most recent 30 from http://mathoverflow.net 2013-05-25T01:31:52Z http://mathoverflow.net/feeds/user/326 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/1684/why-is-the-exterior-algebra-so-ubiquitous/1870#1870 pasquale zito 2009-10-22T13:25:07Z 2009-10-22T13:25:07Z

<p>Somehow, I can't resist (re)formulating an answer in the following minimalistic and tautological way: the exterior algebra appears each time you consider the tensor algebra generated by a vector space (e.g., you want do define a notion of volume for n-parallelepipeds spanned by n-tuples of vectors) and you want to quotient w.r.t. the ideal generated by products of the same vector (you want flat parallelepipeds to have zero volume).</p> <p>It's an "I won't repeat myself" statement which, in its peremptory simplicity, is likely to appear in the early evolutionary stages of many mathematical ideas. </p>

http://mathoverflow.net/questions/500/finite-groups-with-the-same-character-table/504#504 pasquale zito 2009-10-14T18:33:26Z 2009-10-14T18:45:58Z

<p>They're not necessarily equivalent as tensor categories. </p> <p>However, there are examples of finite groups (smallest of order 64) with representation categories which are equivalent as tensor categories but not as symmetric tensor categories (see e.g. <a href="http://arxiv.org/abs/math/0007196" rel="nofollow">http://arxiv.org/abs/math/0007196</a>). In other words, in some cases the same abstract tensor category might be endowed with inequivalent symmetric structures (you can think of these as the pullback of the standard symmetry of the category of vector spaces through inequivalent embedding functors).</p>