Is My Calculation Correct for Topological Action with Veirbein and Levicivita?

[arkajad]

One more thing, When on p. 224 he writes "If we further relabel..." he probably means:

$$\omega_\mu^{\alpha 4}=e_\mu^\alpha$$

$$\phi^4=e^4_4,\,\phi^\alpha=\omega_4^{\alpha 4}.$$

There is no "spin connection". He has connection forms with values in the Lie algebra of the appropriate group.

 

[jinbaw]

There is no "spin connection". He has connection forms with values in the Lie algebra of the appropriate group.

Can you explain more please?

 

[arkajad]

Sure, that is easy for me, so if you will have further question - ask. I am not sure at which level to respond, so let me try this way:

Say, we are constructing a gauge theory of SO(3). So we imagine what is called a principal fibre bundle over base manifold M (our space-time). That is locally we take take the product MxSO(3), except that we are forgetting the origin of SO(3) in the fibers. Selecting (in a smooth way) one origin in each fibre is called "choosing a gauge".
Tangent space at a given point to each fibre is essentially tangent space to SO(3) - thus it is isomorphic to the Lie algebra so(3) of SO(3) - that is antisymmetric 3x3 matrices. If we choose a gauge, then to give a connection form is the same as to give a 1-form on M with values in so(3), which we write as

$${{\omega_\mu}^A}_B$$

If we define

$${\omega_\mu}^{AB}={{\omega_\mu}^A}_C\eta^{BC}$$

where $$\eta$$ is, in our case the Euclidean flat metric, then $${\omega_\mu}^{AB}=-{\omega_\mu}^{BA}$$ - thus $$\omega_\mu$$ have values in so(3).

Spin connection is usually understood in the context of gauging the spin group. In this case that would be gauging SU(2). Its Lie algebra is isomorphic to that of SO(3), but we would represent it normally as 2x2 complex anti-hermitian traceless matrices. And that would be where our connection form would take values for an SU(2) gauge theory.

With this understanding it is not a crime to call an so(3) valued connection form a "spin-connection" (Lie algebras are isomorphic), but it is unnecessarily misleading, because there is no spin around.

If you want me to elaborate more on similar issues - I will try to help as much as I can.

 

[jinbaw]

Actually, I'm very new on this issue. I do understand GR in terms of the metric and affine connection, but when it comes to veirbeins and spin connections i still find difficulties.

What i understand is that the spin connections calculated in chamseddine's paper are actually not the spin connections based on the lie groups, but the "spin connections" based on the lie algebras corresponding to those groups.

Is this correct?

 

[arkajad]

Connections have always values in Lie algebras. Take GR. You are familiar with Levi-Civita connection. It has values in the Lie algebra gl(4), more specifically, it has values in the Lie algebra o(3,1) which is a subset of gl(4). But what it does? At each point of M we gave the set of all orthonormal frames at this point. It is your fiber. The group SO(3,1) acts on these frames (still at the same point of M) - it rotates them. So you have your principal SO(3,1) bundles. What the connection does? It takes an orthonormal frame and transports it along a path in M to an orthonormal frame at a different point of M. How it does it? You solve parallel transport equation using Levi-Civita connection. And so it is in general. You have a fibre, you have group acing on this fibre, you have connection form, with values in the Lie algebra, you have parallel transport equations, you can solve it and transport "rigidly" the whole fibre into the fibre at some other point of the path. Connection form is nothing but a generalization of the Levi-Civita connection.

Normally in GR we use coordinate basis $$\partial_\mu$$ in the tangent space and define connection coefficients through:

$$\nabla_\mu\,\partial_\nu=(\partial_\sigma)\,\Gamma^\sigma_{\mu\nu}$$

I am writing it in this order deliberately, but do not differentiate on the RHS. Then you do not see that you have an o(3,1) valued connection. But if you take instead an orthonormal basis $$e_a$$
and define

$$\nabla_\mu e_a=e_c\,{{\Gamma_\mu}^c}_b$$

then $$\Gamma_\mu$$ matrices are in so(3,1).

 

[arkajad]

Suppose you want to do the gauge theory of SO(10) on M. Instead of playing with tangent space, you put a 10-dimensional vector space $$V_x$$ at each point of M, you endow it with the appropriate SO(10) invariant scalar product (Euclidean in my example), you consider the set of orthonormal bases in each $$V_x$$ and you suppose you have a covariant derivative allowing you to covariantly differentiate vector valued fields $$\phi^A,\, A=1\ldots,10$$. If $$E_A$$ is a 10-beim, then you define

$$\nabla_\mu E_A=E_C{{\Gamma_\mu}^C}_A$$

and because you suppose that your connection preserves the scalar product in the fibers, you find that $$\Gamma_\mu$$ matrices, $$\mu=0,1,2,3$$ are now in so(10) Lie algebra.

Of course at the same time you have the formula

$$(\nabla_\mu \phi )^A=\partial_\mu\,\phi^A+{{\Gamma_\mu}^A}_C\,\phi^C.$$

 

[jinbaw]

Yes. Thanks for making the idea clear. I understood a very big part of what u said, though i know i still have lots of gaps because actually i have never taken a real course neither in topology nor in differential geometry.

I still have a question: although it is mentioned in the title of the thread, but i can't really understand what is exactly meant by the term "topological". can u explain this to me?

 

[arkajad]

Topological in the title means, roughly, that the original action integral you start with is a "topological invariant", that is you calculate some expression (a number) using some connection, but the value of the integral (in case of a compact manifold) is, surprise-surprise, the same for every connection. So it does not depend on which connection you put on your manifold. So, what it depends on? It depends on the "shape of your manifold", whether it has holes or handles or such things. Then, of course, you somehow break this independence, because, after all, you want to get field equations for your connection. But I am not well versed in these tricks.