Let $X$ be a variety defined over $\mathbb{Z}$, $X_{\mathbb{F}}$ be the base change $X\times_{Spec(\mathbb{Z})} Spec(\mathbb{F})$. We say $X$ is of polynomial count if there is a polynomial $f(x)\in \mathbb{Z}[x]$ such that for every finite field $\mathbb{F}_{q}$, $\# X_{\mathbb{F}_{q}}(\mathbb{F}_{q})=f(q)$. Clearly, if $X$ has an affine paving, then it must be of polynomial count. But there are lots of varieties of polynomial count with no affine paving.

For example, a large number of determinantal varieties are of polynomial count. Here the determinant variety is the integral model of the usual one: given $m$ and $n$ and $r<\min(m,n)$, the determinantal variety $Y_{r}$ is the set of all $m\times n$ matrices (over $\mathbb{Z}$) with the determinant of each $(r+1)\times (r+1)$ minors equals zero. The ideal of $\mathbb{Z}[t_{ij}]$ generated by these polynomials is a determinantal ideal. Since the equations defining minors are homogeneous, one can consider $Y_{r}$ either as an affine variety in $mn$-dimensional affine space, or as a projective variety in $(mn-1)$-dimensional projective space, and we take the projective one.

1. My question is: If $X\subset \mathbb{P}^{N}$ is a polynomial count quasi-projective variety, does there exist a chain of varieties $$A_{1}\supset A_{2}\supset \dots \supset A_{n}$$ where $A_{i+1}$ is closed (or locally closed) in $A_{i}$, for $i=1,2,...,n-1$, such that each $A_{i}$ has a affine paving, and $X=A_{1}-A_{2}+A_{3}-\dots (-1)^{n-1}A_{n}$?

If the answer of this question is Yes, I wonder how to give this decomposition of $$\mathrm{Proj}(\mathbb{Z}[t_{ij}]/\det(t_{ij})_{n\times n}),\ 1\le i,j\le n$$ or its affine cone? How can we give the explicit coordinates of this decomposition?

P.S.: One can calculate that the number of closed points of $\mathrm{Proj}(\mathbb{Z}[t_{ij}]/\det(t_{ij})_{n\times n})$ is $f(q)=\sum_{j=0}^{n^{2}-1}q^{j}-q^{n-1}\prod_{i=1}^{n-2}(q^{n}-q^{i})$. One can verify easily that the coefficient of $q^m$ is negative for some $n,m$. This means that in some cases it cannot be a variety which has an affine paving (even though it is of polynomial count), since a variety with an affine paving has $f(x)$ with non-negative coefficients.

2. One more question is: Let $\mathrm{Proj}(\mathbb{Z}[t_{ij}]),\ 1\le i,j\le n$ be the $\mathbb{P}^{n^{2}-1}$ defined over $\mathbb{Z}$, $M$ be a set of some sub square matrices of $(t_{ij})_{n\times n}$, $I$ be the ideal generated by $\det(m),m\in M$. Is this variety $\mathrm{Proj}(\mathbb{Z}[t_{ij}]/I)$ of polynomial count?

Any answer will be welcome. Thanks a lot !

Let $X$ be a variety defined over $\mathbb{Z}$, $X_{\mathbb{F}}$ be the base change $X\times_{\mathrm{Spec}(\mathbb{Z})} \mathrm{Spec}(\mathbb{F})$. We say $X$ is of polynomial count if there is a polynomial $f(x)\in \mathbb{Z}[x]$ such that for every finite field $\mathbb{F}_{q}$, $\# X_{\mathbb{F}_{q}}(\mathbb{F}_{q})=f(q)$. Clearly, if $X$ has an affine paving, then it must be of polynomial count. But there are lots of varieties of polynomial count with no affine paving.

For example, a large number of determinantal varieties are of polynomial count. Here the determinant variety is the integral model of the usual one: given $m$ and $n$ and $r<\min(m,n)$, the determinantal variety $Y_{r}$ is the set of all $m\times n$ matrices (over $\mathbb{Z}$) with the determinant of each $(r+1)\times (r+1)$ minors equals zero. The ideal of $\mathbb{Z}[t_{ij}]$ generated by these polynomials is a determinantal ideal. Since the equations defining minors are homogeneous, one can consider $Y_{r}$ either as an affine variety in $mn$-dimensional affine space, or as a projective variety in $(mn-1)$-dimensional projective space, and we take the projective one.

1. My question is: If $X\subset \mathbb{P}^{N}$ is a polynomial count quasi-projective variety, does there exist a chain of varieties $$A_{1}\supset A_{2}\supset \dots \supset A_{n}$$ where $A_{i+1}$ is closed (or locally closed) in $A_{i}$, for $i=1,2,...,n-1$, such that each $A_{i}$ has a affine paving, and $X=A_{1}-A_{2}+A_{3}-\dots (-1)^{n-1}A_{n}$?

