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12:01 AM
old joke: "A famous maths professor was giving a lecture during which he said "it is obvious that..." and then he paused at length in thought, and then excused himself from the lecture temporarily. Upon his return some fifteen minutes later he said "Yes, it is obvious that...." and continued the lecture."
I'm fond of Qiaochu Yuan's variation on that: "a professor of mine once made an assertion in lecture that I didn't quite see instantly. I asked him "is that obvious?" and he replied "yes." I asked him "is it obvious that that's obvious?" and, after a short pause, he replied "no.""
 
12:38 AM
I heard a variant of that story in which the professor scribbled a tiny picture in the corner of the blackboard, quickly erased it, and said "Yes, it's obvious."
I don't think I have every been guilty of this particular crime.
 
since this is less than or equal to, I can prove any case I want, right?
 
Huh? No.
Go back to your logic lessons.
 
What do you think i mean
maybe I asked my question wrong
 
OK, maybe.
"I can prove any case I want"?
 
Ok, different example:
nevermind, I can't find my other example in my jpegs
 
12:50 AM
Why don't you clarify what that question meant?
Shhhh.
 
shh semic
 
Don't interrupt me when I'm being mean and obnoxious.
 
ok, it says $a \le b +$ epsilon
idk mathjax for epsilon
 
\epsilon
 
12:52 AM
\epsilon, duh
 
whatevs
I can prove where $a
 
less work to try it than to bitch about not knowing it
 
\varepsilon for the cool kids.
Hello everyone.
 
I actually write \varepsilons in real life.
 
12:52 AM
omg ted
 
$\in$
 
though i do like $\varphi$
 
I'm trying to think
 
what did you mean by any way you want bro
just clarify that so you dont go wander off into thinking in a dead end
 
scrap what I just typed
that's not what I'm asking about
I'm proving $a \le b$
 
12:54 AM
indeed, we can read the question. tell us your approach
 
i can prove $a \lt b$ or $a = b$ because it's or, right
or do i have to prove both ways
 
sure, but either case is possible
 
What you're saying does not make sense.
 
ah, that's what you meant
 
so i'm not sure how that helps matters.
 
12:55 AM
How do you "prove it both ways"?
 
that was my question, Ted made me nervous
Like, prove both situations in order for it to be a valid proof
 
if you could prove that $a\leq b+\epsilon$ for all $\epsilon$ implied $a=b$, then you'd be done. same if you could prove that it implied $a>b$.
 
like part a and then part b
 
But either situationn is possible, so that's not really going to help
 
How do you prove EITHER P OR Q happens?
 
12:56 AM
what do you mean ;-;
That's not what I meant
I just asked the question wrong
 
No, you keep thinking wrong.
 
Ok, what do you mean
 
If it were case a) and then case b) you'd be proving P AND Q
 
I was thinking in English, not math
 
I'm tired of excuses.
I'm out
 
12:57 AM
omg
why so aggressive today
sorry ;-;
 
i will give benefit of the doubt and say your question is valid but this is not a fruitful way of going about it
 
ok, what's a better way to go about it
 
there are situations where you prove "either P or Q happens" by arguing "assume (something) blah blah blah then P happens, otherwise blah blah blah then Q happens" but this is not one of em
@CaptainAmerica16 think about it! :)
 
I know, I literally was just asking in a "what's the most acceptable way to answer this question" kind of way.
 
Maybe start by considering a slightly simpler problem. For instance, take the case b=0.
 
1:00 AM
Or maybe I don't know
 
@TedShifrin the complex-valued 1-forms---do they correspond to a real vector bundle?
 
Formulate an answer, then tell us about it
 
Then you want to show that, if $a$ is real and $a\leq \epsilon$ for any $\epsilon>0$, then $a\leq 0$.
 
eh, anakhro?
 
That is, are they sections of a real vector bundle? I have them here defined as $dP + i\,dQ$ for real functions $P,Q$.
 
