User ben - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T23:33:04Z http://mathoverflow.net/feeds/user/2648 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/14910/can-there-exist-two-non-equivalent-equivariant-actions-of-a-group-on-vector-bundl/14936#14936 Answer by Ben for Can there exist two non-equivalent equivariant actions of a group on vector bundle? Ben 2010-02-10T19:20:18Z 2010-02-10T19:20:18Z <p>For an example from Algebraic Geometry, $\mathbb{P}^1_\mathbb{C}$ is a homogeneous space for both $GL(2)$ and $SL(2)$. The trivial line bundle and the determinant line bundle are isomorphic as $SL(2)$-equivariant bundles, and therefore as vector bundles, but are non-isomorphic as $GL(2)$-equivariant bundles.</p> http://mathoverflow.net/questions/10271/what-does-the-typical-non-solvable-group-look-like/10278#10278 Answer by Ben for What does the typical non-solvable group look like? Ben 2009-12-31T14:56:59Z 2009-12-31T14:56:59Z <p>I don't have reasonable access to internet at the moment, but I will edit this and add references when I can.</p> <p>There is an old paper called "Almost Every group is solvable" where one considers a finite group and its jordan holder decomposition. Ignoring all the factors which are cyclic groups, one multiplies the size of the remaining factors and divides by the size of the group. This gives a number which is &lt;=1, and is equal to 1 only for nonabelian simple groups. They show in that paper that the "average" over all groups of this statistic is 0. In other words, most simple composition factors are cyclic abelian groups.</p> <p>I do not know enough about PSL_2(F_p) to say whether this fits the bill (in other words, as p increases, what is the chart of this statistic).</p> http://mathoverflow.net/questions/4841/what-precisely-is-categorification/9432#9432 Answer by Ben for What precisely Is "Categorification"? Ben 2009-12-20T17:58:44Z 2009-12-20T17:58:44Z <p>There are already many good answers to this question given, but I would like to emphasize one aspect of (what it is good for) that hasn't been fully discussed yet.</p> <p>It's all about the morphisms.</p> <p>For example, knowing the knowing the Betti numbers of a topological space is really enough to identify cohomology spaces as vector spaces, but this is uninteresting. What is exciting is that suddenly the theory becomes <em>functorial</em>. There is no notion of a "morphism" from the betti numbers of one topological space to those of another, but having morphisms in cohomology effectively gives rise to all the interesting features one could desire - cup products, etc. In addition, one can now take invariants of morphisms (like traces on homology) instead of just invariants of the spaces themselves.</p> <p>In similar fashion, if one has an additive category with the Krull-Schmidt property, then each element of the additive Grothendieck group uniquely identifies its corresponding object up to isomorphism. It is not in the <em>objects</em> of a categorification where any interesting new information lies, but in the <em>morphisms</em>. Quantum groups had a geometric categorification for some time now, but recent exciting work of Rouquier and Khovanov-Lauda have redescribed this same categorification (see results of Vasserot-Varagnolo). What makes the recent results exciting is that they give an explicit presentation of the morphisms in the category, which was previously not well understood. This has led to a number of new results, but the full implications are still being explored.</p> <p>Categorification is not just a way to find new invariants, it is a way to add new layers of structure.</p> http://mathoverflow.net/questions/7155/famous-mathematical-quotes/9180#9180 Answer by Ben for Famous mathematical quotes Ben 2009-12-17T15:55:23Z 2009-12-17T15:55:23Z <p>Farkas Bolyai to his son Janos, speaking about attempts to study Euclid's Vth postulate on parallel lines:</p> <p>"You must not attempt this approach to parallels. I know this way to its very end. I have traversed this bottomless night, which extinguished all light and joy of my life. I entreat you, leave the science of the parallels alone... I thought I would sacrifice myself for the sake of truth. I was ready to become a martyr who would remove the flaw from geometry and return it purified to mankind. I accomplished monstrous, enormous labors; my creations are far better than those of others and yet I have not achieved complete satisfaction.... I turned back when I saw that no man can reach the bottom of the night. I turned back unconsoled, pitying myself and all mankind.</p> <p>I admit that I expect little from the deviation of your lines. It seems to me that I have been in these regions; that I have traveled past all reefs of this infernal Dead Sea and have always come back with broken mast and torn sail. The ruin of my disposition and my fall date back to this time. I thoughtlessly risked my life and happiness - aut Caesar aut nihil."</p> http://mathoverflow.net/questions/17617/why-are-the-sporadic-simple-groups-huge/17622#17622 Comment by Ben Ben 2010-03-09T20:22:43Z 2010-03-09T20:22:43Z In another way, sporadic finite simple groups are extremely large, compared to our understanding of general groups. A sylow subgroup of size $2^{46}$? No one even knows how many p-groups of size $p^8$ there are, let alone what they are, and the Higman PORC Conjecture is still a conjecture. http://mathoverflow.net/questions/16570/are-there-positive-formulae-for-the-inner-product-between-elements-of-a-lie-algeb/16982#16982 Comment by Ben Ben 2010-03-04T01:06:07Z 2010-03-04T01:06:07Z Uh, sorry, replace $m$ in that formula with the quantum-analog, which keeps track of the lengths of the permutations involved. Similarly, replace $k!$ with $[k]!$. http://mathoverflow.net/questions/16570/are-there-positive-formulae-for-the-inner-product-between-elements-of-a-lie-algeb/16982#16982 Comment by Ben Ben 2010-03-03T17:58:05Z 2010-03-03T17:58:05Z Unfortunately, Ben wants an answer which depends very much on $\lambda$. The formula for the inner product on $U_q^-$ is actually quite easy: $(F_{i_1}\ldots F_{i_k},F_{j_1}\ldots F_{j_l})$ will be zero unless the $i$ sequence is a permutation of the $j$ sequence, and in that case, the answer will be $\frac{m}{(1-t^2)^k}$ where $m$ is the number of permutations sending the $j$ sequence to the $i$ sequence. For example, $(F_i^k,F_i^k)=\frac{k!}{(1-t^2)^k}$. This works because the Shapovalov form is a decategorification of the Hom bifunctor on Khovanov and Lauda's categorification of $U_q^-$.