« first day (1796 days earlier)   

3:17 PM
@BalarkaSen, the indices were wrong in my post.
 
r9m
Drumming my ear drums since morning :D ... this one is more than awesome!! :D
 
@r9m thanks for sharing
 
r9m
@skillpatrol :D
 
Hi Professor @TedShifrin any news?
 
it's a long weekend, @skull
 
3:22 PM
right :-/
They flagged my Happy 4th of July post in the homotopy room :(
 
well, it was trivial up to homotopy
 
r9m
@TedShifrin :P lol ..
 
@TedShifrin have you heard the story about the single added sentence on Einstein's dissertation?
 
nope
 
3:27 PM
In his biography of Einstein, Carl Seelig reports: "Einstein later laughingly recounted that his dissertation was first returned by Kleiner with the comment that it was too short. After he had added a single sentence, it was accepted without further comment."
 
What was that sentence?
 
Dunno
 
Sorta ruins the story :P
 
I just wanted to know how long it was :P
 
@skillpatrol @TedShifrin kofegeek.wordpress.com/2012/05/16/… that explains a lot
 
3:32 PM
Still doesn't answer either of our questions :P
But thanks, Hippa.
 
Heh, thanks for humoring me, @TedShifrin. I told you I couldn't help myself :)
 
More substantive complaints would be better, @pjs.
 
Hey
 
Noted, @TedShifrin
 
@Hippalectryon thanks for the link
 
3:34 PM
So I came across a PDF which says that we should discard formalism and focus on the practical applications of maths
specifically "western formalism"
It basically starts on how early trig. and calculus was invented in India, and then stolen by the Europeans
 
hi @Ted
 
"stolen"
 
hi @Balarka
 
@SohamChowdhury what's the corrected version?
 
and thus the great war to recover trig started
 
3:37 PM
@BalarkaSen any clue about this?
 
clue about what?
 
@BalarkaSen this article about how the europeans stole and ruined trigonometry and calculus, something invented in India
and that we must discard their formalism
hold on I'll link it to you
@Hippalectryon :D
 
I don't think I want to read it right now. But link me, I'll look later.
 
Maybe we should discard that pdf too
 
3:40 PM
maybe, but I'm not qualified enough to comment on it
 
3:58 PM
@BalarkaSen it's up
@VibhavPant lots of angry RSS-isms in there, especially much railing against "Western philosophy"
I doubt Euler stole all his methods from India. :)
 
yep
I really hope this doesn't find its way into indian institutues
 
And even the Wikipedia page calls it the "Madhavacharya-Leibniz series".
Next thing you know, Rajnath Singh's going around saying etale cohomology was discovered in India in the 9th century. Sheesh.
 
haha
 
It's got a very superior tone, and just a bit revisionist.
 
And the end is fully crank-mode philosophy and whatnot.
 
4:07 PM
It's not like Europe intentionally stole from India. From the Muslim world, maybe, and then India by proxy, but it's not like we call them "Indian numerals".
 
yep :/
 
And he apparently doesn't accept the law of the excluded middle.
Oh, yes. He has problems with $\Bbb R$. People, we have found the Indian Wildberger.
 
that part was really wierd
 
"Without efficient techniques of calculation how could the Greeks do science?" Uhh, thinking?
Science in no way requires math.
 
And he's renamed the Grandi series after himself. Weird.
Oh, god.
Anyway, I have work.
 
4:12 PM
What on earth is that @SohamChowdhury
 
"The word “trigonometry” indicates a conceptual misunderstanding: the concepts relate to a circle (chord, jya) not a triangle." ...because you can't draw a right triangle inside a circle.
Like, literally, this guy's gripe seems to be "someone conceptualized a set of ideas in a different way and used a different linguistic tradition to refer to it", which is, well, pointless.
I have to stop reading this or I'll suffer an aneurysm.
 
its like one of those "NASA proves the Bible is true" things, except with a bit more logic
 
I mean, he's not wrong about Eurocentrist history, and that sort of thing. It's just he mixes that with a misunderstanding of cultural diffusion and of the rudiments of mathematics itself.
"This way was first so clearly it's better."
Well, not a misunderstanding of the rudiments of math, but a misunderstanding of "Western understanding".
 
@Fargle this sort of stuff has been gaining some popularity here
that the west has entirely plagiarized Indian mathematics and ruined it with their way of thinking
 
@VibhavPant I can see why, honestly. It appeals to the basest nationalist inside people.
 
4:21 PM
yep
 
And it also reeks of second-option bias.
I don't think any mathematician worth his or her salt would deny the effect Indian mathematics has had on our general tradition, but to present such an east-west dichotomy is to grossly simplify the history of academics in general, and math in particular.
 
@r9m It was all of the meer2kat stuff, I think!
Does anybody know why we need the absolute value of $t_{i+1} - t_{i}$ in the summand of the integral of the step function?
 
