---
title: "Problem Sets: Moduli"
subtitle: "Problem Set 2"
author:
- name: D. Zack Garza
  affiliation: University of Georgia 
  email: dzackgarza@gmail.com 
date: Fall 2022
order: 2
---

# Problem Set 2

## 1

:::{.problem title="1"}
Consider the $\mathrm{SL}_{2}$ action on $X=\left(\mathbb{P}^{1}\right)^{n}$ with a linearized invertible sheaf $L=\mathcal{O}_{X}\left(d_{1}, \ldots, d_{n}\right), d_{i} \in \mathbb{N}$. Define $w_{i}:=\frac{2 d_{i}}{\sum d_{j}}$, so that $\sum w_{i}=2$. Prove that a point $\left(P_{1}, \ldots, P_{n}\right) \in X^{s s}(L)$ (resp. $\left.X^{s}(L)\right) \Longleftrightarrow$ whenever some points $P_{i}$, $i \in I, I \subset\{1, \ldots, n\}$, coincide, one has $\sum_{i \in I} w_{i} \leq 1$ (resp. $<1$ ).
:::

:::{.solution}
Write points in this product as
\[
X \da (\PP^1)^{n} = \ts{ \vector p \da
\begin{bmatrix}
x_0 & \cdots & x_n
\\
y_0 & \cdots & y_n
\end{bmatrix}
}
,\]
corresponding to the $n\dash$tuple $\qty{ \tv{x_0: y_0}, \cdots, \tv{x_n: y_n} }$,
with $\SL_2$ action given by
\[
\SL_2 &\actson X \\
\matt a b c d \cdot \vector p \da 
&\matt a b c d 
\begin{bmatrix}
x_0 & \cdots & x_n
\\
y_0 & \cdots & y_n
\end{bmatrix}
= 
\begin{bmatrix}
a x_0 + by_0 & \cdots & ax_n + by_n
\\
cx_0 + dy_0 & \cdots & cx_n + dy_n
\end{bmatrix}
.\]
We note that the maximal torus acts as
\[
T_{\SL_2} &\actson X \\
\matt t \cdot \cdot {t\inv} \cdot \vector p \da 
&\matt a b c d 
\begin{bmatrix}
x_0 & \cdots & x_n
\\
y_0 & \cdots & y_n
\end{bmatrix}
= 
\begin{bmatrix}
t x_0  & \cdots & tx_n
\\
t\inv y_0 & \cdots & t\inv y_n
\end{bmatrix}
.\]
We identify $X$ with its image (which we'll also denote $X$) under the Veronese embedding $X \to \PP^N$ associated to the ample line bundle $\mcl \da \OO(\vector d)$ where $\vector d \da \tv{d_1,\cdots, d_n} \subseteq \ZZ^n$ viewed as an integer vector.
Writing $D$ for the convex hull of the $d_i$ in $\ZZ^n$, note that every lattice point in $\ZZ^n \intersect D$ defines a monomial, and every point $\vector p \in X$ corresponds to a a collection of lattice points $P_{\vector p } = \ts{ \vector k = \tv{k_1,\cdots, k_n} } \subseteq D \intersect \ZZ^n$ along with a choice of coefficient $\alpha_{\vector k}$ for each $\vector k \in P_{\vector p}$.

The following is an example $D$ and $P_{\vector p}$ when $n=3$ and $\vector d = \tv{3, 5, 4}$:

![](figures/2022-11-22_17-59-23.png)

The three highlighted lattice points are $\vector k_1 = \tv{3,0,0}, \vector k_2 = \tv{0,5,0}, \vector k_3 = \tv{0,0,4}$, $P_{\vector p} \da \ts{\vector k_1, \vector k_2, \vector k_3}$ corresponds to a polynomial
\[
F(x_1, x_2, x_3) = \alpha_1 x_1^3 x_2^0 x_3^0 + \alpha_2 x_1^0 x_2^5 x_3^0 + \alpha_3 x_1^0 x_2^0 x_3^4
.\]

In our situation, lattice points will correspond to monomials
\[
\vector k_{IJ} = x^I y^J \da x_1^{i_1} x_2^{i_2}\cdots x_n^{i_n} \,\cdot\, y_1^{j_1}y_2^{j_2}\cdots y_n^{j_n}
,\]
and so each point in $X$ will correspond to a polynomial
\[
F(x_1,\cdots, x_n, y_1,\cdots, y_n) = \sum_{(I, J) \subseteq D} \alpha_{IJ} x^I y^J 
.\]
where $\sum_{i\in I} i + \sum_{j\in J} j = d_i$.

