@Astyx OK, I get it. I was just wondering how to write that pairing is an isomorphism; I suppose you'd tell me that its an isomorphism for sections over a specified open and so globally an isomorphism.
Remember I am not thinking so you should convince me why something is true. I agree that this Picard group is indeed a group now.
So Cartier divisors are defined as follow on integral schemes: A Cartier divisor is an open cover with rational functions on each open set $(U_i, f_i)$, such that on intersections, the transition function $f_i/f_j$ are invertible regular maps
There apparently are ways to define Cartier divisors in the non integral case (I think you can drop out reduced or something) but things become ugly and you have to define a sheaf of rational functions etc
My brain is mush so I am a little annoyed by not being able to do it on the spot. What is sheaf cohomology? If you have $X$ a space, $F$ a sheaf, a covering $\{U_i\}$ of $X$ by open sets, then define $n$-cochains to be collections $(U_{i_0, \cdots, i_n}, f_{i_0, \cdots, i_n})$ of sections of $F$ over $n+1$-fold intersections of these open sets of the cover.
The boundary map does a messy alternate sum by omitting a single, varying, index. The resulting chain complex has cohomology groups $H^n(X, F)$. Technically, this is Cech cohomology for the cover $\{U_i\}$, by refining over all covers you get sheaf cohomology $H^n(X, F)$, the absolute group.
Now note that there is an exact sequence of sheaves $0 \to O_X^\times \to M_X^\times \to M_X^\times/O_X^\times \to 0$ which gives a long exact sequence in cohomology
@Thorgott I'm trying to figure the same out for bochner-integrals. I.e. for $s \mapsto T_s \in L(U,V)$, $(\int_0^t T_s \, \mathrm{d}s)(u) = \int_0^t T_s(u) \, \mathrm{d}s$ but I don't see directly how linearity solves this. Linearity means two things for me here: $f \mapsto \int f$ is linear and $A(\int f) = \int Af$
I can't remember what the precise valuation is, something like the power required for it to be in the ideal of functions vanishing on the subscheme, or the opposite... something like that
When I taught Riemann surfaces, I taught it to people at the end of a complex analysis course, so I minimized the bundle language. But in the grad diff geo and complex geo courses, of course I did plenty of bundles.
It's not hard to create a line bundle from a divisor by thinking of a Cech cocycle, as Balarka and Astyx have been discussing. Just give the divisor in the open sets of an open cover.
This is a very confusing phrasing. You're looking at germs of regular functions near $Z$ which vanish on $Z$, which is the ring $O_{X, Z}$ if I understand correctly. This is the DVR because $\codim_Z(X) = 1$, $Z$ is closed irreducible, so you can compute what the valuation of $f$ is along this
@BalarkaSen Suppose you have $f = (z-1)^2(z-2)/(z-4i).$ You know the order of the pole is $2$, but let's do it more abstractly. We want to look at $z=1$ so we localise $C[z]$ at $(z-1)$, the ideal. Then the quotient module I mentioned before in this case is $C[z]_{(z-1)}/((z-4i)(z-1)^2) \cong C[z]_{(z-1)}/((z-1)^2).$
@BalarkaSen Now we can form the maximal chain $(0) \subset C[z]_{(z-1)}/((z-1)) \subset C[z]_{(z-1)^2}$ which has length 2 (length is number of inclusions not number of components of the chain). So the order of the zero is 2.
Yeah got it. This is obvious though because the ambient is 1-dimensional, you have a coordinate element $z$, if you localize everything is $z^n$ times a unit
the funny thing is that I can unironically imagine one of our probability professors to just continue a lecture if a student were to explode in the middle of it
So as I was saying in AG we usually work with the Zariski topology, so we're not really considering points but rather prime ideals. In the Zariski topology of an integral ring there is a specific prime that is close to every other prime: the nilradical. This is the generic point. In a more general irreducible scheme, it turns out that you can also define such a point. The interesting property is that localization at this point (read: at this prime) yields precisely the ring of rational functions
For instance with $\mathrm{Spec}\mathbb Z$ (which is integral affine), the localization at $\sqrt{(0)}$ gives $\Bbb Q$
Can anyone tell me a reference or any proof of this line: Any geodesic on any surface must intersect itself transversely if it is not simple or a covering of a simple closed geodesic.