1:02 AM
@BalarkaSen Ok, sorry if it's wrong
 
@anakhro that's not a general complex-valued 1-form - you gave an exact form
 
Oh whoops
I meant $\alpha + i\,\beta$ for (real) 1-forms alpha,beta.
Sorry, I was thinking of my follow up question.
 
Sure, they are sections of $T^*M \otimes \Bbb C$, the complexified cotangent bundle
 
And that's the natural object associated to them? Or is there something more natural?
 
Ok, this is kind of like a rough draft so it's probably not going to have all of the correct wording, but this is what I was thinking: if $a < b + \epsilon$, then $a +\epsilon < b +\epsilon$. Then $(b +\epsilon) - (a + \epsilon) = b-a$. Therefore $a < b$.
 
1:10 AM
@anakhro The way you're writing it is finicky. This is the standard setup setup: If $M$ is a complex $n$-dimensional manifold, use the $z_1, \cdots, z_n, \overline{z_1}, \cdots, \overline{z_n}$ coordinates to decompose $T^*M \otimes \Bbb C = \mathcal{E}^{1, 0} \oplus \mathcal{E}^{0, 1}$ where $\mathcal{E}^{1, 0}$ are $C^\infty$-linear combination of $dz_i$'s and $\mathcal{E}^{0, 1}$ are $C^\infty$-linear combination of $d\overline{z_i}$'s
This decomposition then is invariant under biholomorphic change of coordinates in $M$
I am unsure what you mean by something more natural - similarly the whole battery of forms comes with these $(p, q)$-decompositions in the complex setup and you have a double complex. This turns out to be useful information
 
@BalarkaSen (assuming it's correct) I only showed that $a < b$. That's why I wondering if I should show that a could equal b as well.
 
It's not true that $a < b+\epsilon$ implies $a+\epsilon < b+\epsilon$
 
done
 
that's what I figured, tbh
 
$2 < 1 + 3$ does not imply $2 + 3 < 1 + 3$, there is not much to figure :P
 
1:14 AM
By "is there something more natural" I mean "natural" in the philosophical/pedagogical sense.
 
yeah, i only thought of one example that fit what i was thinking, lol
back to the drawing board. thanks for the help everyone.
 
you're welcome
 
@anakhro Sorry, I don't know what you mean.
 
It may help to think of it like this: If $a\leq b+\epsilon$ for all $\epsilon>0$, then you need to show that $a$ can't be less than $b$.
 
@BalarkaSen that's fine.
I will ask Ted another time.
 
1:17 AM
has gromov published a paper recently
 
@Semiclassical ok
 
G-dawg is steamrolling scalar curvature
That's what his last 80 papers are on
 
unrelated to gromov: can a manifold induce curvature in another manifold
woww nice
 
Curvature is intrinsic to the metric, so you'd then have to induce a metric. This is commonly done through things like the pullback metric and the likes.
 
wait
gromov has writenn 80 papers?
 
1:24 AM
that was joke but idk
 
I'd guess 43
 
Is there an explanation of the monkey on his website
has anyone questioned his choice in banner?
 
self-portrait, @anakhro
 
o u c h
 
according to his own admonition
lol admission
 
1:26 AM
 
autocorrect
 
okay
I finished creating a language that resembles spanish!
@pig
well actually I only have 500 words
 
@anakhro What Balarka said is perfectly fine, although I didn't read it carefully. It's a complex v bundle. Every cx bundle is of course a real bundle.
 
And this is the way that we ought to look at complex-valued 1-forms, or is there a more pedagogical way of viewing them?
Pedagogical given that the learner is familiar with manifolds and vector bundles.
 
1:43 AM
I don't get your point. Complex geometers want $dz$ And $d\bar z$ for lots of reasons. If you want to stick to $dx$ and $dy$ and allow complex coefficients, it's still a complex v bundle.
A complex $n$ manifold is a real $2n$ Manifold.
But you throw away very important structure.
 
