# Tag Info

## Hot answers tagged topology

105

All of physics has two aspects: a local or even infinitesimal aspect, and a global aspect. Much of the standard lore deals just with the local and infinitesimal aspects -- the perturbative aspects_ and fiber bundles play little role there. But they are the all-important structure that govern the global -- the non-perturbative -- aspect. Bundles are the ...

76

Not an expert, but I believe the answer lies in the hairy ball theorem. You see, for a magnetic field to turn charged particles back from a surface, the field must be parallel to the surface, which means that to have a fully confining geometry you must have a smooth, everywhere non-zero, and continuous vector field mapped onto a surface. But the theorem ...

42

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 ...

37

We need to be precise about the phrase the size of the universe. Specifically I'm going to take it to mean the maximum possible separation between any two points. In an infinite universe two points can be separated by an arbitrarily large distance, so if the maximum distance between two points is finite this means the universe must not be infinite. The ...

35

Topology is of fundamental importance even to systems in classical mechanics. The configuration space (or phase space) of a generic classical mechanical system is a manifold and manifolds are topological spaces with some extra structure (e.g. a smooth structure in the case of smooth manifolds). At the very start of any classical mechanics problem, you need ...

35

Basically, this is related to the theory of Riemann surfaces. Such surfaces arise in complex analysis when one tries to extend the complex plane to make multi-valued functions well-defined. Take a simpler example first. Consider the function $f(z) = \sqrt{ z}$. This function is not single valued since $f( e^{2\pi i } z ) = - f(z)$. The usual way this is ...

33

Every knot is the boundary of an orientable surface. Such a surface is called a Seifert surface.$^\dagger$ For any given knot (with a given embedding in 3-d space), the flux is the same through two such surfaces. As usual, the flux can be calculated either by integrating $\mathbf{B}$ over the surface, or by integrating $\mathbf{A}$ around the knot. Figure 6 ...

31

I'm not going to provide a full answer here, because I don't know the answer, but I want to give some statements that illustrate quite nicely the kind of problems one would face when determining topology of anything: We know spacetime is a manifold. That means, locally, it looks just like $\mathbb{R}^4$. That's already a bummer. We can't do jack at one ...

26

Let me first answer your second questions about the physical intuition behind fiber bundles: Fiber bundles ( with compact structure groups) describe internal degrees of freedom such as spin and isospin just as manifolds describe translational degrees of freedom. For example, (a non-trivial fibre bundle is needed to describe the rotation of a neutral spinning ...

26

Yes, a topological defect is a discontinuity that cannot be removed. Let me explain by giving an example similar to a liquid crystal. Fish in a pond I have a two-dimensional pond and I would like to fill it with fish. These fish are longer than they are wide, so when they are densely packed in the pool, they like to point in the same direction as their ...

25

Very loosely speaking the reasoning is this. Imagine a two band system in which the fermi sea has one filled band with Chern number $n$ and another system with $N$ filled bands but also with Chern number $n$. Physically they have the same topological properties (for example the same Hall conductance, edge states and so on), but cannot be deformed ...

24

Yes, if the universe is: flat (zero spatial curvature) has finite mass energy (since we know it is uniform this also means it is bounded. If you drop the bounded es because you don't want to admit uniformity or otherwise, i.e., if it is unbounded, then the answer is clearly no) is simply connected (has what is called a trivial topology) Then it does have ...

23

Well, the simplest case is that some topologies of spacetime may only allow a particular class of metrics. But unfortunately, it usually requires the knowledge of the metric at every point to be quite certain. Here's a few thing we can probably assume about the spacetime manifold : All the usual jazz about manifolds in general relativity (paracompactness, ...