If the answer of this question is Yes, I wonder how to give this decomposition of $$\mathrm{Proj}(\mathbb{Z}[t_{ij}]/\det(t_{ij})_{n\times n}),\ 1\le i,j\le n$$ or its affine cone? How can we give the explicit coordinates of this decomposition?

P.S.: One can calculate that the number of closed points of $\mathrm{Proj}(\mathbb{Z}[t_{ij}]/\det(t_{ij})_{n\times n})$ is $f(q)=\sum_{j=0}^{n^{2}-1}q^{j}-q^{n-1}\prod_{i=1}^{n-2}(q^{n}-q^{i})$. One can verify easily that the coefficient of $q^m$ is negative for some $n,m$. This means that in some cases it cannot be a variety which has an affine paving (even though it is of polynomial count), since a variety with an affine paving has $f(x)$ with non-negative coefficients.

2. One more question is: Let $\mathrm{Proj}(\mathbb{Z}[t_{ij}]),\ 1\le i,j\le n$ be the $\mathbb{P}^{n^{2}-1}$ defined over $\mathbb{Z}$, $M$ be a set of some sub square matrices of $(t_{ij})_{n\times n}$, $I$ be the ideal generated by $\det(m),m\in M$. Is this variety $\mathrm{Proj}(\mathbb{Z}[t_{ij}]/I)$ of polynomial count?

Any answer will be welcome. Thanks a lot !

Let $X$ be a variety defined over $\mathbb{Z}$, $X_{\mathbb{F}}$ be the base change $X\times_{Spec(\mathbb{Z})} Spec(\mathbb{F})$. We say $X$ is of polynomial count if there is a polynomial $f(x)\in \mathbb{Z}[x]$ such that for every finite field $\mathbb{F}_{q}$, $\# X_{\mathbb{F}_{q}}(\mathbb{F}_{q})=f(q)$. Clearly, if $X$ have a affine paving, then must be polynomial counting. But there are lots of polynomial count varieties with no affine paving.

For example, a large number of determinantal varieties are of polynomial count. Here the determinant variety is the integral model of the usual one: given $m$ and $n$ and $r<\min(m,n)$, the determinantal variety $Y_{r}$ is the set of all $m\times n$ matrices (over $\mathbb{Z}$) with the determinant of each $(r+1)\times (r+1)$ minors equals zero. The ideal of $\mathbb{Z}[t_{ij}]$ generated by these polynomials is a determinantal ideal. Since the equations defining minors are homogeneous, one can consider $Y_{r}$ either as an affine variety in $mn-$dimensional affine space, or as a projective variety in $(mn-1)$-dimensional projective space, and we take the projectve one.

1. My question is: If $X\subset \mathbb{P}^{N}$ is a polynomial count quasi-projective variety, does there exist a chain of varieties $$A_{1}\supset A_{2}\supset \dots \supset A_{n}$$ where $A_{i+1}$ is closed (or locally closed) in $A_{i}$, for $i=1,2,...,n-1$, such that each $A_{i}$ has a affine paving, and $X=A_{1}-A_{2}+A_{3}-\dots (-1)^{n-1}A_{n}$?

If the answer of this question is Yes, I wonder how to give this decomposition of $$\mathrm{Proj}(\mathbb{Z}[t_{ij}]/\det(t_{ij})_{n\times n}),\ 1\le i,j\le n$$ or its affine cone? How can we give the explicit coordinates of this decomposition?

P.S.: One can calculate that the number of closed points of $\mathrm{Proj}(\mathbb{Z}[t_{ij}]/\det(t_{ij})_{n\times n})$ is $f(q)=\sum_{j=0}^{n^{2}-1}q^{j}-q^{n-1}\prod_{i=1}^{n-2}(q^{n}-q^{i})$. One can verify easily that the coefficient of $q^m$ is negative for some $n,m$. This means that in some cases it can not be a variety which has an affine paving (even though it is of polynomial count), since a variety with an affine paving has $f(x)$ with non-negative coefficients.

2. One more question is: Let $\mathrm{Proj}(\mathbb{Z}[t_{ij}]),\ 1\le i,j\le n$ be the $\mathbb{P}^{n^{2}-1}$ defined over $\mathbb{Z}$, $M$ be a set of some sub square matrices of $(t_{ij})_{n\times n}$, $I$ be the ideal generated by $\det(m),m\in M$. Is this variety $\mathrm{Proj}(\mathbb{Z}[t_{ij}]/I)$ of polynomial count?

