were objecting to something I said, yes exactly

2:38 PM

hello

can you give me a complete operation hierarchy list?

where are logarithms in operation hierarchy?

@NicolásCastellanos prefiero que el argumento de log sea en parentesis

log(3x) en vez de log 3x

para que no se confunda

now you writeme always in spanish XD

lo prefiere?

no, i can understand english

so you're referring to the pemdas thing?

so people would write sin 3x and mean sin(3x)

2:49 PM

mmm wait

for special functions like these I think you just need to use context

sin x^2 would be more confusing but this should still be sin(x^2)

because (sin x)^2 is usually written sin^2 x

mmm

interesing

sin x+y... now this I see less often

I think people write sin(x+y) more

and, yes, i was saying where is the logarithm in the pemdas thing

and how many operations are there

so you might say it's on similar levels with multiplication

well there's sqrt as well but that one is usually clear because you draw the horizontal line as far as your input

$\sqrt{x+y}$ causes no confusion

so firstly everything is convention

they are not mathematical truths

that's my main issue with the meme "is 9/3(2+1) = 1 or 9"

2:53 PM

i can solve from this form? log, empowerment, root, multiplication, divition, sum, subtrraction?

people pretend like there's something mathematical behind that question

empowerment = exponent

no es palabra

ok

in a way i'm answering your question by saying that it's the wrong question

but if a people groupe solves first exponent that logarithm, their issues will be different if we compare its with the issues of the people that solves first logarithm

because other operators should always come with parentheses

2:54 PM

no?

would exists a international convention for this i thing

think*

can you give me one scenario where this would cause confusion?

mmm

let me create one

does $e^{\log 2}$ have two interpretations?

or $\log e^2$?

ok $\log e^2$ might be confusing

log_2(16) + 5*2

see, the parentheses to the right of log_2 tell you where the input is

2:57 PM

yes, but the important is that if i solve first logarithm, the issue is 29

ah no wait

this case is not different

if i solve first exponent is the same issue

but some cases are different depending to the thing i solve first

no?

Mats Granvik

oeis.org

the page doesnt works

Mats Granvik

Ok thanks. I can't reach the page either.

and... i will solve first logarithms

The page works for me

3:10 PM

forme no

Clarinetist

3:52 PM

Do any of you have tips on how to resign gracefully from an adjunct position, especially since this will have to be done via Zoom?

Long story short, I teach a class that no other faculty in the college have expertise in. They had a hard time filling the position before I arrived. Ideally, I'd like to treat this as a "bye for now, might come back later" type of arrangement, but I don't know if this will be possible. The truth is that I'm burnt out trying to keep up with the technology changes every semester and doing video editing, and I'd like to figure out my long-term priorities.

4:29 PM

Can I bound the L1 norm of two densities by their wasserstein distance?

user2103480

Delta method

In statistics, the delta method is a result concerning the approximate probability distribution for a function of an asymptotically normal statistical estimator from knowledge of the limiting variance of that estimator. == History == The delta method was derived from propagation of error, and the idea behind was known in the early 19th century. Its statistical application can be traced as far back as 1928 by T. L. Kelley. A formal description of the method was presented by J. L. Doob in 1935. Robert Dorfman also described a version of it in 1938. == Univariate delta method == While the ...

dw figured it out

user2103480

You can apply this to g(x,y) = x/y, but I don't know whether the X's being in L^1 suffices here. You might want to apply this after making sure the vector [normalized Sum of X, normalized Sum of Y] converges in distribution. Which, now that I think of it, also has problems since we're using Y^2

It would be a lot more helpful if you included any details about where the problem comes from

these are the full details

the question only includes this information, and says either provide a proof or counterexample

Given two densities $\mu,\nu$ and a compactly supported infinitely differentiable vector field. Does this bound make sense :

4:43 PM

its not a problem from a text , its from a lecture course

$$ \int v(x) (\mu(x)-\nu(x)) dx = \int ( v(y)-v(x) ) d\gamma (y,x) = \int \langle y-x, \nabla v (y) \rangle d\gamma(x,y) + error $$ using taylor, where $v$ is the vector field and $\gamma$ the optimal coupling between the two measures \mu and \nu.

then taking out the supremum of $\nabla v$, and in the error term $\nabla^2 v$, I get that its bounded by $ C \int |x-y|^2 d\gamma(x,y)= W_2^2(\mu,\nu) $ for some constant C.

@Clarinet: Give them as much time as possible in the way of warning. But it may be a burning your bridges sort of thing.

@TedShifrin you were right your way was much easier. ;)

5:10 PM

LOL, @anakhro. Could I have that statement notarized and posted, please? :D

Ted being right? That's a first timer

First time this month.

Sure send me your address and I will send it on a plaque.

Oooh, that sounds tempting.

Ted, I think you've heard "you were right" enough in your life. :P

5:12 PM

@LeakyNun They'd better!

Nah, @anakhro, I'm not that much of an "I told you so" person. Just for you!

What's a simple example of an homogeneous space that is not the underlying space of a topological group?

A sphere @Alessandro

obviously, not $n=1,3,7$.

What's a homogeneous space?

Yeah I imagined you meant $n=2$. Is it easy to see it's not a group?

Euler characteristic has to be $0$ for a group.

$G/H$, @Astyx :)

A space with a transitive continuous (smooth) action.

