For a matter modelling person, the most valuable resource is computing power. For many of us, computing power at hand limits the scale of problems we can solve. There are many national supercomputing facilities for academics. What are the resources available in each country?
Is there any "sandwich" theorem for cardinalities? Say, $A \subset B \subset C$, $|A| = \mathfrak c$ and $|C| = \mathfrak{c}$. Can we say $|B| = \mathfrak c$? I suppose we can, but I'm looking for a proof.
I think I have a messy proof which enables me to state for all $m > 2$ being a prime number:
$$ \sum_{k=2}^{m-1} \pi(2m) - m+2 = \pi(m)+2S(m^2) $$
Where $\pi(x)$ is a function which counts the number of primes $\leq x$ and $S(x)$ is the number of odd square free semi primes less $\leq x$. Can y...
When I went to the AAL concert they had Plini and The Contortionist opening, and the same bassist played with all three of them (I think he was the bassist from the Contortionist but I don't remember exactly)
@BalarkaSen lol that's an accurate description tho
I've got an element $\sigma \in W(K^{\text{ab}}/K)$ and $E_\sigma$ its fixed field. Now $\sigma|_{K^{\text{ur}}}$ acts as $\operatorname{Frob}_K^n$ so $E_\sigma \cap K^{\text{ur}} = K(\mu_{q^n - 1})$. The question is now why $E_\sigma/K(\mu_{q^n - 1})$ is totally ramified. Obviously $[E_\sigma : K] = [E_\sigma : K(\mu_{q^n - 1})][K(\mu_{q^n - 1}) : K]$ and the latter factor is $[k_{K(\mu_{q^n - 1})}:k]$ (because $K(\mu_{q^n - 1})/K$ is unramified).
So why is $[k_{E_\sigma}:k_{K(\mu_{q^n - 1})}] = 1$
halp
maybe there's an obvious reason but I'm not seeing it haha
composition of immersion is immersion by chain rule and the fact that product of no nulltity matrices is no nullity yeah, what is a product of immersions
What is cone of CP^1 and is it an orbifold? I would thicken S^2, and collapse the inner S^2 to a point. So basically, this is a 3-ball with its core crushed. Is this picture correct? Also, if this is an orbifold, what is the local group at the point where the inner S^2 was collapsed (aka. the cone vertex).
No I dont understand the question at all. C(S^2) is topologically B^3 so you're asking for orbifold structures on B^3, which has many -- you can forget how you got the B^3 by coning a sphere because it has no relevance to the question.
@feynhat yes thats one of many orbifold structures on S^2
Okay. When I write R^2/Z_n, would you say its a disk (with trivial orbifold structure), or would you say its a cone with a singular point? I mean R^2/Z_n has an obvious non-trivial orbifold structure.
Well so it's not very explicit, but if you build in $(n,n+1)$ a set which is $\mathbf{\Sigma}^0_n\setminus\mathbf{\Sigma}^0_{n-1}$ and take the union you get a $\mathbf{\Sigma}^0_\omega$ set
that's the standard way of showing $\bigcup_{\alpha<\lambda}\mathbf{\Sigma}^0_\alpha\subsetneq\mathbf{\Sigma}^0_\lambda$ at limit $\lambda$
I'm pretty sure computability people will be able to give you an explicit example in the form "the set of reals which codes some computability related stuff", but I know nothing about this side of descriptive set theory
In $\Bbb R^2$ you can, for every $\xi<\omega_1$, build sets $U$ which are $\Bbb R$-universal for $\mathbf{\Sigma}^0_\xi(\Bbb R)$, meaning that $U\in\mathbf{\Sigma}^0_\xi(\Bbb R^2)$ and that every set in $\mathbf{\Sigma}^0_\xi(\Bbb R)$ can be realized as a slice of $U$. The construction is done recursively so it's somewhat more explicit I guess
usually, your function is either obviously differentiable since built out smooth functions with sum/product/composition or you'll have to go and check by definition
rare cases you can be sneaky and use the IFT to prove something is differentiable
Oh, yes. It cannot be a sphere. I mean look at $\mathbb{CP}^2$. The open set $U_0 = \{[z_0 : z_1 : z_2] : z_0 \ne 0\}$ is a neighborhood of $[1 : 0 : 0]$, but $\mathbb{CP}^2 - U_0$ is homeomorphic to $\mathbb{CP}^1$. We removed a connected neighborhood of $\mathbb{CP}^2$ and its still something which is not contractible, it wouldn't have happened if it were a sphere.
@feynhat CP^2 is R^4 = C^2 with a CP^1 at infinity (whereas there should really be a S^3 at infinity parametrizing all real directions, the point is you only parametrize complex directions)
Yeah, I think that noting that $f$ is a sum/product/composition of smooth function except at discrete points/ lines / planes/ hyperplanes? will get me a lot of "coverage"
Since my work is in computer science, and the problems I solve are non-computable in general, all of this is more about covering as much as we can
The coarsest boundary of a linear space $V$ is the sphere $S_\infty$ at infinity. If $V$ is complex linear space, you have the complex projective space $\Bbb P_\infty$ at infinity. There's a map $S_\infty \to \Bbb P_\infty$ going in the other way coming from the forgetful functor, which is our Hopf map
This situation I think comes up a lot with people who do boundary stuff
Okay, slightly more complicated sanity check: say $f : X \to Y$ is a smooth map whose derivative $df_x$ is a linear iso for all $x$ in a submanifold $Z$, and which is injective on $Z$. Then I am to prove that $f$ is a diffeo on a neighborhood of $Z$.
This is just because if it fails to be a diffeo on such a neighborhood, I can find a point $z \in Z$ and make two distinct sequences that converge to $z$ and which map to the same sequence in $Y$, but that would contradict the inverse function theorem at $z$, right?
Hmm. IFT tells you $f$ is a diffeo onto its image on a neighborhood of each point in $Z$. Glue these neighborhoods together and get a neighborhood of $Z$ on which $f$ is a local diffeo onto its image. This gives the desired conclusion iff $f$ is injective on that neighborhood. Why can we guarantee that..
Well, if it fails, then in particular, it fails somewhere---there must be some $z \in Z$ and distinct sequences $\{a_i\}, \{b_i\}$ in $X$ that converge to $z$ with $f(a_i) = f(b_i)$ by doing nbhd shenanigans
For a matter modelling person, the most valuable resource is computing power. For many of us, computing power at hand limits the scale of problems we can solve. There are many national supercomputing facilities for academics. What are the resources available in each country?
I think I have a messy proof which enables me to state for all $m > 2$ being a prime number:
$$ \sum_{k=2}^{m-1} \pi(2m) - m+2 = \pi(m)+2S(m^2) $$
Where $\pi(x)$ is a function which counts the number of primes $\leq x$ and $S(x)$ is the number of odd square free semi primes less $\leq x$. Can y...
