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12:06 AM
@TedShifrin I didn't express myself very well with what I wrote, but I was trying to say this is the relationship between the Lagrangian and the BH right? So the BH is a map from $\Bbb{R}^{n+1} \mapsto \Bbb{R}^{n+1}$ right? I'm just trying to make sure I understand the basic facts. Presumably then, my $F$ should involve $L$ and write $L$ in such a way that it includes $c$.
But something like $L(c,\lambda,\mathbf{x}) = f(\psi(c)) - \lambda g(\psi(c))$ doesn't make sense.
 
user147690
12:33 AM
@Paul Did you decide on your next post yet?
 
Is it even worth giving a geometric answer to this post? There's already an answer, but it's pretty stuffy.
 
Lol, I was just thinking about that, I think it will be between a sketch of the free product three non trivial groups with an extra relation, is not the trivial group, and the marked group topology. @AlexClark
 
user147690
@PaulPlummer Oh hahaha, take the two votes and combine them. I take it Mike didn't vote then?
 
@Stan, existence of $\psi$ is what you're trying to prove.
 
@pjs36 I would say so, nothing wrong with that, my eyes basically glazed over when I saw that answer
 
user147690
12:37 AM
@PaulPlummer Agreed
 
@AlexClark Well he did not ping me if he did
 
user147690
@PaulPlummer Nice introduction of tropical geometry I have only read a a little bit of it so far, but I will finish it off soon
 
@BalarkaSen Another application of the marked group topology, is that you can use the Baire category theorem to show the existence of some groups.
Looks nice @AlexClark
 
user147690
@PaulPlummer If you looked at it quickly the operations and the tropical additive identity are pretty fun
 
Ah, Sturmfels, what a guy
 
user147690
12:49 AM
@pjs36 I haven't read any of his other work before
 
If I'm not mistaken, he does some really cool geometric/combinatorial probability, for gene sequencing and that sort of thing
I could be mistaken though
 
user147690
That sounds right
 
user147690
Tropical mathematics has interests in phylogenetic trees
 
I actually had this in mind
I like the pair, Pachtor and Sturmfels, because they've found a use for graph associahedra, and I've worked with them a fair amount
 
12:54 AM
hi guys ~ quick combinatorial question: if I have $n$ slots, each of which could have any of $m$ possible values, do I have $nm$ distinct series? Or is it something like nCm or nKm? How many distinct series do I have?
 
user147690
@pjs36 Oh wow thanks for that. I will be doing a research project over the american summer on tropical geometry(although I know essentially none now)
 
@AlexClark Interesting, and you are quite welcome! I know 0 about the subject as well, but I've seen the name a few places
 
user147690
@TedShifrin Were you referring to Eric Stucky?
 
No, he's referring to Eric Auld, someone I don't think youu know.
 
user147690
1:22 AM
@MikeMiller I don't. I went back in the chat history to his first uses of Eric and Eric Stucky turned up
 
nevermind, I have $n^m$
 
user147690
@Paul So with $\Bbb R[x]/(x^2+1)$ I can see I get the form $a+bx\pmod{x^2+1}$ where $x^2\equiv -1$ so I can see how my quotient ring is isomorphic to $\Bbb C$. But As you said $x$ doesn't equal $i$, but has the same 'information' as $i\in\Bbb C$ would. Can I make that idea rigorous? How?
 
Show that there is an isomorphism between the two rings @AlexClark
 
user147690
1:37 AM
@PaulPlummer As in explicitly show it?
 
Yes :)
 
user147690
Will this be easy? Don't give me the answer though
 
Yah, it is pretty easy
 
user147690
Okay sweet, thanks
 
So how'd the tropical geometry thing come about, @AlexClark ?
 
user147690
1:45 AM
@pjs36 The topic or me doing research on it over the american summer?
 
hehe, American summer... yeah, the research
 
user147690
@pjs36 Well Eric Stucky is going to be doing research through CRP(Collabora‌​tive Research Project organisation), and this is the topic they have given - and he has offered people to join him on it.
 
user147690
Well to make sure I am specific I could either say the American summer or the Australian winter, and American summer seems better communication haha.
 