19

You could resumé the importance of cohomology (not only de Rham's) in Physics (and all areas of Applied Mathematics) in a very concise way: if your closed forms are not exact, then cohomology matters. Also, equivalence classes of forms arise in Physics very naturally. Let's see some examples. Take two $n$-forms $A_1$ and $A_2$ and a $(n-1)$-form $B$. If $... 19 For a generic, oriented knot, you can construct an oriented surface which has the knot as it's boundary through the Seifert algorithm. Stokes' theorem says that the flux through any two such surfaces which share the same boundary must be the same. In principle, one could construct a Seifert surface for the trefoil knot, parameterize it, and then evaluate ... 18 This claim is simply wrong. The flat hyperplane is of course infinite, but non trivial topologies can be flat and still finite. The simplest example is the 3-torus, but there are even the Klein bottle and the Hantzsche-Wendt manifold. See for example page 27 of Janna Levin - Topology and the Cosmic Microwave Background, which show you ten different closed ... 18 You don't. Take a set of short permanent magnets. Chain them together. Make a knot out of the chain, and connect the ends. Or form two chains. Make them into linked closed loops. 17 The bible for the mathematical formulation of classical Mechanics, namely Foundations of Mechanics by Abraham and Marsden, defines a hamiltonian system as a triple$(M, \omega, X_H)$where$(M, \omega)$is a symplectic manifold, and$X_H$is the Hamiltonian vector field corresponding to a hamiltonian function$H:M\to\mathbb R. Now, are there typically any ... 17 The relevant Lie group isomorphism reads $$U(2)~\cong~[U(1)\times SU(2)]/\mathbb{Z}_2, \qquad Z(SU(2))~\cong~\mathbb{Z}_2.\tag{1a}$$ In detail, the Lie group isomorphism (1a) is given by $$U(2)~\ni~ g\quad\mapsto\quad$$ $$\left(\sqrt{\det g}, ~\frac{g}{\sqrt{\det g}}\right) ~\sim~ \left(-\sqrt{\det g}, ~-\frac{g}{\sqrt{\det g}}\right)$$ $$~\in ~[U(1)\... 17 The problematic line in your reasoning is in assuming that \nabla \times \vec F=0 implies that \vec F is conservative, i.e. that \vec F = \nabla \phi for some \phi. If this were true, then it would imply that \oint \vec F \cdot \mathrm d\vec r = 0 via the fundamental theorem of vector calculus, but alas it is not. We are only permitted to make the ... 16 Bundles and compactified spacetime A gauge theory cannot be looked at purely locally, it has inherently global features one cannot see locally. The proper mathematical formalization of a Yang-Mills gauge theory is that the gauge field A is a connection on a principal bundle P\to M over spacetime M. However, in practice, it turns out that physicists don'... 16 Consider a pair of points P and Q, say of mass m>0, and suppose that they are constrained as follows. Q stays on the z axis and can freely move along it up to restrictions said below. P can freely rotate around z and is connected to Q by means of an ideal rod (zero mass) of length \ell, and it is also connected to the origin O by ... 15 Let$$\begin{align} \overline{\mathbb{R}^{p,q}}~~:=&~~\left\{y\in \mathbb{R}^{p+1,q+1}\backslash\{0\}\mid \eta^{p+1,q+1}(y,y)=0\right\}/\mathbb{R}^{\times} \cr ~~\subseteq &~~ \mathbb{P}_{p+q+1}(\mathbb{R})~~\equiv~~(\mathbb{R}^{p+1,q+1}\backslash\{0\})/\mathbb{R}^{\times}, \qquad \mathbb{R}^{\times}~~\equiv~~\mathbb{R}\backslash\{0\}, \end{align}\... 15 This is a very good question. The same operator algebra does not imply thatH(J,h)$and$H(h,J)$have the same spectrum. As has been mentioned in Dominic's answer, even the ground state degeneracy is different under the interchange of$J$and$h$($J\gg h$: symmetry-broken two-fold degeneracy, and$J\ll h$unique ground state), therefore it is impossible to ... 15 My favorite reference for these sorts of things that straddle physics and geometry is Frankel's "The geometry of physics". In the chapter on harmonic forms, you will find what he refers to simply as "Hodge's Theorem". It's a little more general than you need, because it applies to general$p$-forms, and you only need functions ($0$-forms). So I'll ... 14 I) OP is interested in totally symmetric covariant$(0,r)$tensor fields $$g\in\Gamma\left({\rm Sym}^r(T^{\ast}M)\right) \tag{A}$$ on an$n$-dimensional manifold$M$. The number of totally symmetric tensor components are $$\begin{pmatrix} n+r-1 \cr r\end{pmatrix} .\tag{B}$$ II) If the manifold$M$is paracompact, we can use the partition of unity to prove ... 14 The short answer is that it depends on which of the symmetries you enforce. More precisely, the simple SSH model has many symmetries, and it is a priori not clear which of these symmetries you consider as 'accidental' and which you consider as 'enforced'. This is a matter of choice. Depending on this choice, the model lands up in different possible symmetry ... 13 Orbifolds are spaces of the type$O = M/G$where$M$is a manifold and$G$is a group acting nonfreely on$M$. That is there are fixed points (or more generally submanifolds of this action); i.e., points$x \in M$such that$G.x = x$for all$G$. These fixed points are called singular points, they have the property that geometrical objects (such as the ... 13 Beyond the polemics, I believe one can still answer this question. The gauge-potential can not be an observable in quantum mechanics, since it is gauge covariant. Quantum mechanics is clear about that. In a first quantised version of QM, you measure only$\left\vert\Psi\right\vert^{2}\$, and it is not affected by a gauge transform. This paragraph does not ...

13

No, because a Lorentz transformation is continuous with a continuous inverse. While an open ball is not mapped to itself, it is mapped to some other open set, in an invertible way. (That a Lorentz transformation is continuous of course follows from that it is linear.)

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