@KhallilBenyattou Strictly speaking, you don't by definition, but I guess they wanted to be extra careful.
 
Thank you, @Fargle! I have another question that I posted yesterday, but not many people were online. It's short, I promise!
yesterday, by Khallil Benyattou
user image
Should it not be $P \cap \left[ u, w \right]$ instead of the open interval, @Fargle?
 
@KhallilBenyattou I don't rightly know. I think either could work--no restrictions are placed upon endpoints of the partition, but if there were restrictions, $P \cap [u,w]$ wouldn't work, I think.
(Why do I always think /intersect is a LaTeX command? Hrgmnblgr.)
 
4:33 PM
\cap
 
Thanks. >_>
Not even thinking.
 
I see what you mean, @Fargle.
Haha, not at all! \cap is one of the least obvious commands in my opinion!
 
I use(d) it all the time! It's just that "\intersect" is always my first instinct. But "\cup" is my instinct for unions. I'm all over the place.
 
r9m
5:09 PM
@KhallilBenyattou :P
 
hi @Rememberme
 
r9m
someone flagged that here in chat room? -_-
 
Hello@BalarkaSen
 
@Rememberme $X$ be a top. space. It's said to be path-connected if for any two points $x_0, x_1 \in X$, there is a "path between them", i.e., a cont. map $f : [0, 1] \to X$ such that $f(0) = x_0$ and $f(1) = x_1$.
 
I think so, @r9m! :-P
 
5:15 PM
Ex : Give an example of a path-conneccted space.
@Rememberme Do the above exercise when you get time, I have an interesting exercise for you afterwards.
hello @Khallil. How is things?
 
Okhay....
 
Not too bad, @Balarka. How've you been? ^_^
 
r9m
@KhallilBenyattou :P lol .. hard to know what made people to flag that ... :P nyway it's better not to reveal all this digging stuff here .. I'm on your fb right?
 
I've been studying algebraic topology. What kind of math have you been thinking about?
 
I'm not sure, @r9m! :-)
 
5:18 PM
@BalarkaSen Did you think about the kurawotski set I was talking about ?
 
r9m
@KhallilBenyattou neither am I .. wait lemme check :P
 
@Remember nah, been busy getting something right.
 
Just been doing some analysis, step functions, regulated integrals, supremum norms etc. @Balarka :-)
 
Galois stuff?@BalarkaSen
 
galois theory/covering spaces, yes.
@KhallilBenyattou cool.
 
5:19 PM
I've only heard good things about topology. Algebra is good fun as well so a mixture seems cool!
 
Both are good. But algebraic topology is of a different flavor than algebra and topology.
 
@BalarkaSen though I haven't proven this yet...
I think R^2 is path connected.... Because you require path connectedness for proving that R and R^2 are not homeomorphic
 
yes, R^2 is path connected. but you now know the defn. of path connectedness, so prove it!
 
Yup....
 
Hint : What does path-connectedness really mean, geometrically?
 
5:24 PM
That there exists a path between them which doesn't have any holes....?(I am really pedantic this time ) @BalarkaSen
 
well, yeah, it just means that you can join any two points in your space by a continuous line ("path")
can you do that with $\Bbb R^2$?
 
r9m
@KhallilBenyattou nah! can't find you on fb man ..
 
Damn, really?
 
r9m
I have 5 weeks of naruto due ... gotta watch em right now
@KhallilBenyattou you are on @Sawarnik imp's fb frnds list right?
 
What does imp's mean, @r9m?
 
5:29 PM
Yes you can .....
If we think of the punctured plane lets say without the point (0,0) then we can still form a continuous line @BalarkaSen(since we are in a plane)
 
r9m
@KhallilBenyattou imp = 'a small, mischievous devil or sprite' (google)
 
Important @KhallilBenyattou
 
I didn't tell you to think about the punctured plane, @Remember
Think about $\Bbb R^2$
First, prove that it is path-connected.
 
I mean a R^2 without a point..... Let that point be (0,0)@BalarkaSen
 
I don't think I have Sawarnik on fb, @r9m.
 
5:31 PM
No, prove it for R^2 first.
 
Can I ask what 'dense' means, @Balarka?
 
Okay...
 
r9m
@KhallilBenyattou oh! wait then .. I need to find you :P
 
The closure equals the set.... I believe @KhallilBenyattou
 
r9m
@KhallilBenyattou dense = a person with thick head :P
 
5:32 PM
Wait...
 
@KhallilBenyattou In real analysis?
 
Oh I forgot to write some important things over there @KhallilBenyattou
 
$A \subset B$ be subsets of $\Bbb R$. $A$ is called dense in $B$ if every point of $B$ is either a point of $A$ or a limit point of $A$.
you know what limit points are, right?
 
r9m
@KhallilBenyattou is your profile pic the reimann zeta function written in white on a black background?
 