> Todo: this is not quite right. If $\alpha_j$ is associated to the embedding along the $d_j$ direction, then the monomial degrees should just sum up to $d_j$.

Indexing these monomials systematically, we can write
\[
F(x_1,\cdots, y_n) = \sum \alpha_j \prod_{i=1}^n x_i^{d_i - k_j} y_i^{k_j}
.\]
When points collide, without loss of generality (using the transitive $\SL_2\dash$action) we can assume that the collision point in $\PP^1$ is $\tv{0: 1}$, so $p\in X$ is of the form
\[
p = 
\begin{bmatrix}
0 & \cdots & 0 & p_{m+1} & \cdots & p_n \\ 
1 & \cdots & 1 & q_{m+1} & \cdots & q_n \\ 
\end{bmatrix}
,\]
where we've written $m$ for the number of colliding points.
We can now compute the weights of the torus action over such colliding points
\[
\lambda(t).F(x_1,\cdots, y_n) 
&= \sum \alpha_j \prod t^{d_i - 2k_j} x_i^{d_i-k_j} y^{k_j} \\
&= \sum t^{w_{ij}} \alpha_j x_i^{d_i-k_j} y^{k_j}, \qquad w_{ij} \da \sum_{i} d_i - 2k_j
.\]
We now need $\mu(x, \lambda) \geq 0$ for semistability, i.e. $\min(w_{ij}) \geq 0$, so $\min(\sum d_i - 2k_j) \geq 0$.
We can maximally destabilize such a quantity by taking $k_j = d_i$ for each $i,j$, and so if the collision set is $\mathcal S$, we require
\[
\sum_{i=1}^n d_i - \sum_{i\in \mathcal S} 2d_i \geq 0 \iff
\sum_{i=1}^n d_i \geq \sum_{i\in \mathcal S} d_i \iff 
{ \sum_{i\in \mathcal S} 2d_i \over \sum_{i=1}^n d_i} \leq 1 \iff
\sum_{i\in \mathcal S} w_i \leq 1 
.\]
:::

## 2

:::{.problem title="2"}
Consider the $\mathrm{SL}_{3}$ action on the set $X=\mathbb{P}^{N}, N=\left(\begin{array}{c}3+2 \\ 2\end{array}\right)-1=9$, parameterizing cubic curves $C \subset \mathbb{P}^{2}$, with a linearized invertible sheaf $L=\mathcal{O}_{X}(1)$. Prove that $C$ is semistable $\Longleftrightarrow C$ has only ordinary double points.
:::

:::{.solution}
We first note that every choice of cubic curve $C\in Y_{3, 2}$ can be represented (after choosing coordinates) by a polynomial 
\[
F(x,y,z) = \sum_{i+j+k=3}a_{ijk}\, x^iy^jz^k = \sum_{i+j+k=3} a_{\vector i} x^i y^j z^k \qquad \vector i \da \tv{i,j,k}
\]
and thus a choice of lattice points $C_P$ in the corresponding weight polytope where each point is labeled with the corresponding coefficient of $F$:

![](figures/2022-11-21_22-18-35.png)

We record the fact that the point $p \da \tv{1:0:0}$ is singular iff $a_{300} = a_{201} = a_{210} = 0$:

![](figures/2022-11-21_22-22-25.png)

Moreover, $p$ is a triple point iff additionally $a_{102} = a_{111} = a_{120} = 0$:

![](figures/2022-11-21_22-25-03.png)

Moreover, all of above holds except $a_{102}$ (the coefficient of $xz^2$) is nonzero, then $p$ is a double point with only a single tangent, and thus not an ordinary double point.
These facts follow from computing the gradients and Hessians which characterize these types of singularities.

