GIT Lecture 5

This lecture ( GIT5 ) was given on 9/4/09.


In the lecture, G is always a finite group.

A worked example of how to apply Molien’s formula is worked out for the case G is the group of symmetries of the cube including those that don’t preserve orientation (so it has order 48).  It can be embedded as a subgroup of SU(2) so in particular it acts on \mathbb{C}^2.  The result is (after some clever factoring) that

P_{R^G}(t) = \frac{1 + t^{18}}{(1-t^{12})(1-t^8)}

Compare this with the example right before the proof of Molien’s formula in the previous lecture.  It follows that R^G is a finitely generated module for a ring of the form \mathbb{C}[f_12, f_8] where the subscripts denote the degree of the element.  In particular the numerator requires the degree 18 peice of the module to be generated by a single element call it f_{18}.  The claim is that there is a relation

f_18^2 = af_{12}f_8^3 + bf^3_{12}

with a,b \ne 0.

Proposition: Let G act on an irred. affine var. X faithfully.  Set R = k[X] and K = Frac(R).

  1. R \supset R^G is an integral extension.
  2. K^G is the field of fractions of R^G
  3. K \supset K^G is a normal extension with Galois group G.

proof: take f\in R, then P_f(T) = \prod_{g \in G} (T - g(f)) \in R^G[T].  This shows 1.  For 2 let f/h \in K^G.  Applying the Renolds operator fixes it so

f/h = \frac{1}{|G|} \sum_{g \in G} g(f)/g(h) = \sum_{g' \in G} \frac{\prod_{g \in G - g'} g(h)g'(f)}{\prod_{g\in G}g(h)}

It can be checked that the far rhs is a ratio of two elements in R^G. Alternatively, let f/h as above.  By 1, h^n + a_1h^{n-1}.... + a_n = 0, so h(h^{n-1} + ... + a_{n-1}) = -a_0, now a_0 is clearly invariant. Now 

f/h = \frac{f(h^{n-1} + ... + a_{n-1})}{h(h^{n-1} + ... + a_{n-1})} = \frac{f(h^{n-1} + ... + a_{n-1})}{-a_0}

now the bottom is invariant so its not hard to show the top is invariant too; that is, f(h^{n-1} + ... + a_{n-1}) is invariant, separating each factor into its fixed and non fixed part it follows that f is invariant, from which it follows that h is invariant.  This has the advantage of not using the Renolds operator and so makes no assumption on the characteristic.


Finally, K is the splitting field of all polynomials P_{f/h}(T) \in K^G[T].QED.

Groups Generated by Reflections

Let V be a vector space over latex k = \bar k or characteristic 0. r \in End_k(V) is called a reflection if it is not the identity, has finite order, and \exists a hyperplane H_r that r fixes.  Let H_r be defined by the linear equations l_r \in V^*.  In some bases can write (for finite dimension n)

r = \left(\begin{array}{cc} \eta & 0 \\ 0 & I_{n-1} \end{array}\right)

Using coordinate x_1, ..., x_n then H_r = V(l_r) = V(x_1).  The next two lectures will be devoted to proving the following theorem:

Thm (reflection theorem): G \subset GL(V), say V.  Then k[V]^G is isomorphic to a polynomial algebra iff G is generated by reflections.

Rmk: This says X//G will be nonsingular.  Further by the above proposition K^G \subset K will be a normal and separable (by char K = 0) hence a finite extension, consequently X, X//G must have the same dimension (i.e. polynomial rings in the same number of variables).

Some preliminaries.  For R = k[V] and s \in G a reflection.  Recall from lecture two that G is a reductive group. Dualzing

s\colon V \to V

s|_{H_s} = id \colon H \to H_s

We get on ring maps

R \leftarrow R \colon s

R/l_s \xleftarrow{id} R/l_s \colon s

so writing f = lf' + f'' it follows that s(f) = s(lf') + f'' = lf'''+g'', so the map

D_s \colon R \to R

f \mapsto \frac{f - s(f)}{l_s}

is well defined and lowers the degree of f by at least 1.  It can also be checked D_s(fg) = fD_s(g) when f \in R^G.

R has a natural grading R = \oplus_i R_i so let I = R\cdot \oplus_{i>0} R_i^G.Here is an important lemma that is used more than once

Lemma: Let g_1,..., g_m, u_1, ..., u_m be homogeneous with g_i \in R^G.  Suppose

\sum_i g_iu_i = 0

and u_1 \not \in I.  Then g_1 \in R^Gg_2 + ... +R^Gg_m.

comments on proof: its done by induction on the degree of u_1.  When it has degree 0 you can divide by it and get the result directly.  The key is that when the degree is \ge 1 the D_s operator described above lowers the degree of an element by 1.  Assuming the result for \deg u_1 = k, and say it fails for some \deg u_1 = k+1, then applying any D_s gives the same hypothesis of the lemma except possibly that D_s(u_1) \not \in I.  But the result should hold given the degree of D_s(u_1), since it fails we must not satisfy the D_s(u_1) \not \in I hypothesis. The rest of the proof is as in the notes :).

Now begins the proof of the theorem.  let f_1, ..., f_r be a minimal set of generators for I ordered so \deg f_1 \le ... \le \deg f_r.  Step 1 is to show for each generator there is i such that df/dx_i \not \in I.  Assuming this fails there is an f_q all of whose partials are in I, but for degree reason df_q/dx_i can only be made from f_1, ..., f_{q-1}.  Now using the Euler identity you can generate f_q from the generators before it, contradicting minimality.

For the next part in the notes, choosing a relation h of minimal degree means in particular, that no t_i factors out of h, i.e. are working with something like x^2y + yz + z^3 vs. x\cdot (poly).  Playing around with this idea you can probably justify the all the partials of h(t_1, ..., t_s) will be lin. indep.


Plugging in gives 0 = h(f_1, ..., f_s), so all its partials are zero.  The notes should read ‘ choose i so that  df/dx_i \not \in I‘ because we want to apply the lemma above.  Note that the h_i are invariant because they are polynomials in the generators of the ring of invariants :).



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