- #1

Jianbing_Shao

- 102

- 2

- TL;DR Summary
- metric compatible equation, parallel propagator, path dependence

If a vector ##V(x)## being transported down a path ##l##, The vector field is described with equation:

$$\partial_\mu V(x)=\Gamma_\mu V(x)$$

The solution of the equation can be described with parallel propagator ##P(x, x_0)##(in mathematics it is also called product integration):

$$V(x)=P(x, x_0)V(x_0) = \mathcal{P} \exp\left(\int_{x_0}^x \Gamma_\mu(x){\rm dx^\mu }\right)V(x_0)$$

If there exists a matrix function ##A(x)## which satisfies the equation:

$$\partial_\mu A(x)=\Gamma_\mu A(x)$$

Then##\Gamma_\mu## can be written as ##\partial_\mu A(x) A^{-1}(x)##. and the parallel propagator constructed with such a connection is:

$$P(x, x_0) = \mathcal{P} \exp\left(\int_{x_0}^x \partial_\mu A(x)A^{-1}(x){\rm dx^\mu }\right)

=A(x)A^{-1}(x_0)$$

So the parallel propagator is path independent, the curvature of ##\Gamma_\mu## is zero.

$$R_{\mu\nu}=\partial_\mu \Gamma_\nu-\partial_\nu \Gamma_\mu+[\Gamma_\mu \ \Gamma_\nu]=0$$

For a matrix function the change of the matrix between two points is path independent. It is very natural get a zero curvature connection from it.

If we use parallel propagator to construct a metric field. From metric compatible equation:

$$∇_ρ g_{μν}=g_{μν,ρ} -Γ_{ρμ}^λ g_{λν}-Γ_{ρν}^λ g_{μλ}=0$$

If the metric ##g_{μν}## moves along a path ##l## between points ##x_0## and ##x##, then the metric can be described with parallel propagator:

$$g(x)=P(x,x_0)g(x_0)P^T(x,x_0)$$

If ##Γ_\mu(x)## the connection compatible with metric ##g(x)## whose curvature is not zero. then parallel propagator ##P(x,x_0)## is path dependent. but because ##g(x)## is a field, So ##P(x,x_0)g(x_0)P^T(x,x_0)## must be path independent. The only possible parallel propagator should take such a form:

$$P(x,x_0)=A(x)P'(x,x_0)=A(x)\mathcal{P}\exp\int^x_{x_0}\Gamma'_\mu(x)dx^\mu$$

Here ##A(x)## is a matrix function. We demand the curvature of ##\Gamma'_\mu(x)## is not zero. ##R(\Gamma'_\mu)\neq 0## and:

$$P'(x,x_0)g(x_0)P'^T(x,x_0)=g(x_0)$$

Then we can get a path dependent parallel propagator and a metric field at the same time. The connection is:##\partial_\mu A(x) A^{-1}(x)+A(x)\Gamma’_\mu(x) A^{-1}(x)##

The metric is only determined by the matrix function ##A(x)##, ##g(x)= A(x) g(x_0)A^T(x)## and has nothing to do with the path dependent part ##P'(x,x_0)##. Then I have some quertions:

1. It seems that for any metric field there exist connections with zero curvature compatible with it. is it true?

2. Although there exist connections compatible with a metric field whose curvature is not 0. But in the definition of the metric field, the path dependent part ##P'(x,x_0)## was canceled. So we can get a metric field (path independent) at the same time a path dependent parallel propagator. Then how can we get a non-zero curvature connection purely from a metric field?