@BalarkaSen Exactly my plans. :)
Hi @AlexClark
 
user147690
Hey @SohamChowdhury
 
1:53 AM
That's a girl's name (but I turned you into a form of alcohol so that's ok)
 
user147690
That's an alcoholic beverage( I was indeed copying you )
 
user147690
Haha
 
So I gave the groups chapter a run-through yesterday, to see what things I knew a little about and just to get a feel for what was there.
I'll start working on it now. What's tropical geometry?
 
user147690
@Paul Are all polynomials of degree greater than $1$ ideals of $\Bbb R[x]$?
 
user147690
@SohamChowdhury Don't know enough yet to give a good explanation(been too busy to get into it)
 
1:55 AM
Mhm.
Okay, I'll be back when I have any problems (which should take ~4 minutes)
 
user147690
@SohamChowdhury Hahaha okay
 
Polynomials are not ideals, ideals are ideals @AlexClark
 
user147690
@PaulPlummer I meant are all polynomial generators ideals
 
user147690
@PaulPlummer I.e. can I take any degree 2 or greater polynomial as $(f(x))$ and it is an ideal of $\Bbb R[x]$
 
1:59 AM
Well yah, given that the definition of $( f(x) )$ is the smallest ideal that contains $f(x)$ @AlexClark
 
user147690
Okay thanks
 
@Alex how did you get the monospace font?
The message in the starboard?
 
user147690
@SohamChowdhury Oh I used the thing attached to ~ on the keyboard '`'. It allows you to add code without it rendering
 
The $$\huge thing
 
This?
 
2:04 AM
this
Oh, yes, I figured it out
 
user147690
Hello
 
you use ticks `
 
Yeah, like SO
Thanks.
 
user147690
Yeah use ticks $\checkmark$
 
user147690
$\checkmark hello \checkmark$
 
2:05 AM
Those are checks
 
user147690
I know haha
 
user147690
It's in the latex
 
You so fancy, Right Hon. Clark
 
user147690
You so fancy, Right Hon. Clark $\color{green}{\checkmark}\quad\left(\color{green}{\frac{10}{10}}\right)$
 
user147690
2:07 AM
hey!
 
user147690
NO!
 
user147690
$\checkmark$ test
 
What? angel face
I'm a good boy
 
user147690
Fortunately you messed up the defining :P
 
2:08 AM
Shouldn't you be proving something is isomorphic to the complex numbers ;) @AlexClark
 
Indeed.
 
user147690
@PaulPlummer Lmao yes, I am easily distracted unfortunately. But first I am reading a page on quotient rings of polynomial rings now that I almost understand it
 
$\square \!\!\!\! \checkmark$ Make @AlexClark feel bad for not doing work.
6
 
user147690
@PaulPlummer $\text{:}\color{blue}{\text{'}}($
 
Now redefine away, @Alex
 
user147690
2:12 AM
Why do I feel like the starboard is in my hono(u)r LOL
 
It was in mine yesterday, I guess
 
$\newcommand{\checkbox}{\square \!\!\!\! \checkmark}$
 
Eh, not working
Let me go back to studying the automorphism groups of groupoids with single elements lol
 
user147690
Okay I'll not talk here for 20 min atleast intentionally(except maybe editting this to laugh at Paul's scaling haha). That too @Soham, I am ruining it
 
Does not scale well with size, $\huge{\checkbox}$ :(
 
2:14 AM
@AlexClark for the beautiful graph that will burn a thousand envious chatters' hearts?
 
I don't think you can usepackage
 
Yeah, I guess
 
Are you trying to make one that does scale?
 
Yeah, off TeX.SE
I need amssymb it seems
And if the fonts were all switched to AMS Euler . . . well, one can dream.
@Paul is that a $\bullet$ under the $1$?
 
