## Group Schemes and Moduli (IV)

Unless there is some specific interest, I think this will be the final post in the series about taking quotients of varieties by actions of group schemes.  Recall that everything is being done over an algebraically closed field, $Spec(k)$.  I’ll discuss what happens in the projective case in this post.  To cover such a topic completely is of course impossible, but hopefully this post will illlustrate some of the major ideas and results.   There are also some more or less completely worked through, hands on, examples to show this subject is still rooted in polynomial computations!

In the following assume that $G$ is a reductive algebraic group scheme acting on an algebraic variety $X$.  Unless stated otherwise, a point will always mean a closed point.  If $L$ is a $G$ linearized line bundle on $X$, then we make the following definition:

Definition:  i) A point $x \in X$ is called semistable (with respect to $L$) if there exists a positive number m and $s \in (L^m(X))^G$ (that is, an invariant global section of the m-th tensor power) such that $X_s = \{ y \in X | s(y) \neq 0 \}$ is affine and contains $x$.

ii) A point $x \in X$ is called stable, if the condition in i) is satisfied, and in addition all orbits in $X_s$ are closed.

iii)  If a point $x \in X$ does not satisfy i), we say it is unstable.

(for those following along in GIT, Mumford also defines other stability notions, but is these which we will focus on).

Let me try to make some sense of this defintion.  Assume that $L$ is very ample and so embeds $X$ equivariantly into projective space $\mathbb{P}(V)$.  Then $L^m(X) = Pol_m(V) / I_m$ where $I_m$ is the subspace of degree m polynomials vanishing on $X$.  At the level of invariants then, we have $(L^m(X))^G = (Pol_m(V) / I_m)^G$.  An $s \in (L^m(X))^G$ can be represented by a polynomial $F$ which is $G$ invariant modulo $I_m$.  This says exactly that $F$ will be constant on orbits of the action.  For a point $x \in X$, we can choose $v \in V$ representing it.   We’ll have that $s(x) \neq 0$ if and only if $F(v) \neq 0$.  So the unstable points can be identified with the set $\{v \in V | F(v) = 0, \forall F \in \bigoplus_{m > 0} Pol_m(V)^G \}$.  For an exercise using this description: show that the unstable points are exactly the ones represented by $v \in V$ such that the origin is in the closure of the orbit $O(v)$.

Denote by $X^{ss}(L)$ and $X^s(L)$ the collection of all semistable (respectively stable) points of $X$.  We are ready to state today’s main theorem:

Theorem:  There is a categorical quotient $f: X^{ss}(L) \rightarrow Y$ such that $f$ is affine and $Y$ is quasi projective.  Furthermore, there is an open set $Y'$ of $Y$ such that $f^{-1}(Y') = X^s(L)$ and the restricted map $f: X^s(L) \rightarrow Y'$ is a geometric quotient.

Remark on Proof:  The proof is actually not so difficult.  To show the existence of $Y$, simply remark that categorical quotients of the various (affine!) $X_s$ exist and can be glued together (by uniqueness of categorical quotients) to give a categorical quotient of $X^{ss}(L)$.  It is only slightly more difficult to show that the result is quasi-projective.  To construct an ample line bundle $M$ on $Y$, first take a trivialization of $L$ on each peice $X_s$.  On the overlap $X_s \cap X_t$, note that the function $s / t$ is $G$ invariant and so by construction is a function on $Y$.  Not going into the details, these transition functions on $Y$ define an ample line bundle.  For the complete proof see Mumford’s GIT.

In fact, when $X$ is projective and $L$ is ample on $X$, then $Y = X^{ss}(L) / G$ is isomorphic to $Proj(R^G)$ where $R = \bigoplus_{m \geq 0} (L^m(X))^G$.  In particular, the quotient is projective.

Let’s work through some examples.

