« first day (3215 days earlier)      last day (121 days later) » 
00:00 - 19:0019:00 - 23:00

12:22 AM
Homomorphisms preserve linear structure, homeomorphisms preserve topological structure. Neither of them have to do with differentiability. That the concepts are different is apparent in the definition. The derivative of a holomorphic function $\mathbb{C}\rightarrow\mathbb{C}$ at a point is a complex number, whereas the derivative of a differentiable function $\mathbb{R}^2\rightarrow\mathbb{R}^2$ at a point is a linear operator.
@Thorgott Thank you :)
Does there exist any connection with diffeomorphism? I have seen this concept in manifolds.
if a map is diffeomorphism then differentiabilty structure same?
At what case differentiability structure same?
1 hour later…
1:46 AM
@N.Maneesh: Yes, when you have diffeomorphic objects, they are "identical" so far as differentiable functions are concerned. If they're $C^k$ diffeomorphic, then we talk about $C^k$ functions. If they're smoothly diffeomorphic, then we talk about smooth functions.
2:08 AM
@TedShifrin sorry. Thank you very much
1 hour later…
3:31 AM
If $A$ be $3\times3$ Matrix with Characteristic polynomial with characteristic polynomial $x^3-3x+a=0$ for what values of $a$, $A$ must be diagonalizable?
$(A) a\in (-2,2)$$ (B)a\in [-10,2] (C)a\in \mathbb R (D) $ No values of $a$
What shall I care about this problem? Should I check the diagonalizability over $\mathbb C$ or $\mathbb R$?
4:05 AM
@Thorgott I thought so.
Please don't empty ping. Ask something if you have to.
I have
I saw your thing above. I don't want to check it in detail, but you can give a cleaner argument by noting that if $x$ is a unique element of order $2$ in $G$, then $\langle x \rangle$ is normal in $G$.
I want you to do that, instead of listing down the multiplication table by hand or whatever
4:21 AM
Is it correct to say that $\lim_{n \rightarrow \infty} a_n = \{x \in \mathbb{Q} \mid \exists n \in \mathbb{N} : \forall m \in \mathbb{N} : m > n \rightarrow a_m > x\}$ ?
Where the right hand side is interpreted as a Dedekind cut
Subhasis do you have any idea on the diagonalizability problem?
@N.Maneesh no. absolutely not. I won't touch other people's problem unless I am pretty confident about it
@user76284 are you concerned about the notational validity of the left hand side?
@N.Maneesh I will come back to it once it learn how to diagonalize
I think irreducibilty come in to play here. Irreducibility of polynomial over a field.
4:56 AM
@SubhasisBiswas I want to express the correct Dedekind cut corresponding to the limit of any Cauchy sequence of rationals.
5:24 AM
@BalarkaSen, Here it goes.

Proof of the statement (you mentioned):

If $a$ be a unique element of order $m$ in $J$, then $a \in Z(J)$.

Proof: Consider the element $jaj^{-1}$.

