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Jeff
Jeff
12:08 AM
Hi folks. Wonder if anyone can offer a clue about what technique to use to integrate
$$\int \frac{5 \sqrt x}{\ln 6x} dx$$.
 
Leaky Nun
Leaky Nun
you can use the technique called "wolfram alpha"
 
Jeff
Jeff
It gives an answer I don't recognize and offers no help with how to integrate it.
 
Ted Shifrin
Ted Shifrin
@Jeff This has no elementary antiderivative.
 
Jeff
Jeff
12:23 AM
@Tec Oh. I was starting to suspect that.
I've never heard of the 'exponential integral' that Wolfram refers to. I'm looking it up now.
 
Semiclassical
Semiclassical
it's a particular definite integral that doesn't have an elementary closed form
the integral you wrote is equivalent to it
(by appropriate change of variables)
 
Ted Shifrin
Ted Shifrin
$\int e^x/x\,dx$
 
Jeff
Jeff
@TedShifrin TY. Got it and "did" it.
What use is the exponential integral?
 
Ted Shifrin
Ted Shifrin
12:57 AM
There are various integrals that show up, like erf (most well-known from probability), $\sin x/x$, etc., but have no elementary expressions. So they get names and have been tabulated numerically.
 
 
3 hours later…
Chris Steinbeck Bell
Chris Steinbeck Bell
3:28 AM
Howdy mathematics. I'm Chris. I'm I need some help in this question regarding discrete quanitties based on time [4041120] this is the code of the question
perhaps if someone has seen this topic before or a similar could help me?. I don't know exactly what to do because the only answer which has appeared there has left me more confused that how i was feeling at the beginning
Thus if someone could enlighten me with a wordy description or maybe a guide on which went wrong I will appreciate the help.
I understand that it might be a simple question but it has got me to think too much and round in circles. Thus could anyone please take a look into question [4041120]
Thanks in advance
 
 
1 hour later…
Leonhard Euler
Leonhard Euler
4:52 AM
N i c e
I wonder who is controlling math
 
User
User
I am reading a paper where the word Curvature(a real number) frequently comes; the paper deals with only Riemannian surface, i.e., dimension 2. Do I assume the author means to say sectional curvature?
I know this is a weird question, but I searched on the word curvature in whole .pdf but got no word like sectional.
 
Ted Shifrin
Ted Shifrin
5:11 AM
You need to learn basics. For surfaces there’s only one curvature.
 
User
User
Yeah, I know this: for a surface sectional curvature of is equal to its Gauss curvature.
 
dc3rd
dc3rd
Oh @TedShifrin I forgot to mention/ask with regards to my reasoning yesterday about the circle being split in three parts. So I came to that conclusion because all three circles had the same length of radius which implies that the three circles are congruent, as a result I could "combine" how the different radii split up the three circles. In other words the way RobJohn drew the picture, but have all three of those angle measures "merged" into one circle...
and from that I could find the "angle" measure for one portion.
A lot more verbose than I wanted to describe it....but yea...
 
 
3 hours later…
robjohn
robjohn
8:23 AM
@TedShifrin do you mean one intrinsic curvature? Otherwise, there's mean curvature, which is different, right?
 
 
1 hour later…
StudySmarterNotHarder
StudySmarterNotHarder
9:43 AM
0
Q: Algebratization of the twin prime conjecture. Twin primes true if and only if constructed group $G$ has infinite order.

StudySmarterNotHarderDefine $g : \Bbb{Q}^{\times} \twoheadrightarrow \Bbb{Z}$ by $\pm\dfrac{p_1\cdots p_n}{q_1\cdots q_m} \mapsto p_1 + \dots + p_n - q_1 - \dots - q_m$. The kernel $K = \ker g$ kind of measures the failure of summation of primes to be unique. Except it infinitely counts one example say $\dfrac{2\cd...

abstract-algebra group-theory homology-cohomology prime-gaps prime-twins
 
user2103480
user2103480
10:20 AM
The hierarchy is clear
Math is only a subfield of forklift operating
 
geocalc33
geocalc33
hmmmm
forklift theory must be rich
 
maths student
maths student
A fair coin is tossed three times. X is the number of heads in the first two tosses and $ Y$ the number of H in the last two tosses.