$\mbox{\ooalign{\checkmark \cr \hidewidth \square\hidewidth\cr}}

\makebox[0pt][l]{\square}\raisebox{.15ex}{\hspace{0.1em}\checkmark}$
 
2:23 AM
$1_\bullet$?
 
Yes I think so
 
Or $1_G$?
 
It is hard to tell with out context, but I am guessing this is suppose to be $\mathrm{Aut} (\bullet)$
where $\bullet$ is the object in you groupoid
Where is this?
 
Nope, I figured it out. $\bullet$ is the group operation. $1_\bullet$ is the id wrt that.
 
Oh you must be looking at a bad scan, because it is $*$
(or a bad printing)
 
2:28 AM
Nope, it is a bullet.
 
That is the notation at the beginning of the chapter, and I have the book right in front of me, it is $*$ (maybe it is a later printing)
 
Yeah, flip a few more pages.
 
That is the operation, I am saying they have $1_*$
 
$*$ is the "lone" object of the groupoid. $\bullet$ is the operation.
 
where that is a $*$
is the object
The subscript is $*$
 
2:30 AM
Must be a later printing. He uses two different symbols.
$*$ at the beginning and $\bullet$ later
 
Yup, exactly, he is using $*$
The subscript on the $1$ is a star
 
In your book?
 
Mine's different.
Probably revised or something. The pdf's clear enough.
 
I can tell it is a star on yours, it is just really hard to tell, and he is using $\bullet$ for the operation
 
2:32 AM
Yes.
Exactly.
Hi @MikeMiller
I get what you mean, yes.
This? $\bullet:\mathrm{Aut{)} \times \mathrm{Aut{)} \rightarrow mathrm{Aut{*)}$
This? $\bullet : \mathrm{Aut}(\star) \times \mathrm{Aut}(\star) \rightarrow \mathrm{Aut}(\star)$
 
What got it to work?
 
\star
Except it's a five-pointed star, but I suppose there's no \makestar where you can control the number of vertices.
@MikeMiller I now understand the universals of products and coproducts (and quotients by an eq. rel), and having done the first chapter is making the second (groups) feel quite a bit easier :)
Haha. Markdown's messing it up.
 
Oh I see it is trying to make things italic
* Does this escape *
$\bullet : \mathrm{Aut} (*) \times \mathrm{Aut} (*) \to \mathrm{Aut}(*) $
 
Woohoo!
I like my five-pointed stars better though, thank you very much
I gotta get back to work now. @AlexClark is making me jealous.
 
user147690
2:45 AM
@SohamChowdhury Oh me, I've been distracted
 
By?
The Australian winter?
 
user147690
@SohamChowdhury Email for the logistics of the research project
 
Oh. Still work.
 
That's pretty nifty, @AlexClark, you'll have to keep me updated on how it goes
 
You can use \ast too
 
2:46 AM
I see
 
user147690
@pjs36 We will be doing weekly videos for it, it seems, so there is that
 
Ooh, that's fun! I'd love to see some polytope stuff work out, but I didn't read his description too much
So I take it you already contacted him, and it's semi-official?
 
user147690
@pjs36 Yep, it will start on June 6th. I will have to do bulk reading before then
 
Nice, very nice indeed
 
user147690
3:01 AM
The only unfortunate thing is it starts a week before my finals, which will make the starting a little harder(which is one of the times where you need to work hardest I have been told[which makes sense])
 
user147690
Hey @TedS
 
Hey @AlexC
 
3:24 AM
Does anyone have any tips on how I can convince my parents to let me go to this math summer program I've been invited to?
It's in the US, so they're like "you can't fly alone" etc.
My dad is ok with it, but it's my mom who's not really willing to let me go.
:(
 
3:38 AM
Make your mom feel bad for ruining your future :D Also what is going to happen on a plane that would not happen if you had someone else with you? It is not like people get targeted on planes, if something bad is going to happen on a plane it is probably going to effect everyone
 
Nope. Here's the thing. I would have to make a stopover at Heathrow. They're afraid I'll miss the flight (i.e. Heathrow to Boston/Hartford)
They're not worried about "bad things" per se.
And there's a lot of general "there's plenty that could go wrong, I won't even bother thinking" going on.
$\hskip -0.7in \uparrow \text{sad}$ :(
 
Well making a flight is not too difficult, you just walk over to the loading area, and don't zone out too much during the layover, and if it is the airlines fault you will get a ticket for a different time
Its not rocket science, or abstract mathematical ideas
 
Yeah, but if you think about all those jet thrusters and so forth, it is pretty much applied rocket science.
 
