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As you mention ncatlab I would bet that you have already revised all this... Looking in the net for old discussions and papers, it seems that the an open question was about defining the open sets beyond 1+1 dimensions. Of course (1+1) has a lot of niceties, I remember Borcherds -with 'd'- exploited very well them. The net of open sets must be consistent ...

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The equivariant moment map has several applications. Its meaning is that it provides an encoding of how the Lie group $G$ acts on the phase space, and it gives you a way to find the observables corresponding to the conserved quantities/generators of the symmetry $G$: It also defines the process of symplectic reduction to a reduced phase space. Given that $$... 0 A helpful starting point is to work backwards from the definition of the Christoffel symbols. Hope that helps. 1 yep, think of  \xi  as a unit vector and replace all instances of it with  \epsilon \xi where  \epsilon  is some small number. Then you will see that those two terms are second order in  \epsilon . 2 I don't think there is any universal "intuition" to tap into, aside from that which comes from practice. You perhaps need to explore different physics texts in the electromagnetic department. I for example loathed Jackson as a learning text: it is comprehensive and useful as a reference for refreshing knowledge, but not good at conveying it. Volume 2 of the ... 0 I'm going to answer this as a non- conformal field theorist, but I have been thinking about this stuff a bit lately. But I do believe I can answer the simplest of your questions: I'm sure an expert will put the following straight if there are mistakes. "So then a cordinate transformation which acts on the metric as a Weyl transformation is a conformal ... 1 Yes. The short answer is you have one action you extremize to get Einstein's Field Equation G_{\alpha\beta}=kT_{\alpha\beta}. Which you can think of as equations of motion for the gravitational metric g_{\alpha\beta}. (They determine the second derivatives of the metric in terms of the matter fields and metric and the first derivatives of the metric.) ... 1 GR can be recast into an equivalent but conceptually quite different form, using teleparallel gravity. This approach introduces the Weitzenboeck connection, which has no curvature, but has torsion. The presence of torsion indicates that gravity is not geometrized. Recall that in GR, we can always choose a locally inertial coordinate system such that the ... 0 I wrote this answer assuming you don't know much GR The analogue of the Electromagnetic field in General Relativity is the metric tensor g_{\mu\nu}. The value of the metric is determined by the mass energy distribution via Einsteins equation, which is analogous to Maxwell's Equations. In the equation below G_{\mu\nu} is a tensor which depends on ... 3 The Gauge Theory of Gravity (GTG) by Lasenby, Doran and Gull has a background spacetime with fields on it. It is basically derived from the same physical principles but as a background theory. It ends up not being the same theory, for instance it doesn't have the same isotropic solutions, and I think it does not allow time travel and such (unlike General ... 0 Sorry to be late to the party, I did an answer yesterday and then had some internet glitch and couldn't post it. I shall dust it off: Can we say that gravity (indirectly) is responsible for motion of electrons around nucleus? No, I'm afraid not. Note that an electron doesn't actually go round a proton (the simplest nucleus) like a planet round a Sun. ... 2 First, we will compute \text{div}\,F. The partial derivatives are given by$${{\partial F}\over{\partial x_i}} = q {\partial\over{\partial x_i}}\left({{x_i}\over{r^3}}\right) = q\left({1\over{r^3}} - {{3r^2 {{x_i}\over{r}} x_i}\over{r^6}}\right) = 0.$$Thus, \text{div}\,F = 0 away from the origin. Consider now a ball B of radius r centered at the ... 3 First some terminology: A non-degenerate 2-form \omega is called an almost symplectic structure. A closed 2-form \omega is often called a pre-symplectic structure. If the 2-form \omega is both non-degenerate and closed, it becomes a symplectic structure. In the non-degenerate case, the closedness condition$$\mathrm{d}\omega~=~0\tag{C}$$is ... 0 The basis change for the Clifford algebra$$ \{\Gamma_A, \Gamma_B\}_+~=~2\eta_{AB} \tag{1.12} $$can be viewed as a linear coordinate transformation in the (complexified) Minkowski space^1 V of dimension 2n. The longitudinal and temporal coordinates go into light-cone coordinates. The transversal coordinates pair up two and two and are transformed in ... 3 Gravity is very, very weak compared to the electromagnetic force. Whatever curvature of space-time is induced inside an atom would be far too weak to hold mass-energy together at that scale. The electromagnetic force is 1/137 the strength of the strong nuclear force, while Gravity is 6 * 10^-39 the strength of the strong nuclear force. Here is a link that ... 3 First, terminology: Symmetry groups are not "defined on domains". Symmetry groups exist in the abstract, and they are then represented on certain spaces. If we have a spacetime manifold \mathcal{M}, then the fields are functions$$ f : \mathcal{M} \to V$$where V is some vector space upon which a representation \rho : \mathrm{SO}(1,3)\to ... 3 A point to emphasize here is that one cannot separate the rest mass of the nucleons and electrons from that of field: rest masses aren't additive in this way and one can only state the rest mass of a system as a whole. The mass-energy of the fields is already built into the rest mass of the system. The rest mass of an electron includes the energy of the ... 0 The simple answer is that if an object falls and accelerates, its X-T graph will be curved so its worldline would be curved so movement of the object will be a curve through space time and we only see space so we see it as a straight line. But we can see curvature of spacetime in Earth revolving around the Sun 2 MS Mohamed et al begin with a "standard vector calculus formulation of the NS equations",$$ \frac{∂ \bf u}{∂ t} - \mu ∆ {\bf u} + ({\bf u} \cdot \nabla) {\bf u} + \nabla p = 0  \nabla \cdot {\bf u} = 0 $$and use the rotational form,$$ \frac{∂ \bf{u}}{∂ t} + \mu \nabla \times \nabla \times {\bf u} - {\bf u}\times(\nabla \times {\bf u}) + \nabla p^d ...

