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2:00 PM
yeah, I looked up what the tensor product is, I'm confused about its relation with the exterior one
@AkivaWeinberger I simply used the Leibniz rule for differentiation. It worked after some toil :)
Hey everyone, quick question. If $A \subset X$ for some topological space $X$, does $A$, not $Cl(A)$, contain any limit points of $X - A$?
because they seem to be operations defined on different spaces to me
If $T_1$ and $T_2$ are $k$ and $\ell$-multilinear maps, instead of the exterior/wedge product there's a much easier tensor product operation which spits out a $k+\ell$ multilinear form $T_1 \otimes T_2$
@anonymous The problem is that you've got an $x$ in the limits and in the integral
so you have to account for both of them
2:01 PM
Defined by $T_1 \otimes T_2(v_1, \cdots, v_k, v_{k+1}, \cdots, v_{k+\ell}) = T_1(v_1, \cdots, v_k) \cdot T_2(v_{k+1}, \cdots, v_{k+\ell})$
@AlessandroCodenotti What's the context, out of curiousity?
@anonymous Oh, never mind -
the Leibniz rule is more general than I thought it was.
You should be fine, then.
Hamiltonian mechanics. The Poincaré-Cartan differential form I was wondering about a few days ago to be precise
@Akiva Yeah, it just occurred to me
You could have functions of $x$ on the limits
2:02 PM
By the way, today is the AMC
@anonymous why?
@AlessandroCodenotti You've probably noticed, but there is this question on MO: mathoverflow.net/questions/96992/…
@AkivaWeinberger IMO, Leibniz almost follows from chain rule. You can clearly see it.
In fact, now is the AMC. So, bye
2:03 PM
@DHMO It is just Leibniz rule of differentiation. You didn't know it ?
@SimplyBeautifulArt The case where there's an $x$ in the limits and in the integral is harder to show from the chain rule, though
unless you use the multivariable chain rule.
@Semiclassical I didn't, I'm looking at the references provided there right now, thanks!
Huzzah for google-fu, then :)
@Alessandro Anyway, now you've got the tensor algebra with the tensor product operation. What you can do now is construct the exterior algebra $\Lambda^k V^*$ as a quotient $\bigotimes^k V^*/\sim$ where $T \sim T'$ if $T(v_1, \cdots, v_i, v_j, \cdots, v_k) = T'(v_1, \cdots, v_j, v_i, \cdots, v_k)$.
@anonymous I see
2:05 PM
Huzzah for integral equations :/
@DHMO integral?
One way to approach that, I'd say, is to first note that $\cos x$ is a solution to $\phi+\phi''=0$. So all that one needs is the second term.
@SimplyBeautifulArt no time
:-) ok
I too am busy
And under this quotient the tensor product can indeed be made into the exterior product, but there's a small subtlety there. If $A_1$ and $A_2$ are two alt. mult. forms, then $A_1 \wedge A_2$ is defined as the "average" of all expressions of the form $T_1 \otimes T_2$ where $T_i$ are the multilinear forms which represent $A_i$'s.
2:08 PM
And that second term is a convolution $-(f\star g)(x)$ with $f(x)=x,g(x)=\phi(x)$.
This is described in detail in G&P, chapter 4
@SimplyBeautifulArt @DHMO May I ask you something? Is a function like $\int_{f(x)}^{g(x)} h(x) dx$ legit? Does it make sense? f(x), g(x) and h(x) are all suppose linear functions of $x$.
No, it doesn't make sense.
2:10 PM
@anonymous no
@Semiclassical Can you elaborate, please?
You can't have integration variables in the limits of integration.
Nope. The integrand should be $h(x,t)$ and you should integrate with respect to $t$.
That's the usual form
It's fine to have independent variables in the integrand and the limits of integration, but you can't have integration variables in the limits.
@DHMO @SimplyBeautifulArt @Semiclassical I get it now. That was a confusion bugging me for past few days. Thanks a lot :)
2:12 PM
Consider what the limits of integration tell you: Integrate $x'$ from $x'=f(x)$ to $x'=g(x)$.
nvm, that made no sense
If $x'$ appeared there, then neither statement would be well-defined.