Any answer will be welcome. Thanks a lot !

Let $X$ be a variety defined over $\mathbb{Z}$, $X_{\mathbb{F}}$ be the base change $X\times_{Spec(\mathbb{Z})} Spec(\mathbb{F})$. We say $X$ is of polynomial count if there is a polynomial $f(x)\in \mathbb{Z}[x]$ such that for every finite field $\mathbb{F}_{q}$, $\# X_{\mathbb{F}_{q}}(\mathbb{F}_{q})=f(q)$. Clearly, if $X$ has an affine paving, then it must be of polynomial count. But there are lots of varieties of polynomial count with no affine paving.

For example, a large number of determinantal varieties are of polynomial count. Here the determinant variety is the integral model of the usual one: given $m$ and $n$ and $r<\min(m,n)$, the determinantal variety $Y_{r}$ is the set of all $m\times n$ matrices (over $\mathbb{Z}$) with the determinant of each $(r+1)\times (r+1)$ minors equals zero. The ideal of $\mathbb{Z}[t_{ij}]$ generated by these polynomials is a determinantal ideal. Since the equations defining minors are homogeneous, one can consider $Y_{r}$ either as an affine variety in $mn$-dimensional affine space, or as a projective variety in $(mn-1)$-dimensional projective space, and we take the projective one.

1. My question is: If $X\subset \mathbb{P}^{N}$ is a polynomial count quasi-projective variety, does there exist a chain of varieties $$A_{1}\supset A_{2}\supset \dots \supset A_{n}$$ where $A_{i+1}$ is closed (or locally closed) in $A_{i}$, for $i=1,2,...,n-1$, such that each $A_{i}$ has a affine paving, and $X=A_{1}-A_{2}+A_{3}-\dots (-1)^{n-1}A_{n}$?

If the answer of this question is Yes, I wonder how to give this decomposition of $$\mathrm{Proj}(\mathbb{Z}[t_{ij}]/\det(t_{ij})_{n\times n}),\ 1\le i,j\le n$$ or its affine cone? How can we give the explicit coordinates of this decomposition?

P.S.: One can calculate that the number of closed points of $\mathrm{Proj}(\mathbb{Z}[t_{ij}]/\det(t_{ij})_{n\times n})$ is $f(q)=\sum_{j=0}^{n^{2}-1}q^{j}-q^{n-1}\prod_{i=1}^{n-2}(q^{n}-q^{i})$. One can verify easily that the coefficient of $q^m$ is negative for some $n,m$. This means that in some cases it cannot be a variety which has an affine paving (even though it is of polynomial count), since a variety with an affine paving has $f(x)$ with non-negative coefficients.

2. One more question is: Let $\mathrm{Proj}(\mathbb{Z}[t_{ij}]),\ 1\le i,j\le n$ be the $\mathbb{P}^{n^{2}-1}$ defined over $\mathbb{Z}$, $M$ be a set of some sub square matrices of $(t_{ij})_{n\times n}$, $I$ be the ideal generated by $\det(m),m\in M$. Is this variety $\mathrm{Proj}(\mathbb{Z}[t_{ij}]/I)$ of polynomial count?

Any answer will be welcome. Thanks a lot !