Find Cov (X,Y)
I find the distribution for X and Y E(X) AND E(Y) but how to find E(XY)
 
Edward Evans
Edward Evans
10:52 AM
@user2103480 lmfao
 
Leaky Nun
Leaky Nun
11:04 AM
if I engineer a coin so that it always lands on the opposite side of the previous flip, is it a fair coin?
what is fair?
is it fair that the big companies get to control the price of the stock market blatantly?
 
bard
bard
Hi! I'm bard, sorry to interrupt: I've posted some days ago the following question, but altough I've recevied some upvotes nobody has answered me. Since I'm pretty confident there's at least an algebraic-geometer here, if you want to gain some rep please take a look, it's a pretty simple problem:

https://math.stackexchange.com/questions/4040038/blow-up-lines-in-reids-pagoda

Sorry again, and have a nice day!
 
robjohn
robjohn
@mathsstudent Just write out all the possibilities:
$$
\begin{array}{ccc|cc}
1&2&3&X&Y\\\hline
0&0&0&0&0\\
0&0&1&0&1\\
0&1&0&1&1\\
0&1&1&1&2\\
1&0&0&1&0\\
1&0&1&1&1\\
1&1&0&2&1\\
1&1&1&2&2\\
\end{array}
$$
$$\sum X=8\quad\sum Y=8\quad\sum XY=10$$
 
geocalc33
geocalc33
11:50 AM
$$\nabla^2f=u\frac{\partial}{\partial u}\left(u\frac{\partial f}{\partial u}\right) + v\frac{\partial}{\partial v}\left(v\frac{\partial f}{\partial v}\right)=0$$
I don't know if anyone likes differential equations, but if anyone can help me solve it that'd be cool
 
user2103480
user2103480
12:02 PM
is that supposed to be the laplacian?
 
geocalc33
geocalc33
12:12 PM
yeah
in a different coordinate system
It's proving to be harder to solve than I thought
 
user2103480
user2103480
... does that matter?
Harmonic function
In mathematics, mathematical physics and the theory of stochastic processes, a harmonic function is a twice continuously differentiable function f : U → R, where U is an open subset of Rn, that satisfies Laplace's equation, that is, ∂ 2 f ∂ x 1 2...
this should be what you're looking for
 
geocalc33
geocalc33
yeah it matters, the solutions will be different in different coordinate systems
 
Eminem
Eminem
Is $\log _3 \lfloor {3^{\frac{1}{n}}} \rfloor = \frac {1}{n}$?
 
user2103480
user2103480
12:28 PM
no, plug in 2
sqrt(3) = 1.7 something so the floor of that is 1 and log_3(1) = 0
 
Eminem
Eminem
Im trying to disprove the following statement:

Given $f,g>1:\mathbb N \rightarrow \mathbb N$ prove\disprove the following:

$f=O(g)\Rightarrow \log_fg=\Omega(1)$

Take $f=2,g= \lfloor3^{1/n} \rfloor$. Let $c=5$, then for sufficiantly large $n$ we have

$f=2 \leq 5\cdot \lfloor3^{1/n} \rfloor=c\cdot g \Rightarrow f=O(g)$

But

$\log_fg=\frac{\log_3g}{\log_3f}=\frac{\log_3\lfloor3^{1/n}\rfloor}{\log_32}=\frac {0}{\log_32}=0\neq\Omega(1)$
 
Leonhard Euler
Leonhard Euler
1:09 PM
I have proved that $$\sum_{m=0}^{\infty}p(5m+4)x^{5m+4}=cx^{4}\frac{\phi(x^{25})^5}{\phi(x^5)^6}$$
How can I prove that $c=5$?
(Ramanujan's identity)
The solution would be easy
In a parallel universe
Or is it easy?
Btw it is an exercise from a book
 
Astyx
Astyx
1:27 PM
evaluate f(x)/x^4 at 0 ?
 
