## ag.algebraische Geometrie – Grothendieck-Teichmüller conjecture and tropization of the curve modules

Abramovich, Caporaso, and Payne (2014) have used the Berkovich analysis of the moduli spaces of stable curves to create functional Tropicalization maps. $$overline {M} _ {g, n}$$to the modular spaces of the expansive tropical curves, $$overline {M} ^ {trop} _ {g, n}$$that are compatible with natural, forgetful, double-sided and adhesive cards.

Grothendieck designed in his Esquisse a program to study the effect of the absolute Galois group Gal ($$overline { mathbb {Q}} / mathbb {Q}$$) on the "Lego-Tëichmuller tower", consisting of profound terminations of basic groupoids of moduli spaces with smooth curves, $$M_ {g, n}$$connected via the same forgetful, clinging and sticky cards.

What effects does the tropicalization result have on the Grothendieck program? Transfers the absolute Galois action the corresponding "tropical tower" of $$overline {M} ^ {trop} _ {g, n}$$ at all in any way and so combinatorially studied?

## Differential Geometry – Can we always find a curve on the manifold whose tangent vector always belongs to a linear subspace?

Suppose we have a smooth manifold $$M$$ and the tangential space of each point $$x in M ​​$$ has a non-empty intersection with a given linear subspace. Could we find a curve? $$M$$ so that the tangent vector of the point on that curve always belongs to this linear subspace?

$$textbf {my try:}$$ If the dimension for the linear subspace is one or two, I think I can find the curve. But I do not know if the dimension is larger than two?

## Vector analysis – Define the path integral if the scalar \$ f \$ and the curve \$ mathbf {c} \$ are in * curvilinear * coordinates

I'm currently doing a multivariable calculus course. I have seen path integrals in the Cartesian coordinate system as the following definition:

Definition.
The path integral of $$f (x, y, z)$$ along the curve $$C$$ is

$$int_C f ds = int_a ^ bf ( mathbf {x} (u)) || mathbf {x} & # 39; (u) || you$$

from where $$mathbf {x}:[a,b] to mathbb R ^ 3$$ is the parametric representation of $$C$$, The definition did not refer to the coordinate system.

Question.
Suppose that $$(x, y, z) = mathbf Phi ( xi_1, xi_2, xi_3)$$ where the
transformation $$mathbf Phi: U to V$$ is sufficiently differentiable and has one
inverse $$mathbf { Phi ^ {- 1}}: V to U$$ from where $$U, V$$ are open subsets of $$mathbb R ^ 3$$,

What would be the definition of a path all over $$C$$ when $$f$$ is a function of curvilinear
Coordinates ($$xi_1, xi_2, xi_3$$) and the curve has been parameterized
curvilinear coordinates as $$mathbf { xi} = mathbf { xi} (u)$$ to the $$a le > u le b$$?

Is the definition consistent with a regular "Cartesian" path? If so, can you derive from the above definition for Cartesian how you would obtain a path integral for a function in curvilinear coordinates?

## Different result when scanning in Epson mode "color negative film" and when scanning in positive mode -> reverse curve in the post content?

Scanning removes the entire orange mask in the color negative film.

Scanning as positive and then inverting after processing will not remove it. The inversion simply changes the orange mask to a deep blue overall picture. NOT bluish, but very deep blue. Then extra work to try to remove it.

This is a difficult task in digital post-processing (not for the process, but for the result) because such extreme color shifts (to remove the strong blue) severely jeopardize digital trimming. A detail can change the colors and lose details.

The scanner can do this as analog (no digital clipping) by simply varying the scan time for each RGB color (the result acts like a correction filter in the light). Scanning color negatives is a bit slower than positive slides, but it's a very good thing.

If you do not say it's "impossible" in the digital realm, you may get an acceptable result, but it's just not the same thing. If you have the scanner, I strongly recommend using it. It is designed for the exact purpose.

These are color negatives, they are the particular problem. Positive transparencies or prints or black and white negatives do not have the orange mask. Therefore, copy methods other than scanning are not excluded (but scanners are good too).

The result of a yellowish occupation is not inherent. It will, however, be much easier to handle than the deep blue otherwise. 🙂 However, there are several factors.

Not every brand of color negative film has the same color as the orange mask, so some scanners offer a variety of film options.
Or the scanner calibration may not be accurate.

Or rather, the film may have been shot with a wrong color white balance (we only had a very small selection of films, but there was a basic choice between film type or flash type or lighting or filters). Often the recording itself needed a better white balance correction. So scan other film negatives in their different lighting situations to see how consistent the yellow cast is. Sunshine outdoors is probably more accurate than indoors.

