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

#### § How does a group act on polynomials?

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

#### § Determinants

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

#### § $S_n$ acting on $\mathbb C^n$ by permuting coordinates.

Polynomials are functions $\mathbb C[x_1, \dots, x_n]$. Symmetric group acts on polynomials by permuting $x_1, \dots, x_n$. What are the invariant polynomials?
• $e_1 \equiv x_1 + x_2 + \dots x_n$
• $e_2 \equiv x_1 x_2 + x_1 x_3 + \dots + x_{n-1} x_n$.
• $e_n \equiv x_1 x_2 \dots x_n$.
These are the famous elementary symmetric functions. If we think of $(y - x_1) (y - x_2) \dots (y - x_n) = y^n - e_1 y^{n-1} + \dots e_n$.
• The basic theory of symmetric functions says that every invariant polynomialin $x_1, \dots x_n$ is a polynomial in $e_1, \dots, e_n$.

#### § Proof of elementary theorem

Define an ordering on the monomials; order by lex order. Define $x_1^{m_1} x_2^{m_2} > x_1^{n_1} x_2^{n_2} \dots$ iff either $m_1 > n_1$ or $m_1 = n_1 \land m_2 > n_2$ or $m_1 = n_1 \land m_2 = n_2 \land m_3 > n_3$ and so on. Suppose $f \in \mathbb C[x_1, \dots, x_n]$ is invariant. Look at the biggest monomial in $f$. Suppose it is $x_1^{n_1} x_2^{n_2} \dots$. We subtract:
\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 $f$. If $f$ is symmetric, Then we can order the term we choose such that $n_1 \geq n_2 \geq n_3 \dots$. We need this to keep the terms $(n_1 - n_2), (n_2 - n_3), \dots$ to be positive. So we have now killed off the largest term of $f$. Keep doing this to kill of $f$ completely. This means that the invariants of $S_n$ acting on $\mathbb C^n$ are a finitely generated algebra over $\mathbb C$. 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 $\mathbb C$. This is the first non-trivial example of invariants being finitely generated. The algebra of invariants is a polynomial ring over $e_1, \dots, e_n$. This means that there are no non-trivial-relations between $e_1, e_2, \dots, e_n$. This is unusual; usually the ring of generators will be complicated. This simiplicity tends to happen if $G$ is a reflection group. We haven't seen what a syzygy is yet; We'll come to that.

#### § Complicated ring of invariants

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

#### § Second order Syzygy

Take $Z/3Z$ act on $\mathbb C^2$. Let $s$ be the generator of $Z/3Z$. We define the action as $s(x, y) = (\omega x, \omega y)$ where $\omega$ is the cube root of unity. We have $x^ay^b$ is invariant if $(a + b)$ is divisible by $3$, since we will just get $\omega^3 = 1$. So the ring is generated by the monomials $(z_0, z_1, z_2, z_3) \equiv (x^3, x^2y, xy^2, y^3)$. Clearly, these have relations between them. For example:
• $z_0 z_2 = x^4y^2 = z_1^2$. So $z_0 z_2 - z_1^2 = 0$.
• $z_1 z_3 x^2y^4 = z_2^2$. So $z_1 z_3 - z_2^2 = 0$.
• $z_0 z_3 = x^3y^3 = z_1 z_2$. So $z_0 z_3 - z_1 z_2 = 0$.
We have 3 first-order syzygies as written above. Things are more complicated than that. We can write the syzygies as:
• $p_1 \equiv z_0 z_2 - z_1^2$.
• $p_2 \equiv z_1 z_3 - z_2^2$.
• $p_3 \equiv z_0 z_3 - z_1 z_2$.
We have $z_0 z_2$ in $p_1$. Let's try to cancel it with the $z_2^2$ in $p_2$. So we consider:
\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 $p_1, p_2, p_3$! This is a second order syzygy, a sygyzy between syzygies. We have a ring $R \equiv k[z_0, z_1, z_2, z_3]$. We have a map $R \rightarrow \texttt{invariants}$. This has a nontrivial kernel, and this kernel is spanned by $(p_1, p_2, p_3) \simeq R^3$. But this itself has a kernel $q = z_1 p_1 + z_2 p_2 + z_3 p_3$. So there's an exact sequence:
\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 $R \equiv k[z_0, z_1, \dots]$ that map onto the invariant ring. We have relationships between the $z_0, \dots, z_n$. This gives us a sequence of syzygies. We have many questions:
1. Is $R$ finitely generated as a $k$ algebra? Can we find a finite number of generators?
2. Is $R^m$ finitely generated (the syzygies as an $R$-MODULE)? To recall the difference, see that $k[x]$is finitely generated as an ALGEBRA by $(k, x)$ since we can multiply the $x$s. It's not finitely generated as aMODULE as we need to take all powers of $x$: $(x^0, x^1, \dots)$.
3. Is this SEQUENCE of sygyzy modulues FINITE?
4. Hilbert showed that the answer is YES if $G$ is reductive and $k$ has characteristic zero. We willdo a special case of $G$ 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.