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\begin{document}
\lecture{21}{November 23, 2005}{Madhu Sudan}{Michael Manapat}
\section{Quantified Statements}
Let $\phi: K^n \to \{\mathrm{T}, \mathrm{F}\}$ be a Boolean function defined as
follows:
\[
\phi(\x_1) = \exists \x_1 \forall \x_2 \exists \x_3 \cdots \mathrm{Q}_w \x_w,
\]
Q a quantifier, such that
\[
f_i(\x_0, \ldots, \x_w) = 0
\]
and
\[
g_j(\x_0, \ldots, \x_w) \neq 0
\]
for $i,j = 1, \ldots, m$, where the $f_i$ and $g_j$ are polynomials of degree at most $d$.
The goal of this lecture is to show how we can generate polynomaisl $P_1, \ldots, P_m$
such that $\phi(\x_0) = \mathrm{T}$ if and only if $P_i(\x_0) = 0$ for all $i$.
We will assume throughout that the ground field $k$ is algebraically closed (and thus
infinite) so, in particular, we will be able to use Hilbert's Nullstellensatz. Note,
however, that the result is fairly trivial if $k$ is a finite field since all functions
on a finite field are polynomial functions.
We begin with the simple case in which $w = 1$, $g_j = 1$ for all $j$, and $\x_0 = 0$.
Then $\phi(\x_0) = \mathrm{T}$ iff there exists an $\x_1$ such that $f_i(\x_1) = 0$
for $i = 1, \ldots, m$ (where $\x_0$ has been abosrbed into the polynomials $f_i$).
By the Nullstellensatz, this happens if and only if there do not exist polynomials
$q_1, \ldots, q_m \in k[\x_1]$ such that $\sum q_i f_i = 1$. There are several results
concerning the bounds of these $q_i$:
\begin{itemize}
\item Mayr and Meyer showed that if such $q_i$'s exist, then $q_i\prime$'s with the
same property exist and have degrees bounded by $d^{n^2}$, which puts the ``Hilbert
Nullstellensatz Problem'' in EXPSPACE (since we can find the coefficients of the $q_i$
by solving a linear system in logarithmic space in the size of the linear system).
\item Brownawell (87) showed, using complex analysis, that if such $q_i$'s exist, then
$q_i\prime$'s with the same property exist and have degree at most $O(md)^n$, putting
the problem in PSPACE.
\item Koll\'ar (88) and Dub\'e (92) proved Brownawell's bound using cohomology and elementary
commutative algebra (respectively).
\end{itemize}
Now instead of fixing $\x_0$ to be $0$, suppose it is an arbitrary vector $\ba$
in $K^n$. For that specific vector $\ba$, we can formulate the question as to
whether polynomials $q_i$ as above exist as a linear system: namely, we seek
a vector $\mathbf{q}$, whose entries are the coefficients of the polynomials
$q_i$, satisfying $M_{\alpha} \mathbf{q} = \mathrm{e}_1$, where $\mathrm{e}_1$ is
the standard basis vector with a $1$ in the first entry and zeroes everywhere else.
The length of $\mathbf{q}$ and the dimension of $M_{\bar{\alpha}}$ will depend on our bounds
for the degrees of the $q_i$---from what we saw above, the linear system will have
dimension $(md)^{n^2}$.
We obtained the matrix $M_{\bar{\alpha}}$ from the coefficients of the polynomials
$f_i$ (after $\alpha$ had been absorbed into those coefficients), but we can instead
produce a similar matrix $M$ with the formal variable $\x_1$ instead of the constant
vector $\bar{\alpha}$. Observe now that the new system $M\mathbf{q} = \mathrm{e}_1$ will
have a solution if there is a nonsingular (square) submatrix $S$ extending from
the first row. For each such submatrix $S$, we have a polynomial $P_S(\x_0) = \det(S)$, and
thus we can say that there exist $q_i$'s such that $\sum q_i f_i = 1$
if and only if $P_S(\x_0) \neq 0$ for some $S$. Negating this, we have the following:
\begin{quotation}
\noindent For a given $\x_0$, there exists an $\x_1$ such that $f_i(\x_0, \x_1) = 0$
for all $i$ iff for all square submatrices $S$ of $M$ as above, $P_S(\x_0) = \det(S) = 0$.
\end{quotation}
Now the number of polynomial constraints will be bounded by $2^{D^n}$, where $D$
is the degree of the polynomials $q_i$. By the bounds above, this means that the number
of constraints is around $2^{d^{n^2}}$, which is quite large. We can reduce the number
of constraints by finding a set of generators for the radical of the ideal generated
by the $P_S$, and we can do this by finding vector space generators for that radical. Now
the $P_S$ have degree at most $d \cdot D^n$, so there are at most $d^{n^2 + 1}$ linearly
independent polynomials in the radical ideal, whence we can take the number of polynomial
constraints to be less than $d^{n^2 + 1}$.
Now that we know how to address the special case above, we can generalize to get
a solution for the problem we originally stated. First, if we want to know if there
is an $\x_1$ such that $f_j(\x_0, \x_1) = 0$ and $g_j(\x_0, \x_1) \neq 0$ for all $j$,
where the $g_j$'s are no longer trivial, we need only determine if there exist
$\x_1$ and $y$ such that $f_j(\x_0, \x_1) = 0$ for all $j$ and
\[
1 - y \prod g_j(\x_0, \x_1) = 0.
\]
Second, given the quantified statement
\[
\exists \x_1 \forall \x_2 \exists \x_3 \cdots \mathrm{Q}_w \x_w,
\]
we can produce an equivalent statement with only existential quantifiers by
writing the statement above as
\[
\exists \x_1 \neg (\exists \x_2 \neg ( \cdots )).
\]
We can then work our way from the inside out inductively, using the solution for the
$w = 1$ case at each step. If $v$, $c$, and $d$ denote the original number of variables,
the original number of constraints, and the original degree of the polynomials, and the
primed $v$, $c$, and $d$ the new values after one iteration, we have
\begin{itemize}
\item $v\prime \leq v + 1$
\item $d\prime \leq d^v$
\item $c\prime \leq (cd)^v$.
\end{itemize}
The only complication arises in how we can (algebraically)
express the condition that the polynomials $P_i$ are not all zero: the negation of $P_1(x) = 0, \ldots, P_M(x) = 0$ holds if and only if there exists a $y$ such that
\[
\sum p_j(x)y^j \neq 0,
\]
which itself holds if and only if there exists a $z$ such that
\[
1 - z \left(\sum p_j(x) y^j \right) = 0.
\]
After these manipulations, the final collection of polynomials we obtain $P_1,\ldots,P_M$,
satisfying the original goals set-forth at the beginning of the lecture, are such that
both $M$ and the degrees of the $P_i$ are bounded by $d^{n^w}$ (with big-O's suppressed).
\section{Computation over the reals}
We briefly state some results on real root-finding:
\begin{itemize}
\item There is a nice algorithm for finding the number of roots of a polynomial
in an interval $[a, b]$ that uses only field arithmetic (the "Sturm sequence").
\item Distinct roots are well-separated (try to show this as an exercise).
\end{itemize}
The existential theory of the reals consists of questions of the following sort: do
there exist $x_1, \ldots, x_n$ such that $P_i(x_1, \ldots, x_n) \geq 0$? Tarski (1950)
showed that the existential theory is decidable, and Collins (70), Conway (87), Kozen and
Reit (87), and Renegar (92) showed that the existential theory is in PSPACE and that the
theory for a constant number of variables is in P.
\end{document}