@Semiclassical If they use the same variable in the limit and the integrand aren't they supposed to mean a different variable? Like writing $x \mapsto \int_0^x f(x) dx$ is considered abuse of notation as it actually means $x \mapsto \int_0^x f(y) dy$.
I'll look at GP and ask you if I have more precise doubts, thanks for your help @Balarka
Sure thing
Right. So $\int_0^x f(x')\,dx'$ is fine, but it wouldn't if both were $x$.
@Semiclassical Agreed!
That's one abuse of notation I refuse to consider myself.
It's vital to distinguish between the function being integrated and the limits of integration.
But if they can never be the same variable there is no room for ambiguity.
One is a dummy variable, the other isn't.
No reason to treat them notationally as though they're the same thing.
2:16 PM
@BalarkaSen Let's say I define a map between (almost) complex manifolds $f:(M,J)\to (N,J')$ to be holomorphic if $Df\circ J=J'\circ Df$. Then do you know how to prove that this is equivalent to $J'(Tf(M))\subset Tf(M)$? One direction is trivial. The other... I don't know.
And when you're doing stuff like convolutions, you definitely need to distinguish them.
Good point
Semi ^ any idea?
@Danu What's $Tf(M)$?
Basically why is a (smooth) submanifold a complex submanifold iff the complex structure stabilizes its tangent bundle?
2:17 PM
@BalarkaSen Tangent bundle of the image
The image may not be a manifold?
I'm basically asking ^^^^
Right, let the image be a submanifold
Got it.
I was hoping this would just be some kind of linear algebra exercise
But I don't see any easy solution
2:20 PM
In what field do you study manifolds? I still haven't encountered them.
Right now, I study almost complex manifolds and in particular their Chern classes
@Danu That sounds almost trivial. Why not identify $f(M)$ with $M$ and $f$ with the inclusion of the submanifold?
So you have the inclusion $i : M \to N$
Now I assume $J$ stabilizes the tangent bundle
Hmm lol
that's kind of almost circular
I think I slightly messed up the wording
Let me go back and check
Right, no actually
I think I'm okay
Can't find any mention to manifolds in any of my books. :I
so assume $J$ stabilizes the tangent bundle---then can you show me how that square commutes?
Take the map to be the inclusion, no problem
2:25 PM
Ok, so $Di$ is the inclusion $TM \to TN$
$J'(Ti(M))$ is precisely $J' \circ Di$
And $J'\circ Di(v)=Di(w)$ for some $w$
Now can you show me why $w=Jv$?
Well, $J$ is an anti-involution of $TN$, right? So let $v = -Jw$
...yes, and then?
2:31 PM
@Danu Sorry, was afk for a while. Ignore that, I misunderstood what you wrote up there.
Hmm, right, that's what we want to show
It's such a stupid little exercise
But maybe it's not as trivial as it looks?
@Danu Yes, it shouldn't be trivial. For 1 dimension it seems like the claiming Cauchy-Riemann equations => holomorphic
Sorry, that's not true. I am thinking of the wrong converse.
Do you know where could I find English translation for Grothendieck's EGA?
I have been reading Eisenbud and Shafarevich for a while, and they mention some interesting papers from EGA.
2:47 PM
@MathWanderer Don't exist, AFAIK.
@Danu Really?
Q: Is an English translation of Grothendieck's EGA available?

Amitesh DattaI have always wondered whether there is an English translation of Grothendieck's EGA (Elements de Geometrie Algebrique) available. Does anyone know whether there is and if so where I can find it? If not, are there English texts that cover similar material to the EGA that you would recommend? (My ...

The consensus seems to be: Learn enough French to read it.
You probably need more than French to read it!
Reading != understanding
2:48 PM
Does EGA provide any insight that cannot gain from Shafarvich or Hartshorne?
Shafarevich's BAG is just classical algebraic geometry; I think Hartshorne covers a lot but I have never read it
@MathWanderer Probably.
Q: The importance of EGA and SGA for "students of today"

user1161That fact that EGA and SGA have played mayor roles is uncontroversial. But they contain many volumes/chapters and going through them would take a lot of time, especially if you do not speak French. This raises the question if a student, like me, should even bother reading EGA. There must nowada...