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When introducing the stress tensor to undergraduate engineers, I note that, if a body is deformed by a stress state, and we imagine it sliced by a plane, then there is a (force) vector acting across (but not normal to) the plane. To fully specify the stress state, you need to cut the body with three planes, each of which can be specified by its normal ...

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A bit late for an answer, but I think I read your question differently to how others have read it. If I understand correctly, you are expecting a totally empty universe (and with lambda = 0) to be flat. I think this is completely accurate (perhaps with the exception of traveling gravitational waves, as mentioned already - if you allow such things in an ...

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The spin connection has a geometric meaning - it is a connection associated to a particular non-coordinate frame, which diagonalizes the metric. Here's how: Let $M$ be our spacetime. The tangent bundle $TM$ may be thought of as the associated bundle to an $\mathrm{SO}(n)$-principal bundle, where the $\mathrm{SO}(n)$ matrices represent ordered orthonormal ...

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I'll elaborate on the things mentioned in the comments. That being said, this will refer mostly to very specific foliations and not much of a general theory. First, since manifolds in GR have a timelike and a spacelike component, it's always worth keeping track of those. Then foliations arise very naturally trying to see whether or not there is a reasonable ...

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The only orthonormal coordinate basis is the Cartesian coordinate basis. The basis vectors for the, e.g., polar coordinate basis are orthogonal but not normalized. That doesn't mean that one can't normalize the polar basic vectors to get the polar unit basis but such a basis isn't a coordinate basis. For the Cartesian coordinate basis, the basis vectors ...

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Technically speaking, manifolds are by definition topological spaces, which resemble locally an inner-product space. Since there are vector spaces (with a dot product) of infinite dimension, then there shall be infinately-dimensional manifolds as well. The infinity of the dimension is not a problem for the tensors as well - each multi-linear function over a ...

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If you have a coordinate system you could move along a coordinate, which indicates some vectors you could use for a basis. These vectors might be orthogonal, that depends on your coordinates (think, does the metric look diagonal in those coordinates)? But even if your coordinates are orthogonal then you still have to pick a magnitude for these vectors. ...

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The name "1-form" is only used because there are 2-forms and 3-forms and so on. An equivalent but probably better name for 1-forms in this case is that of "dual vector." Any vector space will have associated with it the dual space consisting of all linear functions that map vectors into scalars. This is all a dual vector is: a thing that linearly maps ...

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Suppose you have a skewed coordinate system, with vectors that have components $v^i$ defined relative to some basis vectors $\hat e_i$ such that $\vec v = v^i \hat e_i$ in the Einstein summation convention (we sum over any index which is repeated once above, once below). The meaning of a skewed coordinate system is that $\hat e_i \cdot \hat e_j \ne ... 5 Notation$\renewcommand{braket}[2]{\langle #1 | #2 \rangle}$First let us define the term "1-form": A 1-form is a linear function which takes a single vector as it's argument. That is it. It is no more confusing or complicated than that single statement. Consider a vector$v$. Given any other vector$w$we can form the inner product$\braket{v}{w}$. ... 1 Comments to the question (v3): Note that the gamma matrices are covariantly conserved$^1$$$\nabla_{\mu}\gamma^c~=~\omega_{\mu}{}^c{}_b\gamma^b+\frac{1}{4}\omega_{\mu}{}^{ab}[\gamma_{ab},\gamma^c] ~=~0,\qquad \nabla_{\mu}\gamma^{\nu}~=~0,\tag{1}$$ cf. e.g. Ref. 1. Consider the vector current $$J^{\mu}~:=~\bar{\psi}\gamma^{\mu}\psi,\tag{2}$$ where$\psi$... 2 I don't believe there is a general relationship for an arbitrary Lorentzian manifold, such that$R_{abcd}R^{abcd} \propto R$The only relationship that I can think of is the Weyl tensor invariant. Let us denote the Weyl tensor$C_{abcd}$. The quantity you have mentioned above is the Kretschmann scalar, which is denoted:$K = R_{abcd}R^{abcd}\$. One can then ...

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The area of an event horizon is an invariant, so it doesn't matter what coordinate system you use to calculate it. The area is of physical interest because it's proportional to the entropy of the black hole, and in a second law kind of way that means the area cannot decrease.

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