User shripad - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-18T06:41:25Z http://mathoverflow.net/feeds/user/9991 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/90441/non-abelian-divisible-groups/125156#125156 Answer by Shripad for Non-abelian divisible groups Shripad 2013-03-21T11:31:27Z 2013-03-21T11:31:27Z <p>Any connected, compact Lie group has this property. To get convinced about this, take an element $g \in G$ where $G$ is a compact Lie group, then $g$ can be put inside a torus $T$ in $G$, and a torus being a direct product of finitely many copies of circle, $S^1$, the result now follows.</p> <p>Caution: One needs to prove the result that every element of $G$ has an $n$-th root to show that every $g \in G$ can be put in a torus. The proof uses degree argument of a smooth map between manifolds. </p> http://mathoverflow.net/questions/121959/classification-of-tori-of-gl2-up-to-conjugation/121974#121974 Answer by Shripad for Classification of Tori of GL2, up to conjugation Shripad 2013-02-16T07:23:11Z 2013-02-16T07:23:11Z <p>The $n$-dimensional tori up to isomorphism can be described by Galois cohomology, see for instance, Serre's Galois cohomology, or my thesis (http://arxiv.org/abs/math/0409453) for more details. These are given by (equivalence classes of) Galois representations taking values in $GL_n(\mathbb{Z})$.</p> <p>For instance, for real numbers, the Galois group is $\mathbb{Z}/2\mathbb{Z}$ and has three indecomposable integral representations; the trivial 1-dimensional representation, the sign representation and the 2 dimensional permutation representation (where the Galois group operates by permuting a basis). Thus we get three 2-dimensional tori over $\mathbb{R}$; the split one, the restriction of scalars from $\mathbb{G}_m$ from $\mathbb{C}$ to $\mathbb{R}$, and the $SO_2 \times \mathbb{G}_m$. Since the last torus is not embeddable in $GL_2$, we get only two types of tori in $GL_2$ over $\mathbb{R}$. </p> <p>Over $\mathbb{Q}$, the Galois group is much more complicated and hence has many more two-dimensional representations (one corresponding to each quadratic extension). Thus we get many non-conjugate maximal tori in $GL_2$ over $\mathbb{Q}$.</p> http://mathoverflow.net/questions/87053/papers-archives-especially-not-indexed-by-google/87130#87130 Answer by Shripad for papers archives? (especially not indexed by google) Shripad 2012-01-31T12:44:13Z 2012-01-31T12:44:13Z <p>Ulf Rehmann from Bielefeld has a comprehensive list of electronically available journals (with 318 items)</p> <p><a href="http://www.math.uni-bielefeld.de/~rehmann/DML/dml_links.html" rel="nofollow">http://www.math.uni-bielefeld.de/~rehmann/DML/dml_links.html</a></p> <p>As Chandan says, this list could perhaps be added to AMS.</p> http://mathoverflow.net/questions/51484/prime-numbers-dividing-the-orders-of-the-sporadic-groups/86887#86887 Answer by Shripad for Prime numbers dividing the orders of the sporadic groups Shripad 2012-01-28T07:49:48Z 2012-01-28T07:49:48Z <p>A trivial remark: Since the number of these primes is less than the number of sporadic groups, the multiplicative subgroup of ${\mathbb Q}^{\times}$ generated by the orders of sporadic groups is not free on 26 (or 27) generators. :-).</p> http://mathoverflow.net/questions/77985/locally-connected-versus-locally-compact locally connected versus locally compact Shripad 2011-10-13T02:28:25Z 2011-10-14T10:47:26Z <p>In the definition of a locally connected space we demand every neighbourhood of a point to satisfy certain condition whereas for a locally compact space we demand that one neighbourhood be there with the required property. </p> <p>Is there some reason for this difference? Is it so that a compact space needs(?) to be locally compact?</p> <p>Connectedness is a "geometric" property whereas compactness is an "analytic" property. Is that a reason behind such different definitions?