most recent 30 from http://mathoverflow.net 2013-05-25T02:39:06Z http://mathoverflow.net/feeds/user/5048 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/67524/a-paradoxical-decomposition-of-a-group/67697#67697 Sven Raum 2011-06-13T19:53:13Z 2011-06-13T19:53:13Z

<p>According to a note in Grigorchuk's and Sunic's Self-Similarity and Branching in Group Theory, there is a proof not using the matching theorem in the book The Banch-Tarski Paradox by Stan Wagon. By the way, you have to mention that $\Gamma$ is also the union of $E_1,...,E_m,F_1,...,F_n$.</p>

http://mathoverflow.net/questions/58753/center-of-the-algebra-of-bounded-operators/58758#58758 Sven Raum 2011-03-17T14:34:24Z 2011-03-17T14:34:24Z

<p>If I see it correctly, then take a Hamel basis $(v_i)$ with $i \in I$ of the Banach space $X$. Then the coordinate switch $T_{i,j}:v_k \mapsto \delta_{i,k}v_j$ is bounded, because finite dimensional subspaces have a complement in a Banach space. Writing down any $S$ in the centre of $\mathscr{B}(X)$ as a matrix with rows and columns indexed by $I$, commutativity of $S$ and $T_{i,i}$ shows that is must be diagonal and commutativity of $S$ and $T_{i,j}$ shows that all diagonal entries must be the same. Hence $S$ is a scalar multiple of the identity.</p> <p>The essence of the argument is that finite dimensional subspaces are complemented in Banach spaces.</p>

http://mathoverflow.net/questions/56167/hilbert-space-having-all-norms-and-seminorms-continous/56169#56169 Sven Raum 2011-02-21T13:11:12Z 2011-02-21T13:11:12Z

<p>If every seminorm on $H$ is continuous, then every linear functional on H is continuous. Since there are unbounded linear functionals on infinite dimensional Hilbert spaces, $H$ must necessarily be finite dimensional.</p> <p>Actually, the locally convex topology on a vector space that makes all linear functionals continuous, doesn't describe anything new. It's probably useful to reformulate algebraic problems (I remember talking to someone about this), but it doesn't give more structure to the vector space.</p>

http://mathoverflow.net/questions/53251/twist-of-a-group-hopf-algebra/53612#53612 Sven Raum 2011-01-28T12:42:07Z 2011-01-28T12:42:07Z

<p>In one case you will always get a group isomorphic to the group you started with, namely if your cocycle is symmetric. Actually in <a href="http://arxiv.org/abs/1007.1412" rel="nofollow">http://arxiv.org/abs/1007.1412</a> Theorem 3.2 proves that all symmetric cocycles on a cocommutative, compact quantum group are trivial. A quantum group is to be understood in the von Neumann algebraic sense there, but for finite groups this point of view is equivalent to the Hopf algebraic $\mathbb{C}G$.</p>

http://mathoverflow.net/questions/48572/weakly-solid-factors/51850#51850 Sven Raum 2011-01-12T15:38:12Z 2011-01-12T18:18:02Z

<p>Sorry, this was a wrong example. I should have thought about it better.</p>