Unrelated but I just discovered that Barilla (the pasta brand) has a spotify accounts with playlists of lengths equal to the cooking times of different kinds of pasta and I'm not sure what to say about that
@BalarkaSen But Gromov told me that there is no geometry in finite sets
@user2103480 In Bonn there were some amazing Italian places for pizza. I still have to find a really good in Münster since everything's closed more or less apart from take away food
i get it. although i don't think anybody is prioritizing her students. because they're young and mostly healthy, except when they're not. i'm glad i'm not teaching anymore. she has had to deal with a number of students with deaths in the family or household. i'm not built for that
The UK government wasted 22 billion pounds on their failed test and trace app and various corrupt contracts given to friends of the government/party donors
i would love a MS office permanent license. i have to use it for work. one time someone gave me guff about not having it on my home PC (when the office network was down). i shamed them by explaining that IT best practices was not to keep any confidential information on my home computer.
"you don't, like, email yourself word documents at home, do you?"
wow, you weren't kidding. 22 billion. i thought maybe you had just made up an absurdly large number. truth is always stranger than fiction.
is it too late to submit other bids? i will offer a functioning test and trace app to the UK government for 21 billion pounds.
the testing is the hard part. the tracing i'll just build off of grindr. pretty good data there.
sanity check: if $M,N$ are two manifolds and we glue them along a diffeomorphism $f$ of their boundaries, are the images of $H^{2n}(M\cup_fN,\partial M\cup_f\partial N)\rightarrow H^{2n}(M\cup_fN)$ and $H^{2n}(M\cup_fN,\mathring{M}\cup\mathring{N})\rightarrow H^{2n}(M\cup_fN)$ orthogonal with respect to cupping?
I hate your notation so I suggest my own. $M \cup N$ for the union and $S$ for the image of the boundary and $C \cong S \times (-1,1)$ for a collar neighborhood. Notice that the former space can be replaced by $H^{2n}(M \cup N, C)$ at no cost. So these are cochains which vanish on C. The cochains from the other guy are cochains which vanish on $M \cup N \setminus S$.
Notice too that by Barycentric subdivision argument you can replace $C_*(M \cup N)$ with $C_*([M \cup N \setminus S] + [C])$, chains supported in one of these two open factors.
Since your cupped cohomology class vanishes on chains contained in each factor it vanishes on the whole of that second chain complex. Every cycle in $C_*(M \cup N)$ is homologous to one in that chain complex so you are finished.
This ought to be a correct proof that if $X$ is the union of open subspaces $A$ and $B$ then the cup products of $H^n(X, A)$ and $H^n(X, B)$ lie in $H^n(X, A \cup B)$.
Reduce rational number to lowest form $p/q$ and then let total digits of numerator and denominator be $t$, then we map this rational number to $p$ in decimal, then $t$ zeros, then $q$ in decimal concatenated. For example $3/11$ becomes $300011$
@TedShifrin @MikeMiller For any finite set $S \subset \Bbb R^n$ define for any $s \in S$, the isolation radius $\rho(s) = \min\{d(s, t) : t \in S \setminus \{s\}\}$.
Let us denote, for any $\varepsilon \in (0, 1)$, the annulus $\mathcal{A}(\varepsilon, x, y) = S \cap (B(x, \rho(x)/\varepsilon) \setminus B(y, \rho(x)\varepsilon))$. Note that if $\varepsilon$ is closer to $0$, the annulus is "thicker", with incenter $x$ and outcenter $y$.
Fix $0 < \varepsilon < 1$. There exists a uniform constant $C = C(\varepsilon)$ such that for any finite set $A \subset \Bbb R^n$ and for any $k \geq 2$ n…
That is to say, given any finite set $A$, seen from most points of $A$, most of $A$ clutters around a single point of $A$.
there's certainly a surjective map from (integers) x (nonzero integers) to Q. that ought to be enough without specifics of lowest terms or any decimal encoding. it looks like that map might be injective, but proving it seems annoying, given numbers that can terminate in large numbers of zeros and still be parts of fractions in lowest terms. i just wouldn't want to get into analyzing that.