Leonhard Euler
Leonhard Euler
Is f(x) the RHS?
 
Astyx
Astyx
yes
I don't if that works because I don't know anything about $\phi$
 
Leonhard Euler
Leonhard Euler
Euler totient function
The RHS becomes c and the LHS becomes a
Divergent sum
Ramanujan loves divergent sums
 
 
1 hour later…
porridgemathematics
porridgemathematics
2:48 PM
hi, I know this result to be true, but I can't see it using elementary notion of conditional probability, namely, let $X,Y,Z$ be random variables , with $\sigma(Y) \subset \sigma(X)$, then $\mathbb{P}(Z = z | X = x, Y = y) = \mathbb{P}(Z = z | X = x)$ , $(X,Y)_{\ast} \mathbb{P}$ almost surely. This makes total sense if I use the factorization lemma to see it, but suppose I have $\mathbb{P}(X = x, Y=y) > 0$, so these now become elementary conditional probabilities, how can I see it?
uh sorry, I think I also need to say that we are working on a countable state space for these to be elementary conditional probabilities
im not sure if its true if I don't restrict to a countable state space
 
porridgemathematics
porridgemathematics
3:01 PM
hm, I guess one way to see it is that $Y$ is $X$-measurable, so it is of the form $\phi \circ X$ for some $\phi : \mathcal{X} \rightarrow \mathcal{X}$ where $\mathcal{X}$ is the state space, so by asking that $X = x$ and $Y=y$ we are really saying we need $X=x$ and $X \in \phi^{-1}(\{y \})$, and if this happens with nonzero probability then it is equivalent to asking for $X = x$
so conditioning wise, we can forget about asking for the additional $Y=y$
its almost surely redundant extra information
this is still really the factorization lemma way to see it fleshed out, i'm wondering if there is an even simpler way?
 
user2103480
user2103480
So I dont really get youer question. $\Bbb P(A \mid X, Y)(\omega) = \Bbb P(A \mid X)(\omega) = \Bbb E[1_A \mid X](\omega) = g(X(\omega))$, and in the case that $X(\omega) = x$, we say $g(x) = \Bbb P(A\mid X=x)$
the value of Y is determined by the value of x so either X=x and Y=y does not happen or always coincide
The factorization lemma is not really avoidable up to rephrasing
(to my knowledge)
 
porridgemathematics
porridgemathematics
yeah, the determined by the value of x thing essentially comes from the factorization lemma
okay, thats fine
 
robjohn
robjohn
@LeonhardEuler what are $p$ and $\phi$?
In particular, what is $p(4)\phi(0)$?
 
user2103480
user2103480
@porridgemathematics Yes and no
 
porridgemathematics
porridgemathematics
@user2103480 let me rephrase, assuming the state space were working on is $\mathbb{N}$ with the discrete sigma algebra, is there a way to go from $\sigma(Y) \subset \sigma(X)$ to $\frac{\mathbb{P}(Z = z, X=x, Y=y)}{\mathbb{P}(X=x,Y=y)} = \frac{\mathbb{P}(Z = z, X = x)}{\mathbb{P}(X = x)}$, assuming these quotients are well-defined
 
user2103480
user2103480
3:07 PM
the factorization lemma is not the reason, but its a good intuition
 
porridgemathematics
porridgemathematics
without factorization lemma
i.e. just manipulating elementary probabilities via sums
 
user2103480
user2103480
probably, but you see I didn't use the factoriation lemma up there?
The sigma-algebra generated by (X,Y) should be just the one generated by X
 
porridgemathematics
porridgemathematics
yeah, but you used that $\sigma(X,Y) = \sigma(X)$
 
user2103480
user2103480
yeah but that's just the condition
I don't think you need the factorization lemma for that
probably some good sets principle or something
 