We also note that if $\lambda: \GG_m \to \SL_3$ is a 1-parameter subgroup, then $\lambda(t)$ is conjugate to 
\[
\tilde \lambda(t) = \mattt {t^{r_1}} \cdot \cdot \cdot {t^{r_2}} \cdot \cdot \cdot {t^{r_3}}, \qquad \sum_{i=1}^3 r_i = 0
,\]
and thus determines a vector $\vector r \da \tv{r_1, r_2, r_3}\in \ZZ^3$.
The action can then be written
\[
\lambda(t)\cdot F(x, y, z) = \sum_{i+j+k = 3} a_{\vector i}\, t^{\ip {\vector r} {\vector i} } x^i y^j z^k
,\]
and so all weights are of the form $w_{\vector i} = \ip{\vector r}{\vector i} \in \ZZ$.
We note that $C\in Y_{3, 2}$ is unstable iff for every $\lambda$, every weight is negative or every weight is positive, so $w_{\vector i} < 0$ or $w_{\vector i} > 0$ for all $\vector i \in C_P$.
We'll focus on the strictly positive case, since the positive case follows similarly.

$\implies$:
Suppose $C$ is unstable, we will show that $p$ is either a non-ordinary double point, a triple point, or worse.
Pick $\lambda$ and its corresponding $\vector r$ such that all weights $w_{\vector i}$ are positive.
Then in particular
\[
\min \ts{w_{\vector i} \da \ip{\vector r}{\vector i} \st \vector i\in C_P} > 0
.\]

Having strictly positive weights can be phrased geometrically as $\ts{\vector i \st \vector i\in C_P,\, a_{\vector i} \neq 0}$ being contained in the positive half-space corresponding to the hyperplane $H_C \da \vector r^\perp$.
Picking a maximally destabilizing $\lambda$, without loss of generality (changing coordinates if necessary) we can arrange for the lower-left 5 monomials receive non-positive weights:

![](figures/2022-11-21_22-58-07.png)

This forces all of the shaded coefficients except for potentially $a_{102}$ to be zero.
By the earlier remarks, this forces $p = [1:0:0]$ to be singular, and if $a_{102} = 0$ this is a triple point.
Otherwise, if $a_{102}\neq 0$, this yields a double point which only has a single tangent, and is thus not ordinary.
So if $C$ is *not* an unstable curve (i.e. it is semistable), it must have an ordinary double point at worst.

$\impliedby$:
Suppose conversely that $C$ has a triple point or a non-ordinary double point $q$.
Using the transitivity of the $\SL_3$ action, we can move $q$ to $p = [1:0:0]$ and conclude using the singularity criterion above that the following coefficients vanish:

![](figures/2022-11-21_23-58-40.png)

We can now make a specific choice of $\lambda$ that yields the following $H_\lambda$ and gives the remaining coefficients strictly positive weights, allowing us to conclude that $C$ is unstable:

![](figures/2022-11-22_00-00-56.png)

:::

## 3

:::{.problem title="3"}
Give an example showing that Hilbert-Mumford's criterion of (semi)stability for $G \curvearrowright X$ does not hold in general if $X$ is not assumed to be projective. (In other words, produce a counterexample with a non-projective $X$.)
:::

:::{.solution .foldopen}
Consider the following action:
\[
\GG_m &\actson X \da \AA^2 \\
t.\tv{x, y} &\da \tv{tx, ty}
.\]
Thus yields a set theoretic orbit space
\[
\AA^2/\GG_m &= \ts{O_t \st t\in \GG_m} \union \ts{O_x, O_y, O_0} \\ \\
O_t &\da \ts{xy = t \st t \in \GG_m}  \\
O_x &\da \ts{\tv{t, 0} \st t\in \GG_m} = \GG_m . \tv{1, 0} \\
O_y &\da \ts{\tv{0, t} \st t\in \GG_m} = \GG_m . \tv{0, 1} \\
O_0 &\da \ts{0}
,\]
i.e. there is an orbit for each hyperbola $xy=t$, the punctured $x\dash$axis, the punctured $y\dash$axis, and the origin:

![](figures/2022-11-23_14-42-32.png)

We record that the following facts:

- The orbits $O_t$ are all closed with 0-dimensional stabilizers,
- The orbits $O_x, O_y$ are not closed but still have 0-dimensional stabilizers, and
- The orbit $O_0$ is closed but has a 1-dimensional stabilizer $\GG_m$.