Oh, no. I am asking if it's viewing them as sections of $T^*M\otimes\mathbb C$ (as a complex or real bundle) is the way we ought to look at them. I have encountered enough material involving dz and dbar z that I am thoroughly convinced they are important for complex geometry.
I just wonder if I am shying away from some "point" I should be understanding if I view them as sections of a bundle like any normal 1-form (the author of these notes I am reading does not view them as sections of any bundle, but I think it might be because of the target audience).
 
Are difference equations and functional equations the same thing?
 
The “right” way is not to ignore bundles. Indeed, you want vector bundle-valued $(p,q)$-forms, ultimately.
 
Do you know of a good reference on this topic? I figure it would be in something like Kobayashi & Nomizu...
Huybrechts might have it.
 
if you want to stick to curves and no higher than 2-forms, forster is great
doesnt expound on the (p, q)-decomposition that much though
 
1:50 AM
Every book on complex manifolds. Chern, Wells, huybrechts, deMailly.
 
Thanks, I will have a look at some of these and see what I like.
 
Griffiths/Harris, too, of course.
 
Their algebraic geometry book?
 
Yes.
But they have errors, so you'll doubtless bitch about them :)
 
i'm trying to learn a bit about that stuff myself lately, simply because I'm trying to remember how some of the stuff in my thesis worked
and remembering how bad my foundations are there, lol
for instance: I'm trying to convince myself I fully understand the fact that the function $x$ on the elliptic curve $y^2=x^3-x$ has a double zero at $(x,y)=(0,0)$.
 
2:01 AM
Complaining about errors is the best use of time. Though I doubt I will find any myself.
 
(i dunno if that's the right way to say it)
hmm, maybe i do. i'm trying to make sure i can see it in a few ways, and i just clicked on one of them
yea ok
 
@Semiclassical equivalently, does (x, x^3-x-y^2) have a double zero at (0,0)?
 
@Semiclassic Look at the origin on the curve $x=y^2$.
 
right. one obvious statement is just that $x=0$ is tangent to $y^2=x^3-x$ at the origin, so definitely a double zero
or, alternatively, if I perturb $x=0$ to $x=\epsilon$ for sufficiently small $\epsilon$, then there's definitely two roots
but they all come down to $x\sim -y^2$ for points on $x^3-x-y^2=0$ near the origin.
Lubin had a nice answer for how to see it by looking at the projective curve $Y^2 Z=X^3-XZ^2$ (though he did it in more generality)
But I wanted to make sure I could see it in a few ways, like I said
 
@MikeMiller if i recall correctly there's some care required because intersection multiplicity of the zero sets is not order of vanishing of one function on the curve given by the other
one is clearly symmetric whereas the other is not, for one
SemiC's way to see it by perturbing is the right one
one can also alternatively parametrize the curve near the origin, etc
maybe what i said is not correct, i dont remember the subtlety anymore
 
2:17 AM
The projective way of saying it, if I'm remembering right: If we go to the projective curve, then we're instead looking at $X/Z$. But then $$\frac{X}{Z}=\frac{X}{Z}\frac{Y^2 Z}{X^3-XZ^2} = \frac{Y^2}{X^2-Z^2}$$, which obviously(?) has a double zero at $[0,0,1]$ and a double pole at $[0,1,0]$.
 
Ah yes - it cannot be the same: The total intersection multiplicity is 3 but being a function on the elliptic curve the total order of zeroes = total order of poles
IM is not order of vanishing
 
yikes
oh, wait. i'm crediting the wrong person. should have been Álvaro Lozano-Robledo's (link)
 
Something is off about what I am saying lol. Consider a line $\ell$ in $\Bbb{CP}^2$ along with an elliptic curve $E$. Let $f$ be the homogeneous polynomial st. $\ell = \{f = 0\}$. Then $f$ is a function on $E$ whose zeroes are $\ell \cap E$ (unless some zero is at $\infty$ itself) and it will have 3 zeroes $P_1, P_2, P_3$ counting multiplicity
$(f) = [P_1] + [P_2] + [P_3] - 3[\infty]$ as divisors
The point is in elliptic curve addition you treat $[\infty]$ as an absorbing element, so that this translates to $P_1 + P_2 + P_3 = 0$
 
2:33 AM
This is the statement that the Picard group of $E$ is equal to $E$ itself, $\text{Pic}_0(E) = E$
IM is order of vanishing, but I am being uncareful with signs and sometimes zeroes can be attained at $\infty$ - this is a little finicky to say
Whatever man this is an annoying point I always get confused about
 
Is there an abuse of notation there, in using $E$ to denote both the elliptic curve and the group structure that can be placed on it?
 