porridgemathematics
porridgemathematics
and another way of defining things would be, $\mathbb{E}[X | Y = y]$ is the $Y_{\ast} \mathbb{P}$ almost surely defined $\phi : \mathcal{X} \rightarrow \mathcal{X}$ such that $\mathbb{E}[X | Y] = \phi \circ Y$ almost surely
if you use this definition, you can't immediately go from $\sigma(X,Y) = \sigma(X)$ to what we want
because $\mathbb{E}[X | Y = y, K = k]$ is now a map $\phi : \mathcal{X} \times \mathcal{X} \rightarrow \mathcal{X}$
so you inadvertently need to show this is almost surely equal to some map $\psi : \mathcal{X} \rightarrow \mathcal{X}$ (the second component is redundant, essentially)
 
user2103480
user2103480
3:11 PM
@porridgemathematics What do you want/prove?
Except what you stated above there
 
porridgemathematics
porridgemathematics
basically I want to know how to do it in the countable state space case, without explicit use of formal definitions of conditional expectation that apply in the uncountable state space case
 
user2103480
user2103480
if $\sigma(Y) \subset \sigma(X)$ on a countable state space then $Y^{-1}(\{a\}) = X^{-1}(B)$ for some countable set B and then you can probably distribute over that
 
porridgemathematics
porridgemathematics
because in te countable case, you can show conditional expectation exists without any use of the radon-nikodym theorem
you can just define it via elementary conditional probability
ah okay, yeah that does work
nice
 
user2103480
user2103480
yeah and the $B_a$ should be disjoint yada yada
 
porridgemathematics
porridgemathematics
yeah I can see how we can turn this into an argument involving what we've been talking about essentially (in the most general case) distilled to the countbale case
that does answer my question :)
 
user2103480
user2103480
3:15 PM
I dont recall the proof of the factorization lemma but that should also be the reasoning behind it
modulo some fine details like measurability, which I know I messed up when I tried proving it in an exercise
@porridgemathematics ok nice
 
porridgemathematics
porridgemathematics
well $\sigma(Y) \sigma(X)$ immediately gives you $Y^{-1}\{a \} = X^{-1}(B)$ for some countable $B$ and then we can just say $Y = a, X=x$ is the same as $X = x$ and $X \in B$ and if this happens with nonzero probability then we can effectively ignore the $X \in B$ term in our summation
so we don't even need to use the factorization lemma, just the definition of $\sigma(X)$ for some rv $X$
*$\sigma(Y) \subset \sigma(X)$
 
user2103480
user2103480
You don't ever need the factorization lemma if you reprove it everytime, but its quicker to just use it :P
 
porridgemathematics
porridgemathematics
we dont even need to reprove it in the countable case, really
 
user2103480
user2103480
Fair, fair
 
porridgemathematics
porridgemathematics
the proof of it that I know of is just the usual start from indicator functions nonsense
 
user2103480
user2103480
3:21 PM
@porridgemathematics the indicator function basically only consists of the above step
 
porridgemathematics
porridgemathematics
oh yeah haha, thats true :)
its noticing what $\sigma(X) \subset ... $ entails
 
user2103480
user2103480
and then induction of course
 
porridgemathematics
porridgemathematics
yup
 
user2103480
user2103480
I agree that it's hard to see all that at first
I found conditional expectations to be very unintuitive
I still can't say I find them intuitive per se, but I'm used to them now and know some heuristics
How do representants of generators of top cohomologies look like?
For manifolds. Hm. Ah. I know my fault in thinking
we only defined orientability for rings
So I guess a cochain represents the generator of top cohomology of an R-orientable manifold if it maps a chain that represents the fundamental class to the unit in the ring?
Or am I mightily wrong with that
 
user2103480
user2103480
4:23 PM
Cohomology of disjoint unions is just the product of the cohomologies?
This should just work via $C_\ast(\coprod_i X_i) \cong \bigoplus_i C_\ast(X_i)$ and then dualizing
 