Thus $X^s = \AA^2 \sm V(xy)$ is the plane with the axes deleted, and for example $0\in X\sm X^s$ is an unstable point and $\tv{1, 0}, \tv{0, 1}\in X\sm X^s$ are not stable points (and may thus either be unstable or semistable).

Noting that $O_x \sim O_y \sim O_0$ are all orbit-closure equivalent since $0$ is in the closure of $O_x$ and $O_y$, we can separate these orbits by redefining our total space to be $X \da \AA^2\smz$; then $O_x, O_y$ are closed in $X'$ and have 0-dimensional stabilizer and thus points in those orbits become stable for the restricted action $\GG_m\actson X'$.

For example, pick $p \da \tv{1, 0} \in O_x \subseteq X'$, then $p$ is stable by construction.
However, we can now check the Hilbert-Mumford numerical criterion and note that every 1-parameter subgroup $\lambda$ acting with weights $r_1, r_2$ satisfies 
\[
\lambda(t).p = \tv{t^{r_1} 1, t^{r_2} 0} = \tv{t^{r_1}1, 0}
,\]
and in particular always has strictly positive or strictly negative weights, which would otherwise characterize $p$ as an *unstable* point, yielding the desired counterexample.
:::

## 4

:::{.problem title="4"}
Provide a complete VGIT (variation of GIT) analysis for the quotients $\left(\mathbb{P}^{1}\right)^{3} / / \mathbb{G}_{m}$. The line bundle is $L=\mathcal{O}(1,1,1)$. The $\mathbb{G}_{m}$-action is defined as
\[
t .\left(x_{0}: x_{1}\right)=\left(x_{0}: t x_{1}\right), \quad t .\left(y_{0}: y_{1}\right)=\left(y_{0}: t y_{1}\right), \quad t .\left(z_{0}: z_{1}\right)=\left(z_{0}: t z_{1}\right)
\]
The linearization is a lift of this action to the action on the coordinates $w_{i j k}=$ $x_{i} y_{j} z_{k}$ on $\left(\mathbb{P}^{1}\right)^{3}$ embedded into $\mathbb{P}^{7}$ with the 8 homogeneous coordinates $w_{i j k}$. The above equations give an action on the point $\left(w_{i j k}\right) \in \mathbb{P}^{7}$. The linearization is a lift of this action to the point $\left(w_{i j k}\right) \in \mathbb{A}^{8}$.

Determine the following:

(1) The choices for $\mathbb{Q}$-linearizations of $L$ (i.e. linearizations of some $L^{d}, d \in \mathbb{N}$ ).

(2) Chamber decomposition.

(3) For each chamber, the quotient.

(4) For neighboring chambers, the induced morphisms between the quotients.

(5) For each chamber, the sets of unstable and strictly semistable points.

:::

:::{.solution .foldopen}
Todo.
:::

## 5

:::{.problem title="5"}
Let $X \subset \mathbb{P}^{N}$ be a singular projective curve. Suppose that $X$ has $n$ irreducible components $X_{i}$ and that $\left.\operatorname{deg} \mathcal{O}_{X}(1)\right|_{X_{i}}=\lambda_{i} \in \mathbb{N}$. Let $F$ be a coherent sheaf on $X$. Then on an open subset $U_{i} \subset X_{i}$ of each irreducible component it is a locally free sheaf of rank $r_{i}$.

The Seshadri slope of an invertible sheaf $F$ is defined to be
\[
\mu(F)=\frac{\chi(F)}{\sum \lambda_{i} r_{i}}, \quad \text { where } r_{i}=\left.\operatorname{rk} F\right|_{U_{i}} .
\]
By replacing $\mathcal{O}_{X}(1)$ by a rational multiple, one can assume that $\lambda_{i}>0, \sum \lambda_{i}=1$.

1. Let $F$ be a pure-dimensional coherent sheaf on $X$. Prove that $F$ is Hilbertstable (resp. semistable) $\Longleftrightarrow$ for any subsheaf $E \subset F$ one has $\mu(E)<$ $\mu(F)$ (resp. $\leq$ ). (Note in particular, that this definition depends on the polarization $\left(\lambda_{i}\right)$, and there is a Variation of GIT here.)

