# One-point compactification of space of matrices with positive trace

1. Feb 7, 2012

### julian

one-point compactification of space of matrices with non-negative trace

Hi I'm a physicist and my question is a bit text-bookey but it is also part of the proof that the universe had a beginning...so could I ask anyway...You got q which is a continuous function of a 3 by 3 matrix where if any component of the matrix is very large then q is close to the point x, then with the trace of the matrix greater than or equal to zero, the one-point compactification of this space of matrices, where the point at infinity is mapped to x, is compact...please explain. Thanks.

Last edited: Feb 7, 2012
2. Feb 7, 2012

3. Feb 7, 2012

### julian

The book is "HAWKING ON THE BIG BANG AND BLACKHOLES", chapter 1, pg.20. Actually I'm going to sleep now...must look at the stuff you gave tomorrow..thanks for your response julian.

4. Feb 7, 2012

My suggestion would be to give up on Hawking and Penrose sketch of the theorem and look a more precise version, like in

Relativity and Singularities - A Short Introduction for Mathematicians
Jose Natario

http://arxiv.org/abs/math/0603190

and also Hawking and Ellis monograph cited thereof.

5. Feb 8, 2012

### julian

Re: one-point compactification of space of matrices with non-negative trace

I think in "Relativity and Singularities - A Short Introduction for Mathematicians" they are considering a more restrictive form of the singularity theorem and as such dont need to consider this argument of Penrose and Hawking.

As for the topology imposed on the matrices...Penrose Hawking do use the Schwarz's inequality where they define the norm to be || U || = [tr (U^tU)]^{1/2}. And obviously the open balls defined by the norm could be a basis for a topological space...

This Alexandroff thing...it's like how you get the Riemann sphere from the non-compact plane + point at infinity. I'm guessing in general you need some kind of regularity at 'infinity' for the Alexandroff thing to work? I'm thinking that the q of Penrose and Hawking tends to x makes the compactification possibly.

Anyway, version of the Penrose Hawking argument is also in "The large scale structure of space-time", pages 98-99 but doesn't mention one-point compactification although I can see why it would tie in with a compactification. The basic argument is there...If any component of U were large then q would be close to x. Therefore, there is a C and a distance from x such that if any component of U is greater than C then q would lie inside this bounded segment to the future of x. If every component of U was less than or equal C then q would be inside a compact segment (a finite distance in the future of x, given by a q determined by the U's > C?) because the image of a continuous map from a compact space is also compact. Not sure how the fact that the matrices U are symmetric and have non-negative (or non-positive trace depnding on convention) come into the argument though...need to think about it some more.

6. Feb 8, 2012

### julian

Re: one-point compactification of space of matrices with Non-negative trace

When the trace is non-negative the point q lies to the future of x, whereas when it is negative q lies to the past of x...this may mean that the map from the space of all symmetric matrices to q is discontinuous if you did not restrict the trace to be non-positive and a compactification would't follow.

7. Feb 9, 2012

### julian

So for the map to q to be continuous we require that the matrices have non-negative trace - that just imposes what space we need to consider - which us just a simple partition of 6-dimensional Euclidean space. This space has a norm and so is a metric space, making it Hausdorff, and locally compact - this together with the space being non-compact are the necassary and sufficient conditions for the Alexandroff extention to be a compactification. The continuous map q then maps this compact space onto a compact segment of a geodesic with the point at infinity mapped to x.