Yeah, in the right hand side of $\text{Pic}_0(E) = E$, $E$ is the elliptic curve group
 
(i.e. the group is the elliptic curve plus the group law)
right
 
In the left hand side, it's the Picard group of the variety $E$
 
physicists don't get to complain about abuse of notation but i wanted to check
 
2:37 AM
hahah
 
Pig
@BalarkaSen this should be fine as long as the function you pick is actually a local uniformizer (so no multiplicity issue)
 
yeah, @Pig, but I was somehow convincing myself that if we reverse the roles of the function (the other function need not be a local uniformizer for the first one) I would get issues
i dont think this is actually a problem
@Semiclassical Ah lol, the isomorphism $E \to \text{Pic}_0(E)$ is $P \mapsto [P] - [\infty]$ - no need to "forcibly make $[\infty]$ an absorbing element". $P_1 + P_2 + P_3 = 0$ translates to $([P_1] - [\infty]) + ([P_2] - [\infty]) + ([P_3] - [\infty]) = 0$, a coherent group law on both sides
I am an idiot
 
neat
just to make sure I follow the language a bit
$y$ is a local uniformizer for $E:x^3-x-y^2=0$ at the origin, because $y=0$ "obviously" intersects $E$ only once there. (and, similarly, it's a local uniformizer at $(1,0)$ and $(-1,0)$ as well.)
whereas $x$ wouldn't be because of the double zero
as seen by the fact that $x\approx -y^2$ near the origin
i guess the justification for the "obviously" is analogous to before: if you perturb $y=0$ slightly, you still only get one solution near the origin
 
Pig
3:09 AM
yes that's correct
another way to think about it is via local ring
for nonsingular curves, it's known that their local ring at any point is a discrete valuation ring, which effectively means uniformizer exists
for $x^3 - x = y^2$ at the origin, clearly the local ring at origin is generated by $x,y$
but for these two generators, like you said $x \approx y^2$ (another way to put this is $y^2 = x(x-1)(x+1)$ and $(x-1)(x+1)$ is invertible near origin
so the local ring is generated by $y^2, y$ (since $(x-1)(x+1)$ is just unit in the local ring), and so is generated by $y$, and so $y$ must be the uniformizer
 
mmkay. that way of looking at it, unfortunately, is definitely beyond me right now. to the extent i grok this stuff, i suspect it'd be far more classical arguments than modern
 
Pig
meh, i was only giving an argument that $y$ is the uniformizer at origin, but if you already buy that, it doesn't matter
 
right
 
4:05 AM
@Balarka aaaand I finished writing
hahaha
now I'm gonna inject some lovely cancerous red bull into my stomach
 
5:08 AM
@EdwardEvans Cool!
I can give you my email id or something if you want to send it over
 
sure, it's just a handout, and it's in German
lol
dunnit
 
gotit
 
I still have some details to work out but I have a week until my talk so :P
 
5:26 AM
is it possible to elegantly integrate (c mod floor(x)) dx?
 
