§ What is a syzygy?

Word comes from greek word for "yoke" . If we have two oxen pulling, we yoke them together to make it easier for them to pull.

§ The ring of invariants

Rotations of R3\mathbb R^3: We have a group SO(3)SO(3) which is acting on a vector space R3\mathbb R^3. This preserves the length, so it preserves the polynomial x2+y2+z2x^2 + y^2 + z^2. This polynomial x2+y2+z2x^2 + y^2 + z^2 is said to be the invariant polynomial of the group SO(3)SO(3) acting on the vector space R3\mathbb R^3.
  • But what does x,y,zx, y, z even mean? well, they are linear function x,y,z:R3Rx, y, z: \mathbb R^3 \rightarrow \mathbb R.So x2+y2+z2x^2 + y^2 + z^2 is a "polynomial" of these linear functions.

§ How does a group act on polynomials?

  • If GG acts on VV, how does GG act on the polynomial functions VRV \rightarrow \mathbb R?
  • In general, if we have a function f:XYf: X \rightarrow Y where gg acts on XX and YY(in our case, GG acts trivially on Y=RY=\mathbb R), what is g(f)g(f)?
  • We define (gf)(x)g(f(g1(x)))(gf)(x) \equiv g (f(g^{-1}(x))).
  • What is the g1g^{-1}? We should write(gf)(gx)=g(fx)(gf)(gx) = g(fx). This is like g(ab)=g(a)g(b)g(ab) = g(a) g(b). We want to get(gf)(x)=(gf)(g(g1x)=g(f(g1x))(gf)(x) = (gf)(g(g^{-1}x) = g(f(g^{-1}x)).
  • If we miss out g1g^{-1} we get a mess. Let's temporarily define (gf)(x)=f(g(x))(gf)(x) = f(g(x)). Consider(gh)f(x)=f(ghx)(gh)f(x) = f(ghx). But wecan also take this as (gh)(f(x))=g((hf)(x))=(hf)(gx)=f(hgx)(gh)(f(x)) = g((hf)(x)) = (hf)(gx) = f(hgx). This isabsurd as it gives f(ghx)=f(hgx)f(ghx) = f(hgx).

§ Determinants

We have SLn(k)SL_n(k) acting on knk^n, it acts transitively, so there's no interesting non-constant invariants. On the other hand, we can have SLn(k)SL_n(k) act on i=1nkn\oplus_{i=1}^n k^n. So if n=2n=2 we have:
[abcd] \begin{bmatrix} a & b \\ c & d \end{bmatrix}
acting on:
[x1y1x2y2] \begin{bmatrix} x_1 & y_1 \\ x_2 & y_2 \end{bmatrix}
This action preserves the polynomial x1y2x2y1x_1 y_2 - x_2 y_1, aka the determinant. anything that ends with an "-ant" tends to be an "invari-ant" (resultant, discriminant)

§ SnS_n acting on Cn\mathbb C^n by permuting coordinates.

Polynomials are functions C[x1,,xn]\mathbb C[x_1, \dots, x_n]. Symmetric group acts on polynomials by permuting x1,,xnx_1, \dots, x_n. What are the invariant polynomials?
  • e1x1+x2+xne_1 \equiv x_1 + x_2 + \dots x_n
  • e2x1x2+x1x3++xn1xne_2 \equiv x_1 x_2 + x_1 x_3 + \dots + x_{n-1} x_n.
  • enx1x2xne_n \equiv x_1 x_2 \dots x_n.
These are the famous elementary symmetric functions. If we think of (yx1)(yx2)(yxn)=yne1yn1+en(y - x_1) (y - x_2) \dots (y - x_n) = y^n - e_1 y^{n-1} + \dots e_n.
  • The basic theory of symmetric functions says that every invariant polynomialin x1,xnx_1, \dots x_n is a polynomial in e1,,ene_1, \dots, e_n.

