## Group Schemes and Moduli (III)

Finally we’re ready to discuss what sorts of quotients exist when group schemes act on (some) other schemes.  Recall that for simplicity, every scheme/variety/morphism in sight is assumed to be over the spectrum of a fixed algebraically closed field, $k$.  For further simplicity, assume all schemes are of finite type.  In this post, we’ll begin the discussion with the simplest case, that is, the main results involving taking quotients of reductive group schemes acting on affine schemes.  This will be the main motivation for generalizing results to the non-affine case which we also begin to do here.  This theory is well developed and sometimes a bit technical though I’ve tried to avoid anything too difficult below and hopefully the examples illustrate some of what’s going on.  (The main reference is Mumford’s GIT, and proofs of unproven facts can be found there.)

We need one more definition before getting to the main results, the notion of a slightly weaker notion than a geometric quotient, namely:

Definition:  Suppose a group scheme $G$ acts on a scheme $X$.  (Recall this means that there is a morphism $s: G \times X \rightarrow X$ such that ….). A categorical quotient for this setup will be a morphism $f: X \rightarrow Y$ such that

i) The two maps $f \circ p_2, f \circ s : G \times X \rightarrow Y$ are equal.

ii)  And $Y$ is the “smallest” scheme with this property.  In other words, if $g: X \rightarrow Z$ satisfies the same condition as in i), then there is a unique map $h: Y \rightarrow Z$ such that $h \circ f = g$.

A formal exercise shows that geometric quotients (see GSM (I)) are already categorical ones.  The converse is certainly not true, the main difference in the definitions being that orbits need not necessarily be separated in the categorical quotient case.  We will see examples of this below.

The main theorem in the affine case have already been stated and proved by Charles in his talks on Invariants.  I’ll restate it here but certainly won’t reprove it.

Theorem:  If $X = Spec R$ is an affine scheme and $G$ is a reductive group acting on $X$, then a categorical quotient $Y$ exists and is equal to $Spec R^G$, the spectrum of the ring of invariants.  In fact, $Y$ inherits many properties from $X$ (algebraic, noetherian, connected, irreducible, …).  The scheme $Y$ is a geometric quotient if and only if the action is closed (that is to say, each orbit is).

As an example, recall one of the examples from the GSM(I), the action of $\mathbb{G}_m$ on $\mathbb{A}^2$ given by $(x,y) \rightarrow (tx, t^{-1}y)$.  Here there were three types of orbits, the origin, the hyperbolas $xy = c$ and the two axes minus the origin.  Notice that the ring of invariants is isomorphic to $k[xy]$, and the quotient map is given by $\mathbb{A}^2 \rightarrow \mathbb{A}^1$ which sends $(x,y) \rightarrow xy$.  Our theorem asserts this is a categorical quotient and it is clearly not a geometric quotient, the origin and the two axes are collapsed to the origin in the target.

For another example, consider the action of $G = GL(n)$ on itself by the adjoint action, that is $g \cdot x \rightarrow g \cdot x \cdot g^{-1}$.  For any matrix $g$, we can compute the characteristic polynomial $p_g(t) = (-t)^n + c_1(g)(-t)^{n-1} + \ldots + c_n(g)$.  Using this, we can define a $G$ equivariant map (in fact an actual morphism) $c: G \rightarrow \mathbb{A}^n$ which sends $g \rightarrow (c_1(g), \ldots, c_n(g))$.  This is actually a categorical quotient.  To verify this, we need to verify that $\mathcal{O}(G)^G = k[c_1, \ldots, c_n]$.  This is not too hard, I’ll outline the proof.  1) Since any matrix is equivalent to one in Jordan normal form, an invariant function is determined by its values on Jordan matrices.  2)  The set of matrices with diagonal normal form is Zariski dense in the set of all invertible matrices, so an invariant function is determined by its values on diagonal matrices.  3)  Let $S = k[x_1, \ldots, x_n]$ be the ring of functions on the diagonal matrices.  Since diagonal matrices with permuted elements are equivalent under the action of the group, invariant functions must be symmetric in the x’s.  By the Fundamental Theorem of Symmetric Functions, these are polynomials in the elementary symmetric functions $\sigma_i$ in the x’s.  These functions though agree with the functions $c_i$ when restricted to the diagonal matrices.

However, we see in the above example that the quotient is not a geometric one, as there are matrices with different Jordan normal forms that have the same characteristic polynomial.

In fact, there are examples of non-projective algebraic varieties $X$ acted on by finite groups $G$ where the categorical quotient does not even exist in the category of algebraic varieties.  As usual at this point, we’ll simply say the constructions are a little involved (the interested can look at an example Hironaka which is explained in GIT as well as in the Appendix of Hartshorne’s Algebraic Geometry).

The search for geometric quotients continues, but we see from the above examples that even in some simple cases we’ll be out of luck.  Suppose now that $G$ acts on $X$ (maybe no longer affine).  The idea will be to find and characterize open sets $U \subset X$ where geometric quotients do exist, but of course, we get to wade through a little more theory first!

