References:
Mikhailkin math GT/0205011, math AG/0403015
Kopronov etc. math AG/0408311
Gelfond-Kopronov-Zel... (book) pp. 192-197

Summarizing last time:

Venn diagram:
Big circle: bounded rational polytopes in $\R^d$.
Small circle #1: Integer vertices; yielding a toric projective variety
together with a line bundle
Small circle #2: Delzant; yielding a toric symplectic manifold (which
turns out to be Kahler).
Intersection: Smooth toric projective varieties.

Recall the Kodaira embedding theorem: a Kahler manifold is a projective
variety if and only if the Kahler class is integral...

Suppose $M$ is a smooth projective toric variety; then 
  M    \supset (C^*)^d
  |               |
  |               |
  v               v
\Delta \supset \int(\Delta)

where the left map is the moment map, and the
right map is a fibration, and hence (as $\int(\Delta)$ is
contractible) $(C^*)^d=\int(\Delta) \times (S^1)^d$.

Suppose $X \subset M$ is a subvariety.  What is $\mu(X)$?

For instance, if $M=\C P^{n-1}$, then, in homogenous coordinates, 
$\mu(x)=(|x_1|^2-1/n,\dots, |x_n|^2-1/n)$.  To ask for $\mu(X)$ is to ask for 
the "size" of various components of $X$.  But this is qualitative
("your big may be my small").  So we'll start in a setting where
bigness is uncontestable.

Non-archimedean amoebas:

Let $\K=\C[[t]](t^{-1},t^{1/n})$.  This is an algebraically closed field;
elements are of the form
$\psi(t)=\sum_{k=k_0,d \in \N}^{\infty} a_{k/d} a_{k/d} t^{k/d}$.

We have a map $\val:K \to \Q$ given by
$\val(\psi)=-\infty$ if $\psi=0$, $-\inf \{k/d|a_{k/d} \neq 0\}$ if
$\psi \neq 0$.  If we think of $t$ as being "infinitesimal", the
valuation gives us a notion of "size".

Let $f(z) \in \K[z,z^{-1}]$ be a Laurent polynomial over $\K$.
Write $f(z)=\sum_{i=m}^n \psi_i z^i$.  Suppose $m<n$, $\psi_m,\psi_n \neq 0$.
Then $f(\eta)=0$ has $m-n$ solutions $\eta$ in $K^\times=\K-\{0\}$.

What are the sizes of these solutions?

Consider the coordinates $(i,-\val_i \psi_i)$,
draw the vertical line extending up from each, and take the convex
hull $\Gamma_f$ in $\R^2$.  This is called the {\em extended Newton polygon}
of $f$.

Lemma: $\{\val(\eta):f(\eta)=0, \eta \neq 0\}=\{slopes of \partial \Gamma_f\}$.

Why does this work?

Ex: First consider $f(z)=1-\psi z^2$.  This has two solutions, $\eta=\pm
\psi^{-1/2}$.  Note that $\val(\eta)=-1/2 \val(\psi)$.
Moreover, the corresponding polygon will consist of everything above
the line from $(0,0)$ to $(2,-\val \psi)$, which has that slope.

Pf of lemma:  If we replace $f(z)$ by $\xi f(z)$, $\xi \in K^*$,
$\Gamma_f$ shifts up, and so the slopes are unchanged.

If we replace $f(z) by $z^k f(z)$, $\Gamma_f$ shifts to the right, and
so the slopes are unchanged.

If we replace $f(z)$ by $f(\xi z)$ for some $\xi \in \K^\times$, the new
coefficients have valuations $\val(\xi^i \psi_i)=\val(\psi_i)+i\val(\xi)$;
thus all the slopes decrease by $\val \xi$.  On the other hand, the
zeroes get divided by $\xi$.  

Lemma': There is an $\eta$ with $f(\eta)=0$ and $\val(\eta)=0$ iff
$\Gamma_f$ has a horizontal edge.  By the last remark, this statement
implies the original lemma.

Pf: Suppose $\Gamma_f$ has no horizontal edge.  Then there is some coefficient
with strictly largest valuation; thus 
$f(z)=t^r(\alpha z^k+o(1))$ (where $o(1)$ denotes something with only
positive $t$-coefficients).

Take some element in $\K$ with valuation 0; this is $\eta=\beta+o(1)$
for $\beta \in \C^\times$.  Then $f(\eta)=t^r(\alpha \beta^k+o(1))$.  This
is nonzero, as it has nonzero leading-order term.  Hence $f$ has no
zeros with valuation 0.

