Riemann, Ambrose, Singer, Berger

January 27, 2008

The last GR problem displays a deep correlation between the Riemann Tensor, i.e. the Curvature of a Riemannian manifold and the change of a vector field if it is parallel transported around a loop.

The stuff is easy to picture. Take a tangent vector moving around a loop in flat space. When you arrive at the starting point it will still point in the same direction . To change direction something needs to happen along the route. Thats curvature.

Imagine the vector as a ship. A strange ship because it has no rudder… you should not be able to change direction “by hand”. Now you start sailing around the globe. The globe can be seen as a curved manifold (\cong S^2)-all land flooded. Consider 2 different ways.

First, simply sail around the equator (say starting at Somalia, heading for Indonesia). When you arrive at the start you still see the coastline of Indonesia (if it wasnt´ submerged ). That should not be surprising, because what you essentially did was sailing around a flat manifold, the equator (\cong S^1).

Now lets start at Somalia again. This time heading for the north pole. We sail till we ve circled half of the globe. We are now at the equator , but at the opposite side of the earth (somewhere in the pacific ocean ). We close the loop by sailing back along the equator. Our strange ship doesn´t allow us to turn around and so, when we arrive at Somalia, we look at the south pole. A 180 degree rotation.

Different ways would have yield different rotations :

Lets chip in some mathematical notation. Take a smooth  manifold (connected is improtant here, because we want to close loops). At the starting point  choose a tangent vector V. Obviously from the tangent space T_pM at that point. After we terminate our journey we obtain a rotated vector V` from the same tangent space. We can map the two vectors into each other by a GL(n) operation . Therefore we have a group operation on the tangent bundle, generated by parallel transport along any closed loop on the manifold. Its called the Holonomy group.

Parallel transporting around an infinitesimal parallelogram complies with taking the covariant derivative along a vector field \partial_{\mu} , after that along another field \partial_{\nu} and reversing the procedure to come back where we started. Lets calculate this in a map:

\left[D_{\partial_{\mu}},D_{\partial_{\nu}}\right](\partial_{\tau})=\, D_{\partial_{\mu}}(D_{\partial_{\nu}}(\partial_{\tau}))-D_{\partial_{\nu}}(D_{\partial_{\mu}}(\partial_{\tau}))\newline \hspace*{27mm}=\partial_{\mu}\Gamma^{\sigma}_{\nu\tau}\partial_{\sigma}+\Gamma^{\sigma}_{\nu\tau}\Gamma^{\tau}_{\mu\sigma}\partial_{\tau}-(\mu\leftrightarrow \nu)\newline \hspace*{27mm}=\left[\partial_{\mu}\Gamma^{\tau}_{\nu\tau}+\Gamma^{\sigma}_{\nu\tau}\Gamma^{\tau}_{\mu\sigma} -(\mu\leftrightarrow \nu) \right]\partial_{\tau} \newline \hspace*{27mm} = R^{\sigma}_{\tau\mu\nu}

It follows the statement of the Ambrose-Singer Theorem:

“The Lie Algebra of the Holonomy group (at some point p) \mathcal{H}ol_p(\omega) is generated by the values of the curvature tensor R(X,Y)(p) (at that point), as X,Y vary over all possible vector fields (which flows describe closed loops). ”

The \omega stands for the connection.

It is obvious that the Holonomy group of a 1 dimensional manifold is trivial,since \left[\partial_{\mu},\partial_{\nu}\right]=0 . But what can be said about general (semi-)Riemannian manifolds? Parallel transport preserves the metric. Therefore the induced linear transformations on tangent space does it. Hence, the Holonomy group is a subgroup of the isometric transformations. Thats O(n) in the case of a Riemannian manifold. If the manifold is orientable, vector flipping will not occur, so its SO(n)  (Take a look at the tangent frame in the gif and you will agree). In the case of GR it becomes the connected component of the Lorentz group SO(p,q).

The  classification of holonomy groups was done by Berger and is the subject of another theorem. It also includes complex manifolds like Calabi-Yaus, which are popular candidates if string theorists want to compactify a bunch of extra dimensions.

It should be mentioned that the concept of holonomy plays an important role in geometric gauge theory and is a cornerstone of loop quantum gravity.

