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Home , Invariant theory

2 RICHARD BORCHERDS

1. Lecture 1 theory All rings are commutative and have an identity 1 unless otherwise stated. The text is Eisenbud [1]. The fundamental examples of commutative rings are:

• Rings of algebraic integers, such as Z or the Gaussian integers Z[i]. A typical result is that Z[i] is a unique factorization domain. More generally in algebraic number theory we want to know how far a of algebraic integers is from a unique factorization domain: this is measured by the ideal class , a special case of the Picard group of a commutative ring. • Coordinate rings of algebraic varieties For example, the ring k[x, y, z] or the coordinate ring k[x, y]/(y2 − x3 + x) of the affine elliptic curve y2 = x3 − x. Is it a unique factorization domain? The coordinate ring captures the affine elliptic curve completely: for example, the points of the elliptic curve can be recovered as homomorphisms to k. More generally commutative rings correspond to affine schemes. • Rings of invariants. For example, the group of rotations of an icosahedron acts on 3-dimensional space and therefore on the R[x, y, z]. What is the ring of fixed by this action? It obviously contains R[x2 + y2 + z2], but also contains some less obvious polynomials. In fact it is a polynomial ring on generators of degrees 2, 6, 10. The fundamental problem of invariant theory is the following: given a group acting on a , is the ring of invariants finitely generated? This was solved by Hilbert in the case when the group is a reductive group in characteristic 0. We will give his solution (cleaned up a little) Example. Distance in Euclidean space is invariant under rotations and reflec- tions. We can rephrase this as saying the polynomial x2 +y2 +z2 is invariant under the orthogonal group. Moreover any polynomial invariant under this group is a polynomial in x2 + y2 + z2. n Example. Suppose G is the Sn acting on C by permuting the coordinates. Then an invariant is a symmetric polynomial. The ring of symmetric polynomials is the ring of polynomials in the elementary symmetric polynomials Q Σixi,Σi

i n−i is not a polynomial ring. It is generated by the elements zi = x y . There are many relations between these generators: zizj = zkzl whenever i + j = k + l. These relations are called SYZYGIES. Example. Suppose G is the orthogonal group of V = Rn. Then it acts on V m, and has invariants of the form (xi, xj) where xi ∈ V is the i component of an m element of V . A syszygy for m = n + 1 is given by det((xi, xj)) = 0, as if m > n then the m vectors xi are linearly dependent. The second fundamental problem of invariant theory asks whether all syzygies are generated by a finite number. More generally there are second-order syzygies giving relations between first-order syzygies, third order syzygies, and so on, and one can ask if the syzygies of given order are finitely generated. Finally one can ask if there is a bound to the orders of syzygies. These questions were all answered by Hilbert. Example. Binary quantics. Here the group is SL2(C) acting on the n + 1- n 0 0 n dimensional space of coefficients of the binary quantic anx y + ··· a0x y . (This is the nth symmetric power of the 2-dimensional representation of SL2(C).) One invariant is the discriminant of the form, which vanishes exactly when the form has 2 2 2 two equal roots. For example, the discriminant of a2x + a1xy + a0y is a1 − 4a0a2. In one of the triumphs of invariant theory before Hilbert, was able to prove that the ring of invariants of binary quantics are finitely generated. Example. Ternary cubics. These are degree 3 polynomials in 3 variables, so they have 10 coefficients. The ring of invariants in generated by 2 polynomials of degrees 4 and 6, that take a couple of pages to write out explicitly. Ternary cubics give elliptic curves, so these invariants are closely related to the problem of finding the of elliptic curves. (In fact invariants of ternary cubics correspond to level 1 modular forms, and these two invariants correspond to the 2 Eisenstein series E4 and E6.) Invariants can be very complicated to write out explicitly in all but the simplest cases. Salmon’s book given an explicit example of an invariant taking 13 pages to write out explicitly. (Reading 19th century invariant theory papers before Hilbert sometimes gives one the impression that they were competing with each other to produce the most horrendously complicated formulas. They were very good at this: more than one mathematician computing invariants with a long computer calculation has been embarrassed to discover that their results were already known in the 19th century.) It might seem that showing finite generation of syzygies is much harder that showing finite generation of invariants: after all, one first has to find the invariants before one can find the syzygies. Hilbert’s first discovery was that if one can show the invariants are finitely generated, then it automatically follows that the syzygies are finitely generated. The key point is that the invariants form a ring, while the syzygies form an ideal: we have (polynomial ring) → (invariants) where the polynomial ring is generated by our basic invariants, and the syzygies are the kernel of this map. Hilbert’s theorem states: Theorem 1.1. Any ideal of a polynomial ring in a finite number of variables is finitely generated. Note the difference between being finitely generated as an ideal or as a ring (without 1): the ideal (x) of k[x, y] is finitely generated as an ideal, but not as a ring without 1: a set of ring generators is x, xy, xy2, .... As this example suggests, 4 RICHARD BORCHERDS it is often much easier to show something is finitely generated as an ideal than to show something is finitely generated as a ring. A ring such that all ideals are finitely generated is called Noetherian, after . Example. The ring of polynomials k[x1, x2,...] in infinitely many variables is not Noetherian, as the ideal (x1, x2,...) is not finitely generated. Addendum to lecture 1: how does a group acting on a vector space act on polynomials? Theoretical answer: Suppose G acts on sets X and Y . Then it acts on functions f from X to Y by gf(gx) = g(f(x)). In other words any formula remains true when acted on by g ∈ G. 2 Practical answer: suppose G is SL2(C) acting on C with coords x and y. How n n−1 does it act on binary quantics anx + an−1x y + ··· ? Answer: Change x to ax + by, y to cx + dy. We get n n−1 (1.1) an(ax + by) + an−1(ax + by) (cx + dy)

n n−1 n n−1 (1.2) = (ana + an−1na + ··· )x + (something even worse)x y + ··· which will in principle give some explicit but rather horrible matrix describing the action of SL2 on the n + 1 dimensional vector space spanned by an, an−1,....