1.  Let $G = \mathbb{G}_m$ act on projective space $X = \mathbb{P}^n$ by the formula $t \cdot [x_0, \ldots, x_n] = [t^{q_0} x_0, \ldots, t^{q_n} x_n]$ where the $q_0 \leq q_1 \leq \ldots \leq q_n$ are integers.  I should have mentioned last time (when discussing the existence of $G$ linearizations) that there is an exact sequence $0 \rightarrow Ker \rightarrow Pic^G(X) \rightarrow Pic(X) \rightarrow Pic(G)$ where $Pic^G(X)$ denotes the $G$ linearzed line bundles.  This is actually a fairly easy exercise.  Using this sequence then, the kernel becomes identified with the characters $Hom(G, \mathbb{G}_m)$ and in our example this is $\mathbb{Z}$ so that $Pic^G(X) = \mathbb{Z}^2$.  This can be seen directly in the following way:

Any $G$ linearized line bundle must look like $\mathcal{O}(m)$ for some number m.   This defined a $G$ equivariant embedding of $X$ into projective space $\mathbb{P}^N$ (for the appropriate N) by the Veronese embedding with coordinates $x_{i_1\ldots i_m}$ where $i_1\leq \ldots \leq i_m$.  The $G$ action on $\mathbb{P}^N$ has to be $t \cdot x_{i_1\ldots i_m} = t^{q_0 + \ldots + q_n} x_{i_1\ldots i_m}$.  Then a linearization is given by a representation of $G$ in $\Gamma(X, \mathcal{O}(m))$ lifting the action on projective space.  It is defined by the formula $x_{i_1\ldots i_m} \rightarrow t^{-a}t^{q_0 + \ldots + q_n} x_{i_1\ldots i_m}$ for some integer $a$.  The pair $(m,a)$ indexes the $G$ linearized line bundles which we denote $L_{m,a}$.  Raising the bundle to the r-th power corresponds to replacing $(m,a)$ with $(rm, ra)$.  Replacing $L$ by a tensor power of itself does not change the semi-stable locus (exercise!) so we may take $L = L_{1,p/q}$ where by definition, $H^0(X, L^N)^G$ only makes sense if q divides N.  Invariance then means that for any non-zero values of t, we have $F(t^{-a + q_0}x_0, \ldots, t^{-a+q_n}x_n) = F(x_0, \ldots, x_n)$.   Assume now that $q_0 \leq 0$.  If $a \leq q_0$ or $a \geq q_n$, we have from the above formula that $H^0(X, L^n)^G = 0$ for all possible values of N.  This shows that the semi-stable locus is empty for $a$ in this range.  When $a = q_0 = \ldots = q_k < q_{k + 1}$, then we have that $\bigoplus_N H^0(X, L_{1,a}^N)^G = k[T_0, \ldots, T_k]$.  Then the semistable locus is exactly $X - \{x_0 = \ldots x_k = 0\}$ and the quotient is $\mathbb{P}^k$.  As $a$ increases, for example, $q_k < a \leq q_{k+1}$, there are more invariant polynomials and the categorical quotient changes.  For example, if $a = r/s$ then the polynomial $T_0^{sq_{k+1} - r} \cdot T_{k+1}^{sq_0 - r}$ is invariant.  One can show in general that if $a$ stays between two q’s then the quotient does not change, but otherwise certainly does.

2.  In this example, $X = \mathbb{A}^n$ and again $G = \mathbb{G}_m$.  Suppose the action is given by $t \cdot (x_1, \ldots, x_n) = (t^{q_1}x_1, \ldots, t^{q_n}x_n)$.  On affine space, all line bundles $L$ are trivial, so we may write (for the associated space) $L = X \times \mathbb{A}^1$.  Let $L_a$ be the $G$-linearization defined by $t \cdot (x, v) = (t \cdot x, t^av)$.  A section of this bundle $s: X \rightarrow L_a$ may be written $s(x) = (x, F(x))$.  The group $G$ acts on these global sections via $t \cdot s(x) = (x, t^a F(t^{-1} x))$.  So a section defined by the polynomial $F$ is invariant if and only if $F(t\cdot x) = t^a F(x)$.  (Similarly for global sections of $L_a^n$).  When $a = 0$, only the constant polynomials are invariant, so the semistable locus is all of $X$ but the quotient is $Spec( k[x_1, \ldots, x_n]^G = Spec k$.  That is, the quotient is a single point.  However, if $a = 1$, then the semistable locus is $D(x_1) \cup \ldots \cup D(x_n) = \mathbb{A}^n - {0}$.   This can be seen because the invariant functions are $\bigoplus_{m > 0} L_1^m(X) = k[x_1, \ldots, x_n]_{> 0}$.  And then $X^{ss} / \mathbb{G}_m = D_+(x_1) \cup \ldots \cup D_+(x_n)$.  Recall that $D_+(f) = Spec k[x_1, \ldots, x_n]_{(f)}$.  If $q_1 = \ldots = q_n = 1$, this gives the standard projective space $\mathbb{P}^{n-1}$.  However, if the q’s are not all one, but say, still positive, the result is called weighted projective space $\mathbb{P}(q_1, \ldots, q_n)$.  Of course, unlike projective space, it need not be smooth.  For example suppose $\mathbb{G}_m$ acts on $\mathbb{A}^3$ via $t \cdot (u,v,w) = (t \cdot u, t \cdot v, t^2 \cdot w)$.  The quotient on the patch $D(w)$ is given by $D_+(w) = Spec k[u^2/w^2, v^2/w^2, uv/w^2]$.  We’ll leave it as an easy exercise to see that this is not smooth.