3. For a connection with zero curvature, it can be described with a matrix function, so we need ##n^2## parameters to described it. An arbitrary connection has ##n^3## parameters, and the zero curvature condition reduces the number of parameters from ##n^3## to ##n^2##. A Christoffel connection can be described with metric which only has ##\frac{1}{2}n(n+1)## parameters. Then what kind of conditions the Christoffel connection should satisfy then the parameters was reduced from ##n^3## to ##\frac{1}{2}n(n+1)##?

$$\partial_\mu V(x)=\Gamma_\mu V(x)$$

The solution of the equation can be described with parallel propagator ##P(x, x_0)##(in mathematics it is also called product integration):

$$V(x)=P(x, x_0)V(x_0) = \mathcal{P} \exp\left(\int_{x_0}^x \Gamma_\mu(x){\rm dx^\mu }\right)V(x_0)$$

If there exists a matrix function ##A(x)## which satisfies the equation:

$$\partial_\mu A(x)=\Gamma_\mu A(x)$$

Then##\Gamma_\mu## can be written as ##\partial_\mu A(x) A^{-1}(x)##. and the parallel propagator constructed with such a connection is:

$$P(x, x_0) = \mathcal{P} \exp\left(\int_{x_0}^x \partial_\mu A(x)A^{-1}(x){\rm dx^\mu }\right)

=A(x)A^{-1}(x_0)$$

So the parallel propagator is path independent, the curvature of ##\Gamma_\mu## is zero.

$$R_{\mu\nu}=\partial_\mu \Gamma_\nu-\partial_\nu \Gamma_\mu+[\Gamma_\mu \ \Gamma_\nu]=0$$

For a matrix function the change of the matrix between two points is path independent. It is very natural get a zero curvature connection from it.

If we use parallel propagator to construct a metric field. From metric compatible equation:

$$∇_ρ g_{μν}=g_{μν,ρ} -Γ_{ρμ}^λ g_{λν}-Γ_{ρν}^λ g_{μλ}=0$$

If the metric ##g_{μν}## moves along a path ##l## between points ##x_0## and ##x##, then the metric can be described with parallel propagator:

$$g(x)=P(x,x_0)g(x_0)P^T(x,x_0)$$

If ##Γ_\mu(x)## the connection compatible with metric ##g(x)## whose curvature is not zero. then parallel propagator ##P(x,x_0)## is path dependent. but because ##g(x)## is a field, So ##P(x,x_0)g(x_0)P^T(x,x_0)## must be path independent. The only possible parallel propagator should take such a form:

$$P(x,x_0)=A(x)P'(x,x_0)=A(x)\mathcal{P}\exp\int^x_{x_0}\Gamma'_\mu(x)dx^\mu$$

Here ##A(x)## is a matrix function. We demand the curvature of ##\Gamma'_\mu(x)## is not zero. ##R(\Gamma'_\mu)\neq 0## and:

$$P'(x,x_0)g(x_0)P'^T(x,x_0)=g(x_0)$$

Then we can get a path dependent parallel propagator and a metric field at the same time. The connection is:##\partial_\mu A(x) A^{-1}(x)+A(x)\Gamma’_\mu(x) A^{-1}(x)##

The metric is only determined by the matrix function ##A(x)##, ##g(x)= A(x) g(x_0)A^T(x)## and has nothing to do with the path dependent part ##P'(x,x_0)##. Then I have some quertions:

1. It seems that for any metric field there exist connections with zero curvature compatible with it. is it true?

2. Although there exist connections compatible with a metric field whose curvature is not 0. But in the definition of the metric field, the path dependent part ##P'(x,x_0)## was canceled. So we can get a metric field (path independent) at the same time a path dependent parallel propagator. Then how can we get a non-zero curvature connection purely from a metric field?

3. For a connection with zero curvature, it can be described with a matrix function, so we need ##n^2## parameters to described it. An arbitrary connection has ##n^3## parameters, and the zero curvature condition reduces the number of parameters from ##n^3## to ##n^2##. A Christoffel connection can be described with metric which only has ##\frac{1}{2}n(n+1)## parameters. Then what kind of conditions the Christoffel connection should satisfy then the parameters was reduced from ##n^3## to ##\frac{1}{2}n(n+1)##?