# Solutions to exercises

**Exercise 7.1**

Let act on in the usual way. Define a lift of the action to the trivial line bundle as the action composed with . Use the Hilbert-Mumford criteria to find the stable, semistable, and unstable points.

*Proof*

Embed in using . This action extends as the action given by , with the obvious linearisation. To check the stability of the point , because of the Hilbert-Mumford criteria, we only need to consider the weight of this action, and of the inverse 1-PS, as on the line bundle.

Case 1: Suppose . Then is a fixed point of the action. On the line over , acts as multiplication by , so the weight of the action is . So this is an unstable point.

Case 2: Suppose . This is again a fixed point of the action. The weight of the (linearisation) of the action as is , as on the line over this point, . However, the weight of the linearisation action of the inverse subgroup (given by ) is , so this point is unstable.

Case 3: (everything else) The limit as of is . As before, on the line over this point, , so the weight is . We also need to the check the inverse 1-PS, given by , so consider the limit as of . This is . On the line over this point, the action is , so it has weight . Therefore is a stable point.

**Exercise 7.2**

Consider the action of on , . Show that semistable points correspond to injective maps. What is the quotient?

*Proof*

We may represent an element of as an matrix. acts by multiplication on the right; therefore it lets us perform column operations.

If an element has , then we can choose a different basis of in such a way that the matrix representing has a zero column (say the first one). This point is then unstable, since we can consider the 1PS of given (in this basis) by and .

On the other hand, consider an unstable point. There is a bad torus which takes the element to 0 in the limit for . Every torus in can be conjugated into the maximal torus of diagonal matrices. We may then suppose that this torus is given by with and . We may write , where indicate the columns of the matrix representing in the basis "chosen" by the torus. We see that, if we want the limit for to be zero, we need that columns on which the 1PS acts with negative weight (at least ) are themselves zero. So 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 -planes in , .

**Exercise 7.3**

We consider the natural action of on . Every cubic curve in is given by a single degree 3 homogeneous form

*Proof*

In order to apply Hilbert-Mumford criterion, we consider the 1-parameter subgroups of ; they are of the form

We consider as above and we fix the point . We recall that

- p is singular (i.e. there are not terms in and ;
- p is a triple point ;
- p is a double point with a unique tangent .

By Hilbert-Mumford criterion:

F is unstable if and only if

finally, choosing and , we can check that ; so the converse implication also holds.

finally, choosing and , we can check that ; so the converse implication also holds.

We can summarize that:

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