Let $X$ be a variety defined over $\mathbb{Z}$, $X_{\mathbb{F}}$ be the base change $X\times_{Spec(\mathbb{Z})} Spec(\mathbb{F})$. We call $X$ is of polynomial count if there is a polynomial $f(x)\in \mathbb{Z}[x]$, and for every finite field $\mathbb{F}_{q}$, $\# X_{\mathbb{F}_{q}}(\mathbb{F}_{q})=f(q)$. Clearly, if $X$ have a affine paving, then must be polynomial counting. But there are lots of polynomial count varieties with no affine paving. For example, a large number of determinantal varieties (Here the determinant variety is the integral model of the usual one: Give $m$ and $n$ and $r<min(m,n)$, the determinantal variety $Y_{r}$ is the set of all $m\times n$ matrices (over $\mathbb{Z}$) with the determinant of each $(r+1)\times (r+1)$ minors equals zero. The idea of $\mathbb{Z}[t_{ij}]$ generated by these polynomials is a determinantal ideal. Since the equations defining minors are homogeneous, on can consider $Y_{r}$ either as an affine variety in $mn-$dimensional affine space, or as a projective variety in $(mn-1)$-dimensional projective space(Here we take the projectve one).) 1.My question is: If $X\subset \mathbb{P}^{N}$ is a polynomial count quasi-projective variety, is there exist a chain of varieties $$A_{1}\supset A_{2}\supset \dots \supset A_{n}$$ Where $A_{i+1}$ is closed(or local closed) in $A_{i}$, for $i=1,2,...,n-1$. Such that each $A_{i}$ has a affine paving, and $X=A_{1}-A_{2}+A_{3}-\dots (-1)^{n-1}A_{n}$. If the answer of this question is Yes, I wonder how to give this decomposition of $Proj(\mathbb{Z}[t_{ij}]/det(t_{ij})_{n\times n}),\ 1\le i,j\le n$ or its affine cone  ? If it could be, how to give the explicit coordinate of this decomposition  ? P.S. One can calculate the number of close points of $Proj(\mathbb{F}_{q}[t_{ij}]/det(t_{ij})_{n\times n}),\ 1\le i,j\le n.$, that is $f(q)=\sum_{j=0}^{n^{2}-1}q^{j}-q^{n-1}\prod_{i=1}^{n-2}(q^{n}-q^{i})$. One can verify easily that the coefficient of $q^m$ is negative for some $n,m$. This means in some
case it can not be a variety which have a affine paving (Even though it is polynomial counting). Since the affine paving variety has $f(x)$ with non-negative coefficients. 2.One more question is  : Let $Proj(\mathbb{Z}[t_{ij}]),\ 1\le i,j\le n$ be the $\mathbb{P}^{n^{2}-1}$ defined over $\mathbb{Z}$; $M$ be a set of some sub square matrices of $(t_{ij})_{n\times n}$, $I$ be the idea generated by $det(m),m\in M$. Is this variety $Proj(\mathbb{Z}[t_{ij}]/I),(\ 1\le i,j\le n.)$ polynomial counting?
Any answer will be welcome. Thanks a lot !

Let $X$ be a variety defined over $\mathbb{Z}$, $X_{\mathbb{F}}$ be the base change $X\times_{Spec(\mathbb{Z})} Spec(\mathbb{F})$. We say $X$ is of polynomial count if there is a polynomial $f(x)\in \mathbb{Z}[x]$ such that for every finite field $\mathbb{F}_{q}$, $\# X_{\mathbb{F}_{q}}(\mathbb{F}_{q})=f(q)$. Clearly, if $X$ have a affine paving, then must be polynomial counting. But there are lots of polynomial count varieties with no affine paving. 

For example, a large number of determinantal varieties are of polynomial count. Here the determinant variety is the integral model of the usual one: given $m$ and $n$ and $r<\min(m,n)$, the determinantal variety $Y_{r}$ is the set of all $m\times n$ matrices (over $\mathbb{Z}$) with the determinant of each $(r+1)\times (r+1)$ minors equals zero. The ideal of $\mathbb{Z}[t_{ij}]$ generated by these polynomials is a determinantal ideal. Since the equations defining minors are homogeneous, one can consider $Y_{r}$ either as an affine variety in $mn-$dimensional affine space, or as a projective variety in $(mn-1)$-dimensional projective space, and we take the projectve one.

1. My question is: If $X\subset \mathbb{P}^{N}$ is a polynomial count quasi-projective variety, does there exist a chain of varieties $$A_{1}\supset A_{2}\supset \dots \supset A_{n}$$ where $A_{i+1}$ is closed  (or locally closed) in $A_{i}$, for $i=1,2,...,n-1$, such that each $A_{i}$ has a affine paving, and $X=A_{1}-A_{2}+A_{3}-\dots (-1)^{n-1}A_{n}$?

If the answer of this question is Yes, I wonder how to give this decomposition of $$\mathrm{Proj}(\mathbb{Z}[t_{ij}]/\det(t_{ij})_{n\times n}),\ 1\le i,j\le n$$ or its affine cone? How can we give the explicit coordinates of this decomposition?

P.S.: One can calculate that the number of closed points of $\mathrm{Proj}(\mathbb{Z}[t_{ij}]/\det(t_{ij})_{n\times n})$ is $f(q)=\sum_{j=0}^{n^{2}-1}q^{j}-q^{n-1}\prod_{i=1}^{n-2}(q^{n}-q^{i})$. One can verify easily that the coefficient of $q^m$ is negative for some $n,m$. This means that in some cases it can not be a variety which has an affine paving (even though it is of polynomial count), since a variety with an affine paving has $f(x)$ with non-negative coefficients.

2. One more question is: Let $\mathrm{Proj}(\mathbb{Z}[t_{ij}]),\ 1\le i,j\le n$ be the $\mathbb{P}^{n^{2}-1}$ defined over $\mathbb{Z}$, $M$ be a set of some sub square matrices of $(t_{ij})_{n\times n}$, $I$ be the ideal generated by $\det(m),m\in M$. Is this variety $\mathrm{Proj}(\mathbb{Z}[t_{ij}]/I)$ of polynomial count?

Any answer will be welcome. Thanks a lot !