Yes, @r9m!
I do now!
 
r9m
5:41 PM
@KhallilBenyattou ah! found you on google+ and facebook :) I think you have Sawarnik on your g+ circles
 
That must be it, @r9m!
I hardly ever use google+ :-P
 
r9m
@KhallilBenyattou alrighty! added you in both :D
 
So for $A$ to be dense in $B$, every point of $B$ is either in $A$ or $\forall \epsilon > 0$, within an $\epsilon$ of the points of $A$, @Balarka?
 
yes
example : $\Bbb Q$ in $\Bbb R$
 
Does the same apply to two dimensional spaces, @Balarka?
 
5:48 PM
"two dimensional" is a bit vague, but sure.
$\Bbb Q \times \Bbb Q$ is dense in $\Bbb R^2$
 
How can I make two dimensional more precise?
 
Hard question for arbitrary topological spaces : goes back to dimension theory.
 
The dimension is just the cardinality of the number of linearly independent vectors that form a basis of the space, right?
 
Huy
That's a vector space.
 
r9m
@KhallilBenyattou 'within an ϵ' = within a ball of radius $\epsilon$ .. whenever there is a metric definable on $A$ .. else we need the notion of open sets (need to define a topology on $A$ to talk about limit points)
 
5:51 PM
that's dimension in vec. spaces.
wildly different than dimension in top. spaces
@r9m I think he's talking about subsets in R
 
r9m
@BalarkaSen ah! that's fine then
 
What kinds of things do people learn when introduced to topology, @Balarka and @r9m?
(Do metric spaces fall under topology?)
 
how to generalize the notion of "distance" in arbitrary sets
 
Huy
@KhallilBenyattou: Do you know what a topology is yet?
 
r9m
@KhallilBenyattou take a look at Munkres Topology when you have time .. :-)
 
5:54 PM
Not at all, @Huy! :-P
 
and do "analysis with sets"
(not literally correct, as you can't do anything with a barren set. you use the notion of distance.)
 
Huy
I think so, yes, but I think if you're still studying calc1/2 it would serve you better to understand the concepts in the presented way first, then generalizing to topology later.
 
What do calc1/2 entail, @Huy?
 
Huy
@KhallilBenyattou: I don't know, over here it usually contains sequences, series, a bit of topology, continuity, differentiation and integration on $\mathbb{R}^n$ and such.
@KhallilBenyattou: Some universities call such courses analysis, or even real analysis, I don't really know the difference between all the courses offered at many English/American universities, over here we simply have "Analysis 1 / 2".
 
6:03 PM
@anon you're here?
 
ish
 
cool.
 
Yep, I've done analysis 1 and 2! We get an analysis 3 next year as well. It's to do with step functions, regulated functions and I think some norms at the end. (Which is why I asked that question about the supremum norm earlier) @Huy
 
I have a question, but it might take a few seconds. Commutative diagrams are involved.
 
Huy
@KhallilBenyattou: If you have a set $M$ you can basically have a family $\tau$ of subsets of it. This family is called topology if it satisfies certain properties, which are properties open sets in $\mathbb{R}^n$ satisfy. In $\mathbb{R}^n$ you would typically define open balls $B_r(x) := \{p \in \mathbb{R}^n| |p-x| < r\}$ and use those to define what it means to have an open set. In topology you don't always need/want a metric so you just have a topology $\tau$ as the set consisting of open set
s.
@KhallilBenyattou: To be more specific, the properties are that $\emptyset \in \tau$, $M \in \tau$, any union of elements of $\tau$ is an element of $\tau$ again, and lastly any finite intersection of elements of $\tau$ is in $\tau$ again. I'm sure you've proved in $\mathbb{R}^n$ that those statements hold for open sets there, right?
@KhallilBenyattou: These open sets are then used to talk about the concepts of continuity, compactness and connectedness, in basic topology at least. I've never delved much deeper than that basic topology, so I can't really tell you more. :P
 
6:10 PM
The term open set doesn't seem too familiar but it all seems to make some intuitive sense
So open sets generalise the idea of an open interval in $\mathbb{R}$?
 
Huy
@KhallilBenyattou: Really? I guess the way they teach analysis differs at universities a lot.
@KhallilBenyattou: That's a way to think of it.
@KhallilBenyattou: I gave you the properties a topology satisfies in general before. So if open intervals are "open sets", then any union of them are too, as is any finite intersection. And there's also $\mathbb{R}$ and $\emptyset$.
 
