Idea about Einstein Manifold and vacuum Einstein field equation

Question

I have read about einstein manifold online and I saw that it mentioned the condition is equivalent to saying that the metric is a solution of the vacuum Einstein field equations. I also see that Hitchin–Thorpe inequality is the necessarily condition for the compact, smooth, 4-dimensional manifold to have an Einstein metric.

Is this literally mean that in a compact, smooth, 4-dimensional manifold, if the Hitchin-Thorpe inequality is not satisfied, then it does not have Einstein metric and hence there is no solution to the vacuum Einstein field equations for this particular 4-dimensional manifold?

Answer 1

In dimensions larger than two, a metric satisfies the vacuum Einstein field equations (for some value of the cosmological constant) if and only if it is an Einstein metric (see below). However, Einstein's field equations are usually considered for metrics which have signature $(3, 1)$, while the Hitchin-Thorpe inequality refers to Einstein metrics with signature $(4, 0)$, i.e. Riemannian metrics. A priori, a smooth manifold without a Riemannian Einstein metric could have an Einstein metric of signature $(3, 1)$ and hence a solution to the vacuum Einstein field equations.

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The Einstein field equations are

$$R_{\mu\nu} - \frac{1}{2}Rg_{\mu\nu} + \Lambda g_{\mu\nu} = \frac{8\pi G}{c^4}T_{\mu\nu}$$

where

• $R_{\mu\nu}$ is the Ricci tensor,
• $R$ is the scalar curvature,
• $g_{\mu\nu}$ is the metric tensor,
• $\Lambda$ is the cosmological constant,
• $G$ is Newton's gravitational constant,
• $c$ is the speed of light in a vacuum, and
• $T_{\mu\nu}$ is the stress-energy tensor.

In a vacuum, the stress energy tensor is zero, so the Einstein field equations can be rewritten as

$$R_{\mu\nu} = \left(\frac{1}{2}R - \Lambda\right)g_{\mu\nu}.$$

Taking the trace of both sides, you get

\begin{align*} R &= \left(\frac{1}{2}R - \Lambda\right)n\\ R &= \frac{n}{2}R - n\Lambda\\ R - \frac{n}{2}R &= - n\Lambda\\ \frac{2 - n}{2}R &= - n\Lambda. \end{align*}

Provided that $n > 2$, we have

$$\frac{1}{2}R = \frac{n}{n-2}\Lambda.$$

So the vacuum Einstein field equations become

$$R_{\mu\nu} = \left(\frac{n}{n-2}\Lambda - \Lambda\right)g_{\mu\nu} = \frac{n-(n-2)}{n-2}\Lambda g_{\mu\nu} = \frac{2}{n-2}\Lambda g_{\mu\nu}.$$

Therefore $R_{\mu\nu} = \lambda g_{\mu\nu}$ where $\lambda = \frac{2}{n-2}\Lambda$; that is, $g$ is an Einstein metric. Conversely, if $R_{\mu\nu} = \lambda g_{\mu\nu}$, then $g$ is a solution of the vacuum Einstein field equations for the cosmological constant $\Lambda = \frac{n-2}{n}\lambda$.