In this case $x \in Z(G)$. So, for any element $g \in G$


Evidently, $<x>$ is normal in $G$. Here, $Z(G)=<x>$ [since the only possible order of subgroups are $3$ and $2$. If any other element, of order $3$, say $p$ belongs to $Z(G)$, then $p^2=x \implies p^3=e=xp \implies p=x^{-1} $ But, $o(x^{-1}=2$, contradiction ] The quotient group $G/<x>$ exists. But, $o(G/<x>)=o(G)/o(<x>
That's right
@BalarkaSen done?
Sure, sounds like it. You have proved $\Bbb Z_6$ and $S_3$ are the only groups of order $6$ with very general techniques (eg, you can try to classify groups of order $pq$, $p \neq q$ now)
Note that both $\Bbb Z_6$ and $S_3$ have a normal subgroup isomorphic to $\Bbb Z_3$ which, upon quotienting, gives a $\Bbb Z_2$, which is something you were asking yesterday if I recall correctly.
I will try it sometime later. I have to cover a huge amount of syllabus (applied, which I hate with a passion).
@BalarkaSen can't remember. I remember whether $G \cong G'$ and $H \cong H' \implies G/H \cong G'/H'$ in general
turned out to be false, tho
Hah, ok, whatever. There are three scenarios (1) $G \cong G'$, $N \cong N'$, is $G/N \cong G'/N'$? (2) $N \cong N'$, $G/N \cong G'/N'$, is $G \cong G'$? (3) $G \cong G'$, $G/N \cong G'/N'$, is $N \cong N'$?
(1) Take $G = G' = \Bbb Z_4 \times \Bbb Z_2$ and $N = \Bbb Z_2 \times \{0\}$, $N' = \{0\} \times \Bbb Z_2$. (2) $G = S_3$, $G = \Bbb Z_6$, $N = N' = \Bbb Z_3$ (3) $G = G' = \Bbb Z_4 \times \Bbb Z_2$, $N = \Bbb Z_2 \times \Bbb Z_2$, $N' = \Bbb Z_4 \times \{0\}$.
So the answer to all three are "No". The middle one is the most interesting, though.
Here's some abstraction: Suppose $N, G, H$ are groups and $i : N \to G$ and $q : G \to H$ are homomorphisms which fit in the following "diagram" $$0 \to N \stackrel{i}{\to} G \stackrel{q}{\to} H \to 0$$ This sequence of arrows is called a short exact sequence if kernel of any arrow is image of the arrow before it. This in particular implies $i$ is injective, $q$ is surjective, and $\ker q = \text{im}\, i$.
Note that by first isomorphism theorem, $G/i(N) \cong H$.
Here's a classical problem in group theory: Given $N$ and $H$ in a short exact sequence as above, what groups $G$ can fit in the middle? This is called the "Extension problem"
You gave an example where there are two groups that fit in the middle in $0 \to \Bbb Z_3 \to G \to \Bbb Z_2 \to 0$, namely, either $\Bbb Z_6$ or $S_3$.
There are no other groups of order $6$, you proved, so nothing else can fit. These are the only solutions.
5:43 AM
Groups of order 4 yesterday, groups of order 6 today, at this rate it'll get pretty hard in a few weeks :P
I'll get him to classify finite simple groups in a month
Good luck lol
6:08 AM
$p$ and $q$ are prime, right? @BalarkaSen
Yes, of course
This is what I have done so far:
$Z(G)$ cannot be a proper non trivial subgroup of $G$
(It can be, in that case $G$ is abelian EDIT: Oh, proper, ok)
I agree.
@BalarkaSen since, $o(Z(G))=p $ or $q$. So, $o(G/Z)=p$ or $q$. Quotient group is cyclic, which implies the entire group is commutative
6:10 AM
Now, in this case, the exponent is of order $q (>p)$.
By Cauchy's theorem, such an element exists.
But, $q$ and $p$ are distinct primes. So, $p \nmid q$.
What? No. $S_3$ is a non-cyclic group with an element of order $3$.
I removed it lol
So, for a group of order $pq$, the group cannot be commutative
? take the cyclic group
@BalarkaSen I meant non cyclic, and commutative
That requires a proof. Why?
6:19 AM
Previous argument. Since, the non identity elements will be of order $p$ and $q$, and the exponent will have an order of $q$ (the order cannot be higher, since it will then become cyclic). Again, $p \nmid q$, but that is a contradiction since in an abelian group the order of all the elements divide the order of the exponent
And Cauchy's theorem ensures that the must exist an element of order $q$. $G$ being abelian
@BalarkaSen, good so far?
So the classification begins as:

1. Cyclic group of order $pq$

2. non-abelian groups of order $pq$
@SubhasisBiswas Can you explicitly construct an element of order $pq$ in an abelian group of order $pq$?
@BalarkaSen what does "explicitly construct" mean?
i'll try to answer
You mention Cauchy's theorem for example. Try to crank an element of order $pq$ from that.
But, that'd make the group cyclic, right (no problem with that anyway, I'll try to construct)?
Yes, that'd give a constructive proof that every abelian group of order $pq$ is cyclic.
6:28 AM
nice. let me try
6:45 AM
There must exist elements $a$ and $b$ with $o(a)=p$ and $o(b)=q$, $\gcd(p,q)=1$