I see. Since my current research in machine learning will involve tools from the algebraic geometry, I decided to study it. Since I do not have strong background in commutative algebra, I actually started with Eisenbud and Shafarevich.
I only heard that EGA offers a lot of useful theorems
@Danu Haha, I knew that buying all my mathematics textbooks in French was going to be worth it.
You definitely need a lot of commutative algebra to read EGA, or even Hartshorne.
2:53 PM
Yeah, that is why I started with Eisenbud. He does also incorporates some concepts from algebraic geometry. Strangely, Shafarevich did not require much background.
Hold on, what do you mean by "classical" AG?
Algebraic geometry of varieties, mostly.
There's quite a range of opinions in that question
I see. And I believe AG through scheme is "modern"?
What sort of tools from algebraic geometry do you intend to apply?
Not particularly, but schemes are a major part of algebraic geometry
@Danu I can't seem to prove that fact :(
2:57 PM
Me neither.
I guess it's passed the "am I being stupid" check
so I can ask my supervisor
You see, current machine learning started to ask questions about estimating a learning function with polynomials. My interest is how we could find optimal input set, which corresponds to varieties from different estimated polynomials .
To be honest, I am not sure which tools I could apply since I do not know enough theorems about AG.
All I know is that the main structures involve varieties and several-variable polynomials.
I decided to pick up Hartshorne first since I heard that it has vast amount of theorems and results, but it was not readable past first few pages.
Then I switched to books I mentioned. Some colleagues mentioned EGA as great source to pick up results.
what do you guys think of my argument. Let R be a ring and $S \subset R$ be a subset consisting of all elements which are not zero divisors. Prove that every element in $S^{-1}R$ is either a zero divisor or a unit.
Suppose [x/s] is not a zero divisor or the 0 element of the ring. This means x neq 0. Also, we have that property that for all [k/m] neq [0/1] we have $[x/s][k/m] \neq [0/1]$. [x/s][s/x] = [xs / sx]. Let us test if they are in the same class or not xs/sx ~ 1/1 iff there exists t in S such that t(xs - sx) = t(xs - xs) =0 so it is indeed true thus we are done.
@BalarkaSen what do you think of my argument ?
I forget. If an element of a ring isn't a zero divisor, must it be invertible?
Then how is $S^{-1}$ well-defined?
3:08 PM
$S^{-1}R$ is the ring of fractions for the multiplicative set S @Semiclassical
I have a formula to show that a given set of i.i.ds are normally distributed then I want to show a concrete example, however I am stuck on which variables to choose. I have tried splitting up the fraction to define the constants, but that leads me no where. This is what I am facing: mathb.in/126115 So I need to determine c_0 and c_1. I can only picture c_0 = 0, as this will give 0 + c_1*mu = 0 (since mu is 0), but I don't see how that is determined
3:36 PM
Now reaching: Munkres Section 9: AoC
Nvm that's too easy I guess
Answered 20 of 25 questions on the AMC 12. My understanding is that right answers are 6 points, wrong answers are 0 points, and answers left blank are 1.5 points; a score of 100 points or higher is needed to get to the next round.
Assuming I made at most four mistakes, am I guaranteed to pass?
Weird that blank answers can get more marks than attempted ones.
It's to discourage random guesses.
They don't want people passing because of chance.
Oh, is this an MCQ?
3:44 PM
A what?
Multiple Choice Q
Yeah, it's a multiple choice test
Ok, then that makes more sense
Lowest raw mark: ((20-4)*6+1.5*5)=103.5, a bare pass
3:46 PM
I realized towards the end that there were like three questions I forgot about.
@anonymous you know, sin^-1 (cos x) = sqrt(1-x^2) right
but d/dx ln(tan^-1 x) = 1/[(tan^-1 x)(1+x^2)]
and I don't actually know how to continue
@BalarkaSen I was wondering if you could check my argument ?
@Adeek I don't think I want to
oh ok
4:18 PM
@BalarkaSen So I think the "fact" is not true.
A professor pointed out to me that the stability under $J'$ does not depend on $J$ so it would be weird if this is equivalent to a condition involving $J$
@Danu I'm angry at you now!
