# Solutions to exercises

Exercise 7.1

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

Proof

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

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

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

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

Exercise 7.2

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

Proof

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

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

On the other hand, consider an unstable point. There is a bad torus which takes the element to 0 in the limit for $t\rightarrow 0$ . Every torus in $SL_{r}(\mathbb {C} )$ can be conjugated into the maximal torus of diagonal matrices. We may then suppose that this torus is given by $t\mapsto \operatorname {diag} (t^{\lambda _{1}},\ldots ,t^{\lambda _{r}})$ with $\lambda _{1}\leq \ldots \leq \lambda _{r}$ and $\lambda _{1}<0$ . We may write $t.\varphi =t^{\lambda _{1}}(\varphi ^{1},t^{\lambda _{2}-\lambda _{1}}\varphi ^{2},\ldots ,t^{\lambda _{r}-\lambda _{1}}\varphi ^{r})$ , where $(\varphi ^{1},\ldots ,\varphi ^{r})$ indicate the columns of the matrix representing $\varphi$ in the basis "chosen" by the torus. We see that, if we want the limit for $t\mapsto 0$ to be zero, we need that columns on which the 1PS acts with negative weight (at least $\varphi ^{1}$ ) are themselves zero. So $\varphi$ 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 $r$ -planes in $\mathbb {C} ^{n}$ , $\operatorname {Gr} (r,n)$ .

Exercise 7.3

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

$f=\sum _{i+j+k=3}a_{ijk}x^{i}y^{j}z^{k},$ where $[x:y:z]$ are the homogeneous coordinates in $\mathbb {P} ^{2}$ and $SL(3,\mathbb {C} )$ 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 $SL(3,\mathbb {C} )$ ; they are of the form

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

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

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

By Hilbert-Mumford criterion:

F is unstable if and only if

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

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

We can summarize that:

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