@BalarkaSen The fact that $S^2/G$ is a manifold is where you need the orientation-preserving hypothesis, through a case analysis on the structure of the non-free set (eg, that it's 0-dimensional follows because a 1-dimensional fixed set of an oriented action would force everything to be fixed, by thinking of the normal bundle).
To conclude, consider $(S^2)^{\text{irr}} \to (S^2/G)^{\text{irr}}$. The former is a manifold $\Sigma_{0,n}$ for some $n > 0$, and the latter is a surface $\Sigma_{g,k}$ for some $k > 0$, and we have that $\Sigma_{0,n} \to \Sigma_{g,k}$ is a covering map.
That forces $n \geq k$ and $2-n \leq 2-2g-k$; the latter is because you have the inequality for the absolute values, but the two terms have the same parity and 2-n is necessarily negative or we would already be done.
Oh, this isn't a contradiction. Whatever. I thought this was easy.
@TedShifrin Hi. Also bye, now that this is too hard for me.
@BalarkaSen $S^2/G =: X$ is a closed surface. Furthermore, it is orientable, because it is such once you delete the finite set of singular points. Because $G$ acts orientably, if you orient $X$ so that its orientation agrees with that of $S^2$ away from the singular points, we see that the degree of $p: S^2 \to X$ is $|G|$, and in particular, non-zero for all groups. The only closed oriented surface with a map from $S^2$ of nonzero degree is $S^2$, by the classifictation.
Thanks to Marco Marengon for the degree argument.
The previous argument was supposed to be an argument that there are no branched covers $f: S^2 \to \Sigma$ for $g(\Sigma) > 0$, but tried to argue that by just deleting the branch points. This still seems true to me at a glance but for instance $\Sigma_{0,4}$ covers $\Sigma_{1,1}$ so an argument that doesn't pay close attention to the enumeration of boundary curves wouldn't be enough.
@N.Maneesh It means that given the curvature and the torsion of a space curve, you can find a unique curve alpha which will satisfy that curvature and torsion upto discrete isometries of R^3, which are rotations or translations or a combination of both.
Say you have your curve and, consider translating this curve by some value $a$. Can you calculate using your serret Frenet equations what is the curvature and torsion, similarly in case of rotations(rotate the curve using the rotation matrix). This should give you an example.
okay means the curve is unique. curve which satisfy $\K_0$ and $\tau_o$ is either rotation or translation or cobination of both. right? Thank you very much :)
sorry for the awkward language. I understood what do you say. Thank you very much :)@Albas
Yeah, $S^2/G$ being a surface is fine. Fixed points of any element are isolated poles, like I mentioned, which are locally $D^2/\Bbb Z_n$ orbifold points. That's homeomorphic to $D^2$.
If $X$ is a compact Riemann surface of genus more than one, the biholomorphic automorphism group, let's call it $G$, is finite. $X \to X/G$ is a branched covering again by inspecting the fixed loci, which are again isolated $D^2/\Bbb Z_n$ orbifold points. By Riemann-Hurwitz $2 - 2g = |G|(2 - 2g') - \sum_p (r_p - 1)$, or $2g - 2 - \sum_p (r_p - 1) = |G|(2g' - 2)$. So $|G| < (2 - 2g)/(2 - 2g')$. If $g' \geq 2$ then $|G| < g - 1$.
If $g' = 1$ then $2g - 2 = \sum_p (r_p - 1)$. Hm, what do I say with that
$r_p$ divides $|G|$ since it is the order of a stabilizer, thus if we set $q_p=|G|/r_p$, we can factor out $|G|$ and obtain $2-2g=|G|(2-2g'-\sum_p(1-q_p))$ Call $Q=\sum_p(1-q_p)$. If $g'=1$, then there is at least some point that is ramified and hence $Q\neq 0$
in fact, having a ramified point means that $Q \geq \frac{1}{2}$, since all the $q_p$ are $\leq \frac{1}{2}$
If you want the algebraic version, and if you're happy to assume that the automorphism group is algebraic (not too hard, but is an extra step), then the fact that genus \ge 2 curves have finite automorphism group is a consequence of - computing the dimension of the aut group = dim of tangent space at identity - tangent space at identity = vector fields on your curve - for genus \ge 2 curves, there are no non-zero vector fields (Riemann Roch)
I'm confused by something that should be very basic. In the noncommutative geometry class we considered the connection $\nabla^g$ on the Clifford bundle $\Bbb C\ell(T^\ast M)$ ($M$ is some Riemannian manifold with Riemannian metric $g$) and said that $\nabla^g$ is the extension of the Levi-Civita connection on $T^\ast M$. But all the sources I see define the Levi-Civita connection on $TM$ instead?
