your window: boring, flat, no structure, no curvature my window: non-constant sectional curvature, three different algebra structures in (co)homology, high genus, not stably parallelizable
Silly question : can I be sure that my integral converges : f is Lipschitz so a.e differentiable, $\phi$ has compact support, $u_n$ converges to $u$ in L^1. The integral is $$\int u_n \nabla f \cdot \nabla \phi $$ as $n\to\infty$
you should just know what it mean geometrically. the element of H^1(RP^n; Z/2) is generated by the cochain which takes value 1 on smooth, singular 1-simplices in RP^n transverse to RP^(n-1), and 0 elsewhere
if you call this element a, then a^2 actually takes value 1 on smooth, singular simplices in RP^n transverse to RP^(n-2) = intersection of two RP^(n-1)'s in RP^n
computing the cohomology ring of RP^n/CP^n/HP^n is more or less equivalent to calculating that the top Stiefel-Whitney/Chern class of the tautological bundle is a generator
@BalarkaSen to tell the difference: descriptive complexity theory classifies least runtimes by describing the logics in which problems can be formulated
@Astyx "I used an ultrafilter to prove your argument about consumer behaviour wrong"
People really overestimate the usefulness of logic in everyday life. It's the main problem of logic to find logical formalisms that work for problems which were not formalized before
@Astyx “My opponent’s theory has quantifier elimination, so anything he says can be found out to be crap algorithmically, and this is why you should elect me.”
for anyone who might care, when philosophically deriving turing machines, turing actually used a topological argument to argue the number of symbols should be finite
Computing is normally done by writing certain symbols on paper. We may suppose this paper is divided into squares like a child’s arithmetic book… I shall… suppose that the number of symbols which may be printed is finite. If we were to allow an infinity of symbols, then there would be symbols differing to an arbitrarily small extent.* The effect of this restriction of the number of symbols is not very serious. It is always possible to use sequences of symbols in the place of single symbols. Thus an Arabic numeral such as 17 or 999999999999999 is normally treated as a single symbol. Similarl…
Suppose we have on a horizontal timetable
$m\textbf{a}_{rot}=\textbf{F}-\textbf{F}_{corialis}-\textbf{F}_{centrifugal}$
where $\textbf{a}_{rot}$ is the acceleration in a rotating frame. Suppose an ant on the turntable walks from the centre at constant speed. Suppose an ant slips an experiences a...
Does showing that $D_t(X) = 0$ for all curves in $(M,g)$ iff $\nabla_Y X = 0$ for all vector fields Y on M use the fact that M has to be path connected?
$\Bbb{RP}^n$ is just $\Bbb R^n$ where you cannot expect intersections to reduce dimensionally if you push everything to infinity. That is how I think of it.
OK, I am sure you have more experience with projective spaces that I do, so you must be right that my picture is insufficient. I was explaining why the cohomology ring is $\Bbb F_2[\alpha]/(\alpha^{n+1})$, so I was doing mushy topology :)
It doesn't give me any reason @Ted. It just says $(M,g)$ is path connected and asks you to prove that. I can prove that statement without using path connectedness, hence my query as to why the question assumes M as path connected
you should just know what it mean geometrically. the element of H^1(RP^n; Z/2) is generated by the cochain which takes value 1 on smooth, singular 1-simplices in RP^n transverse to RP^(n-1), and 0 elsewhere
if you call this element a, then a^2 actually takes value 1 on smooth, singular simplices in RP^n transverse to RP^(n-2) = intersection of two RP^(n-1)'s in RP^n
To be 100% clear the picture is $D^n$ with $x \sim -x$ at the boundary. All I see from the center of the disk is the interior of the disk $D^n$, but of course I can move around and see more.
This is not a controversial statement. Since when can you see the curvature of Earth while sitting in your room?
@TedShifrin I see, for example, that it is compact. If I move, I come back to a point. I also see that it is nonorientable, because my orientation changes when I come back.
I can "probe" it to gain more information.
Of course, I get all the transition functions by simply moving around.
This is how Thurston describes a lot of 3-manifolds in his book
You have to see it from sitting inside it because otherwise you're screwed.
I am not just sitting in my own room. I can sit in some other room. This is how the Earth was found out to be round, by a ship moving East and coming back
Well, Foucault appears local, but the entire line of longitude spins under your feet (which are off the ground) if you're going to see what happens in 24 hours.
Hmm. I guess the mathematical content of "You can tell what space you are in by probing it by taking long detours from your home and coming back" is "$X$ is determined by $\Omega X$". Which is true, homotopy theoretically. What's a homeomorphism-type counterexample?
I need a quick knock in where to start by finding the limit of: $ 1 / (x * (\sqrt(x^2 +2) - \sqrt(x^2 - 2))) $ - I know the result must be 1/2 , so the denominator's limit needs to be 2, which means the limit of the sqrt term needs to be 2 over x ... and there I am stuck getting to the limit of $(\sqrt(x^2 + 2) - \sqrt(x^2 - 2)$ equals 2/x
So basically \frac{4}{\sqrt{n^2 + 2} + \sqrt{n^2 - 2}} needs to be equal to 2/n ... hold on ... I remember I had MathJax working at some point for me but right now it isnt
... reading the docs of MathJax rn to understand if I need a browser plugin or not
Though I feel like they want to trick their students here with it once being + and once being - to disallow making a clear > or < statement when doing that on $\frac{4}{\sqrt{n^2 + 2} + \sqrt{n^2 - 2}}$
if i try to move up in dimensions, the next interesting case would seem to be roots of $x^5-1=(x-1)(x^4+x^3+x^2+x+1)$. ($x^4-1$ should just reduce to considering $x^2+1$ which is just Gaussian integers.)