In digital work, it would have been better if the white balance of the digital camera had been correct. Normally, however, it is rather mild and can easily be corrected in post-processing.

White balance was the same problem for film as it is today, but the lab that printed the image out of film did a good job before it was corrected for us. But it is NOT corrected in the negative yet.

Do you remember the blue flashlights? Remember, we still got a good result, whether we used it or not. The printing lab has fixed it for us. But in digital or film scans, that's our job right now.

## Algebraic Geometry – Do you show that a given set is a smooth projective curve?

I have recently encountered this problem from the book Riemann surfaces by Simon Donaldson:

To let $$p (z)$$ be a polynomial of degrees $$d$$ with different roots and let
$$X$$ Let be the compact Riemann surface, which is assigned to the functions
$$w ^ 2 = p (z)$$, Show that if $$d = 3$$the natural card of $$X$$ to
$$mathbb {CP} ^ 2$$ has an image that represents a smooth projective curve, but
that this does not apply to $$d> 3$$,

Now I am not sure how to solve it. First and foremost, what exactly is the "natural card" $$X$$ to $$mathbb {CP} ^ 2$$? Is it sending the card? $$(z, w) in X$$ to $$[1:z:w] in mathbb {CP} ^ 2$$? And if that's true, I still have no idea where I would go to solve it. Any ideas?

## Cryptography – What is the current computational speed when performing elliptic curve multiplication?

Hello, I was just looking for some information on how long it would take to crack a private key with a brute force approach in Bitcoin, and I could not find a very good answer on how long it takes to check Whether a particular key (or every single key) is key would work.

Basically, so I'm wondering how long the elliptic curve multiplication process would take to check if a single private key works for a particular public key (on average), thanks 🙂

## A simple proof of the Jordan curve theorem

The one I have is 16 pages long and is for a small exhibition, so I need a slightly shorter one.

Many Thanks

## ecdsa – What is the reason for choosing 2 ^ 256-2 ^ 32-977 for the prime on the secp256k1 curve?

This is probably for others here to answer, but I will try.

We are looking for an elliptic curve that has a cyclic subgroup with a high order (but not higher than `2 ^ 256` since we want to work with private 32-byte keys and the private keys are the order of the elements in relation to the generator point), this particular elliptic curve over this prim field seems to satisfy these characteristics.

The order of the subgroup essentially defines the difficulty for the discrete logarithm. For example, if the subgroup was so small that you could store all the group items on a computer, you could easily calculate them and break any public key that came from that group.

As for the generator point. I think that's an arbitrary decision. Since the group is cyclic (and first order afaik), any other element (except the neutral one) could serve as a generator. However, we need to agree on a fixed generator point for ecdsa to get the same results each time the algorithms are run. For this an element was chosen.

I am not sure if there is more theory why this particular configuration is preferable.

## gnupg – RSA 4096 Vs. ECC curve 25519

When quantum computers become practical, both RSA and ECC are completely destroyed. To be sure in this case, you would need to use post-quantum cryptography. Keep in mind that these algorithms are not as well studied as RSA or ECC, that it would be unwise to fully trust one of them. Instead, use two levels of encryption: a conventional and a postquantum.

What is otherwise safer, algorithms like https://crypto.stackexchange.com/questions/31439/how-do-i-get-the-equivalente-stärke-von-anecc-key can calculate a security level (though this is just the Measuring the processing time and has no bearing on how likely a "break" of either system will be). These calculations seem to indicate that RSA-4096 is slightly stronger than Curve25519.

## Constructing a minimal spanning tree from a curve with shortest paths?

Given a $$n times n$$ shortest path matrix $$D$$, And a complete graphic $$G (D)$$ on $$n$$ Knots where edge $$(i, j)$$ has weight $$D_ {ij}$$, Furthermore, the distance matrix $$D$$ is calculated for a network $$T$$ from $$n$$ Node – d. H. the shortest path distance between two nodes in $$T$$,

My problem is when I calculate a Minimum spanning tree $$T #$$ from $$G (D)$$ for example with Kruskal algorithm becomes the resulting tree $$T & # 39; = T$$?

• I have tried to argue that they are in fact the same because if we consider an advantage $$e$$ was added in one step by Kruskal algorithm then this edge must have been slight edge So it's the shortest route.

I am not sure if this argument is correct. How can I say that? $$T & # 39; = T$$?