§ Proof of elementary theorem

Define an ordering on the monomials; order by lex order. Define x1m1x2m2>x1n1x2n2x_1^{m_1} x_2^{m_2} > x_1^{n_1} x_2^{n_2} \dots iff either m1>n1m_1 > n_1 or m1=n1m2>n2m_1 = n_1 \land m_2 > n_2 or m1=n1m2=n2m3>n3m_1 = n_1 \land m_2 = n_2 \land m_3 > n_3 and so on. Suppose fC[x1,,xn]f \in \mathbb C[x_1, \dots, x_n] is invariant. Look at the biggest monomial in ff. Suppose it is x1n1x2n2x_1^{n_1} x_2^{n_2} \dots. We subtract:
P(x1+x2)n1n2×(x1x2+x1x2)n2n3×(x1x2x3+x1x2x4)n3n4 \begin{aligned} P \equiv &(x_1 + x_2 \dots)^{n_1 - n_2} \\ &\times (x_1 x_2 + x_1 x_2 \dots)^{n_2 - n_3} \\ &\times (x_1 x_2 x_3 + x_1 x_2 x_4 \dots)^{n_3 - n_4} \\ \end{aligned}
This kills of the biggest monomial in ff. If ff is symmetric, Then we can order the term we choose such that n1n2n3n_1 \geq n_2 \geq n_3 \dots. We need this to keep the terms (n1n2),(n2n3),(n_1 - n_2), (n_2 - n_3), \dots to be positive. So we have now killed off the largest term of ff. Keep doing this to kill of ff completely. This means that the invariants of SnS_n acting on Cn\mathbb C^n are a finitely generated algebra over C\mathbb C. So we have a finite number of generating invariants such that every invariant can be written as a polynomial of the generating invariants with coefficients in C\mathbb C. This is the first non-trivial example of invariants being finitely generated. The algebra of invariants is a polynomial ring over e1,,ene_1, \dots, e_n. This means that there are no non-trivial-relations between e1,e2,,ene_1, e_2, \dots, e_n. This is unusual; usually the ring of generators will be complicated. This simiplicity tends to happen if GG is a reflection group. We haven't seen what a syzygy is yet; We'll come to that.

§ Complicated ring of invariants

Let AnA_n (even permutations). Consider the polynomial Δi<j(xixj)\Delta \equiv \prod_{i < j} (x_i - x_j) This is called as the discriminant. This looks like (x1x2)(x_1 - x_2), (x1x2)(x1x3)(x2x3)(x_1 - x_2)(x_1 - x_3)(x_2 - x_3), etc. When SnS_n acts on Δ\Delta, it either keeps the sign the same or changes the sign. AnA_n is the subgroup of SnS_n that keeps the sign fixed. What are the invariants of AnA_n? It's going to be all the invariants of SnS_n, e1,,ene_1, \dots, e_n, plus Δ\Delta (because we defined AnA_n to stabilize Δ\Delta). There are no relations between e1,,ene_1, \dots, e_n. But there are relations between Δ2\Delta^2 and e1,,ene_1, \dots, e_n because Δ2\Delta^2 is a symmetric polynomial. Working this out for n=2n=2,we get Δ2=(x1x2)2=(x1+x2)24x1x2=e124e1e2\Delta^2 = (x_1 - x_2)^2 = (x_1 + x_2)^2 - 4 x_1 x_2 = e_1^2 - 4 e_1 e_2. When nn gets larger, we can still express Δ2\Delta^2 in terms of the symmetric polynomials, but it's frightfully complicated. This phenomenon is an example of a Syzygy. For AnA_n, the ring of invariants is finitely generated by (e1,,en,Δ)(e_1, \dots, e_n, \Delta). There is a non-trivial relation where Δ2poly(e1,,en)=0\Delta^2 - poly(e_1, \dots, e_n) = 0. So this ring is not a polynomial ring. This is a first-order Syzygy. Things can get more complicated!

