No lecture on Thursday... "Today I will engage in one of my favorite pasttime, which is to try and pass as an algebraic geometer." Algebraic varieties: Def'n: A {\em rational polyhedral cone} is some $\sigma \subset \R^n$ defined be finitely many inequalities, $ \geq 0$, \dots, $ \geq 0$, with $v_i \in \Q^n$. To avoid pathologies, we'll assume that $\Sigma$ is $n$-dimensional (i.e., not contained in some hypersurface). To this, we can associate an algebraic variety (by defining the function ring of the variety). Let $A_\Sigma=\C[\Sigma \cap \Z^n]$. Claim: This is a $\C$-\algebra. What is the multiplication? Take $a,b \in \Sigma \cap \Z^n$; these yield monomials $x_a,x_b$; define $x_ax_b=x_{a+b}$. $x_0$ is the unit element. Define $X_\Sigma=\Spec(A_\Sigma)$. $X_\Sigma$ is finite-dimensional, as $A_\Sigma$ is finitely generated (exercise) Example: $\Sigma=(\R^{\geq 0})^n$. If we let $x_1=x_{(1,0,\dots,0)}$, \dots, $x_n=x_{(0,0,\dots,1)}$, then $A_\Sigma=\C[x_1,\dots,x_n]$, and so $X_\Sigma=\C^n$. $\Sigma=\R^n$. Then each of the above things has an additive inverse; thus $A_\Sigma$ is the ring of Laurent polynomials in $x_1,\dots,x_n$, and so $X_\Sigma=(\C^*)^n$. Remark: $X_\Sigma$ carries a natural $(C^*)^n$-action: namely, $(z_1,\dots,z_n)\cdot x_a=(z_1^{a_1},\dots,z_n^{a_n})x_a$, where $a=(a_1,\dots,a_n) \in \Z^n$. General example: Suppose $\Sigma$ is given. Take $\rho_1,\dots,\rho_n \geq1$ integers; define $\Sigma'=(p_1,\dots,p_n)\Sigma$ (i.e., dilate the $i^{\rm th} coordinate by $p_i$). How does this change the corresponding algebraic variety? Take the $(\C^*)^n$-action on $A_{\Sigma'}$ and restrict it to $G=\Z/(\rho_1) \times \dots \times \Z/(\rho_n)$. Then $(A_{\Sigma'})^G=A_\Sigma$. Why? The elements of $A_\Sigma$ in $A_\Sigma'$ will be precisely those where the $i^{\rm th}$ coordinate is divisible by $\rho_i$; thus they will be the ones in which $ The invariant functions on a variety are the functions on a quotient; thus $X_\Sigma=X_{\Sigma'}/G$. So $X_\Sigma$ can easily have singularities of finite order. Ex: Consider the cone in $\C^2$ with both coordinates positive and below the line $y=2x$. If we dilate the $x$-coordinate by a factor of 2, we get the cone with both coordinates positive and below the line $y=x$, which is clearly Lemma: $X_\Sigma$ is smooth iff $X_\Sigma=\C^k \times (\C^*)^{n-k}$ iff $\Sigma=\{ \geq 0, \dots, \geq 0\}$, where $v_1,\dots,v_r$ is a partial basis of the lattice $\Z^n$. Pf: One direction is trivial; the other direction requires more knowledge of algebraic geometry (specifically, knowledge of when varieties are smooth in terms of their rings of functions) than we'll be assuming. Geometry of $X_\Sigma$ $(C^*)^n$-invariant Zariski-closed subsets of $X_\Sigma$<=>$(C^*)^n- invariant pure ideals in $A_\Sigma$<=>subsets $J \subset \Sigma \cap \Z^n$ such that $J+\Sigma \cap \Z^n \subset J$ and $a+b \in J$ implies that $a$ or $b \in J$. How do we get subsets with this property? Choose a face, take monomials that don't belong to that face (the sum of two monomials not belonging to a fact can't belong to the face; if $a+b$ doesn't belong to the face, clearly either $a$ or $b$ doesn't). Similarly for the union of several faces (in fact, this fully describes all such ideals). Lemma: There is a bijective correspondence between $k$-dimensional faces of $\Sigma$ and $k$-dimensional $(\C^*)^n$ orbits of $X_\Sigma$. Given a face, take $J=(\Sigma-F) \cap \Z^n$; this gives the Zariski-closure of the corresponding orbit. How can we realize this as an affine variety? Take $B \in \Sigma \cap \Z^n$ a finite set that generates the semigroup $\Sigma \cap \Z^n$. Note that $B$ must contain $0$, the first integral