Advertisements

Lie derivatives and Diffeomorphism

January 18, 2008

Over the last days I was mulling over the invariance of general relativity under the whole diffeomorphism group. Now, after looking up some textbooks and after some conversations i had with Sandro and Rainer Häußling i come to the conclusion that the subject is not simple at all. Before I bring up my thoughts I should convince myself that its not only my incompetence holding me back from understanding. Here´s a link to a discussion where some much brighter guys struggle with the same problem ; )

The whole thing started with a task about the Lie derivative given on the actual GR problem sheet. So first I will make some comments on the task:

One can derive the isometry group of a given space from the Lie derivative . If the Lie derivative of the metric tensor along some vector field vanishes, that vector field represents a generator of the isometry group. To understand this consider the definition of the Lie derivative :

\mathcal{L}_X T = \displaystyle\lim_{\lambda \to 0} \frac{1}{\lambda} \left( \Phi_{\lambda }^* T - T \right)

Where \Phi_{\lambda }^* indicates the flow of X and acts as a pullback on T. The Lie derivative therefore compares the tensor components evaluated at some point p to the tensor components at another point p\prime, where p and p\prime are in general related by a diffeomorphism.

Back to the metric tensor. It is easy to accept now that the condition \mathcal{L}_Xg=0 corresponds to the statement that a shift of the metric tensor along the flow of X doesn´t change it. The next task is to sort out the diffeomorphism which fullfill this condition. First we write down the local coordinate expression of the lie derivative:

\mathcal{L}_XT^{\mu_1 \dots \mu_r}_{\nu_1 \dots \nu_s} = X^{\lambda}\partial_{\lambda}T^{\mu_1 \dots \mu_r}_{\nu_1 \dots \nu_s}\newline \hspace*{18mm} -T^{\lambda\,\mu_2 \dots \mu_r}_{\nu_1 \dots \nu_s}\partial_{\lambda}X^{\mu_1}-\text{all upper indices}\newline \hspace*{18mm} +T^{\mu_1 \dots \mu_r}_{\lambda\, \dots \nu_s}\partial_{\nu_1}X^{\lambda}+\text{all lower indices}

(The derivation of the above expression can be found in Straumanns textbook).

In case of a fixed solution of Einsteins equations (i. e. Minkowski space) the condition reduces to the famous killing equations

\partial^{\mu} X^{\nu}+\partial^{\nu} X^{\mu}= 0

In both equations the partial derivative can be replaced by the covariant one (the christoffel terms chancel). And because the Levy-Civita connection is metric the (Xg) term always vanishes.

The left hand side of the killing equation is a symmetric 2-tensor A_{\mu\nu} , which leaves 10 possible solutions. Because the Poincare group is the isometry group of the Minkowski spacetime, there are actually 10 vector fields. Minkowski space time is a maximal symmetric space. There are solutions of the Einstein equation with less or even no killing vector fields at all. Beside the generator of transitions, its not obvious how the residual killing vectors look like. The construction can be done in the vielbein formalism. I will tell you where an instruction can be found and present an easy example instead:

{\Bbb R}^2 :\quad ds^2= dx^2+dy^2

A symmetric tensor A_{i j} in 2 dimensions has 3 components and we expects therefore 3 killing vector fields. Two of them are obvious again and the third one also if we take a look at the metric in polar coordinates:

X_1=(1,0)^T\, ,\, X_2=(0,1)^T\, ,\,X_3=(-x , y)^T

So much for the resume. What I want to know is if there is any way to show the diffeomorphism invariance of GR in this context?
My question is: why won´t the equation \mathcal{L}_Xg(x)=0 , not inserting some solution, but the general dynamical metric tensor field in GR spit out the whole diffeomorphism group?

I talked to Florian Scheck today about the problem and he told me the problem is badly posed. I ask for the vector fields which leave every solution of Einsteins equation invariant, which should turn out as no fields at all instead of all vector fields.

Any further input is heavily appreciated.


Hej

January 18, 2008

My name is Martin Bauer, I study physics at the Gutenberg university in mainz . At the time I work on my diploma thesis in Matthias Neuberts group. Our main interest is to squeeze out some measurable results of the Randall-Sundrum model.

Aside from that I assist Florian Scheck , who gives a lecture in general relativity.

I create this blog to add some additional information for the students which attend the GR course, but I will also comment the daily seminar talks and hopefully discuss some ideas (not exclusivly physics) with the few who stumble over this page. And if some of my group members want to add some thoughts they are of course invited.

Have fun and don´t mind my poor english. As long as you get the point…;)