 
3 hours later…
8:04 AM
the ultimate purpose is having something like $\int_{2}^{n-1} \left \lfloor{\left \lfloor{\frac{n}{x}}\right \rfloor - \frac{n}{x} + 1}\right \rfloor dx$, but easily integrable
I don't think there's much variation to this formula that gives a neat integration
 
8:24 AM
any hint for this?
i feel like z1 z2 z3 are conjugate pairs
but i am failed to elaborate it further
i try using the geometrical method too.look like this circles intersect but i am again wrong
 
9:05 AM
@towc Is that not $0$?
$-1\lt\lfloor x\rfloor-x\le0\implies0\le\lfloor x\rfloor-x+1\le1$ and it is only $1$ when $x\in\mathbb{Z}$
That is, $\lfloor\lfloor x\rfloor-x+1\rfloor=0$ unless $x\in\mathbb{Z}$
 
9:31 AM
Hello people
 
Hi feynhat
 
@BalarkaSen I have two questions for you.
 
watchudoin
But you should probably get on those ^ first. lol.
 
listening to weird metal, figuring out weird math
@knight okay
 
Question 1. How to identify a vertical ellipse and hyper bola ? And how equations transform from horizontal to vertical ones
As you see, in the case parabola things are quite clear and easy
 
9:36 AM
im kind of busy with something atm but maybe someone else can help you
 
Okay :-)
 
@feynhat What happened to the matrices whose eigenvalues summed to nothing?
 
Huh?
 
They vanished without a trace
 
@BalarkaSen no
 
9:39 AM
heh.
A few days back Mike mentioned a theorem while talking to Thor. Vector bundle over contractible spaces are trivial.
 
You should still be able to determine if they did something singular of note
 
Let $f_0, f_1 : Y \to X$ be homotopic maps, and $E \to X$ be a smooth vector bundle. $X, Y$ are manifolds and I am only interested in $Y$ compact for now.
Finding a bundle isomorphism between $f^{-1}_0E$ and $f^{-1}_1E$ is same as finding a smooth section for the fibre bundle $\text{Iso}(f_0^{-1}E, f_1^{-1}E)$, right?
 
What do you want to prove
 
$f_0^{-1}E$ and $f_1^{-1}E$ are isomorphic.
 
Check Hatcher, Vector bundles and K-theory
The proof is in the first chapter somewhere
the basic idea is to pull $E$ back to $Y \times I$ by using the homotopy $F : Y \times I \to X$ and trivialize over $Y \times I$
 
9:53 AM
okay. This looks like a somewhat different proof than what I am reading in B&T (maybe the underlying principle is the same). But what they do is is find a section of $\text{Iso} (f_0^{-1}E, f_1^{-1} E)$ over $Y \times t_0$ and then extend it to $Y \times (t_0 -\epsilon, t_0+\epsilon)$.
 
Makes sense, yeah, essentially the same principle.
 
@BalarkaSen I just entered chat but I want to leave already
 
Lol
That's quite uncharacteristic of you, @Alessandro.
 
@BalarkaSen Come on ...
 
9:58 AM
$f : X \to pt$ be the constant map, $\Bbb Z_{pt}$ be sheaf over this point. Denote $\Bbb D_X = f^! \Bbb Z_{pt}$. I would like to compute this guy
So I guess fix an injective resolution $I_\bullet$ of $\Bbb Z_{pt}$ and a $c$-soft flat resolution $K^\bullet$ of $\Bbb Z_X$, so that we're computing $f^!_{K^\bullet} I^\bullet$
 
0
Q: Tangent space isomorphic to derivations taylor series

topologicalorientablesurface$T_p(\mathbb{R}^n)\cong D_p(\mathbb{R}^n)$ All I want to show is that the map: $\phi: T_p(\mathbb{R}^n)\rightarrow D_p(\mathbb{R}^n)$ , given by $\phi(v_p)=D_v=\sum_k v^k \frac{\partial}{\partial{x}^k}|_p.$ is surjective. Let $D$ be a derivation at $p$. Let $[(f,V)]$ be a germ at p (this exi...