2. Prove, however, that if $\chi(F)=0$ then the (semi)stability condition does not depend on a polarization $\left(\lambda_{i}\right)$.

:::

:::{.remark}
You can use the following simple observation. If $\pi: \widetilde{X} \rightarrow X$ is a normalization then $\tilde{X}$ is a smooth curve, so Riemann-Roch is applicable:

$$
\chi(E)=\operatorname{deg}(E)+\operatorname{rank}(E)(1-g),
$$

and the difference of Hilbert polynomials

$$
\chi(X, F(m))-\chi\left(\tilde{X},\left(\pi^{*} F\right)(m)\right)
$$

is a constant.
:::

:::{.solution .foldopen}
We first recall that a sheaf $\mcf\in \Coh(X)$ is Hilbert stable if for every subsheaf $E\leq F$, we have an inequality of reduced Hilbert polynomials $\tilde p_E(n) < \tilde p_F(n)$, and semistability is characterized by replacing $<$ with $\leq$.
Noting that 
\[
p_F(n) \da \chi(X; F(n)) = c_0 n ^{\dim X} = c_0 n + c_1
\]
since $X$ is a curve and consequently $\dim X = 1$.
We have $\tilde p_F(n) = n + {c_0\over c_1}$ and thus $\tilde p_E(n) = n + {d_0\over d_1}$ for some constants $c_i$ depending on $F$ and $d_i$ depending on $E$, and so
\[
\tilde p_E(n) < \tilde p_F(n) \iff {d_0\over d_1} < {c_0 \over c_1}
.\]
Thus it suffices to show that ${d_0\over d_1} = \mu(E)$ and ${c_0\over c_1} = \mu(F)$.
We'll proceed by computing $p_F(n)$ in order to identify what $c_0, c_1$ are in general.

Noting that $X$ may be singular and thus Riemann-Roch won't apply directly, take the normalization $\pi: \tilde X\to X$.
Let $X = \union_i X_i$ be the decomposition of $X$ into irreducible components and let $\tilde X_i$ be their lifts in the normalization, which are all curves with some genera $g_i$.
We now have
\[
p_F(n) 
&\da \chi(X; F(n)) \\ \\
&= \chi(\tilde X, (\pi^* F)(n) ) + c \qquad \text{ for some constant } c \\ \\
&= \sum_{1\leq i\leq n} \chi(\tilde X_i, \ro{ (\pi^* F)(n) }{\tilde X_i} ) + c \\ \\
&= \sum_{1\leq i\leq n} \qty{\deg \ro{(\pi^* F)(n)}{\tilde X_i} + (1-g_i) } + c 
.\]
As an aside, we can compute the degrees inside of the sum as follows:
\[
\deg \ro{(\pi^* F)(n)}{X_i}
&= \deg \ro{F(n)}{X_i} \\
&= \deg \ro{F}{X_i} \tensor \bigoplus_{1\leq j\leq r_i} \OO_{X_i}(n) \\
&= \deg \ro{F}{X_i} + n r_i \lambda_i
.\]
Continuing the above calculation, we have
\[
p_F(n) 
&= \sum_{1\leq i\leq n} \qty{ \deg \ro{F}{X_i} + nr_i \lambda_i + (1-g_i) } + c \\ \\
&= n \qty{ \sum_{1\leq i\leq n} r_i \lambda_i} + \qty{ \sum_{1\leq i\leq n} \deg \ro{F}{X_i} + (1-g_i) + c} \\ \\
&= n \qty{ \sum_{1\leq i\leq n} r_i \lambda_i} + \qty{ \sum_{1\leq i\leq n} \chi(X_i; \ro{F}{X_i} ) + c} \\ \\
&= n \qty{ \sum_{1\leq i\leq n} r_i \lambda_i} + \chi(X; F) 
.\]

Thus $c_0 = \sum r_i \lambda_i, c_1 = \chi(F)$, and ${c_1\over c_0} = { \chi(F) \over \sum r_i \lambda_i} = \mu(F)$.
:::