User moduli - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T07:14:50Z http://mathoverflow.net/feeds/user/7035 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/109084/lie-superalgebra-in-two-dimensions Lie superalgebra in two dimensions Moduli 2012-10-07T17:45:01Z 2013-03-04T20:22:00Z <p>The standard formulation of two dimensional $N=(2,2)$ and $N=(0,2)$ supersymmetry algebras in physics is an explicit one; I am not aware of the underlying abstract Lie superalgebras. Does anyone know what these alegrbras are in terms of Lie superalgebras?</p> http://mathoverflow.net/questions/123178/lie-derivative-of-curvature Lie derivative of curvature Moduli 2013-02-28T02:09:49Z 2013-02-28T03:57:00Z <p>Let $M$ be a Kahler manifold, with Kahler metric $g$. Let $X$ be a holomorphic Killing vector field of $g$, i.e. $L_{X} g = 0$, where $L_{X}$ is the Lie derivative along $X$. Let $R$ be the Riemannian curvature tensor of $g$. Is $L_{X} R = 0$? </p> http://mathoverflow.net/questions/117170/a-partial-differential-equation-on-mathbbcp1 A partial differential equation on $\mathbb{CP}^1$ Moduli 2012-12-24T22:14:02Z 2012-12-25T00:20:33Z <p>Let $f$ be any complex function on $\mathbb{CP}^1$. Denote the local coordinates of $\mathbb{CP}^1$ as $z,\bar{z}$. Does the following equation</p> <p>$\frac{\partial}{\partial z} f = \frac{\partial}{\partial \bar{z}} f = \frac{i}{2r} f$</p> <p>where $r$ is the radius of $\mathbb{CP}^1$, have nontrivial solutions? If yes, then is there an easy example?</p> http://mathoverflow.net/questions/57870/count-the-number-of-homogeneous-polynomials Count the number of homogeneous polynomials Moduli 2011-03-08T18:33:30Z 2011-03-09T07:19:19Z <p>Is there a general way of counting the number of homogeneous polynomials of certain degree in a complex projective space or a weighted complex projective space, mod the ideal generated by some homogeneous polynomials with smaller degrees? </p> <p>As an example, consider the degree 18 homogeneous polynomials in $W\mathbb{P}_{[2,2,2,4]}^3$, mod the ideal generated by two degree 8 homogeneous polynomials $P_1=x^4_1$ and $P_2=x_2^4$, where $x_1$ and $x_2$ are the first and second coordinates of $W\mathbb{P}_{[2,2,2,4]}^3$. I can count the number of equivalent classes by directly listing all of such homogeneous polynomials; I would like to know if there is a more general and efficient way of doing this.</p> <p><em>Edit</em>: to avoid possible confusion, I have replaced "polynomial" by "homogeneous polynomial".</p> http://mathoverflow.net/questions/44420/are-there-non-supersymmetric-and-or-non-calabi-yau-topological-sigma-models/45897#45897 Answer by Moduli for Are there non-supersymmetric and/or non-Calabi-Yau topological sigma models? Moduli 2010-11-13T05:25:36Z 2010-11-13T21:09:53Z <p>I believe that A-model does not require a Calabi-Yau target space. In fact, A-model is well-defined on any almost complex manifold, which was Witten's original construction (Comm. Math. Phys. Volume 118, Number 3 (1988), 411-449). On the other hand, B-model can only be defined on a Calabi-Yau manifold, which follows from anomaly cancelation.</p> <p>In general, topological field theories have many different types (not necessarily supersymmetric). As an example, Chern-Simons theory is topological. Try <a href="http://en.wikipedia.org/wiki/Topological_quantum_field_theory" rel="nofollow">http://en.wikipedia.org/wiki/Topological_quantum_field_theory</a> for some general discussion.</p> http://mathoverflow.net/questions/39772/is-symplectic-reduction-interesting-from-a-physical-point-of-view/43106#43106 Answer by Moduli for Is symplectic reduction interesting from a physical point of view? Moduli 2010-10-21T22:41:48Z 2010-10-21T22:41:48Z <p>Here is a fancy example: Supersymmetry. Rigid N=1 supersymmetric theories in 4 dimension have a natural Kahler structure on the field space. The D-term is precisely a moment map. The moduli space of the theory is the symplectic quotient from this moment map.</p> http://mathoverflow.net/questions/40797/characters-of-kac-moody-algebra-from-orbifold Characters of Kac-Moody algebra from orbifold Moduli 2010-10-01T22:20:55Z 2010-10-01T22:20:55Z <p>In a Wess–Zumino–Witten model on some Lie group G, the character of a particular integrable representation is the same as the specialized character from the corresponding Kac-Moody algebra. Suppose now we have a WZW model on an orbifold, i.e. on G/H where H is a subgroup of the center of G, then how should we compute the WZW characters for this orbifold theory? In particular, are there any relation between the original characters (WZW on G) and the orbifold characters (WZW on G/H)?</p> http://mathoverflow.net/questions/38825/what-kind-of-lagrangians-can-we-have/39185#39185 Answer by Moduli for What kind of Lagrangians can we have? Moduli 2010-09-18T01:25:59Z 2010-09-18T01:42:51Z <p>The most valuable discovery of the 20th century physics is symmetry. When you try to write down a Lagrangian for a system, usually there are two "principles": (1) the symmetries this system has and (2) experience. So no, L is not always T-U. For example, in General Relativity the Lagrangian (or rather Lagrangian density) is R, the scalar curvature of the spacetime manifold. As another example, in WZW models, the Lagrangian has two parts: a "kinetic" term plus a Wess–Zumino term. </p> <p>It is in quantum field theory that all kinds of Lagrangians pop up. What is more interesting is that the best you can do is to write down an “effective Lagrangian” at a certain energy scale, which may (and usually does) get quantum corrections at higher energy scales.