ag.algebraic geometry – How to show analytification functor commutes with forgetful functor?

Also in ME.


Let $k$ be a field complete with respect to a non-trivial non-archimedean
absolute value (so that rigid $k$-space makes sense). Denote $K$ a finite field extension of $k$.

Denote $Xrightsquigarrow X^{mathrm{an}/k}$ the analytification functor from the category of locally of finite type $k$-schemes to the category of rigid $k$-spaces.
Similarly there is an analytification functor $Xrightsquigarrow X^{mathrm{an}/K}$ over $K$.

There is a well-defind forgetful functor $S:Xrightsquigarrow X$ from $K$-schemes to $k$-schemes ($S$ represents schemes) and a forgetful functor $R:Yrightsquigarrow Y$ from rigid $K$-spaces to rigid $k$-spaces ($R$ represents rigid).

Let $X$ be a locally of finite type $K$-scheme. I believe that $S(X)^{mathrm{an}/k}cong R(X^{mathrm{an}/K})$ as rigid $k$-spaces. The universal property induces a canonical map $R(X^{mathrm{an}/K})to S(X)^{mathrm{an}/k}$ but I cannot show it is an isomorphism. A proof or reference would be nice.

p.s. the idea comes from proving absolute/relative Frobenius morphism commutes with analytification, but I first need to make sure the maps have the same source.

geometry – point of intersection of two lines in barycentric coordinate system

I am looking for an efficient way to determine the point intersection of two lines which go through a triangle (face) of a 3D triangular surface mesh.

For both lines I know the two points at which they intersect with the edges of a triangle (face).
Denoted $P_A^1, P_B^1$ for the first line and $P_A^2, P_B^2$ for the second line (see Figure below).

Based on my investigations my preferred approach would be to:

  1. Transform the the points of intersection for both lines ($P_A^1, P_B^1$ and $P_A^2, P_B^2$) into barycentric coordinates resulting in $B_A^1, B_B^1$ and $B_A^2, B_B^2$.
  2. Define two lines: $L_1$ which goes through $B_A^1, B_B^1$ and $L_2$ going through $B_A^2, B_B^2$ based on the two-point form defined in section 4.1.1. of this document
  3. Determine the point $B_I$ as the barycentric coordinates where $L_1$ and $L_2$ intersect based on the equations in section 4.3 of this document.
  4. Transform $B_I$ back into the cartesian coordinate system to have its 3D coordinates, denoted as $P_I$.

Unfortunately steps 2 and 3 do not lead to meaningful results (i.e., the intersection points fall outside the triangle) and I have doubts that I am applying equations in section 4.1.1. and section 4.3 properly.

Even when using a simple example be based on the upper illustration in this figure here by defining $B^1_A=(1,0,0), B^1_B=(0,1/2,1/2)$ and $B^2_A=(0,1,0), B^2_B=(1/2,0,1/2)$ I cannot determine $B_I$ correctly as $B_I=(1/3,1/3,1/3)$ with the above procedure.

Simple illustrative example

dg.differential geometry – Mean Gaussian Curvature from non-unit vector

Pg.248 of “Textbook in Tensor Calculus and Differential Geometry” by Prasun Nayak.


Let us suppose that $lambda_{h|}^i$
is not a unit vector and therefore, the mean curvature $M_h$ in
this case is given by

$$M_h=-frac{R_{ij}lambda_{h|}^ilambda_{h|}^j}{g_{ij}lambda^i_{h|}lambda^j_{h|}} tag{1}$$


A proof was not given so I am trying to derive this.

For when $lambda_{h|}^i$ are components of an orthornormal set of vectors
$$M_h=sum_{k}K_{hk}=sum_kfrac{lambda_{h|}^llambda_{k|}^ilambda_{h|}^jlambda_{k|}^rR_{lijr}}{lambda_{h|}^llambda_{k|}^ilambda_{h|}^jlambda_{k|}^r(g_{lj}g_{ir}-g_{lr}g_{ij})}=sum_kfrac{lambda_{h|}^llambda_{k|}^ilambda_{h|}^jlambda_{k|}^rR_{lijr}}{(1)cdot(1)-0}$$
$$=lambda_{h|}^llambda^j_{h|}R_{lijr}g^{jr}=-lambda_{h|}^llambda^j_{h|}R_{lj}$$

where $sum_klambda^i_{k|}lambda^r_{k|}=g^{ir}$.

Now for when $lambda_{h|}^i$ are not components of an orthonormal set of vectors, succumbing to the same approach gives

$$M_h=sum_k K_{hk}=frac{lambda_{h|}^llambda_{h|}^j R_{lijr}}{(g_{lj}lambda_{h|}^llambda_{h|}^j)g_{pq}}sum_kfrac{lambda_{k|}^ilambda_{k|}^r}{lambda_{k|}^plambda_{k|}^q} tag{2}$$
It is not obvious to me how $(2) to (1)$. Modifying the definition of $M_h$
$$M_h=frac{sum_klambda_{h|}^llambda_{k|}^ilambda_{h|}^jlambda_{k|}^rR_{lijr}}{(g_{lj}lambda_{h|}^llambda_{h|}^j)sum_k g_{pq}lambda^p_{k|}lambda_{k|}^q}=-frac{R_{ij}lambda_{h|}^ilambda_{h|}^j}{g_{ij}lambda^i_{h|}lambda^j_{h|}N} tag{3}$$

which is the nearest I can get to $(1)$, but with an extra factor $frac{1}{N}$.


What’s gone wrong? Is there a way to derive $(1)$ or is $(1)$ a definition itself?

ag.algebraic geometry – K-equivalence => isomorphism of Chow motives?