 
Uh, $D_X(U) = f^{!}_{K^\bullet}(I^\bullet)(U) = \text{Hom}(f_! i_! i^* K^\bullet, I^\bullet)$, where $i : U \to X$ is inclusion
By Hom I really mean the Hom-complex
But $f_! i_! i^* K^\bullet = \Gamma_c(X; i_! i^* K^\bullet)$ is the full space of compactly supported global sections, which is the same as $\Gamma_c(U; K^\bullet)$
So that's the dualizing sheaf explicitly, $\Bbb D_X(U) = \text{Hom}^\bullet (\Gamma_c(U; K^\bullet), I^\bullet)$. This is independent of the choices of $K^\bullet$ and $I^\bullet$ once we pass to the derived category, so it's a well-defined complex of sheaves upto quisms
 
I don't get how they fatten the domain of the section though. First we cover $Y \times t_0$ by finitely many trivializing open covers. By tube lemma we can get a fatter strip. Then, they say that this section is also a section of $\text{Hom}(f_0^{-1}E, f_1^{-1})$, and as the fibers of $\text{Hom}(f_0^{-1}E, f_1^{-1})$ are Euclidean spaces the section can be extended to the fatter domain.
what does fiber being Euclidean spaces have to do with extending the domain?
 
Let $X\subseteq\mathbb{R}^k$, $Y\subseteq\mathbb{R}^m$ and $f\colon X\rightarrow Y$. We want to declare when such a map ought to be called differentiable and there are two possible notions, both of which may feel natural. We call $f$ differentiable if, for each $x\in X$, there is an open neighborhood $U$ of $x$ and a differentiable (in the usual sense) extension $\hat{f}\colon U\rightarrow\mathbb{R}^m$ of $f$. We can also do the same thing, but additionally require that $\hat{f}(U)\subseteq Y$. Both of these definitions yield a notion of morphisms in a category where objects are subsets of
 
@feynhat You have some vector bundle on $X \times I$ which is trivial on $X \times \{0\}$ and you want to extend that to a trivialization on $X \times [0, \varepsilon)$, right?
 
10:13 AM
Right.
@feynhat This is garbage.
 
This sounds finicky. I am more used to trivializing on tubes of the form $U \times I$ for some open $U \subset X$
Your tubes are going the other way
That is very strange
 
Exactly. I am pulling back by the maps, when I should be pulling back by homotopy, and the projection $Y \times I \to Y$.
Ignore that message.
 
nvm what I said doesn't make sense
 
Okay. Let $f$ be the homotopy and suppose $f_t^{-1} E$ is a bundle $E'$. So, we can pullack $E'$ over $Y \times I$ and also pullback $E$ (by the homotopy) over $Y \times I$.
Therefore $\text{Iso}(f^{-1}E, \pi^{-1}E')$ has a section over $Y \times t$.
This is also a section of $\text{Hom}(f^{-1}E, \pi^{-1}E')$, right?
*over Y x t
 
10:28 AM
Shit so, cohomology of $\Bbb D_X$ is $\text{Hom}(H^\bullet_c(X; \Bbb Z), \Bbb Z) \oplus \text{Ext}(H^\bullet_c(X; \Bbb Z)[1]; \Bbb Z)$
Am I just going to get Poincare duality if I take sheaf cohomology of $\Bbb D_X$?
 
Exercise A.1.22. The fundamental group of English is generated by the 26 letters
a, . . . , z, subject to all relations w = u where w and u are letter sequences that are
pronounced identically in at least some contexts. E.g., gh is trivial since it is silent
in some words, and ir = er because of the words “bird” and “her”.
Prove that the fundamental group of English is trivial.
 
Classic Artin exercise
OK, yeah, $\Bbb D_X \cong \mathcal{O}_X[n]$ for a manifold $X$, where $\mathcal{O}_X$ is the orientation sheaf
 
Now I cover $Y \times t$ with trivializing opens sets of $\text{Hom}(f^{-1}E, \pi^{-1}E')$, to get a tube around $Y \times t$. What I don't get is how can we extend this section to this tube.
 
This is just PD
Nuts
 
I pronounce abcdefghijklmnopqrstuvwxyz silently, qed
 
10:34 AM
I have a question
 
@BalarkaSen great
@BalarkaSen wait why is he doing the french one in english and the english one in french
???
 