point along each edge, and maybe some others. This gives rise to an embedding of $X_\Sigma$ into $\C^{|B|}$, namely through $\C[B] \to A_\Sigma$. Since $B$ generates, this is surjective, and thus yields a closed embedding. To get explicit equations: take a relation $\sum n_i b_i=0$, where the $b_i \in B$, $n_i \in \Z$; this gives the relation $x_{b_1}^{n_1}\dots x_{b_k}^{n_k}=1$. Some of the $n_i$ might be negative; move these to the other side of the equation to get a polynomial relation. So it's a subvariety defined by monomial equations... Let $(C^*)^n$ act on $\C^{|B|}$ by the weights $b_{ij}$, $b_i \in B$, $j \in \{1,\dots,n\}$. Then the embedding $i:X_\Sigma \to \C^B$ is equivariant and $i(X_\Sigma)$, and $i(X_\Sigma)$ is the closure of $(C^*)^n(1,\dots,1)$ (in either Zariski or ordinary topology...) The correspondence between faces and orbits is given by $F$ corresponds to $\{x \in i(X_\Sigma) \subset \C^{|B|}$ such that $x=(x_b)_{b \in B}$, where $x_b \neq 0$ iff $b \in F$. So this gives us some affine toric varieties. Projective toric varieties: Take $P \subset \R^n$ a bounded convex polyhedron with integer vertices. Associate to this a graded ring $R_P=\bigoplus_{k \geq 0} R_P^k$, where $R_p^k=\C[kP \cap \Z^n]=\C[P \cap (1/k) \Z^n$. Multiplication: $\R_P^k \times \R_P^l \to \R_P^{k+l}$ is given by addition of points (in the second view, by weighted averages of points). Define $X_P=\Proj(\R_P^k)$. "What the hell is that?" In practical terms, $X_P$ is a variety carrying a holomorphic line bundle $L_P$ such that $\Gamma(X_p,L_P^{\otimes k})=R_P^k$ for sufficiently large $k$. Ex: The triangle with vertices $(0,0)$, $(1,0)$, $(0,1)$. $R_P=\C[x_a,x_b,x_c]$; all 3 are of degree 1 and there are no relations (by the existence of barycentric coordinates...); So $X_P=\C P^2$, the line bundle is $\O(1)$. Ex: The segment with vertices $(0,0)$ and $(0,1)$. Similarly, we get $\C P^1$. Now, consider the segment $[-1,1]$. Three integer points; we have the relation $x_0^2=x_{-1}x_1$. So $\R_P=C[x_0,x_{-1},x_1]/(x_0^2=x_{-1}x_1)$. So $X_P$ is the variety defined by $x_0^2=x_1x_{-1}$. This is a smooth conic in $\C P^2$, and thus is isomorphic to $\C P^1$. $L_P=\O_{\C P^1}(2)$... There is a natural $(\C^*)^n$-action on $(X_p,L_p)$ given by $(\C^*)^n$ acting on $\R_P$ as before. Let $A \subset P \cap \Z^n$ be a subset containing all vertices of $P$. This gives rise to a closed embedding $X_p \to \P(\C^{|A|})$, dual to $C[A] \to R_P$ by taking every point to its corresponding monomial. This isn't surjective, but it's "essentially surjective" (i.e., $R_P/\C[A]$ is nilpotent). This turns out to be good enough. If we consider the $(C^*)^n$-action on $\P(\C^{|A|})$ given by the coordinates of the points in $A$, then the embedding $i_A$ is equivariant. Lemma: There is a bijective correspondence between orbits of the $(\C^*)^n$ action and faces of $P$. (considering $X_P \subset \P(\C^{|A|})$, $x_a \neq 0$ if $a \in A \cap F$, $X_a=0$ if $a \in A-F$). In particular, the vertices of $P$ correspond to fixed points. Lemma: $i_A(X_P)$ is the closure of $(C^*)^n(1,1,dots,1)$ in $\P(\C^|A|)$. Lemma: Let $v \in P$ be a vertex. Then a Zariski-n'hood of the corresponding fixed point in $X_P$ looks like $X_\Sigma$, where $\Sigma$ is the cone over $P$ at $v$. Cor: $X_p$ is smooth iff $p$ is Delzant. In fact, if $A=\{vertices\}$, consider $X_P \to \P(\C^A) \to \R^n$, where the second map is the moment map of the $S^1$-action; its image is precisely $P$. So we get the same symplectic manifold in this way as previously. The condition that such a manifold is algebraic is precisely that it have integer coordinates. Reference: Gelfond-Kapranov-Zellkinski, Ch. 5