Let $o(ab)=k$. $(ab)^{pq}=e \implies pq|k$. $(ab)^{qk}=a^{qk}=e \implies p|qk \implies p|k$. Similarly, $q|k$. So, $pq|k$. Therefore $o(ab)=pq$.
Looks good.
@BalarkaSen wish some pretty girl said that to me :'(
Ok, how much depth do I have?
In group theory. What are my weaknesses
Here's a random detour (but connects back to everything, I promise): Suppose $H, N$ are subgroups of a group $G$ such that $N$ is moreover normal. Consider $HN = \{hn : h \in H, n \in N\}$. Check that $HN$ is a subgroup of $G$.
Suppose $G = HN$, $H \cap N = \{e\}$ and $H$ is also normal in $G$. Then prove that $G \cong H \times N$.
$N$ is normal, right?
6:52 AM
Let me try. Looks tough
It's not hard.
no hints rn
@SubhasisBiswas You're doing pretty good. I'd advice not to evaluate yourself and just enjoy the math though
It's not worth evaluating if you're doing well or not at the end of the day, there's just bound to be things you're good at and things you're bad at - who cares? It's the math that's interesting
@Balarka if I remove the basepoint from the reduced suspension of $X$, do I just get $X\times \Bbb R$?
@BalarkaSen, very true though.
@AlessandroCodenotti where is this question from , I mean, the topic?
6:56 AM
@AlessandroCodenotti Reduced suspension of $S^1$ is $S^2$, and removing the basepoint gives a disk, not a cylinder!
Topology but I'm asking it to solve a doubt I have in noncommutative geometry concerning K-theory of $C^\ast$-algebras
@BalarkaSen isn't it $S^2$ with two poles identified?
No! $\Sigma X$ is $X \times [0, 1]$ with $X \times \{0\} \sqcup \{x_0\} \times [0, 1] \sqcup X \times \{1\}$ collapsed to a point, isn't it?
Ah right I see
If you just collapsed $X \times \{0\} \sqcup X \times \{1\}$ to a point you'd get, in case of $X = S^1$, a sphere with north and south pole identified.
Agreed, thanks
Urgh why is $X\times\Bbb R$ called the suspension of $X$ in ncg
7:01 AM
Rofl what
I guess that's because you want the suspension of the $C^\ast$-algebra $A$ to be $C_0(\Bbb R,A)$
Weird, why is that the correct analogue?
Do you get some natural isomorphism between $K_*(A)$ and $K_*(\Sigma A)$ that increases degree by $1$, or something?
Dunno, the professor just wanted to define higher K-theory very quickly to show us "it can be done", but we're not goong to use it
I never understood higher K-theory even in the topological context to be honest
@BalarkaSen is this sarcasm or is there really something that even you don't understand?
7:16 AM
I don't understand any topological K-theory
What's $K^0(S^1)$?
There is a unique nontrivial rank $k$ bundle on $S^1$ upto stable equivalence for every $k$: the Mobius bundle
So uh, $\Bbb Z$?
all of those various moebius bundles of different ranks are stably equivalent I mean
$\Bbb Z$ seems reasonable, but what is the Möbius bundle?
No, $\Bbb Z_2$, right? Mobius \sum Mobius is trivial
@Alessandro Upto isomorphism there are exactly two rank $k$ bundles on $S^1$, one is the trivial guy and another is the twisted guy, exactly like in $k = 1$
Let $M$ be the Moebius line bundle on $S^1$, then the rank $k$ Moebius bundle is isomorphic to $M \oplus \underline{\Bbb R}^{k-1}$
Seems reasonable
So all these various Moebius bundles are equivalent upto direct sum with a trivial bundle
That means there are exactly two elements in $K^0(S^1)$, the class of the trivial bundle, and the class of the Moebius bundle
So it is $\Bbb Z_2$ I think
Which makes sense, because $M \oplus M \cong \Bbb R^2$ which is trivial
7:32 AM
Uh. I'll try to understand this from the $C^\ast$-algebras point of view in half a hour or so
Let me know if you get $\Bbb Z_2$
Hello. I am trying to figure out how this formula should be evaluated (which result is vector [w1,w2,w3]) :
RT is a (transposed) matrix
I am not asking what the formula does or the meaning each variable (this is OK)
What is not clear to me is those values inside round parentheses, should I treat this as matrices (and so multiply several matrices together to evaluate the formula) ?
I have to understand $C(S^1)$ first though
I checked Hatcher, and apparently what I'm doing is the reduced K-group. $\widetilde{K}^0(X)$ is the group of stable equivalence classes of vector bundles over $X$ equipped with direct sum