$p : E \to X$ be a covering map. Then for some $x \in X$, the fiber $p^{-1}(x)$, by definition, is the pullback of the diagram $$\require{AMScd}
\begin{CD}
x \times_X E @>>> E\\
@VVV @VVV \\
x @>{i}>> X
\end{CD}$$ Where $i$ is the inclusion map of the point. I tried drawing the corresponding diagram for fields, and it looked like this $$\require{AMScd}
\begin{CD}
K \otimes_k \overline{k} @<<< K\\
@AAA @AAA \\
\overline{k} @<{i}<< k
\end{CD}$$ where $i$ is again the inclusion (the algebraic version of a "point", by the discussion we have had previously) and the leftmost map is the inclusion
Note that the second diagram is a pushout, as is natural : the corresponding things in field theory seems to get inverted (i.e., there should be a contravariant correspondence, if any)
And I am taking pushouts as rings : fields don't have pushouts (I think?)
 
Huy
@KhallilBenyattou: In $\mathbb{R}^n$ you don't have open intervals anymore, which is why we take open balls as I defined earlier in case you didn't have them yet. Then, in $\mathbb{R}^n$, you say a set $O$ is open if for any point $x \in O$, there exists an $\epsilon > 0$ such that $B_\epsilon(x) \subset O$. In words, no matter where you are in $O$, you can always put a (small) ball around yourself and that ball is still fully in $O$. Can you see why this holds for open intervals ($\mathbb{R}$)?
 
So : does it make sense to say $K \otimes_k \overline{k}$ is the correct analogy for fibers?
 
6:18 PM
iunno
good idea though
 
I don't know why it should. In my opinion, fibers should be a subset of $\mathsf{Hom}_k(K, \overline{k})$ : "fibers" of a "point" $k \hookrightarrow \overline{k}$ of $k$ should also be a "point" $K \hookrightarrow \overline{k}$.
I am confuzzled by all these possible analogies.
 
I hadn't known about open balls before, @Huy!
(Sorry for the late reply!)
 
Huy
@KhallilBenyattou: Well now you know!
@KhallilBenyattou: Open intervals are open balls, at least in $\mathbb{R}$. :)
 
@Huy It holds for open intervals in $\mathbb{R}$ by the completeness axiom, right?
 
@anon oh well. thanks for looking, though!
 
Huy
6:20 PM
Urm, not sure, I never studied these kinds of axioms too much.
 
@KhallilBenyattou Somewhat indirectly, yes.
 
Huy
@KhallilBenyattou: But if you take for example an interval $(0,1)$ and look at the point $x = \frac{3}{4}$, it should be quite obvious that any ball with radius $r < \frac{1}{4}$ centered around $x$ will be completely in the interval. The same kind of argument can be made for the general case.
 
@BalarkaSen note that $\hom_k(K,\bar{k})$ is $K^*\otimes_k\bar{k}$ (in Vect), so maybe think about directions of things if you're getting $K$ when you want $K^*$
 
If $x \in (a,b)$, choose $\epsilon > 0$ such that $x - \epsilon > a$ and $x + \epsilon < b$, then $(x - \epsilon, x + \epsilon) \subset (a,b)$.
 
@anon hmm. interesting point.
 
6:23 PM
Yep, this makes sense @Huy and @Fargle!
 
although yeah, Aut(p) acts on the lifts of x->X just as Gal(K/k) acts on lifts of k-embeddings of K into k^alg.
 
We are free to make such a choice because $\Bbb R$ is a continuum--there are real numbers between any real numbers.
 
That's the word I was looking for! Continuum :-P
 
Good! Topology's very odd and abstract at first, but using the mental example of open intervals/open balls in $\Bbb R^n$ as the general idea of open sets is pretty helpful. (Though a bit limiting, when it comes time to define more general topologies.)
 
6:49 PM
hey guys
I think it is helpful to visualize some concepts in topology
 
7:10 PM
if you have a polynomial with real coefficients, say $f(x)$, and you let $w$ be a fake number such that $w \neq 0$ but $w^2 = 0$ then $f(x+w) = f(x) + wf'(x)$
 
How is that even possible, @Samuel?
 
how is what possible?
 
$w \neq 0$ but $w^2 = 0$ ...
 
oh, well certainly no such real number exists
but if you pretend one does then everything works
the intuition should be that $w$ is really tiny
 
one can invent a new algebra containing the real numbers and a thing w for which w^2=0 but w is nonzero. in abstract algebra we would be talking about the "dual numbers" R[w]/(w^2)
 
7:15 PM
yep yep yep
I just like that it's not so hard to phrase this at the level of a good high school student
things one can say at that level are somehow more true to me, whatever that means
 
and if you have a new number v with v^3=0 but v^2 not 0, then f(x+v)=f(x)+f'(x)v+[f''(x)/2]v^2.
 
yeah I was thinking about that too, but I didn't work it out
 
the whole taylor series for f(x+h) is valid in the two-variable polynomial ring R[x,h]
(since f is a polynomial, its derivative will always eventually vanish, so the taylor series is always a finite sum)
 
and I suppose that's the logical conclusion of that then
nice
 

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