# Is de Sitter the only vacuum with positive CC

1. May 20, 2015

### wabbit

Assuming a spacetime with zero Weyl curvature and an Einstein tensor proportional to the metric, is it true that in a finite neighborhood of any point, that spacetime must be isometric to a de Sitter vacuum, or are there other possible solutions, and if so how are they classified?

Thanks

2. May 20, 2015

### fzero

Einstein tensor proportional to the metric implies constant scalar curvature. Then vanishing of the Weyl curvature implies that the space is maximally symmetric $R_{mnpq} = k ( g_{mp}g_{nq} - g_{mq}g_{np})$ and therefore of constant sectional curvature. Therefore such a space is, by a conformal rescaling of the metric, equivalent to one of the model constant curvature spaces. For signature $(1,d-1)$ and positive cosmological constant, this is indeed de Sitter.

3. May 20, 2015

### wabbit

Thanks - you say "by a conformal rescaling", so it isn't isometric, only conformally equivalent ? I must say conformal transformations isn't something I am really familiar with.

4. May 20, 2015

### fzero

The issue is that the denominator of the formula for sectional curvature involves precisely the same contractions that correspond to the Riemann tensor of a maximally symmetric manifold. So we can rescale $g' = e^{2\sigma(x)} g$ without changing the sectional curvature. With this relation we say that $g'$ is pointwise conformal to $g$. If there is a diffeomorphism that pulls $g'$ back to $g$, then we say that the metrics are conformally equivalent and there is a genuine isometry. I think this is a stronger condition than the assumptions warrant.

It is probably overkill and yet might not even answer all questions that you might have, but the most specific reference I know of is Besse, Einstein Manifolds. Some results are discussed in the first few pages of this lecture.

5. May 21, 2015

### wabbit

Ah, the situation seems more complex than I thought - will check these, thanks for the references.