The unreduced thing is this $\oplus \Bbb Z$
8:03 AM
Let $f:[0,1]\to\Bbb R$ be a differentiable function such that $f'(x)\ne0$ for all $x\in(0,1)$. Then which of the following statements is false?
a) $f$ monotone. b) $f$ one-to-one. c) $f$ uniformly continuous. d) None of these.
I think d is the answer, right? because e.g. there are monotone as well as non-monotone functions with non-zero derivatives, and so on for b and c as well.
Sorry! It should be c, since because of intermediate value theorem for derivatives, hence either $f'(x)>0$ or $f'(x)<0$ for all $x\in(0,1)$, so monotone, and one-to-one.
Am i correct?
8:51 AM
$f$ must be continuous, right?
continuous on a compact interval means uniformly continuous
9:06 AM
9:17 AM
@SubhasisBiswas, thank you. I am sorry that I wrote $f:[0,1]\to\Bbb R$, the question I have says $f:(0,1)\to\Bbb R$
9:42 AM
Then consider $f(x)=1/x$
@Silent You're completely correct in your argument, but that has nothing whatsoever to do with (c) :P
It's (a) and (b).
@BalarkaSen option (c) is wrong
that's the answer
Ah, which of them are false.
Weird but ok
I suppose "None of these is true" is also false, because certainly (a) and (b) are!
So (d) is wrong as well :P
10:01 AM
@BalarkaSen plis
To show that $f:(0,1) \to \mathbb{R}$ is not uniformly continuous on $(0,1)$, we pick the sequences $\{1/n\}=(u_n)$ and $\{\frac{1}{n+1}\} =(v_n) $. Now, $\{|u_n-v_n|\}$ converges to zero, but, $|f(u_n)-f(v_n)|=|1| \nrightarrow 0$
$\sum a_n$ is a convergent series of positive terms. What can you say about the convergence of $\sum\frac{ \sqrt{a_n}}{n}$? Since, $\sum a_n$ is a convergent series of positive terms. For $\epsilon=1$, There is a natural number corresponding to it. such that $a_n <1$. So, $\sqrt{a_n}<1$. I am not able to use comparison test. please help me.
Looks like Cauchy-Schwarz to me
yes. you are correct
10:17 AM
Does that answer your question then? :)
Hi everyone
Can I ask a probability stats question here?
I tried at Basic Maths but the only person there left
"Just ask; don't ask to ask"
If someone wants to answer they will
If the standard error of the mean is 10 for N = 12, what is the standard error
of the mean for N = 22?
I know that standard error is nothing other than the standard deviation of the sampling distribution
yes. All kind of stuff hidden in some corner of my brain:? I should take chat to my exam hall :/
I also know that sd of sampling distribution is equal to the population sd/square root of population
However the answer I am getting is completely different from what they gave
Any help would be appreciated
10:29 AM
@BalarkaSen Going off this post, the answer to the question does turn out to be affirmative as a consequence of Hahn-Banach; math.stackexchange.com/a/2440188/422019
Nice. It makes sense, Hahn-Banach is precisely what you need to extend bounded linear functions on subspaces to the whole normed space.
10:48 AM
@Balarka turns out $C^\ast$-algebras are too hard and I don't know how to figure our their K-theory explicitely in any nontrivial case
By Serre-Swan $K^0(S^1)=K_0(\mathcal C(S^1))$, so one only needs to classify finitely generated projective $\mathcal C(S^1)$-modules up to isomorphism, which sounds awful
We also have a different definition of $K_0(A)$ for a $C^\ast$-algebra based on stable equivalence of projections in $M_\infty(A)$, which doesn't sound much easier to compute...
@AlessandroCodenotti By Serre-Swan again those correspond to vector bundles over $S^1$ again: If you trust me there are exactly two such things for every rank $k$, then you only have to figure out what the modules corresponding to those are. For the trivial bundle, the corresponding module is free of rank $k$. Now what would be the space of sections of the Mobius bundle?
Let's try $k = 1$. Mobius line bundle on $S^1$.
Is it possible to explicitly write down what the $C(S^1)$-module $\Gamma(S^1, \text{Mob})$ is?
I don't know, maybe?
Let $u : C(S^1) \to C([0, 1])$ be the map obtained from dualizing $[0, 1] \to S^1$ which maps the interval diffeomorphically to the upper hemicircle.
Consider $\Gamma(S^1, \text{Mob}) \otimes_{C(S^1)} C[0, 1]$, where $C[0, 1]$ is considered as a $C(S^1)$-module by the ring homomorphism $u$.
