@BalarkaSen I'm flying to Israel tonight, which means that we'll end up being only 2.5 hours apart instead of the 9.5-hour time difference we have now. I thought you'd find that interesting.
How do you make the connection that because $\mathbb{Z^+}$ is infinite, there must exist an infinite amount of prime factors and consequently prime numbers?
I'm trying to prove that for any $N \in \mathbb{Z^+}$, there exists only finite many integers $n$ with $\phi(n) = N$ (i.e. finite amount of numbers that have $N$ numbers relatively prime to them)
I started by addressing the lower bound of $\phi(n)$ as the set of primes $P = \{0<p\leq n~|~\phi(p)...
that suggests a nice (if rather open-ended) question, actually. What are some infinite subsets of the natural numbers which unexpectedly contain an infinite number of primes?
Hint 2: Every number can be written as a product of primes (why?). So if I give you that finite list of primes, all you have to do is find something that's not a product of things on that list to show that my list is incomplete.
The Bunyakovsky conjecture (or Bouniakowsky conjecture) stated in 1857 by the Russian mathematician Viktor Bunyakovsky, asserts when a polynomial in one variable with positive degree and integer coefficients should have infinitely many prime values for positive integer inputs. Three necessary conditions are
the leading coefficient of is positive,
the polynomial is irreducible over the integers, and
as runs over the positive integers, the numbers should be relatively prime. (Thus, the coefficients of should be relatively prime.)
Bunyakovsky's conjecture is that these three conditions ar...
it mentions on that page, btw: "That n^2+1 should be prime infinitely often is a problem first raised by Euler, and it is also the fifth Hardy–Littlewood conjecture and the fourth of Landau's problems." so quite a pedigree.
@Krijn My algebraic geometry prof let me borrow his copy of Fulton's intersection theory for a few days. Going a bit off-course shortly to read that :)
Well, read chapter 1. No way I am going to be able to read the whole book without more background, and I don't want to anyway.
I wonder if PVAL saw the counterexample to his question I came up with.
It's not even trivial. If you have a purported finite set of primes which appear in the prime factorization of every integer, just take a prime outside that finite set. (primes are infinite, so you can always do that).
This contradicts existence of any such "fundamental primes".
@datalava It's not phrased very well, but it's enough, at least if the OP invokes the theorems that I assume they invoke. Pointwise convergence of the ($C^1$) functions $(f_n)$ plus uniform convergence of the derivatives $(f_n')$ implies that the limit function is continuously differentiable, and its derivative is the limit of $f_n'$.
@Krijn Oh sure there are libraries. But prof gave it to me from his own good will, so I took it :) After all library books don't compare with owned books, especially is the owner is an expert on the topic it's about.
@DanielFischer Heh... I need to change my window size so that I can click on that link (the flag, star, link icons block it for me normally). I was thinking that it might be that, but if others did not notice anything, it might just be something with my system.
I guess the military-industrial complex was offended by the implication that they are of fundamentally no value to the world; I think that's a flaggable offense.
So suppose I have a smooth manifold $M$. 0-forms on $M$ are just real valued smooth functions on $M$. $1$-forms are smooth sections of the cotangent bundle, i.e., maps $M \to T^*M$ which sends a point to a covector of the contagent space $T_xM$. $k$-forms in general are smooth sections of the antisymmetric tensor product $\bigwedge T^*M$ of the cotangent bundle.
@MikeM: In semi-seriousness, your response is why I spent more energy on teaching (and book writing) and less on research over the last half of my career.
I am not sure what answer you want to the "important" bit. I can form a cochain complex out of the vector space of all $k$-forms and compute it's cohomology. That's an invariant of the smooth manifold, so seems like we shouldn't ignore that.
beyond that, it's interesting in the same way that a crossword or a jigsaw puzzle is interesting. it's a satisfying intellectual challenge, but i don't hold any illusions about it actually mattering
@Balarka: But there is a good question. Your first answer was very abstract vector-bundle-ish. What is special about the exterior powers of $T^*M$ is that you do have $d$. Why do we have $d$ in this case and not in general?