Well making a flight is not rocket science, maybe building a plane is sort of rocket science, but it is not like he has to build a plane when he gets to Heathrow
Plus I am sure there is cool stuff going on where your at, or closer to home @SohamChowdhury
 
user147690
4:08 AM
This place is an amazing source of humor hahahahaha
 
user147690
@BalarkaSen This is not maximal, since when generating from $(x-1)$ we get $(x-1)(x-2)$ in the ideal, and then generate everything from that, and hence $\langle(x-1)(x-2) \rangle$ is inside of $\langle (x-1)\rangle$
 
user147690
@PaulPlummer Is this right, or did you just tell me how to find maximal ideals?
 
@PaulPlummer Not really, no. I wish.
 
user147690
@SohamChowdhury Wow those descriptions are awesome
 
And these too
@AlexClark I know :'(
Strange attractors, Galois theory, knots . . . I can't even . . .
And Stephen Wolfram's visiting this year.
 
user147690
4:25 AM
Wait what the, Harvard and MIT are just down the road from eachother?
 
Yeah.
MIT's brochure calls it "Hahvahd" lol
On the other side of a bridge, I think.
 
user147690
Ahhhh harvard wanted to merge, MIT did not, eventually they almost merged but the supreme judicial court put an end to it
 
The bridge's length was once measured in Smoots (just google that, you'll lol hard when you find out what that is)
 
user147690
@SohamChowdhury Hahahaha
 
user147690
"The 182.2-smoot mark is accompanied by the words "Halfway to Hell" and an arrow pointing towards MIT. "
 
user147690
4:30 AM
Well I better get back to work, talk later
 
Mhm okay
 
 
1 hour later…
user147690
5:34 AM
@Paul This is where I am at so far with the presentation(starting at 'The Talk:')
 
@Soham: I know Andy Soffer. He's great.
 
6:31 AM
@SohamChowdhury I am not sure how much algebra are in the first four chapter of Aluffi, though.
 
@BalarkaSen How much did you do?
 
@MikeMiller Any plans to write up the second part of the blogpost?
@AlexClark Yes, your proof is ok.
 
@BalarkaSen Ei, @BalarkaSen. How much algebra had you done pre-topology?
First four chapters: basically quite a bit of group theory and intro rings/modules I've heard the last chapter is a bit hurried (homological algebra, although I guess I need to finish modules before I have any idea of what that means), so I'll probably look at a different book.
 
@SohamChowdhury I knew some galois theory when I started topology, for example.
 
Then I think I'll finish the first five or six chapters.
7 is fields.
 
6:35 AM
you should either do some analysis or know algebra thoroughly well before studying topology. otherwise you won't understand enough motivational examples.
 
Okay.
Do you know homological algebra?
 
a bit.
most of the knowledge comes from studying singular homology.
 
I'll try doing the first 8 chapters then, which should take me a while.
 
but there is surely more to it than a bit of diagram chasing.
 
@BalarkaSen Your knowlege of HA?
 
6:37 AM
what's HA?
 
I'll try finishing the first groups chapter this week.
@BalarkaSen homological algebra is a big phrase
 
ah. well, homological algebra is mostly motivated from the geometric version of homology :)
 
which is a part of topology, I guess?
 
algebraic topology, yes.
 
@Balarka: Not today.
 
6:39 AM
okay, I'm helping my mom with the book she's writing, see you
bbl
oh, she's going out, I'm free now
@BalarkaSen, what are you doing right now?
 