Conversely, if $\Gamma_f$ has a horizontal edge, $f(z)=z^s t^r(g(z)+o(1))$,
where $g(z)$ is a polynomial in $\C$ with at least 2 nonzero
terms.  Now, there is an $\alpha \in C^\times$ such that
$g(\alpha)=0$.  Look for a solution of $f(\eta)=0$, where
$\eta=\alpha+o(1)$.

(Hensel's lemma: this is possible; correct it order by order).

We want to generalize this to Laurent polynomials in several variables.

Combinatorial terminology:

Let $A \subset \Z^d$ be finite, not contained in any affine hyperplane.  
Take a function $v:A \to \Q$.

Let $P=\{convex hull of A\}$ (the Newton polytope).

Define $\Gamma$, the extended Newton polytope, to be the convex hull
of the set of pairs $(k,r) \in \R^{d+1}$, where $k \in A$, $r>v(k)$.

We can write $\Gamma=\{(k,r) \in \R^{d+1}:r \geq w(k)$, where $w:P \to
\R$ is a piecewise affine function.

This induces a decomposition $P=P_1 \cup P_d$ such that $w|P_i$ is
affine and the vertices of each $P_i$ are contained in $A$.

Define $w^*:(\R^d)^* \to \R$, the Legendre transform of $w$.  This
will replace the previous notion of "slope".

We set $w^*(l)=\min_{k \in P} \{<l,k>-w(k)\}=\min_{k \in A} \{l(k)-v(k)\}$.

What does $w*$ "look like"?

Fix $l$; in $\R^{d+1}$, consider the hyperplane $\{(s,k)|<l,k>-w^*(l)=s$.
Claim: this affine subspace barely touches $\Gamma$.  Why? there is at least
one point $k$ such that $w^*(l)=<l,k>-v(l)$...

You can think of this as a definition of the Legendre transform: each
linear functional $l$ defines a family of affine hyperplanes in $\R^{d+1}$,
vertically shifted from each other.  $w^*(l)$ defines the smallest 
shift that will cause the hyperplane to intersect $\Gamma$.

Note that $w^* is piecewise affine.  Define $\Omega=\{l \in (\R^d)^\times|
w^* is {\em not} smooth\}$.  Then $\Omega$ is a piecewise affine complex of
dimension $(d-1)$.

Example: Take a triangle with one interior point, such that $v=1$ at
each vertex, and $0$ at the interior point.  What behavior does the
Legendre transform have?

The extended Newton polytope looks like a pyramid under a prism

So $(\R^2)^*$ decomposes into four regions; three of these (corresponding
to the vertices) are unbounded; the other one (corresponding to the
interior point) is a bounded triangle, with rays from its vertices bounding
the three unbounded regions.

Structure of $\Omega$:

$\l \in \Omega$ iff the graph of $<l,->:\R^d \to \R$ is parallel to
an edge of $\partial \Gamma$.

The geometry at $\infty$ of $\Omega$:  Let $k$ be a vertex of $P$.  Consider
$N_k \subset (\R^d)^*$: the {\em normal cone of $k$}, defined by
$\{l \in (\R^d)^*|<l,->:P \to \R assumes a maximum at $k$\}$.

Take for each $k$ a $b_k \in N_k$; then, "at $\infty$", $\Omega$
is "asymptotic" to $(\R^d)^* \cup_k (b_k+N_k)=\Omega'$ in the sense of
Hausdorff convergence:

The Hausdorff topology for subsets of $S^{d-1}$ is defined by $A,B \subset
S^{d-1}$ have $\dist(A,B)=\max\{sup_{a \in A} d(a,B),\sup_{b \in B} d(A,b)\}$.

Thm: $\lim_{r \to \infty} \frac{1}{r}\Omega \cap S^{d-1}=\lim_{r \to 
\infty} \frac{1}{r} \Omega' \cap S^{d-1}$.  Note that this limit is independent
of $b_k$, and depends only on $P$.

Moreover, this is a cell decomposition of $S^{d-1}$ which is dual to the
decomposition given by $P$.

Thm: Suppose $f$ is a Laurent polynomial over $\K$ in $d$ variables; then
$f=\sum_{k \in \Z^d} \psi_k {z_1}^{k_1}\dots z_d^{k_d}$.

Let $A=\{k:\psi_k \neq 0\}$, let $v:A \to \Q$ such that $v(k)=-\val(\psi_k)$.

Consider the map $\Val:(\K^\times)^d \to \Q^d$ which is $\val$ on each
coordinate.  Then:

Thm (Kapranov): $\overline{\Val({\eta|f(\eta)=0\})}=\Omega$.

This is called the {\em non-archimedean amoeba}.

Think about why this gives back the 1-dimensional statement...