This is a $C[0, 1]$-module now; I have extended scalars. This should be free.
Extension of scalars of a projective module is projective, yeah?
11:02 AM
@BalarkaSen Uhm I'm not sure actually, but it sounds very believable
Tensor product commutes with direct sum
DoesnÄt it
@ÍgjøgnumMeg Oh of course, I always think about the lifting property, but direct summand of a free module is a much nicer definition of projective often enough
Let's work it out. $f : A \to B$ ring hom, $P$ projective $A$-module, so $P \oplus Q$ is free.
@Alessandro Right
$(P \otimes_A B) \oplus (Q \otimes_A B) \cong (P \oplus Q) \otimes_A B$, free again
Great, so the thing above is a f.g. projective $C[0, 1]$-module. Serre-Swan of course tells you all such things are free, but is there a purely algebraic proof?
I doubt. You'd need machinery.
Anyway, do the same thing with the lower hemisphere map $l : C(S^1) \to C[0, 1]$ to get another free $C[0, 1]$-module. Let's call these two things $\Gamma \otimes_u C[0, 1]$ and $\Gamma \otimes_l C[0, 1]$.
There should be a way to recover $\Gamma$ from these two free modules.
11:09 AM
Hmmm it looks like just working in the topological setting is much nicer than trying to translate the topological argument to an algebraic one :P
$$\require{AMScd}\begin{CD}C(S^1) @>{u}>> C[0, 1] \\ @V{l}VV @VVV \\ C[0, 1] @>>> C(\{0, 1\})\end{CD}$$
This diagram commutes, where the maps $C[0, 1] \to C(\{0, 1\})$ are just restriction to the boundary
$C(S^1)$ is a pullback in CRing, basically.
I have two free $C[0, 1]$-modules over the top-right and bottom-left corners of this diagram, obtained from extending scalars of $\Gamma$ over the top-left corner.
Ok, extend scalars of this free $C[0, 1]$-module to get a free module over $C(\{0, 1\})$. Then we have a corresponding pullback diagram of modules over this pullback diagram, and the whole cube commutes.
Some argument here should tell you $\Gamma$ is uniquely determined by this free module over $C(\{0, 1\})$.
But $C(\{0, 1\})$ is $\Bbb R\oplus \Bbb R$. There are exactly two free modules over rank $k$ over $\Bbb R\oplus \Bbb R$ upto isomorphism, yeah?
One is $\Bbb R^k \oplus \Bbb R^k$ where both copies of $\Bbb R$ in $\Bbb R \oplus \Bbb R$ act by $c \cdot v = cv$ and another is when one acts by $c \cdot v = cv$ and the other acts by $c \cdot v = -cv$. These are not isomorphic as $\Bbb R \oplus \Bbb R$-modules.
Something like that.
@AlessandroCodenotti Modulo details, this tells you there are two f.g. projective modules of rank $k$ over $C(S^1)$ :D
11:33 AM
This looks awful but I can believe it works
This is just a hack algebrification of the clutching construction
12:31 PM
why isn't there any integer solution to $a^a=b!-b$ ?
$(1,0)$ certainly is a solution
I'd say it looks like a^a grows much faster than b! - b
so I guess you'd only have finitely many possibilities for solutions or smth
@Alessandro I didn't give you a correct description of the gluing. There is only one module upto isomorphism over $\Bbb R \oplus \Bbb R$ (modules over $A \times B$ are really nothing but pairs $(M, N)$ where $M$ is an $A$-module and $N$ is a $B$-module). That wasn't the right thing to say.
The point I think is two different free rank $k$ modules on $C[0, 1]$ extend by scalars to two different free rank $k$ modules on $\Bbb R \oplus \Bbb R$; they are isomorphic but there's a $\text{GL}_k(\Bbb R) \times \text{GL}_k(\Bbb R)$'s worth of isomorphisms.
If this pair of elements of $\text{GL}_k(\Bbb R)$ have determinants of the same sign, $\Gamma$ if free. Otherwise it's not.
13 mins ago, by Mathphile
why isn't there any integer solution to $a^a=b!-b$ ?
this seems quite similar to Brocard's problem
12:48 PM
Yo what's up?
Mentioned this in Washington but turns out class groups give nice ways to solve certain Diophantine equations
Oh yeah I dunno anything about that
So consider for example $y^2 = x^3 - 54$. You write that as $x^3 = (y+3\sqrt{-6})(y-3\sqrt{-6})$. Well, if we have a prime ideal $\mathfrak{p}$ which divides both $(y+3\sqrt{-6})$ and $(y-3\sqrt{-6})$, then it divides their sum, which contains (and thus divides) $(6\sqrt{-6}) = (\sqrt{-6})^3$. By unique factorization in a Dedekind domain, $\mathfrak{p} = \sqrt{-6}$