3.  Let’s work out one more special case where the weights of the $G = \mathbb{G}_m$ action are not all positive.  Suppose that $G$ acts on $\mathbb{A}^4$ via $t \cdot (u,v,w,x) = (tu, tv, t^{-1}w, t^{-1}x)$.  When (with notation as in example 2) $a = 0$, as above all points of $X$ are semi-stable.  It is not difficult to see that the ring of invariants is $k[uw, ux, vw, vx]$ which is isomorphic to $k[t_1, t_2, t_3, t_4] / (t_1t_4 - t_2t_3)$.  It is a quadric cone in $\mathbb{A}^4$.  Even more interestingly, take $a = 1$.  We’ll leave it as an exercise to show that the semistable locus is $\mathbb{A}^4 - V(u, v)$.  This set is covered by the opens $U_1 = D(u)$ and $U_2 = D(v)$.  Then $\mathcal(O)(U_1)^G = k[u,v,w,x]_{(u)} = k[u,v,w,x]_0[v/u]$ (similarly for the other open).  I claim that the quotient of the semi-stable locus is a closed subvariety in $\mathbb{A}^4 \times \mathbb{P}^1$ given by $BT_1 - AT_3 = 0, BT_2 - AT_4 = 0, T_1T_4 - T_2T_3 = 0$ where $(A,B)$ are homogeneous coordinates on $\mathbb{P}^1$.  The last set of claims which is not difficult to verify are as follows: 1. The stable locus actually agrees with the semistable locus (as it did in the above examples!) so the quotient is a geometric one.  2.  In this case it is smooth.  3.  There is a map from the quotient $Y_1$ when $a = 1$ to the quotient $Y_0$ when $a = 0$.  4.  In fact, that morphism is a blowing up of $Y_0$ at the conical point, that is, the origin.

There is more to the theory of taking and identifying quotients.  In the above examples, identifying the semi-stable and stable locus was not so difficult, but as the examples become more complicated, this process becomes a bit harder.  There are tools to do this (namely the numerical criterion), and it is fun and still hands on.  Also as the above examples illustrated, varying the linearization changes the (semi) stable locus and so also the quotient.  Is there a systematic description of how they are related, if indeed they are?  The third example is supposed to be an illustration of this phenomenon.  Also one can leave the category of schemes to take “finer” quotients, but that requires more foundational work.  Unless there is overwhelming interest though, perhaps I’ll stop here in this subject for now.

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### 3 Responses to Group Schemes and Moduli (IV)

1. Jason Starr says:

How does the semistable locus change under blowing up? If you start with a projective X for which the semistable locus is dense, then the rational map X – – > X//G can be “regularized” by blowing up X. Is there a way to see what to blow up by investigating the unstable locus?

2. Matt DeLand says:

That’s an interesting question — one I don’t know the answer to. I’ll think about it though.

I do know that in many cases, two quotients X//G (with respect to different linearizations), are birational and related by (often calculatable) flips. (This is work due to Michael Thaddeus: J. American Math Soc. 1996 no 3).

3. Jason Starr says:

I had one naive idea, but now I realize it is wrong. I thought that perhaps, after replacing X by a blowing up and replacing the original invertible sheaf by some new invertible sheaf, one could arrange for the semistable locus to equal all of X. But since the morphism X_{ss} –> X//G is affine, if the fiber dimension is positive, X_{ss} can never be projective (the fibers are closed subsets of X_{ss} which are affine).