# Solutions to exercises

Exercise 7.1

Let ${\displaystyle \mathbb {C} ^{*}}$ act on ${\displaystyle \mathbb {C} ^{n}}$ in the usual way. Define a lift of the action to the trivial line bundle as the ${\displaystyle \mathbb {C} ^{*}}$ action composed with ${\displaystyle \lambda \mapsto \lambda ^{-1}}$. Use the Hilbert-Mumford criteria to find the stable, semistable, and unstable points.

Proof

Embed ${\displaystyle \mathbb {C} ^{n}}$ in ${\displaystyle \mathbb {P} ^{n}}$ using ${\displaystyle z\mapsto [z:1]}$. This ${\displaystyle \mathbb {C} ^{*}}$ action extends as the action given by ${\displaystyle \lambda \cdot [z_{0}:\cdots :z_{n}]=[\lambda z_{0}:\cdots :\lambda z_{n-1}:\lambda ^{-n}z_{n}]}$, with the obvious linearisation. To check the stability of the point ${\displaystyle z=[z_{0}:\cdots :z_{n}]}$, because of the Hilbert-Mumford criteria, we only need to consider the weight of this action, and of the inverse 1-PS, as ${\displaystyle \lambda \rightarrow 0}$ on the line bundle.

Case 1: Suppose ${\displaystyle z_{n}=0}$. Then ${\displaystyle z}$ is a fixed point of the action. On the line over ${\displaystyle z}$, ${\displaystyle \lambda }$ acts as multiplication by ${\displaystyle \lambda }$, so the weight of the action is ${\displaystyle 1>0}$. So this is an unstable point.

Case 2: Suppose ${\displaystyle z=[0:\cdots :0:1]}$. This is again a fixed point of the action. The weight of the (linearisation) of the action as ${\displaystyle \lambda \rightarrow 0}$ is ${\displaystyle -n}$, as on the line over this point, ${\displaystyle \lambda \cdot (0,\dots ,0,x)=(0,\dots ,0,\lambda ^{-n}x)}$. However, the weight of the linearisation action of the inverse subgroup (given by ${\displaystyle \lambda \mapsto \lambda ^{-1}}$) is ${\displaystyle n}$, so this point is unstable.

Case 3: (everything else) The limit as ${\displaystyle \lambda \rightarrow 0}$ of ${\displaystyle \lambda \cdot z=[\lambda z_{0}:\cdots :\lambda z_{n-1}:\lambda ^{-n}z_{n}]}$ is ${\displaystyle [0:\cdots :0:1]}$. As before, on the line over this point, ${\displaystyle \lambda \cdot (0,\dots ,0,x)=(0,\dots ,0,\lambda ^{-n}x)}$, so the weight is ${\displaystyle -n<0}$. We also need to the check the inverse 1-PS, given by ${\displaystyle \lambda \mapsto \lambda ^{-1}}$, so consider the limit as ${\displaystyle \lambda \rightarrow 0}$ of ${\displaystyle \lambda ^{-1}\cdot z}$. This is ${\displaystyle [z_{0}:\cdots :z_{n-1}:0]}$. On the line over this point, the action is ${\displaystyle \lambda ^{-1}(xz_{0},\ldots ,xz_{n-1},0)}$, so it has weight ${\displaystyle -1}$. Therefore ${\displaystyle z}$ is a stable point.

Exercise 7.2

Consider the action of ${\displaystyle SL_{r}(\mathbb {C} )}$ on ${\displaystyle \operatorname {Hom} (\mathbb {C} ^{r},\mathbb {C} ^{n})}$, ${\displaystyle r. Show that semistable points correspond to injective maps. What is the quotient?

Proof

We may represent an element of ${\displaystyle \operatorname {Hom} (\mathbb {C} ^{r},\mathbb {C} ^{n})}$ as an ${\displaystyle r\times n}$ matrix. ${\displaystyle SL_{r}(\mathbb {C} )}$ acts by multiplication on the right; therefore it lets us perform column operations.

If an element ${\displaystyle \varphi \in {\mathcal {M}}_{r\times n}(\mathbb {C} )}$ has ${\displaystyle \operatorname {rk} , then we can choose a different basis of ${\displaystyle \mathbb {C} ^{r}}$ in such a way that the matrix representing ${\displaystyle \varphi }$ has a zero column (say the first one). This point is then unstable, since we can consider the 1PS of ${\displaystyle SL_{r}(\mathbb {C} )}$ given (in this basis) by ${\displaystyle t\mapsto \operatorname {diag} (t^{-r+1},t,\ldots ,t)}$ and ${\displaystyle \lim _{t\rightarrow 0}t.\varphi =0}$.

On the other hand, consider an unstable point. There is a bad torus which takes the element to 0 in the limit for ${\displaystyle t\rightarrow 0}$. Every torus in ${\displaystyle SL_{r}(\mathbb {C} )}$ can be conjugated into the maximal torus of diagonal matrices. We may then suppose that this torus is given by ${\displaystyle t\mapsto \operatorname {diag} (t^{\lambda _{1}},\ldots ,t^{\lambda _{r}})}$ with ${\displaystyle \lambda _{1}\leq \ldots \leq \lambda _{r}}$ and ${\displaystyle \lambda _{1}<0}$. We may write ${\displaystyle t.\varphi =t^{\lambda _{1}}(\varphi ^{1},t^{\lambda _{2}-\lambda _{1}}\varphi ^{2},\ldots ,t^{\lambda _{r}-\lambda _{1}}\varphi ^{r})}$, where ${\displaystyle (\varphi ^{1},\ldots ,\varphi ^{r})}$ indicate the columns of the matrix representing ${\displaystyle \varphi }$ in the basis "chosen" by the torus. We see that, if we want the limit for ${\displaystyle t\mapsto 0}$ to be zero, we need that columns on which the 1PS acts with negative weight (at least ${\displaystyle \varphi ^{1}}$) are themselves zero. So ${\displaystyle \varphi }$ is not injective.