Thorgott
Thorgott
@user2103480 yes, H^n is dual to H_n since H_{n-1} is torsion-free for closed orientable manifolds
@user2103480 yes
 
user2103480
user2103480
ok nice
you mentioned that example of computing the cross product of the generators of first cohomology of spheres to get a generator of second cohomology of the torus
I mean I get how it should work
$(\mathrm{pr}_1^\ast[S^1]^\ast \cup \mathrm{pr}_2^\ast[S^1]^\ast)(T^2) = \langle S^1, T^2|_{(0,1)} \rangle \cdot \langle S^1, T^2|_{(1,2)} \rangle = \langle S^1,S^1 \rangle \cdot \langle S^1,S^1 \rangle = 1_R$
In very suggestive notation
But to actually obtain this I need to work with an explicit chain that represents the fundamental class of the torus, right?
So that restricting that chain directly yields two circles
 
Thorgott
Thorgott
yeah
 
user2103480
user2103480
ugh
not gonna do that
although it should be possible with two singular simplices, or probably one if one is smart
gotta be careful about orientation though
Okay, no, it's just the obvious thing
 
Thorgott
Thorgott
4:48 PM
yeah
what's the easiest example of an open manifold that's not an interior?
 
user2103480
user2103480
is that directed at me
I dont even know what that means
 
Thorgott
Thorgott
general question
 
Balarka Sen
Balarka Sen
@Thorgott Take the Loch Ness Monster
 
Thorgott
Thorgott
as in surface with infinite genus?
 
Balarka Sen
Balarka Sen
yes
 
Alessandro Codenotti
Alessandro Codenotti
5:15 PM
Is it obvious that if $P$ is a prime ideal and $p\in P\setminus P^2$, then $p^n\in P^n\setminus P^{n+1}$?
 
Balarka Sen
Balarka Sen
Are your prime ideals maximal by any chance
 
Alessandro Codenotti
Alessandro Codenotti
Yes, I'm in a Dedekind domain
 
Balarka Sen
Balarka Sen
You may be using Nakayama's lemma
Yeah so that's it
If $p^n$ is zero in $P^n/P^{n+1}$ then $p$ is zero in $P/P^2$
 
Alessandro Codenotti
Alessandro Codenotti
I'm not seeing it
 
Balarka Sen
Balarka Sen
Or some variation. Let me say it correctly.
 
Alessandro Codenotti
Alessandro Codenotti
5:17 PM
I have no intuition for Nakayama
 
Thorgott
Thorgott
this is some end-theoretic obstruction, I wager
 
Alessandro Codenotti
Alessandro Codenotti
@Astyx 4Z is not prime though
 
Astyx
Astyx
Lol I should read
 
user2103480
user2103480
french primes
 
Alessandro Codenotti
Alessandro Codenotti
There's a long tradition of French primes starting with Grothendieck's 57
 
Balarka Sen
Balarka Sen
5:27 PM
@Alessandro $P/P^2$ is 1-dimensional as a $A/P$-vector space, so actually $P = (p) + P^2$. Now multiply by $P$ again to get $P^2 = (p)P + P^3 = (p^2) + P^3$, etc etc
I am using full Dedekindness
Ugh, this is not telling you much, sorry. It's something along these lines, one second
I am not an algebraist
 
Astyx
Astyx
The localization at P is a DVR, and you're stating that the valuation of p^n is n times the valuation of p
 
Balarka Sen
Balarka Sen
Ah ok, the point is if $p^n \in P^{n+1}$ then $P^n = P^{n+1}$
Now you use Nakayama to boil it down to $P = 0$
Ok, there you go
 
Alessandro Codenotti
Alessandro Codenotti
@BalarkaSen wait where did you get this from?
 
