User moduli - MathOverflowmost recent 30 from http://mathoverflow.net2013-05-22T07:14:50Zhttp://mathoverflow.net/feeds/user/7035http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/109084/lie-superalgebra-in-two-dimensionsLie superalgebra in two dimensionsModuli2012-10-07T17:45:01Z2013-03-04T20:22:00Z
<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-curvatureLie derivative of curvatureModuli2013-02-28T02:09:49Z2013-02-28T03:57:00Z
<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-mathbbcp1A partial differential equation on $\mathbb{CP}^1$Moduli2012-12-24T22:14:02Z2012-12-25T00:20:33Z
<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-polynomialsCount the number of homogeneous polynomialsModuli2011-03-08T18:33:30Z2011-03-09T07:19:19Z
<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#45897Answer by Moduli for Are there non-supersymmetric and/or non-Calabi-Yau topological sigma models?Moduli2010-11-13T05:25:36Z2010-11-13T21:09:53Z<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#43106Answer by Moduli for Is symplectic reduction interesting from a physical point of view?Moduli2010-10-21T22:41:48Z2010-10-21T22:41:48Z<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-orbifoldCharacters of Kac-Moody algebra from orbifoldModuli2010-10-01T22:20:55Z2010-10-01T22:20:55Z
<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#39185Answer by Moduli for What kind of Lagrangians can we have?Moduli2010-09-18T01:25:59Z2010-09-18T01:42:51Z<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#30139Answer by Moduli for complexified kahler form Moduli2010-07-01T04:19:05Z2010-07-01T17:13:19Z<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-curvatureComment by ModuliModuli2013-02-28T15:13:50Z2013-02-28T15:13:50Z@Deane Yang: Thanks for your clear explanation.http://mathoverflow.net/questions/117170/a-partial-differential-equation-on-mathbbcp1/117175#117175Comment by ModuliModuli2012-12-28T20:57:25Z2012-12-28T20:57:25ZWait, $z+\bar{z}=2x$, 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#117175Comment by ModuliModuli2012-12-25T00:35:11Z2012-12-25T00:35:11ZIndeed, this seems to be the only nontrivial solution. http://mathoverflow.net/questions/117170/a-partial-differential-equation-on-mathbbcp1Comment by ModuliModuli2012-12-24T23:43:42Z2012-12-24T23:43:42ZWell, I guess I should say nontrivial solutions.http://mathoverflow.net/questions/88406/meaning-origin-of-seiberg-witten-equations-invariants/88489#88489Comment by ModuliModuli2012-02-18T04:47:23Z2012-02-18T04:47:23ZThe 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-polynomialsComment by ModuliModuli2011-10-23T05:22:57Z2011-10-23T05:22:57ZFair enough. I should have said that $A$ is in fact a vector space. Modified the question.http://mathoverflow.net/questions/77192/chern-simons-functionalComment by ModuliModuli2011-10-05T12:57:12Z2011-10-05T12:57:12ZGood points, guys.http://mathoverflow.net/questions/57870/count-the-number-of-homogeneous-polynomials/57874#57874Comment by ModuliModuli2011-03-09T05:00:11Z2011-03-09T05:00:11ZI 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#45897Comment by ModuliModuli2010-11-13T21:08:55Z2010-11-13T21:08:55ZA-model was discovered ealier than B-model. The original A-model paper (in whihc the name "A-model" has not been invented yet) is the one I gave. I am guessing the paper you are reading is the classic one "Mirror Manifolds And Topological Field Theory" (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#43106Comment by ModuliModuli2010-10-22T02:15:20Z2010-10-22T02:15:20ZRelated 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#43106Comment by ModuliModuli2010-10-22T02:13:40Z2010-10-22T02:13:40ZJ. 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-orbifoldComment by ModuliModuli2010-10-14T15:51:26Z2010-10-14T15:51:26ZIn 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#39185Comment by ModuliModuli2010-09-18T15:33:32Z2010-09-18T15:33:32ZGood point. But energy in GR has never been well defined, so it is tricky to talk about "energy" 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.