</p> http://mathoverflow.net/questions/28541/complexified-kahler-form/30139#30139 Answer by Moduli for complexified kahler form Moduli 2010-07-01T04:19:05Z 2010-07-01T17:13:19Z <p>This B field is from string theory. When you quantize bosonic, type II or Heterotic string theories, you will find a massless quantum state which is a second rank tensor on spacetime. The antisymmetric part of this tensor is the B field (the symmetric part is the graviton, and the trace part is the daliton). When you compactify superstring theories on Calabi-Yau 3-fold, the moduli space of this Calabi-Yau space is interpreted as massless fields in the four dimensional effective theory. Remeber the B field is massless, which means it shoud be part of the moduli space. This is the "physical" reason why we consider the complexified kahler form.</p> http://mathoverflow.net/questions/123178/lie-derivative-of-curvature Comment by Moduli Moduli 2013-02-28T15:13:50Z 2013-02-28T15:13:50Z @Deane Yang: Thanks for your clear explanation. http://mathoverflow.net/questions/117170/a-partial-differential-equation-on-mathbbcp1/117175#117175 Comment by Moduli Moduli 2012-12-28T20:57:25Z 2012-12-28T20:57:25Z Wait, $z+\bar{z}=2x$, so the solutions is $ae^{i(z+\bar{z})/2r}=ae^{ix/r}$, which is ill defined at $x\rightarrow \infty$. http://mathoverflow.net/questions/117170/a-partial-differential-equation-on-mathbbcp1/117175#117175 Comment by Moduli Moduli 2012-12-25T00:35:11Z 2012-12-25T00:35:11Z Indeed, this seems to be the only nontrivial solution. http://mathoverflow.net/questions/117170/a-partial-differential-equation-on-mathbbcp1 Comment by Moduli Moduli 2012-12-24T23:43:42Z 2012-12-24T23:43:42Z Well, I guess I should say nontrivial solutions. http://mathoverflow.net/questions/88406/meaning-origin-of-seiberg-witten-equations-invariants/88489#88489 Comment by Moduli Moduli 2012-02-18T04:47:23Z 2012-02-18T04:47:23Z The identification between the ultraviolet region and the infrared region, I think, is from the fact that we have a TFT, which means the theory is scale invariant and hence UV=IR. http://mathoverflow.net/questions/78865/kernel-of-a-linear-map-about-polynomials Comment by Moduli Moduli 2011-10-23T05:22:57Z 2011-10-23T05:22:57Z Fair enough. I should have said that $A$ is in fact a vector space. Modified the question. http://mathoverflow.net/questions/77192/chern-simons-functional Comment by Moduli Moduli 2011-10-05T12:57:12Z 2011-10-05T12:57:12Z Good points, guys. http://mathoverflow.net/questions/57870/count-the-number-of-homogeneous-polynomials/57874#57874 Comment by Moduli Moduli 2011-03-09T05:00:11Z 2011-03-09T05:00:11Z I used Macaulay2 as suggested, and I got 60. I suppose we need to add s(n-16). Thank you guys. http://mathoverflow.net/questions/44420/are-there-non-supersymmetric-and-or-non-calabi-yau-topological-sigma-models/45897#45897 Comment by Moduli Moduli 2010-11-13T21:08:55Z 2010-11-13T21:08:55Z A-model was discovered ealier than B-model. The original A-model paper (in whihc the name &quot;A-model&quot; has not been invented yet) is the one I gave. I am guessing the paper you are reading is the classic one &quot;Mirror Manifolds And Topological Field Theory&quot; (hep-th/9112056). If so, it was pointed out in that paper that A-model can be defined on almost complex manifolds, while B-model can only be defined on a Calabi-Yau manifold, which follows from anomaly cancelation. http://mathoverflow.net/questions/39772/is-symplectic-reduction-interesting-from-a-physical-point-of-view/43106#43106 Comment by Moduli Moduli 2010-10-22T02:15:20Z 2010-10-22T02:15:20Z Related to this, in a gauged supergravity theory, we do not have a symplectic quotient anymore; instead we have a GIT quotient. http://mathoverflow.net/questions/39772/is-symplectic-reduction-interesting-from-a-physical-point-of-view/43106#43106 Comment by Moduli Moduli 2010-10-22T02:13:40Z 2010-10-22T02:13:40Z J. Bagger and E. Witten, “The gauge invariant supersymmetric nonlinear sigma model,” Phys. Lett. B118 (1982) 103–106. J. Bagger and J. Wess, “Gauging the supersymmetric sigma model with a Goldstone field,” Phys. Lett. B199 (1987) 243–246. http://mathoverflow.net/questions/40797/characters-of-kac-moody-algebra-from-orbifold Comment by Moduli Moduli 2010-10-14T15:51:26Z 2010-10-14T15:51:26Z In particular, how can one decomposite the representations of affine G into representations of affine G/H? This should be the first step of the question. http://mathoverflow.net/questions/38825/what-kind-of-lagrangians-can-we-have/39185#39185 Comment by Moduli Moduli 2010-09-18T15:33:32Z 2010-09-18T15:33:32Z Good point. But energy in GR has never been well defined, so it is tricky to talk about &quot;energy&quot; in GR. You can say R is like a potential energy by some sort of analogies, but personally I do not think this is a good argument.