Balarka Sen
Balarka Sen
Because by calculation above $P^n = (p^n) + P^{n+1}$
 
Alessandro Codenotti
Alessandro Codenotti
Oh ok I see
 
Balarka Sen
Balarka Sen
5:40 PM
All these quotients are 1-dimensional, because Dedekind domains are 1-dimensionals, basically
 
Alessandro Codenotti
Alessandro Codenotti
At this point one can also use the $\bicap P^n=\varnothing$ to conclude instead of Nakayama then I guess
 
Balarka Sen
Balarka Sen
s'pose so
 
user2103480
user2103480
does suspension do anything complex on cohomology rings
for connected spaces
If not then I'll try proving that
 
Balarka Sen
Balarka Sen
It destroys the ring structure usually
 
user2103480
user2103480
oh sheesh
so trivial everything
okay I'll try to show that
 
Balarka Sen
Balarka Sen
5:44 PM
Yup
That's actually a Hatcher exercise IIRC good to show it
 
Alessandro Codenotti
Alessandro Codenotti
I see, thanks
 
user2103480
user2103480
still trying to find a slick proof that for connected $X, Y$, where $(X,x_0), (Y, y_0)$ are good pairs, the sum of the maps induced by the collapsing maps gives an isomorphism $H^k(X;R) \oplus H^k(Y;R) \rightarrow H^k(X \vee Y;R)$
 
Thorgott
Thorgott
you can give the inverse explicitly
it's trivial to calculate it's an inverse on one side, but it's necessarily a two-sided inverse then, because M-V will tell you that the map is an iso
 
user2103480
user2103480
*path connected

LES goes in the wrong direction, didn't find one via universal coefficient SES, meh. Maybe there's some algebra I'm overlooking? Both collapsing maps are retractions and muh
 
Balarka Sen
Balarka Sen
No need, just consider the pair $(X \sqcup Y, x_0 \sqcup y_0)$
This will also tell you the ring structure agrees
 
user2103480
user2103480
5:50 PM
@Thorgott yeah that'd be the LES map I assume
@BalarkaSen that's the purpose
(it's a three step exercise)
but isnt there some diagram I can draw to avoid explicit calculation
@BalarkaSen or does this not refer to the LES
 
Thorgott
Thorgott
not sure which LES you're looking at, tell me what the inverse map is explicitly
 
Alessandro Codenotti
Alessandro Codenotti
@user2103480 by any chance do you know what the Scott rank of a countable model is?
 
user2103480
user2103480
$H^k(X \vee Y;R) \rightarrow H^k(X \sqcup Y;R) \rightarrow H^k(\text{pt};R)$
@AlessandroCodenotti was that the choice business?
 
Alessandro Codenotti
Alessandro Codenotti
which choice business? I don't think so
 
user2103480
user2103480
Defining cardinalities unambiguously when choice fails
Then no
 
Alessandro Codenotti
Alessandro Codenotti
5:55 PM
that's Scott's trick
 
user2103480
user2103480
ah nah that was scotts trick. close enough lol, it's about ranks and scott
So no, don't know it
 
Thorgott
Thorgott
ok, but that map has an easier description
 
user2103480
user2103480
it's the tuple of inclusions
I am still too lazy to calculate, but I'll do it now then
 
Thorgott
Thorgott
yes, that's what I wanted to hear
 
Ted Shifrin
Ted Shifrin
@robjohn Yes, of course, but no embedding of $M$ as a surface in $\Bbb R^3$ was provided. So, no.
Isn't @Thor supposed to be studying physics today?
 
Thorgott
Thorgott
6:13 PM
I am, actually
 
Ted Shifrin
Ted Shifrin
OK, just checking up on you.
@Thorgott You mean interior of a manifold with boundary?
 
Balarka Sen
Balarka Sen
@Thorgott This case is much easier. But in general, yes.
 
Ted Shifrin
Ted Shifrin
Howdy, a @Balarka.
 
