Are Euclidean solutions also solutions of Einstein's equations?

In Special relativity we have the metric $$(+---)$$. But in General relativity we have a metric tensor $$g$$. In the equations themselves there doesn't appear to be anything that tells you what the signature of space-time must be.

Hence are 4D Euclidean spaces also solutions to the equations? What about spaces with signature $$(++--)$$?

Is it just the starting conditions that determine the signature of space-time?

• The whole premise of the thing involves local reference frames appearing to be Minkowski frames, which requires the standard signature. – Jerry Schirmer Jan 24 at 17:08
• @Jerry yes, I'm talking mathematically if the solutions can include any signature. Or to put it another way, would Einstein's equations in a Universe with a different signature be the same? – zooby Jan 24 at 18:38
• you can mathematically write down any metric $g_{ab}$ you want with ten arbitrary functions, and then go and compute $R_{ab} - \frac{1}{2}R\, g_{ab}$, say that that is your stress energy tensor. It just won't really mean anything. – Jerry Schirmer Jan 24 at 20:57

Hence are 4D Euclidean spaces also solutions to the equations?

Yes.

What about spaces with signature (++−−)?

Yes, they can also be solutions.

Is it just the starting conditions that determine the signature of space-time?

Yes, or the equations of state of the matter fields, which come into the field equations through the stress-energy tensor. The ways that we normally characterize the behavior of matter fields in our $$+---$$ world would not make sense in a world with a signature like $$++++$$ or $$++--$$.

I think the answer to this question depends on what one understands by a solution to Einstein's equations. For simplicity consider the vacuum equations $$Ric(g)=0$$ What is meant by a solution? If one is looking for a Ricci flat metric, then there is no restriction to the signature and the answer would be 'yes'. That is probably how most physics texts talk about solutions. But there is no restriction to the dimension either, so metrics of signature $$(-,-,+,+,+,+,+,+)$$ can also be solutions.

On the other, I think, that the solution to the equation should include the manifold itself. In other words to find a solution means to find a four dimensional manifold with a metric of signature $$(-,+,+,+)$$, which satisfies the equation. Then the answer would be 'no'.

• On the other, I think, that the solution to the equation should include the manifold itself. I don't understand what you mean by this. There is no way to specify a solution without either explicitly or implicitly specifying the manifold. On the other, I think, that the solution to the equation should include the manifold itself. In other words to find a solution means to find a four dimensional manifold with a metric of signature (−,+,+,+), which satisfies the equation. This makes it sound like the signature is part of the manifold, which isn't true. – Ben Crowell Jan 26 at 16:03