User moe hirsch - MathOverflow

Answer by Moe Hirsch for Can we actually find any fixed points with Brouwer's theorem?

2012-08-31T14:59:16Z

The paper "Exponential lower bounds for finding Brouwer fixed points"

Addendum by original poster: It was non-trivial to find a copy of this great paper of Hirsch, Papadimitriou and Vavasis. It does answer my general question quite clearly: finding Brouwer fixed points is exponentially hard in the worst case no matter what algorithm you use. Here is a link to this paper for all those who are interested and don't want to run into many, many pay-walls. I will take it down in a few days. -VN

Answer by Moe Hirsch for Ways to prove the fundamental theorem of algebra

2012-08-08T18:09:01Z

Two very short proofs, mostly topological, that a nonconstant polynomial map $f:{\bf C} \to \bf C$ is surjective (joint work with Robert Palais):

(1) Complex analysis shows that $f$ is an open map (images of open sets are open). A standard estimate, $|f(z)|\to\infty$ as $|z|\to\infty$ , implies $f$ is also a closed map (images of closed sets are closed). Thus $f(\bf C)$ is an open, closed, nonempty subset of the connected space $\bf C$ , therefore $ f(\bf C)=\bf C$ .

The openness of $f$ is nontrivial, but it can be replaced by elemntary algebra and topology:

(2) The set $K$ of roots of $f'$ is finite. The inverse function theorem shows that the set $A:=f({\bf C})\setminus f(K)$ is open, with finite boundary $A'=f(K) \setminus A$ because $f$ is closed. Thus $A$ has closure $\bar A = f({\bf C})$. Since a finite set cannot disconnect the plane, $\bar A = \bf C$ .

A nice feature of these proofs is that they have straightforward (but not trivial) generalizations to higher dimensions:

Theorem: Every nonconstant, closed, holomorphic map between connected, complex n-dimensional manifolds, is surjective.

Answer by Moe Hirsch for Is there a Poincare-Hopf Index theorem for non compact manifolds?

2011-12-05T19:56:20Z

Suppose $M$ has empty boundary. Let $U\subset M$ be an open set with compact closure whose topological boundary contains no zero of the continuous vector field $X$ on $M$. Suppose $X$ is smooth and hence generates a local semiflow $f_t$ , $t \geq 0$.

For sufficiently small $t>0$ the map $f_t$ $\colon U \to M$ is defined and has a "fixed point index" $I(f_t, U)$ (see A. Dold, ``Lectures on Algebraic Topology,'' Die Grundlehren der matematischen Wissenschaften Bd. 52. Springer, New York (1972)). It can be shown that the integer $i(X,U):=I(f_t, U)$ is independent of $t$ and $U$, and is stable under perturbation of $X$.

If $X$ is not smooth, approximate it by a sequence of smooth fields $X_j$ and define $I(X,U):= \lim_{j\to \infty} I(X_j,U)$.

If $X$ has only finitely many zeros in $U$ and none on $U\cap \partial $, then $I(X, U)$ is the sum of their Poincare-Hopf indices.

If $M$ has nonempty boundary, this work if at every boundary point $p$ there is an integral curve $u\colon [0,\epsilon)\to M$ with initial point $u(0)=p$.

Orbit spaces of finite groups acting on projective varieties

2011-07-22T22:26:40Z

Let $V/G$ be the orbit space of a finite group $G$ of automorphisms of a complex projective variety $V$. Is $V/G$ a projective variety?

Example: $V/G$ is the space of sets in complex projective $n$-space $P$, of cardinality $\le k$. Here $V=P\times\dots\times P$ ($k$ factors) and $G$ is the permutation group on $k$ letters.

Comment by Moe Hirsch

2012-09-07T18:50:23Z

The paper is in J. Complexity 5 (1989), no. 4, 379--416. It shows that any algorithm based on function evaluations has a worst-case complexity that is exponential in both the number of digits of accuracy and the dimension of the ambient Euclidean space.

Comment by Moe Hirsch

2011-07-24T17:31:00Z

The natural map $V\to V/G$ should be a morphism of projective varieties.