§ Second order Syzygy

Take Z/3ZZ/3Z act on C2\mathbb C^2. Let ss be the generator of Z/3ZZ/3Z. We define the action as s(x,y)=(ωx,ωy)s(x, y) = (\omega x, \omega y) where ω\omega is the cube root of unity. We have xaybx^ay^b is invariant if (a+b)(a + b) is divisible by 33, since we will just get ω3=1\omega^3 = 1. So the ring is generated by the monomials (z0,z1,z2,z3)(x3,x2y,xy2,y3)(z_0, z_1, z_2, z_3) \equiv (x^3, x^2y, xy^2, y^3). Clearly, these have relations between them. For example:
  • z0z2=x4y2=z12z_0 z_2 = x^4y^2 = z_1^2. So z0z2z12=0z_0 z_2 - z_1^2 = 0.
  • z1z3x2y4=z22z_1 z_3 x^2y^4 = z_2^2. So z1z3z22=0z_1 z_3 - z_2^2 = 0.
  • z0z3=x3y3=z1z2z_0 z_3 = x^3y^3 = z_1 z_2. So z0z3z1z2=0z_0 z_3 - z_1 z_2 = 0.
We have 3 first-order syzygies as written above. Things are more complicated than that. We can write the syzygies as:
  • p1z0z2z12p_1 \equiv z_0 z_2 - z_1^2.
  • p2z1z3z22p_2 \equiv z_1 z_3 - z_2^2.
  • p3z0z3z1z2p_3 \equiv z_0 z_3 - z_1 z_2.
We have z0z2z_0 z_2 in p1p_1. Let's try to cancel it with the z22z_2^2 in p2p_2. So we consider:
z2p1+z0p2=z2(z0z2z12)+z0(z1z3z22)=(z0z22z2z12)+(z0z1z3z0z22)=z0z1z3z2z12=z1(z0z3z1z2)=z1p3 \begin{aligned} & z_2 p_1 + z_0 p_2 \\ &= z_2 (z_0 z_2 - z_1^2) + z_0 (z_1 z_3 - z_2^2) \\ &= (z_0 z_2^2 - z_2 z_1^2) + (z_0 z_1 z_3 - z_0 z_2^2) \\ &= z_0 z_1 z_3 - z_2 z_1^2 \\ &= z_1(z_0 z_3 - z_1 z_2) \\ &= z_1 p_3 \end{aligned}
So we have non-trivial relations between p1,p2,p3p_1, p_2, p_3! This is a second order syzygy, a sygyzy between syzygies. We have a ring Rk[z0,z1,z2,z3]R \equiv k[z_0, z_1, z_2, z_3]. We have a map RinvariantsR \rightarrow \texttt{invariants}. This has a nontrivial kernel, and this kernel is spanned by (p1,p2,p3)R3(p_1, p_2, p_3) \simeq R^3. But this itself has a kernel q=z1p1+z2p2+z3p3q = z_1 p_1 + z_2 p_2 + z_3 p_3. So there's an exact sequence:
0R1R3R=k[z0,z1,z2,z3]invariants \begin{aligned} 0 \rightarrow R^1 \rightarrow R^3 \rightarrow R=k[z_0, z_1, z_2, z_3] \rightarrow \texttt{invariants} \end{aligned}
In general, we get an invariant ring of linear maps that are invariant under the group action. We have polynomials Rk[z0,z1,]R \equiv k[z_0, z_1, \dots] that map onto the invariant ring. We have relationships between the z0,,znz_0, \dots, z_n. This gives us a sequence of syzygies. We have many questions:
  1. Is RR finitely generated as a kk algebra? Can we find a finite number of generators?
  2. Is RmR^m finitely generated (the syzygies as an RR-MODULE)? To recall the difference, see that k[x]k[x]is finitely generated as an ALGEBRA by (k,x)(k, x) since we can multiply the xxs. It's not finitely generated as aMODULE as we need to take all powers of xx: (x0,x1,)(x^0, x^1, \dots).
  3. Is this SEQUENCE of sygyzy modulues FINITE?
  4. Hilbert showed that the answer is YES if GG is reductive and kk has characteristic zero. We willdo a special case of GG finite group.
We can see why a syzygy is called such; The second order sygyzy "yokes" the first order sygyzy. It ties together the polynomials in the first order syzygy the same way oxen are yoked by a syzygy.

§ References