User reza rezazadegan - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T01:34:02Z http://mathoverflow.net/feeds/user/22721 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/109287/decorations-in-szabos-combinatorial-spectral-sequence Decorations in Szabo's combinatorial spectral sequence Reza Rezazadegan 2012-10-10T10:00:56Z 2013-04-22T07:53:15Z <p>Szabo in <a href="http://arxiv.org/abs/1010.4252" rel="nofollow">http://arxiv.org/abs/1010.4252</a> gives a combinatorial candidate for what an explicit calculation of the spectral sequence of branched double covers should yield. In other words he gives a conjectural combinatorial model for HF-hat of branched double covers of links. </p> <p>However the input for his algorithm is not a bare link diagram; it's a "dcorated" diagram. The decoration is a choice of orientations for the arcs which connect the two strands of the link at crossings. Szabo sais in the paper that these decorations are analogues of the extra structure given by Heegaard diagrams and almost complex structures in Heegard-Floer homology. But I don't understand how.</p> <p>So, my question is, what is the analogue of Szabo's decorations in Heegaard-Floer homology? (Since the two theories are expected to be isomorphic, there should be such an analogue.) I suspect this has to do with Lipshitz's cylindrical reformulation but am not sure. By the way, this whole story is over Z/2 so signs are not involved.</p> http://mathoverflow.net/questions/126006/is-it-true-that-the-geodesics-on-son-and-sun-are-closed Is it true that the geodesics on SO(n) and SU(n) are closed? Reza Rezazadegan 2013-03-30T12:36:23Z 2013-03-30T13:21:23Z <p>I mean for the bi-invariant metric (but actually any metric would work). In this metric geodesics are translates of 1-parameter subgroups so we need only to show that $exp(t X)$ for any X in the lie algebra is a closed curve. Then we can use the standard forms (like the Jordan form) for matrices. My source of doubt comes from the fact that these groups don't seem to be on the list of Riemannian manifolds with periodic geodesic flow.</p> http://mathoverflow.net/questions/106717/sarkars-maslov-index-formula Sarkar's Maslov index formula Reza Rezazadegan 2012-09-09T09:12:18Z 2012-09-18T02:11:35Z <p>I have difficulty understanding Sarkar's maslov index formula in symmetric products from <a href="http://arxiv.org/abs/math/0609673" rel="nofollow">http://arxiv.org/abs/math/0609673</a>.</p> <p>If $D$ is an $n$-sided region with corner points $p_1,\ldots, p_n$ then it reads like</p> <p>$\mu(D)= \mu_{p_1}(D)+\mu_{P_n}(D)+ e(D)-g(n-2)/2+\sum_{ 1&lt; i&lt; j \leq n} \partial_j(D) \cdot \partial_i(D)$ </p> <p>where $\mu_{P_i}$ and $e$ are the point and euler measures. </p> <p>First of all it seems to treat $P_1,P_n$ different from other points. Secondly he says the euler measure of an $n$ sided region is $1-n/4$ which looks different from what Lipshitz says (which takes the accute and obtuse corners into account). Last but not least I don't understand the definition of the last term. It is defined by moving the two sides in 4 different directions in such a way that no endpoint of one is on another and then taking intersection points. It's not clear for me what these 4 directions are and why the result is not always zero.</p> http://mathoverflow.net/questions/92546/maslov-index-and-heegard-floer-homology/93392#93392 Answer by Reza Rezazadegan for Maslov index and heegard floer homology Reza Rezazadegan 2012-04-07T07:20:01Z 2012-04-07T07:20:01Z <p>Knowing that the maslov index is equal to the "expected" dimension of the moduli space of the disks is helpful too. For example in the figure on the right side of page 17, you can see that a holomorphic disk with boundary on $\alpha$ and $\beta$ can have "cuts" along the dashed lines, i.e. can send part of the boundary of the disk to the dashed lines. Since the length of these two "cuts" are variable, the moduli space is two dimensional and therefore the Maslov index is two. This also shows you that the moduli space is not compact. (Why?) </p> http://mathoverflow.net/questions/126006/is-it-true-that-the-geodesics-on-son-and-sun-are-closed/126009#126009 Comment by Reza Rezazadegan Reza Rezazadegan 2013-03-31T06:05:10Z 2013-03-31T06:05:10Z Thank you guys. I got the answer to my original question. However what if we mod out by the maximal torus? For example $U(n)/T^n$ is a flag variety and as the answe to to this Mathoverflow question <a href="http://mathoverflow.net/questions/7750/geodesics-on-a-grassmannian" rel="nofollow" title="geodesics on a grassmannian">mathoverflow.net/questions/7750/&hellip;</a> explains, the geodesics on it are given by $exp(tX)$ where $X$ lies in a complement of the Lie algebra of the maximal torus. http://mathoverflow.net/questions/126006/is-it-true-that-the-geodesics-on-son-and-sun-are-closed/126009#126009 Comment by Reza Rezazadegan Reza Rezazadegan 2013-03-30T14:25:57Z 2013-03-30T14:25:57Z OK, thanks. But the round metric on $SU(2)\cong S^3$ has closed geodesics.