Note how the triviality of the homophonic group of English is established in French, and that of French is established in English
Lmao u got tricked
 
leitmotiv = leitmotif ????
On the
other hand, it appears that the analogously de ned
group for Japanese (written in katakana) is free on
46 generators.
lol
 
RIP
 
10:41 AM
I doubt that actually
since ケイ (ke-i) and ケエ (ke-e) should be pronounced the same
 
why?
 
or maybe that's only for hiragana?
 
Let $e_{i_p}$ denote the standard basis for $T_p(\mathbb{R}^n)$. Theres a vector space isomorphism between $T_p(\mathbb{R}^n)$ and $D_p(\mathbb{R}^n)$, where $D_p$ is the set of derivations at $p$, with isomorphism $\phi$. $\phi: T_p(\mathbb{R}^n)\rightarrow D_p(\mathbb{R}^n)$ is given by $\phi(v_p)=D_{v_p}=\sum_k v^k\frac{\partial}{\partial{x}^k}|_p$. So, those partial derivatives are a basis for $D_p(\mathbb{R}^n)$.


Lee sais that we may write $v_p\in T_p(\mathbb{R}^n)$ as $v_p$ $=\sum_iv^ie_{i_p}=\sum_iv^i \frac{\partial}{\partial{x^i}}|_p$
 
it's definitely true for hiragana
けい (ke-i) is pronounced with a long "e"
 
English and French are both phonetically hopeless. I wonder if this is actually a good invariant for phonetic complexity
 
10:43 AM
in katakana I think you would use ケー instead
 
It should be a largish free group in my native tongue
 
the last sentence is a bit sloppy, no?
 
sad Chinese noises
 
@LeakyNun I have no clue actually, will it be massive for Chinese? Probably, right?
 
well if you quotient by homophones then you essentially get one generator for every possible syllable
 
10:45 AM
Lmao
 
with potentially more relations due to one character sometimes having more than one pronunciation
 
Crap
 
every character is one syllable
 
@topologicalorientablesurface we identify them via this canonical isomorphism
 
If you want a phonetically consistent language, try german
 
10:48 AM
you want to think about tangent spaces as derivations when you do abstract manifolds, so this is instructive to do
if you want a phonetically consistent language, try latin
 
okay this identification stuff i don't completely understand," identification" in the sense above, is not really formal, right?
@Thorgott?
 
@Astyx it still isn't free on 26 generators
 
Ah, that's true
 
it means whenever we have an element of one of these spaces, we simultaneously think of it as an element of the other space via the given isomorphism
 
10:52 AM
@Thorgott i was gonna suggest sanskrit
 
we just don't write $\phi(v_p)$ instead of $v_p$ every time, because that would get very messy
 
lorem ipsum dolor sit amet morbit bio
 
I mean, to start with, the fundamental group of german would be on 30 generators
but 4 of them are completely superfluous
 
bis/Biss; das/dass; bunt/Bund; Heer/her; isst/ist; kannte/Kante; Lehre/Leere; Leib/Laib; Lied/Lid; Mahl/mal; man/Mann; Meer/mehr; seit/seid; fiel/viel; war/wahr; wider/wieder
well ok but ä = e
@Thorgott which 4?
 
10:54 AM
then I should be more careful with calculations that require me to use that identifcation
 
äöüß
 
why are ö and ü superfluous?
 
but ä=ae. ö=oe, ü=ue and ß=ss
 
eh...
Fundamental » All languages » German » Terms by lexical property » Terms with homophones German terms that have one or more homophones: other terms that are pronounced in the same way but spelled differently....
 
0
Q: Is * associative?

sai tenOn a set A ,define a binary operator * such that $∀x,y ∈ A$, we have $x*(x*y)=y$ $(y*x)*x=y$ Is * associative? Note:* is commutative because We have $x=x$ $<=>y*(y*x)=[x*(y*x)](y*x)$ $<=>[y*(y*x)]*(y*x)=\{[x*(y*x)](y*x)\}(y*x)$ $<=>y=x*(y*x)$ $<=>x*y=x*(x*(y*x))$ $<=>x*y=y*x$ Could yo...

I don't understand whats the binary operator here
 
10:59 AM
$\ast$
 
@LeakyNun Nice notes
 

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