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1

In order for a theory to present stable monopole solutions it has to satisfy three requirements: i) It has to have the topological conditions, generally showed as non trivial second homotopy group of the vacuum manifold. ii) It has to satisfy a quantization condition $$e^{ieQ_m}=\mathbb 1,$$ where $Q_m$ is the (non-Abelian) magnetic charge. This is a ...


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Your intuition that the Einstein equations are equations for the metric tensor, not for the manifold is mostly on the right track, but the details are wrong. That core bit of intuition is best phrased, I think, as saying that the Einstein equations are local equations for the geometry of the manifold. That is, they tell you that, whatever manifold your ...


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A conical intersection in an multidimensional parameter space of dimension $M$ is an $M-2$ dimensional hypersurface. (In your example $M$ should be $3N-6$ after the removal of the center of mass translations and rigid rotations). This is because every energy constraint is an $M-1$ dimensional hypersurface and the conical intersection is $M-2$ dimensional ...


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Well, by a process of elimination, if you have your time orientation (a vector field $\tau^\mu$), and you have two equivalence classes, for $g(X,\tau) > 0$ and $g(Y, \tau) <0$, the only remaining possibility for a third class is that $g(Z,\tau) = 0$. But any vector tangent to a timelike vector will be spacelike. It can be proven thusly : if you have ...


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You misunderstood the meaning of $H^{\bullet}_{S^1}(U_1) = \mathbb{C}[\Omega]$. This does not mean $H^i_{S^1}(U_1) = \mathbb{C}[\Omega]\forall i$, it means that the cohomology ring (with multiplication given by the cup product) is given by the polynomial algebra in one element that has degree 2. Translated back into the individual degrees $H^i$ this ...


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The first thing that must be said is that the question is not really specific enough: Applications to what exactly are you looking for? To me, a book on algebraic geometry and mirror symmetry, and how it relates to mirror symmetry as physicists know it, is very relevant and interesting. However, I have the feeling that this is not exactly what you're looking ...


3

If you remove a closed subset from a valid spacetime, then the result is a valid spacetime. You can put the Lorentzian metric on the original spacetime, then simply restrict it to the part that you keep. So it's still a 4d manifold without boundary. It's still Lorentzian and so on. You could even imagine the original manifold as some charts and transition ...



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