To motivate the theory, we return to the affine case and consider the following Lemma:

Lemma:  Suppose that $X = Spec R$ is acted on by an affine algebraic group $G = Spec W$.  Then there exists an equivariant embedding of $X$ into $\mathbb{A}^n$ such that $G$ acts on $X$ through a linear representation.

Sketch of Proof:  Suppose that $f_1, \ldots, f_n$ generate the k algebra $R$.  The Lemma follows by looking at the collection of $G$-translates of the $f_i$‘s and taking the linear space they span.  One need only remember that if the action is given (dually) by $\mu : R \rightarrow R \otimes W$, if we write $\mu (f_i) = \Sigma r_{i,j} \otimes w_{i,j}$ then the $G$ invariant subspace spanned by the $G$-translates of $f_i$ is finite dimensional and is contained in the span of the finitely many $r_{i,j}$‘s.

We attempt to follow a similar path when $X$ is quasi-projective.  Recall that a map to projective space $X \rightarrow \mathbb{P}^n$ is given by a line bundle $L$ and a collection of n + 1 global sections of $L$.   We define a lifting of an action of $G$ on $X$ to an action of $G$ on $L$:

Defintion:  Suppose $s: G \times X \rightarrow X$  is an action of an algebraic group scheme on $X$ and $L$ is an invertible sheaf on $X$.  We abuse notation and identify $L$ with its total space, that is, the line bundle $\mathbb{V}(L)$ with its natural projection $\pi: L \rightarrow X$.  Then a $G$ linearization of $L$ is an action $s': G \times L \rightarrow L$ such $\pi \circ s' = s \circ (id \times \pi)$ and such that the zero section of $\pi$ is $G$-equivariant.

Alternatively, one can rephrase this definition to say that there exists an isomorphism of invertible sheaves $s^* L \rightarrow pr_2^*L$ which “satisfy the cocycle condition”.  Since we won’t use this formulation, we’ll leave it to the reader to work this out (or see Mumford’s GIT 1.3).

So as not to go too far out of the way, we’ll state some important facts about G-linearizations:

Fact 1:  Suppose that $G$ is a connected affine algebraic group, and $X$ is a normal algebraic variety, then there is some positive integer n such that $L^n$ admits a $G$ linearization.

Fact 2:  Suppose that $G$ is an irreducible affine algebraic group acting on a quasi projective $X$.  Then there exists a $G$-equivariant embedding $X \rightarrow \mathbb{P}^n$ where $G$ acts on $\mathbb{P}^n$ through a linear representation $G \rightarrow GL_(n+1)$.

Example:

Suppose that $G = PGL(n+1)$ acts on $X = \mathbb{P}^n$ in the natural way (call the action $\sigma$).  The group $G$ sits inside $\mathbb{P}^N$ where $N = n^2 + 2n$ as the complement of the hypersurface $D$ given by the vanishing of the determinant in the variables $x_{ij}$.   The action is the restriction of the rational map $s: \mathbb{P}^N \times \mathbb{P}^n \rightarrow \mathbb{P}^n$ given by matrix multiplication.  This map is not defined at any point $(A,v)$ such that $A \cdot v = 0$.  The restriction of the first projection of $\mathbb{P}^N \times \mathbb{P}^n$ to this set of points, call them $Z$, maps onto $D$.  Since $Z$ has codimension 2 or more in $\mathbb{P}^N \times \mathbb{P}^n$, it must be the case that $\sigma^*\mathcal{O}(1)$ is the restriction of a line bundle on $\mathbb{P}^N \times \mathbb{P}^n$ (exercise).  The formula for the action though, shows then that this bundle must be $pr_1^*(\mathcal{O}(1)) \otimes pr_2^*(\mathcal{O}(1))$.  If the bundle $\mathcal{O}_{\mathbb{P}^n}(1)$ admits a linearization, then $\sigma^*\mathcal{O}(1) \cong pr_2^*\mathcal{O}_{\mathbb{P}^n}(1)$ and so  $pr_1^*\mathcal{O}(1)$ would be trivial.  This can’t be the case though because $\sigma^*\mathcal{O}(1)$ and $\mathcal{O}_{\mathbb{P}^N}(1)$ restrict to the same bundle over $\mathbb{P}^N / D$.  The bundle they restrict to though is a generater of $Pic(\mathbb{P}^N / D)$ which is isomorphic to $\mathbb{Z}/(n+1)\mathbb{Z}$ (exercise).  Thus we see that $\mathcal{O}_{\mathbb{P}^n}(1)$ does not admite a linearizaiton for this action.  The same argument though shows that $\mathcal{O}_{\mathbb{P}^n}(n+1)$ does admit a linearization!

When $X$ admits an action by a reductive group $G$, the idea will be to cover $X$ by affine invariant open sets $U_i$, take the quotient $U_i \rightarrow V_i$ and glue together the results.  Of course, such a naive cover won’t always exist!  Instead, we’ll find such a cover of an open set of $X$, and this really is the best we can hope for.  The construction of that open subset will depend upon a parameter, namely the choice of a $G$-linearizized line bundle $L$.  But that will come in the next installment!