# Tag Info

2

Comments to the question (v4): By definition, the Lagrangian form $\mathbb{L}$ of Chern-Simons (CS) theory (wrt. a Lie algebra valued one-form gauge field $A$) is a CS form, i.e. the CS action reads $$S[A]~=~\int_M\mathbb{L}.$$ The exterior derivative $\mathrm{d}\mathbb{L}$ of a CS form is (also by definition) the Lie algebra trace of a polynomial of the ...

5

The rule of the game is to use $A$ and $F=dA$ to write a topological action, and in $d+1$-space time dimension you need to come up with a gauge-invariant $d+1$-form which can then be integrated over the manifold to give you the action. Such an action does not depend on metric at all. Take $U(1)$ gauge field as an example. In $2+1$, the only thing you can ...

3

In terms of a metric $g_{ab}$, the Riemann curvature tensor is given by, $$R^a_{bcd} = \partial_c \Gamma^a_{db} - \partial_d \Gamma^a_{cb} + \Gamma^a_{c e} \Gamma^e_{d b} - \Gamma^a_{de} \Gamma^e_{c b}$$ and consequently changing $d$ does not change the formula, though of course the actual numerical measures of curvature may change. Nevertheless, there are ...

1

As you can find on google, or in any book of supersymmetry, the number of components of a spinor in dimension $d$ is $2^{[d/2]}$. Where $[d/2]$ denotes the integer part of $d/2$. In certain dimensions you can impose a further property: the Majorana condition on your spinor, that reduces further of a factor of $2$ the number of independent (real) components. ...

2

this may or may not be a misunderstanding, but there is no centre of the universe. Imagine the universe as being the surface of an expanding balloon, with all the galaxies and stars on the surface, being stretched away from each other. Just as there is no centre, for example, of the Earth's surface, there is no centre of the universe. If by 'centre' you mean ...

3

Good question. The main difference is that we cannot freely move in time or in other words, we and everything else are moving together. Other than that, I think nobody can say for sure yet if the answer is 1, 2, 3 or something else. The important thing to realize is that time as a 4th dimension is used to make models or theories of reality. Compare that to a ...

0

Expand your line element and obtain the metric $g_{ij}$. It is of the form $$g_{ij}=\delta_{ij}-n_in_j$$ where $n=\langle \frac3{13}, \frac4{13}, \frac{12}{13}\rangle$ and so $n_in^i=1$ What you have now is a projection operator (because $g_{ij}g_{jk} = \delta_{ik}$, check it symbolically) which does this: It takes any 3D vector $v$ and gives you its ...

1

It is not a space-time because it is not Lorentzian. It is actually Riemannian. This exercise may be from a general relativity book, but is in fact a geometry question. So I take it that the question is to show that it represents a two dimensional space. But since it is in the general relativity tag one can be smart and guess the following. Consider the ...

0

Having a zero column in a diagonalization is bad (since the metric would be degenerate), but also bad would be if somehow it looked like $$dl^2=dx^2+dy^2+dz^2$$ or $$dl^2=-dx^2-dy^2-dz^2.$$ S you also want to avoid the metric being positive definite or negative definite. For more dimensions you'd also want to worry about having two spatial directions and ...

1

Things are not difficult if you put yourself in the right perspective. "Curvature" is a mathematical concept. Like many mathematical concepts the word may sound like something a taxi driver in London or Jakarta may have heard, but it is really just a "formula". Likewise: "Work" may evoke ideas of money, strikes, unions, customers..., but in physics it is ...

0

In the representation theory of Virasoro algebra there is a mathematically strict theorem: http://en.wikipedia.org/wiki/Goddard%E2%80%93Thorn_theorem for bosonic string theory, and similarly for superstring theory. Intuitively the critical dimensions come from zeta function regularization.

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In GR,we are told that matter tells spacetime how to curve and spacetime tells matter how to move in fact.So the dimension of the spacetime of the universe is four.

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The simplest way I can describe this is: 1 dimension If you're on a main street and are talking with your friend on the phone, telling them where on that street you are. There are only 2 directions to decide from at any moment, up or down the street. 2 dimensions You're in a city, talking to your friend where to meet you. They are on a different street and ...

1

In supergravity theory, 11d have 3 dimensional scalar object and 6 dimensional anti-symmetric tensor object. This allows that M-theory have M2 brane and M5 brane only. (By studying on following question, D branes Ns brane and p-branes i found some interesting paper, supermembrane, arXiv:9611203, It seems to me that section an "brane scan" describes ...

2

There are models where the extra dimensions don't need to be curled up. The main issue with extra dimensions is, 'why don't the particles/fields we interact with travel in those directions?' We have extremely good limits on standard model particles (electrons, photons) travelling in extra dimensions. However, it is possible to imagine a string inspired ...

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I don't think its a weakness in any sense. Because in all string theories, $10+1$ dimensional Lorentz transformations ARE a symmetry of the action itself. However not only in order to agree with phenomenology, but also as an attempt (not completely successful so far) to reproduce the entire structure of the standard model interactions, string theory ...

5

No, it's not possible. The other galaxies we see are to radically different. Additionally, if we are the surface volume of hyperspace, then the universe should be closed. Our best estimates and observations indicate it's flat. Let me address both of these in more detail. As for the other galaxies. First of all, there's the Andromeda galaxy. That is a galaxy ...

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