An old conjecture of Bondal-Orlov-Kawamata predicts that K-equivalent varieties are D-equivalent, see Kawamata’s paper for definitions. In particular this applies to birational Calabi-Yau varieties. See this MO question for the Bondal-Orlov formulation of the conjecture and the known cases. These conjectures are mostly open, and considered to be an important bridge between derived categories and birational geometry.

Derived categories and Chow motives play the role of universal cohomology theories, in noncommutative, and commutative worlds respectively. Do we expect that K-equivalence implies isomorphism of rational, or even integral Chow motives?

Example. Integral Chow motives of varieties related by a standard flop are isomorphic: paper by Q. Jiang. This was the motivation for the question.

Remarks. In many cases, D-equivalence is known to imply isomorphisms for rational Chow motives, but not for integral ones: examples can be found among K3 surfaces. It is also known that D-equivalence does not imply K-equivalence, even for rational surfaces.

geometry – Prove the given polytope is tetrahedron

So I have to prove that given a 3-dimensional compact polytope such that every two vertices are adjacent, then it is a tetrahedron.

Somehow I’m able to visualise from the fact that every two vertices are adjacent and sees that it fits to the diagram of tetrahedron. Like it is a polyhedron composed of four triangular faces, six straight edges, and four vertex corners from this it is suffice that every other vertices is adjacent to each other. But I’m not quite sure how to prove it.

Could you please help me in proving the above statement.

Cheers.

sg.symplectic geometry – Computing Gromov Witten invariant of 4 lines in CP^3

I’m trying to understand what the number of genus 0 curves through four lines in $mathbb{C}P^3$ is i.e $Gr_{0,4}^{mathbb{C}P^3, L}(PD(L),PD(L),PD(L),PD(L))$ where $L$ is the class of a line $mathbb{C}P^1 subset mathbb{C}P^3$.

I can see that $sum_{i=1}^{1} deg(PD(L)) = 16$ and dim $mathcal{M}_{0,4}(mathbb{C}P^3,L) = 16$. Hence the Gromov Witten invariants are not trivially zero.

However, I’m struggling to actually compute this number. Could someone help me with how to proceed in computing this?

ag.algebraic geometry – The image of a curve under the multiplication endomorphism of its Jacobian

Let $X$ be a complex smooth projective curve of genus $ggeq 2$. Embed $X$ in its Jacobian
${rm{J}}(X)cong{rm{Div_0}}(X)/{rm{Div_p}}(X)$ where ${rm{Div_0}}(X)$ is the group of degree zero divisors and ${rm{Div_p}}(X)$ is the subgroup formed by principal divisors. The embedding can be in the form of for instance
$$pmapsto text{the divisor }p-p_0text{ modulo }{rm{Div_p}}(X)$$
where $p_0$ is an arbitrary base point. What kind of curve the image of $Xsubset {rm{J}}(X)$ under the multiplication by $n$ map ($n>1$ an integer) $(n):{rm{J}}(X)rightarrow{rm{J}}(X)$ is? Since the endomorphism is locally biholomorphic, I guess the image is smooth if $(n)$ restricts to an injective map on $X$. In general, what kind of singular curves may arise in this way? In what situations $(n)$ sends $X$ to another smooth curve inside ${rm{J}}(X)$?

Here is a special case that I can (almost) analyze: Suppose $nleq g$. Then distinct points $p,p’in X$ have the same image under $(n)$ iff $np-np’$ is a principal divisor and this happens only when $p$ and $p’$ are Weierstrass points. Thus the restriction of $(n)$ is generically one-to-one and so the image is singular if there exists a principal divisor of the form $np-np’$ (e.g. when $X$ is of the form $y^n=f(x)$ with $f$ square-free).

probability distributions – Geometry interpretation of any continuous random variable

Given a continuous random variable $X$ with the cdf $F_X(x)$, I want to know whether there exists a random variable $mathbf{Z}$ uniformly distributed in a geometry region $mathscr{Z}_n$ in $mathbb{R}^n$ such that any one-dimensional marginal distribution $F_{mathscr{Z}_n}^1$ of $mathbf{Z}$ is close to the distribution of $X$. In addition, an asymptotic perspective is also welcome. For example, it is also meaningful to know whether $F_{mathscr{Z}_n}^1=F_X$ is possible if $n$ tends to $+infty$.


Some useful facts are given as follows:

  1. A normal distribution can be approximated by a uniform random variable distributed in a hypersphere.
  2. A negative exponential distribution can be approximated by a uniform random variable distributed in a simplex.

This question has been asked in enter link description here yesterday.

ag.algebraic geometry – the map on divisor class groups induced by restriction to a toric subvariety

Let $X$ be a (say, complex) toric variety acted upon by a torus $T$ and defined by a fan $Sigma$ in the cocharacter lattice $N=mathrm{Hom}(mathbb{C}^times, T)$, and let $M$ be the character lattice. For any cone $sigma in Sigma$ put $M(sigma) = sigma^perp cap M$, $N(sigma) = mathrm{Hom}(M(sigma), mathbb{C}^times)$. There is a natural projection $N to N(sigma)$. Then the closure of the orbit corresponding to $sigma$ has the structure of a toric variety with respect to the quotient torus with the cocharacter lattice $N/N(sigma)$ and given by the fan $Star(sigma)$ consisting of the images in $N(sigma)$ of the cones of $Sigma$ containing $sigma$. Note that the closed embedding $X_{Star(sigma)} to X$ is generally not a toric morphism, since the dense toric orbit of $X_{Star(sigma)}$ does not intersect the dense toric orbit of $X$.

My question is: is there a way to describe the restriction map $mathrm{Cl}(X) to mathrm{Cl}(X_{Star(sigma)})$ in terms of the fans $Sigma$ and $Star(sigma)$?