User moe hirsch - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T15:45:12Z http://mathoverflow.net/feeds/user/16671 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/105883/can-we-actually-find-any-fixed-points-with-brouwers-theorem/106044#106044 Answer by Moe Hirsch for Can we actually find any fixed points with Brouwer's theorem? Moe Hirsch 2012-08-31T14:59:16Z 2012-08-31T19:12:10Z <p>The paper "Exponential lower bounds for finding Brouwer fixed points"</p> <p><em>Addendum by original poster:</em> 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. <a href="https://dl.dropbox.com/u/74531549/HPVBrouwer.pdf" rel="nofollow">Here</a> 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</p> http://mathoverflow.net/questions/10535/ways-to-prove-the-fundamental-theorem-of-algebra/104294#104294 Answer by Moe Hirsch for Ways to prove the fundamental theorem of algebra Moe Hirsch 2012-08-08T18:09:01Z 2012-08-08T18:09:01Z <p>Two very short proofs, mostly topological, that a nonconstant polynomial map <code>$f:{\bf C} \to \bf C$</code> is surjective (joint work with Robert Palais):</p> <p><strong>(1)</strong> Complex analysis shows that <code>$f$</code> is an open map (images of open sets are open). A standard estimate, <code>$|f(z)|\to\infty$</code> as <code>$|z|\to\infty$</code>, implies<code>$f$</code> is also a closed map (images of closed sets are closed). Thus <code>$f(\bf C)$</code> is an open, closed, nonempty subset of the connected space <code>$\bf C$</code>, therefore <code>$f(\bf C)=\bf C$</code>.</p> <p>The openness of <code>$f$</code> is nontrivial, but it can be replaced by elemntary algebra and topology:</p> <p><strong>(2)</strong> The set <code>$K$</code> of roots of <code>$f'$</code> is finite. The inverse function theorem shows that the set <code>$A:=f({\bf C})\setminus f(K)$</code> is open, with finite boundary <code>$A'=f(K) \setminus A$</code> because <code>$f$</code> is closed. Thus <code>$A$</code> has closure $\bar A = f({\bf C})$. Since a finite set cannot disconnect the plane, <code>$\bar A = \bf C$</code>.</p> <p>A nice feature of these proofs is that they have straightforward (but not trivial) generalizations to higher dimensions:</p> <p><strong>Theorem</strong>: Every nonconstant, closed, holomorphic map between connected, complex <em>n</em>-dimensional manifolds, is surjective.</p> http://mathoverflow.net/questions/52023/is-there-a-poincare-hopf-index-theorem-for-non-compact-manifolds/82731#82731 Answer by Moe Hirsch for Is there a Poincare-Hopf Index theorem for non compact manifolds? Moe Hirsch 2011-12-05T19:56:20Z 2011-12-05T19:56:20Z <p>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 <code>$f_t$</code>, $t \geq 0$.</p> <p>For sufficiently small $t>0$ the map <code>$f_t$</code>$\colon U \to M$ 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 $X$. </p> <p>If $X$ is not smooth, approximate it by a sequence of smooth fields $X_j$ and define $I(X,U):= \lim_{j\to \infty} I(X_j,U)$. </p> <p>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.</p> <p>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 $u(0)=p$.</p> http://mathoverflow.net/questions/71027/orbit-spaces-of-finite-groups-acting-on-projective-varieties Orbit spaces of finite groups acting on projective varieties Moe Hirsch 2011-07-22T22:26:40Z 2011-07-23T19:57:45Z <p>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?</p> <p>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. </p> http://mathoverflow.net/questions/105883/can-we-actually-find-any-fixed-points-with-brouwers-theorem/106044#106044 Comment by Moe Hirsch Moe Hirsch 2012-09-07T18:50:23Z 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. http://mathoverflow.net/questions/71027/orbit-spaces-of-finite-groups-acting-on-projective-varieties Comment by Moe Hirsch Moe Hirsch 2011-07-24T17:31:00Z 2011-07-24T17:31:00Z The natural map $V\to V/G$ should be a morphism of projective varieties.