From the above discussion it is clear that there are no strictly semistable points (i.e. all semistable points are stable). The quotient is the Grassmannian of ${\displaystyle r}$-planes in ${\displaystyle \mathbb {C} ^{n}}$, ${\displaystyle \operatorname {Gr} (r,n)}$.

Exercise 7.3

We consider the natural action of ${\displaystyle SL(3,\mathbb {C} )}$ on ${\displaystyle \mathbb {P} ^{2}}$. Every cubic curve in ${\displaystyle \mathbb {P} ^{2}}$ is given by a single degree 3 homogeneous form

${\displaystyle f=\sum _{i+j+k=3}a_{ijk}x^{i}y^{j}z^{k},}$
where ${\displaystyle [x:y:z]}$ are the homogeneous coordinates in ${\displaystyle \mathbb {P} ^{2}}$ and ${\displaystyle SL(3,\mathbb {C} )}$ induces an action on the space of cubic curves. We want to study stable, semistable and unstable cubics.

Proof

In order to apply Hilbert-Mumford criterion, we consider the 1-parameter subgroups of ${\displaystyle SL(3,\mathbb {C} )}$; they are of the form

${\displaystyle \lambda _{abc}:t\mapsto \left({\begin{array}{ccc}t^{a}&0&0\\0&t^{b}&0\\0&0&t^{c}\end{array}}\right)}$
with ${\displaystyle a,b,c\in \mathbb {Z} }$, ${\displaystyle a+b+c=0}$ and ${\displaystyle a\geqslant b\geqslant c}$. The action is explicitly given by
${\displaystyle \lambda _{abc}(t).f=\sum _{i+j+k=3}t^{-(ai+bj+ck}x^{i}y^{j}z^{k}}$
and we denote by ${\displaystyle \mu (f,\lambda _{abc})={\textrm {max}}\{ai+bj+ck|a_{ijk}\neq 0\}}$. According to Hilbert-Mumford criterion, we will now study the sign of ${\displaystyle \mu (f)}$.

We consider ${\displaystyle f}$ as above and we fix the point ${\displaystyle p=[1:0:0]}$. We recall that

• p is singular ${\displaystyle \iff a_{300}=a_{210}=a_{201}=0}$ (i.e. there are not terms in ${\displaystyle x^{3},x^{2}y}$ and ${\displaystyle x^{2}z}$;
• p is a triple point ${\displaystyle \iff a_{300}=a_{210}=a_{201}=a_{120}=a_{111}=a_{102}=0}$;
• p is a double point with a unique tangent ${\displaystyle z=0\iff a_{300}=a_{210}=a_{201}=a_{120}=a_{111}=0}$.

By Hilbert-Mumford criterion:

F is unstable if and only if

{\displaystyle {\begin{aligned}\exists (a,b,c)\left.\right|\mu (f,\lambda _{abc})<0\\\iff &\exists (a,b,c)\left.\right|{\textrm {max}}\{ai+bj+ck|a_{ijk}\neq 0\}<0\\\Longrightarrow &a_{300}=a_{210}=a_{201}=a_{120}=a_{111}=0\\&since\quad 3a,2a+b,a+2b,2a+c\quad and\quad a+b+c\geq 0\\\end{aligned}}}
And this happens if and only if F has a triple point or a double point with a unique tangent

finally, choosing ${\displaystyle a=3,b=-1}$ and ${\displaystyle c=-2}$, we can check that ${\displaystyle {\textrm {max}}\{ai+bj+ck|a_{ijk}\neq 0\}<0}$; so the converse implication also holds.

{\displaystyle {\begin{aligned}{\textrm {fisnotstable}}\iff &\exists (a,b,c){\textrm {suchthat}}\mu (f,\lambda _{abc})\leqslant 0\\\iff &\exists (a,b,c){\textrm {suchthat}}{\textrm {max}}\{ai+bj+ck|a_{ijk}\neq 0\}\leqslant 0\\\Longrightarrow &a_{300}=a_{210}=0{\textrm {necessarilyand}}\\&{\textrm {if}}a_{201}=0,{\textrm {thenfissingularinp}}\\&{\textrm {if}}a_{201}\neq 0,{\textrm {then}}a=b{\textrm {and}}c=-2a\\&{\textrm {hence}}{\textrm {max}}\{a(i+j-2k)|a_{ijk}\neq 0\}\leqslant 0{\textrm {i.e.}}\forall a_{ijk}\neq 0,i+j-2k\leqslant 0\\&{\textrm {hencenecessarily}}a_{120}=0\\&{\textrm {itimpliesthat}}f=zg,{\textrm {forsomegofdegree2}}\\&{\textrm {thenfhasasingularpoint}}\end{aligned}}}

finally, choosing ${\displaystyle a=2,b=-1}$ and ${\displaystyle c=-1}$, we can check that ${\displaystyle {\textrm {max}}\{ai+bj+ck|a_{ijk}\neq 0\}\leqslant 0}$; so the converse implication also holds.

We can summarize that:

• f is stable ${\displaystyle \iff }$ f is smooth
• f is strictly semistable ${\displaystyle \iff }$ f has double points (one or more) with distinct tangents
• f is semistable ${\displaystyle \iff }$ f has at worst ordinary double points
• f is unstable ${\displaystyle \iff }$ f has a triple point or a double point with unique tangent.