Add Lecture 5 on Feb 7th

By Rodrigo Horruitiner (rmh322)

lecture-notes/lecture-notes.tex

@@ -17,6 +17,21 @@
 		-A\\
 		-I
 \end{bmatrix}}
+\newcommand{\bb}[1]{{\mathbb{#1}}}
+\newcommand{\mc}[1]{{\mathscr{#1}}}
+\newcommand{\tens}[1]{%
+	\mathbin{\mathop{\otimes}\displaylimits_{#1}}%
+}
+\newcommand{\fk}{\mathfrak}
+
+\newcommand\restr[2]{{% we make the whole thing an ordinary symbol
+		\left.\kern-\nulldelimiterspace % automatically resize the bar with \right
+		#1 % the function
+		\vphantom{\big|} % pretend it's a little taller at normal size
+		\right|_{#2} % this is the delimiter
+}}
+\DeclareMathOperator{\Aut}{Aut}
+\DeclareMathOperator{\Inn}{Inn}
 \DeclareMathOperator{\Hom}{Hom}
 \DeclareMathOperator{\Spec}{Spec}
 
@@ -26,6 +41,7 @@
 \usepackage{dsfont}
 \usepackage{enumerate}
 \usepackage{graphicx}
+\usepackage{tikz-cd}
 \usepackage{subcaption}
 
 
@@ -45,7 +61,7 @@
 \theoremstyle{definition}
 \newtheorem*{Problem}{Problem}
 \newtheorem*{Def}{Definition}
-\newtheorem{Eg}{Example}
+\newtheorem{Eg}{Example}[section]
 
 \theoremstyle{remark}
 \newtheorem*{Remark}{Remark}
@@ -709,4 +725,163 @@ \section{Lecture 04}
 Note that $A_\sigma = \CC[\sigma^\vee \cap M] \subseteq \CC[M] = \CC[t_1^{\pm 1}, \dots, t_n^{\pm 1}]$ and $\CC[t_1^{\pm 1}, \dots, t_n^{\pm 1}]$ is a domain.
 Thus $A_\sigma$ is a domain, $I_\sigma$ is prime, and $X_\sigma$ is irreducible.
 