In particular, $\mathfrak{p}$ is self-conjugate under the Galois action, so the it divides $(y+3\sqrt{-6})$ with the same order as it does $(y-3\sqrt{-6})$. In particular, that must be a multiple of $3$
But if a prime ideal only divides one of the two, ofc the order is also a multiple of $3$, so $(y+3\sqrt{-6}) = I^3$ for some ideal $I$. But here's the kicker: the class number of $\mathbb{Q}(\sqrt{-6})$ is 2, so the square of any ideal is trivial in the class group. If the cube is also, then the ideal itself is trivial, aka principal
12:55 PM
So you can actually write $y+3\sqrt{-6}$ as the cube of an element of $\mathbb{Z}[\sqrt{-6}]$, and solve it out to find the solutions
I like this
Using this kinda argument my NT pset had us find solutions to $y^2 = x^3 - 20$, $y^2 = x^3 - 54$, and $y^2 = x^3 - 56$, so that was a fun time
Actually @Balarka how is $K^n(X),n\geq 1$ defined for a topological space $X$?
Nobody knows the definition of anything other than $K^0$
Class group of a number field is the set of fractional ideals with the equivalence relation $(a)I \sim (b)J$, right?
12:58 PM
@AlessandroCodenotti $K^{-n}(X) = K(\Sigma^n X)$ and I don't know what happens for positive degree
Some weird thing
^evidence of what I said
K-theory is fucked up man
I get so excited when NT is in chat
@Daminark Fractional ideals in a Dedekind domain are 1-dimensional projective modules, I think, so the ideal class group is actually the 0-th K-group of the domain
1:01 PM
At some point I'd like to check it out (probably been saying that for almost 2 years now but still)
$(a)I \sim (b)J$ is really stable equivalence
@BalarkaSen Ah, ok
If you have a Dedekind domain $R$ the class group is just $\mathrm{Pic}(\mathrm{Spec}(R))$
Ya, becaue 1-dimensional projective modules over $R$ are line bundles over $\text{Spec}(R)$
1:03 PM
I'll say that even though mornings aren't my friend so I've had trouble going to lectures, the psets alone for my NT class have gotten me excited about NT again. For some time it was like yeah I had fun with it and gotta learn it properly
Now I'm like YOOOO
@Daminark so excited to go and fanboy NT in Heidelberg
@ÍgjøgnumMeg shit what the fuck dawg
Also recently I've finally been convinced that AG is cool, day before yesterday Venkatesh gave some lectures about stable homology of symplectic groups over $\mathbb{Z}$. Didn't really understand it too well but it has finally converted me
I'm crossing my fingers for an iwasawa theory seminar in heidelberg this semester
Oh no, it's too late to save you now
1:04 PM
rip dami
ur a social outcast now
D: I was kidding I swear
AG? N e r d s h i t emirite? sweats
anything without algebraic before it is for nerds
Solution: ANT and AT :D
Analytic number theory and Analytic topology?
Actually funny thing along those lines, at one university I was considering there was a guy who, among other things, works on "topological modular forms"
Which sounds trippy
1:08 PM
sounds fit
i think topological modular forms is a cohomology theory which makes me doubt if it has anything to do with either topology or modular forms
this is something u get when u search for topological modular forms
These look like some of the diagrams Behrens showed us lmao
I remember there being some 192-periodic thing
how do people even do algebraic topology these days
everything is an Adams spectral sequence for homotopy groups of a ringed spectrum
like holy cow
1:17 PM
Lol, yeah that's pretty true. Though I guess there seem to be folk who still do something like topology. I've heard Hopkins, also one guy I'm interested in talking to a bit does algebraic topology of varieties
Yeah there's definitely still a lot of classical things
I just can't comprehend how people learn and actively use modern algebraic topology in their day to day lives
It's so hard
1:55 PM
Does anyone know any definition of a measurable space via operators? Like e.g., the definition of topological space via closure operators.
@ÉricoMeloSilva did you get a room assignment?
I think I got the New GC. Is that good or bad?
2:47 PM
How far do I need to go before I understand topology
I mean, being able to see things beyond mere definitions
3:22 PM
Q: Evaluate limit.