But good to hear that
It was stated so matter-of-factly in Salamon's paper that I didn't think it would ijnvolve any specifics
he made it sound like a well-known easy result
I guess it has to come from the specifics of my map
4:34 PM
@Danu That makes sense. What's true is that if the almost complex structure on the ambient manifold preserves the tangent bundle of the submanifold, then the submanifold admits an almost complex structure under which it's an almost complex submanifold of the ambient manifold.
But that's obvious.
@DHMO No, it is much simpler than what you think
Remember than arcsin(y)+arccos(y)=pi/2
And $e^{\ln(\tan^{-1}x)}=\tan^{-1}(x)$
Can you do it after this ?
@anonymous oh god i didnt notice this
I think it's trivial then
Same with me. I deleted the problem coz I thought it will be too easy for you :)
I have some more problems...if you are interested I can give you
(On definite integration)
Can you find the reduction formula for this $I_n$^ ?
4:49 PM
what is ln^n x?
log(x) raised to the power n
Integration by reduction formula in integral calculus is a technique of integration, in the form of a recurrence relation. It is used when an expression containing an integer parameter, usually in the form of powers of elementary functions, or products of transcendental functions and polynomials of arbitrary degree, can't be integrated directly. But using other methods of integration a reduction formula can be set up to obtain the integral of the same or similar expression with a lower integer parameter, progressively simplifying the integral until it can be evaluated. This method of integration...
simple integration by part
yes, okay bye :)
So basically, no choice if the sets in the collection has nontrivial intersections
AC for pairwise disjoint families of sets is equivalent to AC over ZF
5:01 PM
5:14 PM
AC proofs are often nonconstructive, is it because the choice function is hard to specify explicitly in general?
If you can write it explicitely then you don't need AC to prove that this family of sets admits a choice function
I see
@Secret if $F$ is your family of sets you can take $F'=\{A\times\{A\},A\in F\}$ as a new disjoint family, use that version of AC to get a choice function and then construct one on $F$ using the one on $F'$
(While AC$\implies$ AC for pairwise disjoint families is obvious)
yeah munkres use similar ways to define something similar to the set $F'$ by using order pairs of one of the set in the collection and its elements in order to "partition" each element in the sets in the collection in terms of which set they belong, even if there are two sets that are not disjoint
I am not used to thinking in that path of logic though probably because my thinking level is too focused on the elements in each set in the collection, and thus don't realise we can group them up by using distinct ordered pairs
5:30 PM
@DanielFischer Hi, If we take $\Bbb{R}$ with the usual topology then $\{x_n, n\ge 1\}\cup \{l\}$ where $x_n$ converges to $l$ is compact. I was wondering if the same property hold for $\Bbb{R}$ with the topology generated by the intervals $[a,b)$. For exemple $\{1/n, n\ge 1\}\cup \{0\}$ is not compact. But can we prove that is never compact for any sequence ?
Is this proof correct? texpaste.com/n/ftgmk2pt
I can copy it here, but is a bit long.
I suspect the proofs on the chain of length $\mathfrak{c}$ that we all went through last week might be actually a case we found some explicit choice function for $\mathscr{P}$(countable).
That is, whether it is the dedekind cut on the rationals or the 10-adic like construction on the naturals (to produce one of the sets of supernatural numbers) to map every real to every subset, each infinite subset in the collection seemed to be indiced uniquely by the aforementioned two constructions, thus acting as a choice function for $\mathscr{P}$(countable)
@Topologicalife I don't understand your "then", second line
@BalarkaSen Agreed on both counts.
@JeSuis there is only one 'then' and it is on the first line.
What do you not understand?
I show that $f(2^a + 2^b) = f(2^b(2^{a-b} +1)) = (-1)^b [(-1)^{a-b}+1] = (-1)^a + (-1)^b$
5:43 PM
the second equality $f(2^b(2^{a-b} +1)) = (-1)^b [(-1)^{a-b}+1]$
From the definition, it follows that $f(2^a) = (-1)^a$
yeah I agree, but why did you get the equality ? ok $f(2^{a-b}+1)=(-1)^{a-b}+1$ but are you using somethinf as $f(mn)=f(m)f(n)$ ?