Let there exist a multivariable function $f(\mathbf x)$, where it's domain is $D\subseteq \mathbb R^n$. Let there be an open set $S \subset D$. It is given that $\partial f/\partial x_i$ is defined at every point $P \in S$, for all $i$ from $1$ to $n$, where $x_i$ are the $n$ orthogonal coordinates forming up the $\mathbb R^n$ space. From the above given information, can we conclude that $f$ is continuous at every point $P\in S$? I think yes, we can.
Yeah. If you scale $v$, then you just scale the parametrization; the image of the curve remains the same.
This is why you can rule out the covering of a simple closed geodesic case.
(That just means going around a single closed geodesic 100 times)
@FakeMod This is not true. Consider $f(x, y) = (xy)/(x^2 + y^2)$ for $(x, y) \neq (0, 0)$ and $f(0, 0) = 0$. Around any neighborhood of the origin, both the partial derivatives exist.
@BalarkaSen Oh, I see. If I apply one extra constraint that the partial derivatives must be bounded, would it then allow me to deduce the continuity of $f$.
@BalarkaSen But in case of single variable calculus, the point where the derivative of the function is not continuous, that either has to be a point of non-differentiability or a point which doesn't belong to the domain. I am not able to think of any case, where a single variable function is differentiable at a point but it's derivative isn't continuous.
@FakeMod i can only recall similar ones. they have $x^2$ to get differentability at zero, and then some bounded function of ${1 \over x}$ to get the wild derivative.
The defintion of rational equivalence on a variety $X$ is as follows, for two cycles $A_0$ and $A_1$, I say both of them are rationally equivalent there exists a cycle in $\Bbb{P}^1 \times X$ such that the fiber at $t_0$ is $A_0$ and the fiber at $t_1$ is $A_1$. Very naively this suspiciously looks a lot like two things, one a "rational" homotopy and also like a bordism between two manifolds. One even has a transversality type result here.
Where you say that two cycles are transversal upto a rational equivalence
So intersections make sense in the Chow ring. But I was asked to be extremely vary in thinking of this equivalence as homotopy or bordism, why is that? Is there an issue in thinking of this as a $\Bbb{P}^1$ homotopy?
Why is it a homotopy? Homotopy of what map? It's a family of cycles indexed over $\Bbb P^1$
Rationally equivalent cycles are "cobordant" as a special case because you can take a path joining $t_0$ and $t_1$ in $\Bbb P^1$ and take preimage
But even this is fidgety because these are singular manifolds
At least that proves that $CH_* \to H_{2*}$ is a well-defined map for varieties over $\Bbb C$, because you can triangulate the cycles and then rational equivalence gives a simplicial cobordism or whatever between the cycles; $\pi_X : \Bbb P^1 \times X \supset \pi_{\Bbb P^1}^{-1}(\gamma) \to X$ where $\gamma$ is the path.
@copper.hat @BalarkaSen From our older discussion, I wonder whether the following function $$f(x, y) = x^2\sin\left(\frac{1}{x}\right) + y^2\sin\left(\frac{1}{y}\right)$$ differentiable at $(0,0)$? I am not sure about this because both $\partial f/\partial x$ and $\partial f/\partial y$ are discontinuous at $(0,0)$, but the functions $$g(x, y) = x^2\sin\left(\frac{1}{x}\right) \quad \text{and}\quad h(x,y) = y^2\sin\left(\frac{1}{y}\right)$$ are differentiable...
so their sum ($g(x,y) + h(x,y) = f(x,y)$) must also be differentiable.
I am looking for books or to state of the art papers about current the development trends for a strong-AI.
Please, do not include opinions about the books, just refer the book with a brief description. To emphasize, I am not looking for books on applied AI (e.g. neural networks or the book by No...
Neat little fact I'm surprised I never saw before: If $X$ is compact metric, $f\colon X\to Y$ is continuous and $f(X)$ is Hausdorff, then $f(X)$ is metrizable
You could also state it as, in your last example, 3x+4 = 9 or -(3x+4) = 9. The minus can go on either side of the equation---but it has to apply to the whole side, so 3x - 4 = 9 isn't true if |3x+4| = 9 is true.
You would need instead to say that 3x+4 = 9 or -3x - 4 = 9, which is the same as saying 3x+4 = 9 or 3x+4 = -9.