user114873Let $f : \mathbb R \to \mathbb R$ be differentiable at $x = a$. Evaluate: $$ \lim_{n\to \infty}\large[{f(a +\frac{1}{n^2})}+{f(a +\frac{2}{n^2})}+...+{f(a +\frac{n}{n^2})}-nf(a)] $$ Answer: $\ $ $\ \frac{1}{2}f'(a)$ My attempt: $$ \lim_{n\to \infty}\large[{f(a +\frac{1}{n^2})}-f(a)+{f(a...

I have doubt in learnmore answer
$=\sum _{i=1}^n [f(a+\frac{k}{n^2})-f(a)]{\frac{k}{n^2}}$

Note that $\lim_{n\to \infty }\dfrac{[f(a+\frac{k}{n^2})-f(a)]}{\frac{k}{n^2}}=f^{'}(a)\to (1)$
$\sum _{i=1}^n \frac{[f(a+\frac{k}{n^2})-f(a)]}{\frac{k}{n^2}}{\frac{k}{n^2}}$ instead of $=\sum _{i=1}^n {[f(a+\frac{k}{n^2})-f(a)]}{\frac{k}{n^2}}$
Yeah, there is a missing denominator
How did he first take the limit and then add $\frac{1+2+...+n}{n^2}$? in the limit
@SubhasisBiswas If you have done chapter 2 Rudin, exercises and all, that's quite a lot of point-set topology for starters.
There are other flavors of topology of course
Is $\lim_{n\to \infty}\sum _{k=1}^n \frac{[f(a+\frac{k}{n^2})-f(a)]}{\frac{k}{n^2}}{\frac{k}{n^2}}=$ $\sum _{k=1}^n \lim_{n\to \infty}(\frac{[f(a+\frac{k}{n^2})-f(a)]}{\frac{k}{n^2}})\lim_{n\to \infty}{\frac{k}{n^2}}=f'(a)\sum _{k=1}^n \frac{k}{n^2}$
Did he deduce by this logic?
But this justification is invalid, I think.
okay I got it. what he did.
3:40 PM
That would be very wrong. Those steps are pulling some $n$'s out of the limit where they stop making sense. Not to mention the number of summands goes to infinity, so that interchange would be non-trivial, if valid. The justification in the answer is given in the "Added:" section.
@Thorgott Thank you
hey guys what's the chrome extension that turns strings into Latex?
for this chat
I don't know of any chrome extension, but you can use the javascript that is linked in the infobox on the top right.
Damn it. I downloaded it for my laptop but I forgot the name :(
This might not be yours; but it should be fine:
3:56 PM
@MaximilianJanisch That's not the one, but I can Tex any website with that? :o
I'm using the Brave browser, it works anyway
$\sum_{i=0}^n i^2 = \frac{(n^2+n)(2n+1)}{6}$
$$\sum_{i=0}^n i^2 = \frac{(n^2+n)(2n+1)}{6}$$
Thanks Maximilian
4:16 PM
You’re welcome :D
the extension doesn't Tex Twitter though :unbridled rage:
Why would you want to TeX on Twitter? A single line of equations probably exceeds the character limit there
I want TeX on Pornhub damn it!
4:22 PM
It works on Whatsapp web too! you just improved my life by a lot Maximilian :D
If $f \in \Bbb{C}[x_1,...,x_n]$ is nonconstant, does $f$ have at least one root?
Doesn't that follow from the FTA?
I tried induction and using the FTA, but it isn't obvious.
FTA says its true for $\Bbb{C}[x]$.
Well, pick an $x_i$ such that there is a monomial term in $f$ containing $x_i$ with non-zero coefficient. Then keep all the other $x_j,j\neq i$ constant (and non-zero). This restriction yields a non-constant polynomial in $\mathbb{C}[x]$ and you can apply FTA.
4:33 PM
Why must there be such a monomial term? What do you mean by "keep all other $x_j$" constant? Do you mean, evaluate them at some nonzero complex number?
If all monomials had a zero coefficient, $f$ would be constant. I'm not sure if evaluation is the correct term in this case, but you can think of it as a kind of partial evaluation, yes.
Oh, slight oversight, I of course mean the non-constant monomials*
1 hour later…
5:50 PM
@RyanUnger ive heard bad things about the new gc overall, best thing anyone has said to me is “it wasn’t that bad”