Write +- p/q in lowest terms. Write p and q in binary. Then write the integer whose first digit is 9 if +, 8 if -, then is p written in binary, then the digit 5, then q written in binary.
@MikeMiller actually no to that question, i dont know much about all that things, only our teacher asked us just like that to think why rational is countable
In [8] H. Namazi, P. Pankka, and J. Souto found the lemma useful to show a certain non-degeneracy of the distributional limit of random Riemannian manifolds with a curvature bound.
@BalarkaSen I gotta learn more about cameron-martin spaces next semester to really tell the story, but one day you'll be hit by a wall about gaussian measures on hilbert spaces and how there's a commutative diagram of maps, for some covariance operator Q and a predictable operator-valued process H
Collection of Gaussians -> Collection of Brownian Motions -> collection of stochastic integrals over projections | | |
I would say there is no eye-independent statement; you have to pick a point (at random, maybe), place an eye there, and ask for some accumulation point wrt that eye's horizon.
Certainly, because you can imagine the rest of the points beyond your visibility range accumulating towards some point in your horizon. But for a random choice of an eye, this accumulation point is also unique.
I attended a talk by a probabilist who kept using this lemma and hyperbolic geometry to prove whatever he wanted to prove and was amazed that they understand geometry better than me.
I'm shattered by the thought to be honest
@user2103480 wtf is this diagram
dude you should teach me (a) what a Gaussian free field is and (b) what a Schramm-Loewner evolution is
I do not know enough stochastic geometry for either
I also absolutely did not know a Brownian motion is conformally invariant but that makes sense in retrospect
So [obvious anachronism] you study C(X) and then invent C* algebras as the relevant properties of these. Then you notice there are plenty of noncommutative C* algebras. For instance, bounded operators on Hilbert space.
Then you pretend you are doing topology because you were when you were working with the commutative ones.
IIRC the first example of a "noncommutative space" that Connes talked about is attached to a foliation of a manifold, so a real geometric object. Somehow the noncommutativity is about the way the leaves accumulate onto each other in some confusing way.
If you take foliation by $\alpha$-slope lines in $T^2$ you get the "$\alpha$-noncommutative torus", the algebra $\Bbb C \langle T, S\rangle/\langle TS - ST = e^{i\alpha} T\rangle$ or something liek this.
Through some K-theory junk one can actually use this gadget to classify the foliations of $T^2$ by lines up to diffeomorphism.
(These things are equivalent in the relevant sense iff $\alpha_1, \alpha_2$ are rationally dependent, I believe, so that the foliations by $\alpha_1, \alpha_2$ lines are equivalent iff those are rationally dependent.)
I believe I agree with you that classical results should be enough (thoughh I'm not certain which you mean, did he classify diffeomorphisms up to smooth conjugacy????)
@BalarkaSen No, you tell me. What structure is used to recover Omega in the case of planar domains?
I'm just saying that if $T_\alpha$ means torus with foliation by lines of slope $\alpha$, then $T_\alpha = \text{Susp}(R_\alpha)$, but $R_\alpha$ is not conjugate to $R_0$ unless $\alpha = 0$.
Identity is its own conjugacy class
I'm sure this must be old and well-known though, this fact I've stated
The question is a bit ambiguous, but we can interpret it as one about computability and decidable equality. Let $\mathbb{Q}$ be the set of rational numbers, $\mathbb{Q}^+$ be the set of positive rational numbers, and $\mathbb{R}$ be the set of real numbers. A computable real number is a real numb...
But you gotta wait like 5 months, I dont have enough PDE power to handle the laplacian/sobolev space stuff yet. We had gaussian measures on hilbert spaces in the SPDE course but we swept the details under the rug. Next semester there's a course on "gaussian measures in infinite dimensions & invariant measures for PDE" and I think it'll be about much of the stuff
if I have a symmetric bilinear form on a vector space $V$ and a direct (not necessarily orthogonal) sum decomposition $V=V_1\oplus V_2$ and furthermore the form is zero when restricted when $V_2$, is the signature of the form equal to its signature when restricted to $V_1$?
this is supposedly true, but I can't even do linear algebra anymore
