Topological examples will be artificial but take the sheaf of holomorphic functions on the Riemann sphere as an example of where Mike's situation happens
Basically the "canonical example" is that X is a space, O_X(U) = C(U, R) or perhaps C(U, Z), and F is the sheaf of sections of either a real line bundle or a bundle of Abelian groups
Eg R -> S^1 is a bundle of Abelian groups --- I know how to add fiberwise --- and is a C(U, Z)-module where global sections are trivial
@MikeMiller Given a compact Riemannian manifold, can you take lots of points $x_1, \cdots, x_N$, and call the cells the set of points which are closer to $x_i$ than any of the rest of $x_j$'s the $i$-th cell? Seems like geodesic convex polytopes inside $M$
So you can lift it up by exponential map and get an actual polytope, subdivide it a little more, and project
@ShaVuklia I don't know your precise background. I suggest trying to pick up some standard examples of ringed spaces and sheaves --- each "worse-behaving" than before --- as test cases.
Also I'm interested in knowing if the "spaces of triangulations on $D^k$" (I can define this as a simplicial set maybe) is indeed weak homotopy equivalent to a point
Information-theoretic view of Bayesian learning
I once heard that the problem of approximating an unknown function can be modeled as a communication problem. How is this possible?
Yes, this is indeed possible. More precisely, there is an information-theoretic view of Bayesian learning in neural...
@MikeMiller Hm, ye, well, since you said it doesn't hold and you kind of convinced me with your example, I'm good. The thing I have to show is that if $\mathcal F$ is a sheaf of $\mathcal O_X$ modules, then the dual $\mathcal F^\vee$ is a sheaf as well. The restriction maps were easy, but I was failing at locality.
The reason I asked about these generators is because I managed to show that if for a morphism $\eta$ between the sheaves $\mathcal F\vert_U$ and $\mathcal O_U$, whenever you have a cover $U_i$ of $U$, that $\eta(U_i)=0$ implies that $\eta(U)=0$, however, now I still want to conclude that if $V\subset U$ open, then $\eta(V)=0$.
That's why I was hoping that $\eta(s)\vert_V$ would generate $\mathcal F(V)$, but that doesn't seem to hold. We were also allowed to use that $\mathcal F$ is locally free of finite rank, but they also told us that might not be necessary for the proof, so we were trying to show it without that assumption. I guess you mentioned you have no time, so feel free not to respond to this I guess (which anyone always is ofc)
@ShaVuklia Hm, just a second. The dual sheaf $F^*$ is the sheaf $U \mapsto \text{Hom}_{O|_U}(F|_U, O|_U)$, where the latter is the $O|_U$-module of sheaf homomorphisms, correct?
If you have shown that some element in $\text{Hom}_{O|_U}(F|_U, O|_U)$ is zero, then definitionally it's restriction as an element of $\text{Hom}_{O|_V}(F|_V, O|_V)$ is going to be zero, because that's what being a zero homomorphism of sheaves means
I feel like the confusion might be that you're conflating the sheaf homomorphism with it's induced homomorphism of global sections (and this is a confusing point for sure)
@TedShifrin Hi, I was wondering, do you know what the critical points of $\text{dist}^2(x, y) : M \times N \to \Bbb R$ would mean geometrically, if $M, N \subset \Bbb R^n$ are embedded submanifolds? I seem to be getting that if $(x, y)$ is a critical point, and $\ell$ is the secant joining them, $T_x M \subseteq T_y N + \ell^\perp$
Let's say I have a probability space $(\Omega, \mathcal{F}, P)$ and a random variable defined on it, say $X$, which is equal to $0$ and $1$ with probabilities $p_0 > 0, p_1 > 0$ respectively, with $p_0 + p_1 = 1$.
I have two questions:
I know the notation for $X$ would be something like $X: (\Om...
@Clarinetist Why B? It would be a map X : (Omega, F, P) -> ({0, 1}, M) where M = sigma({0}, {1}), correct, which is in fact {{}, {0}, {1}, {0, 1}}, the full power set.
The domain is the probability space X is defined on. If you want to omit the probability function and say it's a map of measure spaces, X : (Omega, F) -> ({0, 1}, M) is fine.
@BalarkaSen Oh, thanks for that. So basically when we say we're defining a random variable on a probability space, we assume the domain is $(\Omega, \mathcal{F})$ (in this siutation).
@ShaVuklia It seems to me there's no real reason to talk about U: if you prove that if eta: F -> G is a sheaf map so that eta|_{U_i} = 0 for a cover U_i of X, then eta = 0, one obtains your result by specifying to F|_U, G|_U, and eta|_U
@Clarinetist Yeah. That is to say, you declare "all possible outcomes of all possible experiments you would do from now on are in F, a subset of the power set of Omega"
and then I also have to keep in mind when I'm talking about $\sigma(\{\{0\}, \{1\})$, my universal set is actually going to be $\{0, 1\}$ as opposed to $\Omega$ like it usually is
So if V < X is any open set, you get that eta|_{V cap U_i} = 0 by hypothesis; since V is covered by the V cap U_i by the result you already proved eta|_V = 0.