Thorgott
Thorgott
@TedShifrin yeah
 
Balarka Sen
Balarka Sen
There are famous four papers of Frank Quinn which solved the problem "Which noncompact manifolds are interiors?" completely.
Ends of Maps I-IV
 
Thorgott
Thorgott
6:25 PM
I know this was also studied by Siebenmann in his thesis
 
Balarka Sen
Balarka Sen
I forgot that actually.
 
robjohn
robjohn
@TedShifrin . o O ( Thor's mom )
 
Ted Shifrin
Ted Shifrin
:)
 
Balarka Sen
Balarka Sen
Topology of ends is an extremely deep branch of mathematics. For example, Stallings is famous for his theorem that a contractible PL $n$-manifold is PL-homeomorphic to $\Bbb R^n$ iff it is simply connected at infinity.
PL can be ignored due to later developments in mathematics.
 
Ted Shifrin
Ted Shifrin
I'd rather have ends of loaves of baguette. :)
 
Balarka Sen
Balarka Sen
6:30 PM
Ah, also, $n \geq 5$.
$n = 4$ is Freedman, I suppose.
 
Thorgott
Thorgott
yeah, Siebenmann actually gives a complete classification in dimensions >5
this preceded the work of Quinn you mention
 
Balarka Sen
Balarka Sen
Smart guy
I tried to read Quinn's paper in some occasions, but he's an extremely messy mathematician
Everything's all over the place. It's a very technical theory
 
Ted Shifrin
Ted Shifrin
Odd to hear a @Balarka complaining about lack of neatness :)
 
Thorgott
Thorgott
@BalarkaSen is this related to Stalling's ends of groups theorem or are they just both by Stallings and both on ends
 
Balarka Sen
Balarka Sen
Haha, I suppose he thought about ends a lot
We can try to read Stalling's paper. If I recall correctly it's short and neat
 
Thorgott
Thorgott
6:34 PM
Siebenmann's thesis looks pretty readable, but it's far beyond my background
@BalarkaSen what's the name
 
Balarka Sen
Balarka Sen
@TedShifrin I'm not a mathematician, so I don't have to be neat :)
 
Ted Shifrin
Ted Shifrin
You sound like you want to be qualified to be a doctor because your handwriting is so messy.
 
Balarka Sen
Balarka Sen
If I was actually communicating mathematics as a job I'd try to be neat
But I am not so I have no obligation to be
 
Ted Shifrin
Ted Shifrin
Oh, engulfing shows up.
 
Thorgott
Thorgott
man, I don't know any PL topology
 
Ted Shifrin
Ted Shifrin
6:36 PM
Me neither.
 
Balarka Sen
Balarka Sen
It's a good place to start
 
Ted Shifrin
Ted Shifrin
 
user2103480
user2103480
Ah, I think I now have the same problem I had last time. The pairs are $(X,x_0)$ and $(Y,y_0)$. I send $([\varphi], [\psi])$ to $(i_X^\ast p_X^\ast([\varphi]) + i_X^\ast p_Y^\ast([\psi])$ and want to show the last thing is zero. $i_X^\ast p_Y^\ast = (p_Y i_X)^\ast = (X \rightarrow \{y_0\})^\ast$.

I think I understand something wrong. I try to calculate using a representant $\psi$, on cochain level. Then for an arbitrary sum of simplices $\sum_i a_i \sigma_i $, I get $ i_X^\ast p_Y^\ast \psi(\sum_i a_i \sigma_i) = \sum_i a_i\psi(\Delta^k \rightarrow y_0)$ and duh is there an immediate reaso…
 
Balarka Sen
Balarka Sen
@TedShifrin Neat ^
 
user2103480
user2103480
fucc formatting
 
Edward Evans
Edward Evans
6:40 PM
Ithinkiunderstandsomethingwrong
 
Thorgott
Thorgott
is chat broken again
ok no
constant map induces 0 on homology in degrees >0
cause it factors through H(pt)
you can do this calculation explicitly, but it will just be the same one as when you calculated the homology of a point
 
user2103480
user2103480
I tried to do the factoring but somehow went wrong
thanks I'll try again
 
Ted Shifrin
Ted Shifrin
@EdwardEvans What's your point?
 