I mean, the metric gives a canonical isomorphism. Levi-Civita connection explicitly depends on the metric, so I don't see what you want to accomplish by getting a canonical connection
the reason a metric gives you a canonical isomorphism is just the same reason that a non-degenerate bilinear form gives you a canonical isomorphism $V \to V^*$
@loch Ok, geometrification of your argument: $G$ is a Lie group (it's the subgroup of SL2(R) preserving the $\pi_1 \Sigma_g$-action, so I guess you argue that means it's a closed subgroup of SL2(R)). The Lie algebra is the space of vector fields on $\Sigma_g$ which is 0-dimensional by Poincare-Hopf
That'd prove discreteness and since $\Sigma_g$ is compact, $G$-orbit of a point is a discrete hence finite set, and stabilizer of a point is a discrete hence finite subgroup of $SO(2)$ (= stabilizer of a point in $\Bbb H^2$) so $G$ is finite.
@Mathein Let me rewrite. $2 - 2g = |G|(2 - 2g') - \sum_{p \in X} (r_p - 1)$ where the sum runs over branch points of $X \to X/G$ on the range $X$. $G$ acts on these branch points, so we pick representatives $p_1, \cdots, p_k$ from the $G$-orbit and let the branching index at $p_i$ be $e_i$ and the size of the the $G$-orbit of $p_i$ be $n_i$.
Then $\sum_p (r_p - 1) = \sum_i n_i (e_i - 1)$, and since $n_i e_i = |G|$, factoring out a $G$ gives $\sum_i (1 - 1/e_i)$. So $2 - 2g = |G|((2 - 2g') - \sum_{i = 1}^k (1 - 1/e_i)$). If $g' = 0$, $k = 3$ and $e_1 = 2, e_2 = 3, e_3 = 7$ then $2 - 2g = |G|(2 - 1/2 - 2/3 - 6/7) = -|G|/42$, so $|G| = 84(g - 1)$.
This is the maximal case. The Hurwitz surface is a branched cover over $\Bbb P^1$ with three branch points in the codomain with branching index $2$, $3$ and $7$
Consider the triangle group $\langle a, b, c | a^2, b^2, c^2, (ab)^2, (bc)^3, (ac)^7 \rangle$. This is generated by reflections along an abstract geodesic triangle with angles $\pi/2, \pi/3, \pi/7$, which embeds in $\Bbb H^2$ by Gauss-Bonnet reasons. So it's discrete subgroup of $\text{SL}_2(\Bbb R)$, and the von Dyck subgroup is $\langle x, y | x^2, y^3, (xy)^7 \rangle$ which is the subgroup of $SL_2(\Bbb R)$, orientation-preserving symmetries of the $(2, 3, 7)$ triangle tessellation
Let's call this $\Gamma$. Then I claim $\Bbb H^2/\Gamma$ is the Hurwitz surface.