+\newpage
+\section{Lecture 5}
+Recall some notation and observations from last time:
+\begin{align*}
+A_\sigma := \CC[\sigma^\vee \cap M]
+X_\sigma = \Spec A_\sigma = \bb{V}(I_\sigma) \subseteq \CC^r
+\end{align*}
+if $\sigma^\vee \cap M = \langle m_1, ..., m_r \rangle$ then the natural projection $\varphi: x_i \mapsto m_i$ forms an exact sequence
+\begin{center}
+	\begin{tikzcd}
+		0 \arrow[r] & I_\sigma \arrow[r] & {\CC[x_1,...,x_r]} \arrow[r, "\varphi"] & A_\sigma \arrow[r] & 0.
+	\end{tikzcd}
+\end{center}
+
+Recall from last time that $I_\sigma$ is a prime ideal, so that $X_\sigma$ is an irreducible (reduced) variety. What is $I_\sigma$?
+
+\begin{Proposition}
+$I_\sigma$ is generated by $\langle x_1^{\alpha_1} \cdots x_r^{\alpha_r} - x_1^{\beta_1} \cdots x_r^{\beta_r} : \forall i (\alpha_i, \beta_i \geq 0, \sum_i \alpha_i m_i = \sum_i \beta_i m_i) \rangle $.
+\end{Proposition}
+\begin{proof}
+
+($\supseteq$) $\varphi(x_1^{\alpha_1} \cdots x_r^{\alpha_r}) = t^{\sum_i \alpha_i m_i}$ and $\varphi(x_1^{\beta_1} \cdots x_r^{\beta_r}) = t^{\sum_i \beta_i m_i}$
+
+($\subseteq$) Note that $x^\alpha - x^\beta \in I_\sigma \iff \sum_i \alpha_i m_i = \sum_i \beta_i m_i$. Note as well that if $J := \langle t^{m_1} - x_1, ..., t^{m_r} - x_r\rangle \subseteq \CC[t_1, ..., t_n x_1, ..., x_r]$ (where $M = \ZZ^n$ is generated by $e_1, ..., e_n$, and we let $t_i := t^{e_i}$) then $I_\sigma = J \cap \CC[x_1, ..., x_r]$.
+
+In order to see this, recall the theory of Groebner bases. Choose the lexicographic monomial order $t_1 > t_2 > ... > x_r$. Let $G$ be the Groebner basis of $J = \{g_1, ..., g_N\}$ in this order. Then $J \cap \CC[x_1, ..., x_r]$ is generated by the $\{g_i : g_i \in \CC[x_1, ..., x_r]$. It is easy to see that $G$ is generated by binomials.
+\end{proof}
+\paragraph{Products (exercise)}
+Suppose $\sigma$ is a cone in $N$, and $\sigma'$ is a cone in $N'$. Then 
+\begin{enumerate}
+	\item $\sigma \times \sigma'$ is a cone in $N \oplus N'$
+	\item $(\sigma \times \sigma')^\vee =$ ???
+	\item $A_{\sigma \times \sigma'} =$ ???
+	\item Show that $X_{\sigma \times \sigma'} \simeq X_\sigma \times X_{\sigma'}$ ( $= X_\sigma \times_{\Spec \CC} X_{\sigma'}$ )
+\end{enumerate}
+
+\begin{Eg}
+Let
+
+$\sigma = 0 \subset N = \ZZ^r$
+
+$\sigma' = \langle e_1, ..., e_s \rangle \subseteq N' = \ZZ^s$ ($\sigma' = N'_\RR$)
+
+then $X_\sigma \times X_{\sigma'} = (\CC*)^r \times \CC^s$.
+\end{Eg}
+
+The plan for this session is to cover some local properties of $X_\sigma$. This includes \textit{points}, the \textit{zariski tangent space}, the \textit{torus action} and the \textit{cone-orbit correspondence} in the affine case.
+
+
+\paragraph{Points on $X_\sigma$}
+If $X$ is a variety (or scheme) then a (closed) point of $X$ is ``really'' a morphism  $\Spec \CC \to X$ (understanding $\Spec \CC$ as a point).
+
+Unraveling this, we have a map $\varphi: \Spec \CC \to \Spec \CC[\sigma^\vee \cap M]$, so that the data of $\varphi$ is corresponds one to one with the data of a map $\varphi^\#: \CC[\sigma^\vee \cap M] \to \CC$, an \textit{evaluation map}. Such a map corresponds one to one with a semigroup homomorphism $\rho: \sigma^\vee \cap M \to \CC^\times$ (i.e. $\rho(m_1 + m_2) = \rho(m_1) \rho(m_2)$ whenever $m_1$ and $m_2$ are in $\sigma^\vee \cap M$, and $\rho(id) = 1$).
+
+In other words, we have a correspondence 
+\begin{center}
+	\begin{tikzcd}
+		\text{points of } X_\sigma \arrow[rr] &  & {\Hom_{\text{semigroup}} (\sigma^\vee \cap M, \CC)} \arrow[ll, "1 \text{ to } 1"'].
+	\end{tikzcd}
+\end{center}
+
+\begin{Def}[Distinguished point]
+$p_\sigma \in X_\sigma$ is the point corresponding to 
+\begin{align*}
+\sigma^\vee \cap M &\to \CC \\
+m &\mapsto \begin{cases}
+1 \text{ if } m \in \sigma^\perp \\
+0 \text{ if } m \not \in \sigma^\perp
+\end{cases}
+\end{align*}
+\end{Def}
+Note that if $\sigma$ is full dimensional then this map sends $m$ to $1$ if and only if $m = 0$. This is a semigroup homomorphism since if $m_1, m_2 \in \sigma^\vee \cap M$ then $m_1 + m_2 \in \sigma^\perp \iff m_1, m_2 \in \sigma^\perp$.
+
+\begin{Proposition}
+	Let $\tau \leq sigma$ (i.e. $\tau$ is a face of $\sigma$). Let $u \in \sigma^\vee \cap M$ s.t. $\tau = \sigma \cap u^\perp$. Then $S_\tau = S_\sigma + \ZZ_{\geq 0} \cdot (-u)$. 
+\end{Proposition}
+\begin{proof}
+Exercise.
+\end{proof}
+
+\begin{Def}[Principal affine open sets]
+If $X$ is an affine variety, and $f \in A(X)$, then $U_f := X \setminus V(f) \subset X$ is called a \textbf{ principal (affine) or basic open set of $V$}.