i got a room in OGC
i heard the gc is generally kind of bad
6:02 PM
Guys, here's a question
$g(x) =max(x, x^{-1})$, $x>0$, then $g(c)g(c^{-1})=?$, $c>0$
I am getting $g(c)g(c^{-1})=c^2$, when $c>1$, $=c^{-2}$ when $c<1$ and $=1$ when $c=1$
That is true. You can also simply note that $g(c^{-1})=g(c)$ for $c>0$.
since $\max\{a,b\}=\max\{b,a\}$ amirit?
Yes, and $(c^{-1})^{-1}=c$ of course
Different question: Let $g\colon\mathbb{R}\rightarrow\mathbb{R}$ be continuous and $\delta>0$. Why is there a sequence $(x_n)_{n\in\mathbb{Z}}$ s.t. $x_n\rightarrow\pm\infty$ when $n\rightarrow\pm\infty$ resp. and $|g(x)-g(x_n)|\le\delta$ for all $x_n\le x\le x_{n+1}$ and $n\in\mathbb{N}$ (this essentially says that $g$ can be approximated by step functions with up to countably many steps). I tried using continuity locally and then use that $\mathbb{R}$ is Lindelöf, but it didn't work out
6:14 PM
that's a great question. Where did you get it?
is $\delta$ arbitrary?
@ÉricoMeloSilva bad how
I got it from a proof I'm reading where this is stated as true without further comment. And yes, $\delta$ is an arbitrary positive real.
@RyanUnger iunno no one gave me specifics bc i didn’t ask much about ngc but ppl generally told me they’re bad in the way that living in a dorm as an adult is bad
@Thorgott maybe use the fact that continuous functions on compact sets are uniformly continuous
@Thorgott break up the entire real line into many pieces of compact sets and use that in a compact interval, a cont function is uniformly con
@LeakyNun wow, I also came up with this :)
6:26 PM
@SubhasisBiswas great minds think alike
@LeakyNun but I'm noob
don't drag yourself down to my level.
maybe I'm also a noob
it's gonna be dangerous
as far as I think, we can modify the Lindelof property by changing the countable sub cover of an open covering into a compact-covering (idk whether the term is valid).
I just had the same idea as well after Leaky's tip. I was considering compact intervals earlier, but even though I considered a limiting construction on increasing compact sets , the much more elegant thought of just splitting up the real line took a second attempt. Thanks :)
@ÉricoMeloSilva well sure
just as long as its not haunted
6:31 PM
@Thorgott can we proceed along my line of thought?
I need some go-ahead
A compact cover can be given explicitly: $\mathbb{R}=\bigcup_{k\in\mathbb{Z}}[k,k+1]$
@RyanUnger it looks nice and newish
We define $x_k =k$
ogc looks like it would be haunted tho
@ÉricoMeloSilva my advisor at UTK lived in NCG when he went
in 82
it's not new
6:33 PM
i did say newish
The terms of the sequence is to be adjusted accordingly as $\delta$, right?
all the other buildings obv look hella old
Yes. If one sequence worked for all $\delta$, we would have a contradiction (or $g$ constant).
@Thorgott, I came up with something
It might need some correction
6:53 PM
Well, the statement is already proven with the above considerations, isn't it
First of all, we cut out a portion of the real line, say, $I=[-a,a]$.

Pick an arbitrary $\delta>0$. Now, $ I$ being compact, we get an $\eta(\delta)=\eta>0$ such that $|x_i -x_j|< \eta \implies |g(x_i)-g(x_j)|< \delta$ .

Clearly, for $ x \in [x_i,x_i +\eta -\epsilon] \implies |g(x_i)-g(x)| < \delta$ where $\epsilon \to 0^+$.

We fix $x_0=0$ and let $x_n=x_{n-1}+\eta - \epsilon$.
@Thorgott I am trying to be more formal here
00:00 - 19:0019:00 - 23:00

« first day (3215 days earlier)      last day (121 days later) »