</p> http://mathoverflow.net/questions/76835/subgroups-with-the-same-number-of-roots-that-the-group/76848#76848 Answer by Shripad for subgroups with the same number of roots that the group. Shripad 2011-09-30T12:25:52Z 2011-09-30T12:25:52Z <p>Removing a random edge and keeping the nodes do not give you a subgroup. The subgroups you mention are obtained by adding one node (and a few edges) and then by removing an inner node. Perhaps you should see Borel-Siebenthal's paper on maximal subgroups of maximal rank in compact Lie groups' for more on this process. </p> http://mathoverflow.net/questions/14574/your-favorite-surprising-connections-in-mathematics/71314#71314 Answer by Shripad for Your favorite surprising connections in Mathematics Shripad 2011-07-26T13:48:19Z 2011-07-26T13:48:19Z <p>Root systems, which are completely combinatorial objects have a lot to do with topological objects, such as compact Lie groups, and linear algebraic objects, such as Lie algebras. Not just that, they classify semisimple ones among them!</p> http://mathoverflow.net/questions/14574/your-favorite-surprising-connections-in-mathematics/71241#71241 Answer by Shripad for Your favorite surprising connections in Mathematics Shripad 2011-07-25T15:57:17Z 2011-07-25T15:57:17Z <p>The amazing connection between $\eta$-identities and affine root systems, due to Macdonald and further elaborated upon by Kac! These identities encompass the jacobi triple product identity, Euler's pentagonal number identity and many others. And these have connections to Complex simple Lie algebras.</p> http://mathoverflow.net/questions/34424/number-of-finite-simple-groups-of-given-order-is-at-most-2-is-a-classification/70645#70645 Answer by Shripad for Number of finite simple groups of given order is at most 2 - is a classification-free proof possible? Shripad 2011-07-18T16:58:05Z 2011-07-18T16:58:05Z <p>Emil Artin proved in 1955 in two papers that the above mentioned examples are the only instances of non-isomorphic finite simple groups having the same order. He proved the result only for the groups that were known till then. As new groups were being discovered Jacques Tits took the responsibility of checking that there were no such further cases. For an exposition of this, one may look in Kimmerle and others, Proc. London Math. Soc. 60(3) (1990) 89–122'.</p> <p>So, indeed the classification is used to some extent.</p> http://mathoverflow.net/questions/90441/non-abelian-divisible-groups/125156#125156 Comment by Shripad Shripad 2013-03-26T03:56:10Z 2013-03-26T03:56:10Z Yes Misha, that will also do. But I thought that the knowledge of (even existence of) Riemannian metric is difficult than the basic cohomology theory of compact manifolds. :-). http://mathoverflow.net/questions/121959/classification-of-tori-of-gl2-up-to-conjugation/121974#121974 Comment by Shripad Shripad 2013-02-18T06:59:11Z 2013-02-18T06:59:11Z Thanks Will, your comment elaborates my answer very well. Jeremy, by complicated for $\mathbb{Q}$ I meant only in comparison with the case over $\mathbb{R}$. http://mathoverflow.net/questions/57337/when-should-a-supervisor-be-a-co-author/57452#57452 Comment by Shripad Shripad 2012-05-29T05:37:49Z 2012-05-29T05:37:49Z @ Gil, non-related to the question possibly but I sometimes write (s)he. But then there is also the problem of his versus her. http://mathoverflow.net/questions/91696/are-torsors-over-unipotent-groups-trivial Comment by Shripad Shripad 2012-03-20T12:28:15Z 2012-03-20T12:28:15Z Just adding to Jason's comment that over perfect fields the torsors are indeed trivial. http://mathoverflow.net/questions/84255/how-about-the-lie-algebra-over-commutative-ring/84284#84284 Comment by Shripad Shripad 2011-12-26T17:34:09Z 2011-12-26T17:34:09Z He perhaps means it to be `associative'. http://mathoverflow.net/questions/80056/using-slides-in-math-classroom/80098#80098 Comment by Shripad Shripad 2011-11-06T11:23:25Z 2011-11-06T11:23:25Z ice fried chicken cream? a bad joke perhaps. http://mathoverflow.net/questions/77985/locally-connected-versus-locally-compact/78034#78034 Comment by Shripad Shripad 2011-10-14T07:24:02Z 2011-10-14T07:24:02Z Thank you Terry for the explanation. Now I guess I understand the modifier &quot;locally&quot; better.