+\end{Def}
+
+\begin{Corollary}
+If $\tau \leq \sigma$, and $u$ is as above, then $A_\tau = (A_\sigma)_{t^{u}}$ and so $X_\tau \subseteq X_\sigma$ is a principal open set. 
+\end{Corollary}
+Note: If $\tau \leq \sigma$ is a face, we have $p_\tau \in X_\tau \subset X_\sigma$. In other words, we have distinguished points $p_\tau \in X_\sigma$ for all faces $\tau$ of $\sigma$.
+
+\paragraph{Example}
+Consider the cone $\sigma$ pictured below in red:
+
+\begin{center}
+	\includegraphics[scale=0.2]{pic/toricvar_feb7_1}
+\end{center}
+
+We want to understand the points $p_\sigma, p_{\tau_1}, p_{\tau_2}, p_0$.
+\begin{enumerate}
+	\item $p_\sigma(m) = \begin{cases}
+	1 \text{ if } m = 0 \\
+	0 \text{ if } m \neq 0 \\
+	\end{cases}$
+
+	so that $p_\sigma = (0,0) \in \CC^2_{uv}$.
+	\item $p_{\tau_1} = (1,0)$
+	\item $p_{\tau_2} = (0,1)$
+	\item $p_{0} = (1,1)$
+\end{enumerate}
+We have inclusions on open sets given by the diagrams:
+\begin{center}
+	\begin{tikzcd}
+		& X_{\tau_1} = \CC^* \times \CC \arrow[rd, hook] &  \\
+		(\CC^*)^2 = X_0 \arrow[rd, hook] \arrow[ru, hook] &  & X_\sigma = \CC^2 \\
+		& X_{\tau_2} = \CC \times \CC^* \arrow[ru, hook] & 
+	\end{tikzcd}
+\end{center}
+
+\paragraph{Zariski tangent space / nonsingularity}
+Recall: if $X \subseteq \CC^n$ is an affine variety, $A(X) = \CC[x_1, ..., x_n]/I$. If $p \in X$ is a (closed) point: $p = (p_1, ..., p_n) \in X \subseteq \CC^n$, we have an exact sequence 
+\begin{center}
+	\begin{tikzcd}
+		0 \arrow[r] & \mathfrak{m}_p \arrow[r] & {\CC[x_1, ..., x_n]/I} \arrow[r, "ev_p"] & \CC \arrow[r] & 0
+	\end{tikzcd}
+\end{center}
+where $\mathfrak{m}$ denotes the maximal ideal of $p$ in $X$, and $ev_p$ is the evaluation map sending each $x_i$ to $p_i$ (in the language of sheaves, $\mathfrak{m}_p = \mathfrak{m}_{X,p} = \mathcal{O}_{X,p}$).
+
+We define the \textbf{Zariski cotangent space} at the point $p$ as $T^*_{X, p} := \mathfrak{m}_p / \mathfrak{m}_p^2$, a finite dimensional $\CC$-vector space. Similarly, the \textbf{Zariski tangent space} is $T_{X,p} := (\mathfrak{m}_p / \mathfrak{m}_p^2)^*$, the dual vector space.
+
+(Note: A basic fact from any basic reference, for example the first chapter of Hartshorne, states that $\dim_\CC T_{X,p} \geq \dim_p X$, where $\dim_p X$ is the Krull dimension of $\mathcal{O}_{X,p}$. Also, $X$ is nonsingular at $p$ exactly when equality is achieved.) \\
+
+Our situation is the following. We have a pointed cone $\sigma$ in $N$, a corresponding toric variety $X_\sigma$, and a distinguished point $p_\sigma \in X_\sigma$, along with other distinguished points $X_\tau$. The following questions arise:
+\begin{enumerate}%[label=(\arabic*)]
+	\item How do we identify $T_{X_\sigma, p_\tau}$, or at least its dimension?
+	\item How do we study other points of $X_\sigma$?
+	\item How do we identify the singular locus, or set of singular points.
+\end{enumerate}
+
+\paragraph{Proposition}
+\begin{Proposition}
+$\dim T_{X_\sigma, p_\sigma} = \mathcal{H}$, where $\mathcal{H}$ is the Hilbert basis of $\sigma^\vee \cap M$.
+\end{Proposition}
+\begin{proof}
+Recall that $p_\sigma(m) = \begin{cases}
+0, m \neq 0 \\
+1, m = 0 
+\end{cases}$. What is $\mathfrak{m}_{p_\sigma}$? 
+
+We know that $A_\sigma = \bigoplus_{m \in \sigma^\vee \cap M} \CC \cdot t^m$ so $\mathfrak{m}_p = \bigoplus_{m \neq 0} \CC \cdot t^m \subset A_\sigma$. We also have $\mathfrak{m}_p^2 = \bigoplus_{m \text{ not irreducible}} \CC \cdot t^m$, so that 
+\begin{align*}
+\mathfrak{m}_p / \mathfrak{m}_p^2 = \sum_{m \in \mathcal{H}} \CC \cdot t^m \implies \dim \mathfrak{m}_p / \mathfrak{m}_p^2 =  | \mathcal{H} | 
+\end{align*}
+and the result follows.
+\end{proof}
+
+\begin{Remark}
+	Assume $\sigma$ is a pointed cone in $N$. What is $\dim X_\sigma$? We can get a torus $T = (\CC^*)^n \subseteq X_\sigma$ an inclusion of an open subset, where $n$ is the rank of $N$. So the field of rational functions on $X_\sigma$ satisfies $K(X_\sigma) = K((\CC^*)^n) = \CC(t_1, ..., t_n)$, and this is the dimension of $X$. 
+\end{Remark}
+Suppose now $X_\sigma$ is smooth, that is, nonsingular, and once again that $\sigma$ is a pointed cone of maximal dimension. Then $\dim T_{X, p_\sigma} = \dim X = n$, so the Hilbert basis $\mathcal{H}$ of $\sigma^\vee \cap M$ satisfies $|\mathcal{H}| = n$. Because $X$ has maximal dimension, it has at least $n$ extremal rays. But $|\mathcal{H}| \geq \# \textit{ extremal rays } \geq n$, so that the number of extremal rays is exactly $n$.
+
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