User ben krause - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T03:47:36Z http://mathoverflow.net/feeds/user/11057 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/52902/proving-theorems-by-using-functions-with-fixed-points/52987#52987 Answer by Ben Krause for Proving theorems by using functions with fixed points. Ben Krause 2011-01-23T19:04:39Z 2011-01-23T19:16:41Z <p>I'm not sure if this is what you had in mind, but counting fixed points seem to come up often in elementary group theory, particularly in arguments involving group actions. In this setting, not only must you pick the right function (homomorphism of $G$ into an appropriate permutation group) but you also have to pick the correct "domain" (a suitable group $G$).</p> <p>For example, one way to show that all $p$-Sylow subgroups of a group are conjugate involves counting fixed points of $p$-Sylow groups under conjugation by other $p$-Sylows; a simpler (and cuter!) example, is the proof that every group of size divisible by $p$ has an order-$p$ element. To the best of my memory, this (standard) proof is found in Hungerford: Suppose $G$ is a group with $p \mid |G|$, and let $U = { (g_1,\dots,g_{p-1},x): (g_1\cdot \dots \cdot g_{p-1})\cdot x =1_{G} }$, i.e. the set of all $p$-tuples of elements in $G$ whose product is the identity. Since $x$ is uniquely determined by the $g_i$, $|U| = |G|^{p-1}$, so $p \mid |U|$ as well. Now, letting $Z/pZ$ act on $U$ by cyclic permutation yields a fixed set with size divisible by $p$, but greater than one, for at least one non-trivial element $(g_1,\dots,g_{p-1},x) \in U$ which is invariant under cyclic permutation, i.e. some $g_1=\dots=g_{p-1}=x \neq 1_{G}$. Consequently, we have $x^p = 1_{G}$, as desired.</p> http://mathoverflow.net/questions/48771/proofs-that-require-fundamentally-new-ways-of-thinking/49077#49077 Answer by Ben Krause for Proofs that require fundamentally new ways of thinking Ben Krause 2010-12-11T20:57:29Z 2010-12-11T20:57:29Z <p>How about Rabinowitsch's proof of the Nullstellensatz?</p> http://mathoverflow.net/questions/29252/geometric-proof-of-the-vandermonde-determinant/48039#48039 Answer by Ben Krause for Geometric proof of the Vandermonde determinant? Ben Krause 2010-12-02T12:01:36Z 2010-12-02T12:01:36Z <p>A professor of mine brought this question up during a seminar the other day, and I offered a "peudo-geometric" answer, interpreting the Gaussian elimination proof in terms of volume preserving shear maps $\tau_{\lambda} \in End(R^n)$ of the form $(x_1,x_2,\dots,x_n) \mapsto (x_1-\lambda,x_2-\lambda x_1,\dots,x_n-\lambda x_{n-1})$ for $\lambda \in R$ scalar. Essentially, this argument boiled down to the fact that Gaussian row-reduction operations are "well-behaved" with respect to volume. I wonder if this cuts it?</p>