@MikeMiller ah, I see. well, we're all waiting for it eagerly :)
nothing, studying.
 
user147690
 
I seriously don't know.
 
user147690
@BalarkaSen Had you seen this before?
 
6:43 AM
The hell.
No, I hadn't.
 
That guy was talking to me. He invited me to a room.
I left after a while. He must have found a private room I made a while back to try and continue the convo.
not really "private"
 
@AlexClark Studied anything more on inverse limits, then?
 
Is there any function that sort of "tends" to the sine function for $x \rightarrow \infty$ like this?
In a similar way, I mean.
 
user147690
@BalarkaSen No, but I think I have ideals and quotient rings mostly done, so I will get back into that very soon
 
cool.
 
user147690
6:48 AM
@Balarka first off, hopefully this is all right: math.stackexchange.com/a/1295009/233746
 
i will tell you a bit about solenoids when you get back :) they're very cool stuff.
 
user147690
But mostly, what the hell is he talking about when he says in the comment: "The curly subscript is unfamiliar." Is he talking about the letter $n$?
 
He probably means the homomorphism symbol.
 
I don't know. It'd be pretty weird if he's unfamiliar with english letters :P
ah, the isomorphism symbol. possibly.
 
user147690
I was thinking that, but it isn't a subscript
 
6:51 AM
^ It's the only curly thing
He's made one of those two mistakes. Now choose.
 
user147690
But an $n$ is curly, so it succeeds in being curly and a subscript(hence by the art of deduction...)
 
@SohamChowdhury (1+1/x)sin(x)?
 
@anon Seems right.
 
it is right. 1 + 1/x tends to 1.
 
Any $f(x) \sin(x)$ will work, then, if $\lim_{x\rightarrow \infty} f(x)= 1$, right?
 
6:55 AM
so the whole thing tends to sin(x).
@Soham $\lim_{x \to \infty} f(x) = 1$
 
Yeah. I actually wanted to write $1+f(x)$ at first, I guess.
 
bah, there's not enough questions to answer in MSE anymore.
 
user147690
@BalarkaSen Well the one I just answered has turned into 5 questions in my comments hahaha(still no upvote and he accepted the other answer)
 
I hate it when OP doesn't even care about the answer I posted.
 
user147690
@BalarkaSen I actually wasn't even looking for a question to answer, it was just there and I went in and answered a comment question he had
 
7:09 AM
@BalarkaSen do you have any pictures of the Cayley graph of a free group with $>2$ generators?
 
yes.
 
@Soham It's in the wiki page. I can't write it down just right now.
Cayley graphs are good stuff.
 
Is it some sort of 3D fractal-ish thing?
Which wiki page?
 
no, it can be embedded in R^2, but not isometrically, as the (gromov) growth of free groups are exponential
wiki page of cayley graphs
(nudge to @AlexClark : the whole solenoid business of mine started from looking at a sensible way to find a notion of cayley graphs for profinite groups, i.e., inverse limit of groups)
 
user147690
7:13 AM
@BalarkaSen And now you are stuck doing them forever :)
 
oh, @Solenoid, you mean cayley graph of free group of > 2 generators, not n = 2.
 
@Solenoid?
 
user147690
Hahaha you are so stuck that you tagged solenoid
 
well, yes, you still can do that, but a picture would be wacky.
 
They're in your head . . .
 
7:14 AM
you don't have to worry about them, @Soham.
 
@BalarkaSen I thought it would look like some kind of Menger sponge in the shape of a tetrahedron.
 
and no, they aren't the solenoids you have in your car.
 
I know that.
 
nope you can just draw the cayley graph in R^2.
have you ever tried to draw the cayley graph for F_2?
 
@BalarkaSen $F_2$ being?
 
7:15 AM
you can do the same with F_n with $n \geq 3$
free group on two generators
 
the plus-ish thing?
Oh.
Yes.
 
you can do the same with higher free groups.
but once again, the picture can be complicated.
 
