and the vector \(\bar x = \begin{pmatrix} x\\ y\\ \end{pmatrix}\) This gives \(M \bar x = \begin{bmatrix}
x \\
-y \\
\end{bmatrix}\text{.}\) Thus for the polynomial \(f(\bar x) = f(\begin{bmatrix}
x \\
y \\
\end{bmatrix}) = x+y\) we have \(f(M\bar x) = f(\begin{bmatrix}
x \\
-y \\
\end{bmatrix})= x-y\text{.}\)
Definition9.1.1.
\(G \leq GLm(\mathbb{K}), |G| < \infty\text{,}\) then \(G\) is a finite matrix group. (In other words if \(G\) is a group of actions under which \(\mathbb{K}\) remains invariant under then it is smaller or equal to the total amount of group actions \(GLm(\mathbb{K})\) that would keep the polynomial invariant. AND the \(|G|\) is finite then \(G\) is a finite matrix group?)
NOTE: An action of a finite group \(G \curvearrowright \mathbb{K}^n\) given a realization of \(G\) as a finite matrix group.
Notation \(\bar x = (x_1, x_2,..., x_n)\text{,}\) with \(R = \mathbb{K}[x_1,x_2,...,x_n]\)
Definition9.1.2.
\(G\) is a finite matrix group within \(GLm(\mathbb{K})\) when? \(f\in \mathbb{K}[x_1,x_2,...,x_n]\) is invariant under the action of \(G\) if and only if \(f(A\bar x) = f(\bar x)\text{,}\)\(\forall A \in G\text{.}\)
Ex. \(f(\bar x)=x\) and \(f(\bar x) = x +y^2\) in \(\mathbb{K}[x_1,x_2,...,x_n]\) is invariant under \(C_2 = \langle\begin{bmatrix}
1 \amp 0 \\
0 \amp -1 \\
\end{bmatrix} \rangle\) However \(f(\bar x)=x+y\) is not. What are others?
Definition9.1.3.
\(R^G : = \{f \in R \, | f(A\bar x) = f(\bar x), \forall A \in G\} \subseteq R\) is the invariant ring for the action of \(G\)
Show this is a subring.
How does on find generators for \(R^G\text{?}\)
Is \(R^G\) even finitely generated?
Work through Hilbert’s proof.
Subsection9.1.3Reynolds Operator
Idea: "Averaging" over the action of \(G\) we get an invariant
\(\implies\) NDB : The ring of invariants is generated in degrees \(\leq |G|\)
Note: This is a computational tool! We can apply \(R_G\) to all the finitely many monomials in degrees \(\leq |G|\) to get generators for \(R^G\text{.}\) Exercise: Try this for \(C_4\) ... show!
Subsection9.1.5Hilbert Ideal
Note: In general for \(\{ f_1,..., f_s\} \subseteq \R\text{,}\)\(\{f_1,...f_s\}\) and \(\R\) can be quite different objects Exercise?
Theorem9.1.6.
Let \(J_G = R(R^G)_t\text{,}\) ideal generated by all positive degree invariants. If \(J_G = (f_1,...,f_s)\) and \(f_i\in R^G, \,\, \forall i\) (apply \(R^G\) if it is not), then \(R^G = \mathbb{K}[f_1,...f_s]\)
Subsection9.1.6Presentations
Definition9.1.7.
Definition: Let \(S = \mathbb{K}[f_1,...f_s] \subset R\text{.}\) A presentation of \(S\) is a map,
such that \(\frac{T}{ker(\phi)} \cong S\) With the syzygies of \(f_i\)’s giving the presentation ideal.
Proposition9.1.8.
(Elimination Theory): In \(S \bigotimes \mathbb{K}[u_1,...,u_s] = \mathbb{K}[x_1,...,x_n,u_1,...u_s]\) consider the ideal,
\begin{equation*}
I = (u_i - f_x(\bar x) | \, \langle f_i\rangle = S
\end{equation*}
Then,
\begin{equation*}
ker \phi = I \cap \mathbb{K}[u_1,...,u_s]
\end{equation*}
Algorithm9.1.9.
Compute a Groebner Basis \(G\) for \(I\) with elimination order for the \(x\)’s. Then, \(G \cap \mathbb{K}[y_1,...y_s]\) is the Groebner Basis for \(ker \phi\)
Subsection9.1.7Graph of Linear Actions
Definition9.1.10.
Let \(G \leq GL_n(\mathbb{K}), \,\, G\curvearrowright \mathbb{K}^n =:V, \,\, |G|\infty\text{.}\) For \(A\in G\) consider,
for \(\mu_i : d_i^{th}\) primitive root of unity and \(i \in [x]\text{,}\)\(j \in [n]\text{.}\) And encoded in the weight matrix \(W = (\omega_{ij})_{ij} =
\begin{bmatrix}
x_1 \amp \cdots \amp x_n \\
\vdots \amp \ddots \amp \\
x_n \amp \amp
\end{bmatrix}\)
Theorem9.1.13.
\(\bar x^{\bar \beta} \in R^G \iff W_{\bar \beta}\cong (0,...,0)\) for zeros being the weight of \(g_1\) acting on \(\bar x^{\bar \beta}\) and being modulo \(d_i\text{.}\)
Note: We can examine all monomials \(|\bar \beta| \leq |G|\) and sort them by their weight \(W\bar \beta\text{.}\) The ones with weight \(\bar 0\) will be invariant!
Question: Does this work for monomials in the exterior algebra?