5:14 PM

@Astyx A space such that for all pair of points $x,y$ there is an automorphism of the space mapping $x$ to $y$

oh nice

Topological groups are homogeneous because (right) multiplication by a fixed element is an homeo

@TedShifrin hmm

Kind of silly since I don't think it matters anyway, but the subgroup of $Aut\mathbb H^+$ which fixes $i$ is easily found to be the fractional linear transformations of the form $\frac{z + tan\theta}{1 - z\tan\theta}$, $\theta\in\mathbb R$. This exercise asks me to show that it is the case for $\theta/2$ instead of just $\theta$, and so I was wondering if there is a particular reason we would want a half angle?

In the smooth category, I have a nowhere-zero vector field. In the continuous category, use left (right?) multiplication by an element and compute Lefschetz number.

Is every manifold homogeneous or am I missing something?

5:18 PM

Definitely not, @Astyx.

connected manifold*

then yes

Alessandro's definition isn't quite right.

manifold implies no boundary

You're defining two-point homogeneous, not homogeneous the way algebraists and geometers use it.

5:19 PM

Manifolds have even a stronger homogeneity property, you can have $n$-tuples of points instead of using two of them

If you say every manifold meets your definition, then obviously there are manifolds that are not topological groups.

Your definition is not commonly accepted.

@TedShifrin Uhm that's the standard meaning of homogeneous in general topology, I didn't know it's also used in different ways

Look up homogeneous space on wiki.

Or any differentiable manifolds book.

BigSocks

@AlessandroCodenotti so an $\infty$-groupoid with objectwise left inverse?

@BigSocks up to (weak?) homotopy equivalence, I guess

BigSocks

5:22 PM

n e a t

@TedShifrin Hm, so the general topology concept is the special case in which $G=\mathrm{Homeo}(X)$

@Astyx @alessandro Here you go.

Right, but the topology wiki agrees with me, I think it's just one of the many instances of the same word having two meanings in maths (luckily they are at least related meanings this time)

I wouldn't want @Astyx to be miseducated! :P

Why doesn't the group action of automorphisms of the top manifold qualify for your definition Ted?

5:25 PM

Ted's definition is more general, he doesn't require the full homeo group

But if his def is more general, then every connected manifold is homogeneous? Or did you not claim it wasn't Ted?

I am used to thinking of it in the context of a finite-dimensional Lie group acting by isometries, I confess.

Any geometer who refers to a homogeneous space means that.

when i get bored i spin in my office chair. it's a finite dimensional lie group acting by isometries.

I'd like to see you demonstrate more than a small subset of that group!~

or a representation of that group, i guess.

5:28 PM

Lie groups are too nice, I mostly think about non locally compact groups

Oh I see, a homogeneous space is the data of a space AND the acting group

i guess i could see why a point-set topology oriented wiki would use the two point version. still seems weird to me and i don't even use homogeneous spaces.

and the action, not just the group if we want to be very precise

Right yes

Wow, @leslie backed Ted up for once.

Two plaques in one day :D

5:31 PM

apparently Word lets you type equations in latex now. my wife needed to use it. except, the editor does all of this autocorrecting to your markup. garbling it. so it's somehow even worse than the equation editor. i was able to turn some but not all of the autocorrecting off.

Word has had a TeX equation editor since the 80s, I think.

David Penney (of Edwards and Penney calculus/ODE fame) typed the books in Word back in the day.

i guess they've brought it to the foreground in more recent versions, and maybe even sidelined their old kludgy thing.

Of course, he eventually became a total devotee of TeX.

I don't remember if he ever did LaTeX.

there ought to be a option in Word preferences that says "my latex markup is perfect, please do not modify it in any way ever"

it seems to be aesthetically motivated, throwing in a lot of extraneous spacing markup, and \left and \right when it doesn't actually change the size. putting the widehat in \widehat{a}b over both a and b to improve the visibility of the wide hat. thanks, Word.

Hello everybody

5:36 PM

Um ... Worse than a bad human editor.

Why use Word for latex?

it's worse than me writing what i want and pasting in a picture of it taken on my phone.

my wife is a sociologist. her collaborators use Word exclusively. but apparently some of her equations are breaking the equation editor. i can't imagine why, what a functional piece of software that was. hence, latex.

Karim Mansour

Hi everyone

hello

in honor of the day, i am continuing to drink green tea. i drink it fairly weak. maybe at 64 oz for the day.

5:42 PM

I hate green tea

Maybe have some kimchi with it.

Simone: Well, it isn't chianti.

@TedShifrin I live in Siena, I can say that confidently XD

I hope I get back to visit Italy one of these years. And France. And Croatia. And New Zealand ...

OK, be back later. Have a good day/evening, all.

by the way we're tackling the vax campaign it'll be a while

have a good day ted

Actually now I am stuck because $-1/z$ does not seem to be of the form $\frac{z + \tan\theta}{1 - z\tan\theta}$.

5:47 PM

you should hear what green tea says about you.

XD

i didn't like it for a long time. but i think i was drinking too much stuff that made me feel unhealthy (sugary juices, things with synthetic sugars, soda). green tea is where i have ended up. also other herbal "teas." black tea winds me up a little too much.

I think it tastes like mowed down grass in boiling water

and I tried good quality imported fresh chinese green tea

sencha teas definitely do. some of the other ones are milder.

if i had a bottle of red wine i would probably be drinking that.

red wine

goes to my head

makes me forget that I

Still love her so