I should be able to cook up the appropriate branched cover $\Bbb H^2/\Gamma \to \Bbb P^1$. Should be very obvious, let me think
iirc, from some remarks in my modular forms course: If $\Gamma(7)$ is the kernel of $\Gamma=\mathrm{PSL}_2(\Bbb Z) \to \mathrm{PSL}_2(\Bbb F_7)$, then $\Bbb H^2/\Gamma(7)$ is the Klein quartic and we obtain a branched cover $\Bbb H^2/\Gamma(7) \to \Bbb H^2/\Gamma$, where $\Bbb H^2/\Gamma \cong \Bbb P^1$ via the q-invariant
Right, the j-invariant gives a branched cover. $j$-invariant is $\text{PSL}_2(\Bbb Z)$-invariant hence also $\Gamma(7)$-invariant, so that descends to a branched cover from the affine Klein quartic to $\Bbb P^1$
@Mathein Oh yeah so can you tell me more about that? Here's what I understand: a complex torus is built as $\Bbb C/(\Bbb Z + \tau\Bbb Z)$ where $\tau$ is a complex number with $\text{Im}(\tau) > 0$. The Teichmuller space of elliptic curves then is $\Bbb H^2$, the parameter space of that $\tau$. $\text{SL}_2(\Bbb Z)$ is the stabilizer subgroup of $\text{GL}_2(\Bbb R)$ acting on lattices in $\Bbb R^2$, so the moduli space is $\Bbb H^2/\text{SL}_2(\Bbb Z)$.
Is this how you think about it? I have also seen it as the "space of isogeny classes of elliptic curves", since $j$-invariant determines isogeny.
How to think about $X(n) = \Bbb H^2/\Gamma(n)$ in this formalism?
I guess formally the Teichmuller space is the space of marked complex structures. A $\tau \in \Bbb H^2$ automatically gives a marked complex structure on $T^2$ by quotienting $\Bbb C$ by $\Bbb Z + \tau\Bbb Z$. Now the moduli space is the Teichmuller space upto reparametrizations on the domain, which are isotopy-classes self-diffeomorphisms of $T^2$ - that's the mapping class group $\text{MCG}(T^2) = \text{SL}_2(\Bbb Z)$.
So $Moduli = Teich/\text{MCG} = \Bbb H^2/\text{SL}_2(\Bbb Z)$
So the $n$-torsion of $\Bbb C/(\Bbb Z + \tau \Bbb Z)$ is always a free $\Bbb Z/n\Bbb Z$-module of rank two. Consider the moduli space $S(N)=\{[\Bbb C/\Lambda_{\tau},(\tau/N+\Lambda_{\tau},1/N+\Lambda_{\tau})] \mid \tau \in \Bbb H\}$, then you can prove that $\Gamma(N)$ acts freely and transitively on this, via some computations involving the Weil pairing.
$S(N)$ consists of elliptic curves which are "$N$-enhanced" in the sense that we choose a basis for the $N$-torsion
the important arithmetico-geometric consequence of this is that, since $X(N)$ is the moduli space of a moduli problem that makes sense over $\Bbb Q$, it is defined over $\Bbb Q$ as a variety! Thus if you take the étale cohomology of it, it comes with an action of $\mathrm{Gal}(\overline{\Bbb Q}/\Bbb Q)$
that's an important step in associating a Galois representation to a modular form, which is crucial to even making sense of Taniyama-Shimura (aka the modularity theorem)
Diamond/Shurman "A First Course in Modular Forms" has lots on this (though not the étale cohomology part), it's a very geometric approach to modular forms
I'm only in the process of understanding the details of this myself for the bachelor thesis
Probability distribution over A conditioned to S = s
It's neither over A nor S. It's the "slice of A given by S = s". The correct probability space over which conditional distributions are defined is a quotient measure space
P(A | S) isn't a probability distribution, it's a family {P(- | S = s)}_s of probability distributions for each s in S. But yes, that terminology sounds fine.
@BalarkaSen Do you have a very quick argument that every element of $H_1(S;\Bbb F_2)$ is represented by an embedded curve, where S is a surface? I thought to just surger out double points but it's not obvious that doesn't change the homology class.
I see. OK, let me think of something cleaner maybe.
Dualize to an element of $H^1(S; \Bbb F_2)$ and take the classifying map $S \to \Bbb{RP}^N$ where $N$ is large. Make it transverse to $\Bbb{RP}^{N-1}$ and pullback to get an embedded curve in $S$. I bet this represents the original class.
Let $M$ be a Riemannian manifold with Riemannian metric $g$. Let $d_g(x,y)$ be the induced metric. What's the regularity of $d_p(x)\colon M\to\Bbb R$, $d_p(x)=d_g(p,x)$?