@MikeM: You'd think they'd put those meetings in the evening and not in the middle of the day.
Yes, @Balarka. Of course, Stokes's Theorem is the answer to a lot of these questions. But to me a lot of the motivation comes from differential geometry (and I mean geometry, not manifolds).
Sure, @Semiclassic, but isn't that just window-dressing?
@Ted: No, this is a good decision. Usually they are held in the union office at 5PM on a weekday. Nobody comes but leadership. We've moved them to mid-day (when people are still on campus) to the sculpture garden. This means some people have conflicts, but it also means more people will probably come.
By the way, that point-set question I left for ForeverMozart is available to all: Any idea of a normal space $X$ for which $X\times [0,1]$ is not normal? Surely that can't be.
(In general, the product of normal spaces needn't be normal, of course. But $[0,1]$?)
@Semiclassical You say this as if it's obviously necessary to doing anything, but there are a number of mathematicians who do alright by working more-or-less invariantly.
@TedShifrin Can you elaborate without revealing what you mean when you say "why do we have $d$ in this case and not in general"? In particular, can you explain what you mean by "general"? Just trying to understand the question a bit better.
@TedShifrin Oh of course I agree. I am going back home 23rd, and going to study forms then. I was merely getting a general overview before plunging it :)
@TedShifrin So what does $d$ really do? Suppose let's just look at $\Omega^0(M) \to \Omega^1(M)$. Then $d$ eats a smooth function $f$ on $M$ and spits out {1-form which eats a vector field $X$ on $M$ and spits out directional derivative $X(f) : M \to \Bbb R$}.
Yes, @Balarka. And if you have a general vector bundle $E$, you could ask about directional derivatives of sections of $E$ (and you could multiply sections by smooth functions, too).
@MikeM: Seems the key point (as it often is) is that $(\text{im}\, T)^\perp = \ker(T^*)$ and the companion formula.
oh. i think i asked @mike this already, but not sure about you @ted. is the Hamilton-Jacobi equation something you've ever run into? i'm curious as to how obscure that topic is to non-physicists.
Oh yeah. I can still take a vector field on $E/M$, which is a smooth section $X : M \to E$ and take a continuous function $f$, directional derivate along each $X(p)$. Hmm.
So what does really go wrong? Interesting. I'll ponder a bit more.
I didn't read it carefully, @MikeM. But long paragraphs with tons of formulas are hard to read. So better to emphasize the key point and then put in details.
@BalarkaSen can you explain to me why is $\bar{f}_{*} : H_0(RP^n,Z_2) \rightarrow H_0(RP^n,Z_2)$ is an isomorphism where $\bar{f} : RP^n \rightarrow RP^n$ defined as $[x] \mapsto [f(x)]$ where f is a odd map between $S^n$ and itself.
@BalarkaSen In your original case, you take a function and spit out a 1-form. In your case, what's the input and what's the output? (At least, what should it be.)
In the mathematical field of general topology, a Dowker space is a topological space that is T4 but not countably paracompact. They are named after Clifford Hugh Dowker.
The non-trivial task of providing an example of a Dowker space (and therefore also proving their existence as mathematical objects) helped mathematicians better understand the nature and variety of topological spaces. Topological spaces are sets together with some subsets (designated as "open sets") satisfying certain properties. Topological spaces arose as generalization of the open sets of spaces studied in elementary mathematics...
Karim: Here's a specific question. Consider the 2-fold covering map $f\colon S^n\to \Bbb RP^n$. What is $f_*\colon H_0(S^n,\Bbb Z/2) \to H_0(\Bbb RP^n,\Bbb Z/2)$?
@Adeek Isn't the generator of that group the homology class of a point? Unless I'm missing something, the fact that $\bar f$ sends a point to a point means that it sends the generator to itself