But it has been closed. Although I had read the guidelines and so on, I made the necessary edits and it has remained closed. As indicated the subject isn't easy for me and given the source from where this problem was obtained and indicated the part which I'm struggling isn't that enough?.
It makes me sad because I think its a good problem, but who knows, maybe I'm mistaken. But I don't want to be left ignored or maybe my problem is discriminated due being too simple?. Can someone help me here?. As I indicated, from the looks of it, seems like a congruency but I don't know which sort of construction can be used in order to fit using euclidean postulates.
If there's a good samaritan help out there I would appreciate to give it a look.
@ChrisSteinbeckBell: If you can't figure out how to draw the picture, I'm not sure why you should even be working on the question. What can you not figure out how to draw?
I have never understood why people find these sorts of problems intriguing. Just seems like random stuff to me, even though I'm good at geometry.
Yup. It is just algebra, but eventually you get $HM=MC=AB$. ... But it's an idiotic multiple choice question, because by logic the only conceivable answer is $30^\circ$.
I wonder how $37^\circ$ (or $53^\circ$) could show up, trigonometrically speaking.
@ChrisSteinbeckBell: If you don't know it well, start by taking a right triangle and dropping a perpendicular to the hypotenuse. See how you end up with two small triangles each similar to the original right triangle. This is used repeatedly in geometry.
Yes, by definition of compatible, that derivative is invertible at each point.
@Astyx: So my conclusion is that this "geometry" problem is totally contrived to make the algebra work out. I cannot for a second see why it would be geometrically of interest.
@Astyx Really? This actually has a wonderful application in elementary mechanics. If you have a (uniform) ladder slipping down a wall, its center of mass travels along a circle.
"The fact that $M^m$ is compact and of class $C^2$ implies that there exists a number $\rho>0$ so small that no $(\nu-1)$-sphere of radius $\rho$ tangent to $M^m$ encloses a point of $M^m$." Here, $M^m$ is a compact submanifold of $\mathbb{R}^{\nu}$. Why is this true? I guess the point geometrically is that spheres of sufficiently small radius are more curved than the submanifold so that locally at the point of tangency, all other points of the submanifold have higher distance from the radius, but I don't have any actual arguments.
Yes, @Thor, you're exactly right. Locally, it's convexity. Globally, you've chosen the radius small enough to keep away far-away points. Did you try a proof by contradiction?
So the point geometrically should be that each point in a tubular neighborhood has a unique closest point on $M$.
@Huy I'm well aware that space filling curves are not injective (as it's clear by looking at them on compact intervals), but they are a geometrically clear reason for why an injection C->R exists, and the other injection is obvious, so you can conclude
I have $f:X\to Y$ a dominant morphism of integral schemes of finite type over a field, I want to prove the dimension of the fiber of the generic point of Y is dim X - dim Y
@Alessandro It's interesting to figure out at how many points the space filling curves must fail to be injective. And can you do it with just 2-to-1 or will there have to be some points that are 3-to-1 or 4-to-1, etc.
I've hit a bit of a wall with a result. I'm trying to show that if a ring $R$ is of strong finite type (for any ideal $I\subseteq R$, there exists a f.g. ideal $B\subseteq I$ and a natural number $n\in\mathbb{N}$ such that $x^n\in B$ for all $x\in I$), then it satisfies the ascending chain condition on radical ideals. So far I've shown that for any chain of ideals $I_i$, then the chain $\sqrt{I_i}$ is the same as the chain $\sqrt{B_i}$.
I think I want to show that the chain of $B_i$ stabilize, because each $B_i$ is finitely generated, but I don't know how (or if) that would work
Ah, I may have figured it out. I simply have to say that $\bigcup_iI_i$ has a corresponding $B$ that contains $\bigcup_iB_i$ and is finitely generated. Therefore the $B_i$ chain must stabilize. (right?)
You should be able to prove that for small enough $\epsilon>0$, the $\epsilon$-neighborhood of the zero section of the normal bundle embeds in $\Bbb R^\nu$. This will imply your result.
This is a special case of the tubular neighborhood theorem.
(But you can write the map down in Euclidean space very easily. Just take $(x,v) \rightsquigarrow x+v$.)
@robjohn: Are there some abstract considerations for any space-filling map? For Hilbert and Peano one has coordinates to work with, basically.
Information-theoretic view of Bayesian learning
I once heard that the problem of approximating an unknown function can be modeled as a communication problem. How is this possible?
Yes, this is indeed possible. More precisely, there is an information-theoretic view of Bayesian learning in neural...