@MikeMiller I guess the content of my previous argument was $H_1(S; \Bbb F_2)$ is PD-isomorphic to $H^1(S; \Bbb F_2)$ which classifies double covers hence line bundles over $S$, and you can take zero set of a generic section to get a submanifold of $S$ which would represent the original class.
Seems a little annoying to check the various naturality ("the hexagon commutes because lol")
The proof I have in mind passes through the classifying map again
(cont.) ok so having open sets being hereditary countable also means there are neighbourhood systems that contains Frechet filters. But then we have a problem because taking points from frechet filters to form a sequence and said sequence does not really converge to anything since they are non principle filters
And what do you think of this notation $\pi_{\mathbb{s}}(A \mid S) = \{ \pi_{\mathbb{s}}(A \mid s_1 = S), \dots, \pi_{\mathbb{s}}(A \mid s_{|S|} = S)\}$?
@nbro There's no literal difference. A set of probability distribution just sounds weird, while a family of probability distributions parametrized over S sounds more meaningful to my ears, that's all.
Could you help, please. I need the information about the ultrafilters, namely, any ideas how one can see that they exist and a proof of the fact that for any ultrafilter every sequence on a compact topological space has a limit. I hope these basic facts can be collected somewhere in a popular for...
Suppose $(x_n)_n$ is a bounded sequence of complex numbers, there must exist a accumulation point, say $x_0$, thus we can find a free ultrafilter $\mathcal{F}$ on $\Bbb N$ such that $\lim_{\mathcal{F}}x_n=x_0$.
Can we find a free ultrafiler $\omega$ on $\Bbb N$ such that $\lim_{\omega}x_n\not \...
So, I was giving an answer to a question I had asked, in the context of reinforcement learning, and I wanted to definite and formalize precisely the concepts of stochastic and deterministic policies. So, here's the answer I gave based also on your advices: https://ai.stackexchange.com/a/12275/2444. Please, have a look at it and tell me if it is precise enough and if it sounds good to you, even though you might have no idea of what RL is.
Ok, basically, every deterministic process is specified by some rule to evolve from any given state to a future state, correct?
An interesting subset of deterministic process are those that are chaotic, which given slightly different intitial conditions as starting states, you can apply the deterministic policy and their state evolution will exponentially diverge
I don't have a strong background on the terminology to know whether they are correct, but the explanation flows to me and seemed to satisfy what I vaguely knew about deterministic and some stochastic process that are governed by pdfs (such as brownian motion kind of dynamics).
If $f(x,y)$ is polynomial in two real variables and with real coefficients and we want to discuss number of intersection points of a line L with a curve $f(x,y)=0$ then we can suppose, WLOG, that L passes through the origin. But how to prove that?
As far as I can tell, there is a difference that here we have only the assumption that the curve is continuous (not smooth). But since you say that it is "a special case of a recent American Mathematical Monthly problem", perhaps you could have the reference to AMM and it might be interesting for the OP.
@TedShifrin I've had a nice discussion with Balarka on the Klein quartic. Riemann surfaces are really fascinating, I still haven't got around to studying them in detail
though I'll be doing a bit on modular curves for my bachelor thesis
@MartinSleziak: Thanks. Yes, as you pointed out, I pilfered from the Math Monthly. I first worked it out using differentiability and then a very gifted student figured out the "right" proof using none.
All you need to end up using is that a circle is uniquely determined by 3 (non-collinear) points.
You guys ever have a situation where you've been operating for a couple of months under the assumption that an algorithm you wrote provided you with accurate data, just to discover that it apparently failed in one or more respects?
Suppose $(x_n)_n$ is a bounded sequence of complex numbers, there must exist a accumulation point, say $x_0$, thus we can find a free ultrafilter $\mathcal{F}$ on $\Bbb N$ such that $\lim_{\mathcal{F}}x_n=x_0$.
Can we find a free ultrafiler $\omega$ on $\Bbb N$ such that $\lim_{\omega}x_n\not \...