@Secret you don't need AC to get a choice function on $P(\Bbb N)$ or any other family of well ordered sets, just pick the minimum from each
but the rationals are not well ordered as there are subsets of them that decreases indefinitely?
5:49 PM
Is $ w = e^{y ln(z) } $ equivalent to $ f(x,y,w) = e^{ y ln(z) } - w $ ?
? @Maks
Or do you mean the powerset of Q is well ordered?
Every set in the powerset of N is well ordered
@JeSuis I'm practicing partial derivatives
All the exercises have f(x,y) = ... or f(x,y,z) = ...
And I have two which say differentiate w = ...
How am I supposed to differentiate $w = e^{y ln(z)}$ ?
@Maks You want to differentiate as the function of which variables ?
5:54 PM
y,z and I guess w
But I dont know, I know it has 3 variables
so you get a function from $\Bbb{R}^3$ to $\Bbb{R}$
That's what I did
On top, with f(x,y,w)
No wait, there are uncountably many sets with indefinitely decreasing sequences, and they are of order type $\omega$ in the rationals, thus P(Q) cannot be well ordered in the standard ordering
so what is the problem ? Just write the derivative as matrix ?
@JeSuis yes, cause it holds.
Just skip that step, I still have to prove it, but it holds.
6:08 PM
I was just doing some basic number theory questions and I can't quickly see how to prove that for every prime number p >3 it holds that 24 | p^2 - 1. How does one prove that again?
@Helios Every prime number p>3 is of the form $6n+1$ or $6n-1$. Use it.
Ah yep! Thank! :)
Welcome :)
6:47 PM
Fun fact: $\int_0^\pi\frac{x^n(\pi-x)^n}{n!}\sin(x)\operatorname d\!x$ is in $\Bbb Z_n[\pi]$
(meaning, it's of the form $a_0+a_1\pi+\dotsb+a_n\pi^n$ for integers $a_i$.)
@PVAL I don't know.
It's bounded below by $0$ and above by $\frac{{\rm constant}^n}{n!}$.
That's enough to prove that $\pi$ is irrational, after not too much further work.
(I think ${\rm constant}=\pi/4$, actually)
(I mean, it's strictly bounded below by $0$. It's positive.)
7:02 PM
You could easily fit Niven's proof into an exercise at the end of a textbook chapter.
Hey. I need help with a cardinality proof.
Q: Proof that $|A-\{x\}| = n-1$ using bijection

JessyunBourneIf $A$ is finite with $|A|=n \geq 1$ and $x \in A$, I need to prove that $A - \{x\}$ is finite with $|A-\{x\}| = n-1$, where $|\cdot|$ is the cardinality. I know that there are proofs for $|A-\{x\}| = n-1$ which rely on the result $A = (A-\{x\})\cup \{x\} $, but I was wondering if there was a wa...

I don't fully understand your restrictions, if you have a bijection $f: \{1,..,n\}\to A$ and $f^{-1}(x)=i$, let $\tau: \{1,..,n\}\to\{1,..,n\}$ be the map that swaps $i$ and $n$, then the inclusion $\{1,...,n-1\}\to\{1,..,n\}$ followed by $\tau$ and then $f$ is an injection into $A$ as a composition of injections, then you can verify that any $y\in A$, $y\neq x$ has a preimage that lies in $\{1,..,n-1\}$ under these maps
7:18 PM
@PaulPlummer You just internalize it eventually. Eg it's not inherently obvious that $\alpha \wedge d\alpha = 0$ means integrable, so you do the computation, but after that the intuition about why $\alpha \wedge d\alpha \neq 0$ pointwise means "completely non-integrable" makes more sense.
I was asked to prove that if p_1, ..., p_t are different primes then these are linear independent given that the coefficients are rational.
Excuse me that message got sent accidentally, It was supposed to be longer

I was asked to prove that if p_1, ..., p_t are different primes then these are linear independent given that the coefficients are rational. Now, I know that if for every x in Q, there exists a sequence of primes n_p such that
x= sign(x) \product p^{n_p}. I know I can use this here, yet I am not sure how to start the proof.
@MikeMiller @BalarkaSen I will probably be participating in a seminar going through Milnor's book on Morse theory next semester. I may have to choose what part to present on tomorrow. What would you recommend?