Edward Evans
Edward Evans
The formatting was broken lol
 
user2103480
user2103480
okay yeah I tried not hard enough with the factoring. Reminder to myself not to write $X \subset Y$ instead of $X \hookrightarrow Y$
 
Balarka Sen
Balarka Sen
7:40 PM
This is annoying. I wanted to prove the tower law for relative different but I'm having trouble with one of the inclusions.
 
Thorgott
Thorgott
for relative different?
 
Balarka Sen
Balarka Sen
$E/L/K$ be a tower, $\delta_{E/K} \supset \delta_{E/L} \delta_{L/K}$ is clear. Easier to do with inverse, let $\alpha \in \delta_{E/L}^{-1}$, $\beta \in \delta_{L/K}^{-1}$ then $\text{tr}_{E/K}(\alpha \beta \mathcal{O}_E) = \text{tr}_{L/K} \text{tr}_{E/L} (\alpha \beta \mathcal{O}_E) = \text{tr}_{L/K} (\beta \text{tr}_{E/L}(\alpha \mathcal{O}_E)) \subseteq \text{tr}_{L/K}(\beta \mathcal{O}_L) \subseteq \mathcal{O}_K$
This shows $\delta_{E/L}^{-1} \delta_{L/K}^{-1} \subseteq \delta_{E/K}^{-1}$
What about the other direction lol
This stuff is so confusing
So I want to prove $\delta_{E/K}^{-1} \subseteq \delta_{E/L}^{-1} \delta^{-1}_{L/K}$, or alternatively $\delta_{E/L} \delta_{E/K}^{-1} \subseteq \delta^{-1}_{L/K}$, which I imagine will be easier
 
user2103480
user2103480
is there any subject except logic that you dont study
 
Balarka Sen
Balarka Sen
Any $\alpha \in \delta_{E/L}$ multiplies with (anything which pairs with anything in $\mathcal{O}_E$ and spits out something in $\mathcal{O}_L$) and spits out something in $\mathcal{O}_E$
This is logic bro
Look at those recursive statements
 
user2103480
user2103480
lOgIC oF sPacEs
@BalarkaSen basically programming
 
Balarka Sen
Balarka Sen
7:48 PM
But anything in $\delta^{-1}_{E/K}$ pairs with anything in $\mathcal{O}_E$ and spits out something in $\mathcal{O}_K$, which is actually a subring of $\mathcal{O}_L$, so I guess $\alpha$ multiplies with such a thing and spits out something in $\mathcal{O}_E$
This is useless
Maybe $\delta_{L/K} \delta_{E/K}^{-1}$
Any $\alpha \in \delta_{L/K}$ multiplies with (anything which pairs with anything in $\mathcal{O}_L$ and spits out something in $\mathcal{O}_K$) and spits out something in $\mathcal{O}_L$
But elements of $\delta^{-1}_{E/K}$ are such a thing because $\mathcal{O}_L$ is a subring of $\mathcal{O}_E$, so $\delta_{L/K} \delta^{-1}_{E/K} \subseteq \mathcal{O}_L$.
Isn't that nonsense
What is happening so confusing
@EdwardEvans I have decided that I would quit too if I was a Masters student in ANT
 
Edward Evans
Edward Evans
literally
 
Balarka Sen
Balarka Sen
Help lol
 
Edward Evans
Edward Evans
i never actually learned about the different lmao
 
Balarka Sen
Balarka Sen
How do you show $\delta_{E/K} \subseteq \delta_{E/L} \delta_{L/K}$
Lmao
So fucked up man
 
Edward Evans
Edward Evans
I'd usually just default to Lukas
 
Thorgott
Thorgott
7:56 PM
that's just transitivity
 
Edward Evans
Edward Evans
but he disappeared again
 
Balarka Sen
Balarka Sen
@Thorgott Ok how
 
Thorgott
Thorgott
there's like E/K on the left and E/L/K on the right
transitivity
 
Balarka Sen
Balarka Sen
Lol
as usual, thanks for being of no help
 
Thorgott
Thorgott
ontological proof by symbolics
 
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