There's no such picture on the wiki page.
 
i know, i thought you meant free group on 2 generators instead of > 2
 
That's in my book
 
7:17 AM
ok, i gotta go.
good luck drawing wacky cayley graphs.
a better exercise is to draw cayley graphs of finite groups, really.
 
8th page isn't loading. The one with the sweet stuff.
Yes, tetrahedron!
@BalarkaSen ^
 
7:31 AM
@SohamChowdhury It can be embedded in $\Bbb R^3$ like a tetrahedron. I never said it cannot be.
I just said that it can be embedded in $\Bbb R^2$ too.
 
And that was what intuitively made sense to me.
I don't even know what embedding means. Something like a projection?
 
For example, replace the $+$-shaped fractal for $\Gamma(F_2)$ by a star-shaped one.
You'll get back $\Gamma(F_3)$
 
$\Gamma$ = Cayley graph of, right?
 
Yes.
I drew a lot of Cayley graphs when I first learned about it. The topological theory behind them is very nice.
I can't stop myself from talking about it : if you look at three arbitrary non-colinear points in the Cayley graph of $F_2$, you'll see that the triangle made out of joining them by the paths in the graph is something like a very deflated euclidean triangle.
 
@BalarkaSen I guess you love Cayley graphs a lot. Sort of like what happened to me when I learned about generating functions. :)
 
7:39 AM
This indicates that there might be some kind of notion of hyperbolicity behind $\Gamma(F_2)$. The formalization of this idea leads you to the beautiful branch geometric group theory of Gromov's.
 
There's a little of that in the pdf I linked.
Don't you have projects to do?
 
Yes, but you have to know a bit about topology to understand the pdf.
What I said above is a rough idea you can make sense out of.
What kind of project?
 
ICSE shit
 
I am not in ICSE. I am a WBBSE guy.
 
Baah. None of that nonsense, right?
 
7:41 AM
nope. loads of free time.
 
One eight of Gauss circle problem is closely related to calculating integer sided acute triangles with largest side n.
 
Our EVS teacher wants a project with ~30 bar/pie charts.
I really want to go tell her, "Do you realise that this is *EVS*?" (obviously in a condescending tone)
But that's only one. Last year we did around 12 in the year. :/
 
That's silly, really.
 
Yeah.
I'm glad it's over.
 
user147690
@BalarkaSen So then in $\Bbb R[x]$ all maximal ideals are <polynomials with complex roots> or <degree 2 polynomials>
 
user147690
7:49 AM
I meant generated by
 
user147690
Hence the lazily adding <> around them
 
Well, they're the irreducible polynomials. Not all ideals generated by degree 2 polynomials are maximal, as you have already seen.
$x^2 - 3x + 2$, say.
 
user147690
Oh sorry I meant degree 1
 
Sure, then what you say works.
@AlexClark Now that you're at it, can you classify maximal ideals of $\Bbb Z[x]$?
 
user147690
@BalarkaSen Yes, one second
 
user147690
7:54 AM
@BalarkaSen All degree one polynomials and all polynomials with complex or irrational roots
 
$(x^2 + 1)$ is not a maximal ideal of $\Bbb Z[x]$
'Cause $\Bbb Z[x]/(x^2 + 1) \cong \Bbb Z[i]$ is a ring, not a field.
 
user147690
But it has complex roots?
 
user147690
Oh woops(that^ is precisely why it is a counter example)
 
How does having complex roots mean the ideal generated by it is maximal?
Right, this one's a wee bit hard. I'll just give one hint : maximal ideals of $\Bbb R[x]$ were all principal ideals. It might not necessarily be so for arbitrary rings.
 
user147690
So $\Bbb R[x]$ is a PID because $\Bbb R$ is a field
 
user147690
7:57 AM
But for $\Bbb Z$ being a ring, it isn't necessarily that strong
 
user147690
Wait the ring of integers are a PID
 
Yes, $\Bbb Z$ is a PID
 
user147690
So it is something to do with being a polynomial ring
 
mutes himself
I'll let you ponder :)
 
user147690
Okay haha
 

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