I dunno.
The really cool thing in that book is the application to loop spaces.
Is that in part IV?
I dunno.
7:33 PM
So how do you know it's in there? :P
@s.harp were you talking to me?
@s.harp I am so confused by what Omnomnom is saying.
I don't know what explicit bijection to pick to map $A$ to $\mathbb{N}_{n}$. Apparently I'm supposed to explicitly define one.
There is nothing like this in my notes, we did nothing like this in class, and I'm so incredibly confused.
What does it mean for a set $A$ to have cardinality $n$?
It means that you can find a bijection between $A$ and the first $n$ natural numbers.
7:39 PM
strictly speaking that there exists such a bijection
So if $|A|=n$ there exists a bijection $f: A\to \Bbb N_n$, and you can work with that
Well, we're told that $A$ is finite and $|A|=n \geq 1$, and we're told that that's what cardinaoty means, so I don't see that we don't know that there is one.
Yes, I know, but he's making it sound like I need to come up with some explicit function! And I don't know how to do that.
If I know that there exists $f:A\to \mathbb{N}_{n}$, how do I know that there exists either of those other two bijections he mentions, and how can I even prove that they are bijections??
You don't need to come up with an explicit function, because you do not know what $A$ looks like, $A$ could be $\{\text{apples},\text{bananas},\text{monkey}\}$
Okay then what about the last question I asked you above?
Ok, so the first bijection he mentions is $g: \Bbb N_n -\{f(x)\} \to \Bbb N_{n-1}$
Consider the map $\Bbb N_{n}\to \Bbb N_{n}$ that switches $n$ with $f(x)$
@s.harp $\mathbb{N}_{n} \to \mathbb{N}_{n}$ just maps the first n integers to the first n integers. It doesn't "switch" anything.
7:44 PM
Call this map $\tau$, if you restrict $\tau$ to $\Bbb N_n -\{f(x)\}$ you find it is a bijection onto $\{1,..,n-1\}$
How do I prove that though?
$\tau(k)=\begin{cases}n & k= f(x)\\ f(x) & k=n \\ k&\text{else}\end{cases}$
@Danu Because I read the book but don't remember what the parts are.
@s.harp why didn' you just say so?
Thats what I meant with $\tau$ switches $n$ and $f(x)$
7:46 PM
@MikeMiller OK. Thanks anyways :)
Can you explain how that map works with specific numbers?
Well, it depends specifically on what $f(x)$ is and what $n$ is
For example if you have $n=10$ and $f(x)=7$, $\tau(1)=1,..,\tau(6)=6, \tau(7)=10,\tau(8)=8,\tau(9)=9,\tau(10)=7$
@Danu: There are all sorts of good things in there for you. You could do the Lefschetz Theorem. You could do section 4 with symmetric spaces and Bott periodicity. Plus Jacobi fields and relations between topology and curvature in section 3.
So now I'm trying to figure out how all of this fits back in terms of A and $\mathbb{N}_{n}$
7:53 PM
If $|A|=n$ there exists a bijection from $A$ to $\Bbb N_n$. Let $f$ be such a bijection, let $\tau: \Bbb N_n\to \Bbb N_n$ be the bijection that swaps $n$ with $f(x)$. If you denote the inclusion $I:\Bbb N_{n-1}\to\Bbb N_n$, then the map
$$f^{-1}\circ \tau\circ I: \Bbb N_{n-1}\to A-\{f(x)\}$$
is well defined and a bijection
@s.harp what is the inclusion $I$?
So, it's the identity
so $I:\{1,...,n-1\}\to\{1,..,n\}$
no because the codomain is not the same as the domain
I've never heard that terminology before
7:54 PM
what terminology?
If you have a set $X$ and a subset $Y\subset X$ there is an "obvious" map from $Y\to X$ that sends every element to itself, this is called the inclusion
$I$ is not a bijection, though.
No, it is an injection
So, the composition of an injection and a bijection is a bijection?
7:58 PM
One needs to verify that the map up there is a surjection
No, it isn't in general. But for these specific maps it will be the case
The composition of injections is always an injection, so the map up there is an injection

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