12
$\begingroup$

It is well known that there exists three and four manifolds that do not admit an Einstein metric, but I wonder if this question is still open for manifolds of dimension higher than four. That is, does anyone know of a compact n-manifold, $n>4$, that does not admit an Einstein metric?

$\endgroup$

2 Answers 2

14
$\begingroup$

No one knows this. Here is a citation from Gromov's beautiful article: http://www.ihes.fr/~gromov/topics/SpacesandQuestions.pdf

Page 19.

Following Alex we (I speak for myself) are lead to the pessimistic conclusion that there is no chance for a distinguished $g_{best}$ for $n\ge 5$ and "natural" metrics, e.g. Einstein $G$, $Ri(g)=\lambda g$ for $\lambda<0$, must be chaotically scattered in the vastness of $G$ with no meaningful link between geometry and topology (This does not preclude, but rather predicts, the existence of such metrics, e.g. Einstein, on all $V$ if dimension $\ge 5$: the problem is there may be too many of them)

$\endgroup$
4
$\begingroup$

In general, for dimensions $n>5$ we don't have topological obstructions to the existence of Einstein metrics (like as Hitchin-Thorpe inequality when $n=4$).

However, there exist compact homogeneous Riemannian manifolds with no $G$-invariant Einstein metrics. In this case, one works with cosets $M=G/K$ of a compact Lie group $G\subset {\rm Isom}(M)$ and the Einstein equation $Ric = c \cdot g$ is written with respect to a $G$-invariant Riemannian metric $g$ on $M=G/K$. For such a metric the Einstein equation reduces to a polynomial system and positive real solutions correspond to invariant Einstein metrics.

In particular, let $\frak{g}=\frak{k}\oplus\frak{m}$ be a reductive decomposition for $M=G/K$ (such an orthogonal decomposition always exists for a homogeneous Riemannian manifold $(M=G/K, g)$). For simplicity, let us assume that the tangent space ${\frak{m}}\cong T_{o}G/K$ decomposes into $q$ pairwise inequivalent isotropy summands (i.e., irreducible $Ad(K)$-submodules)

$ \frak{m}= \frak{m}_1 \oplus \frak{m}_2 +\oplus ..... \oplus \frak{m}_q. $

Then any $G$-invariant Riemannian metric is given by

$ g = < , > = x_1\cdot (-B)|_{\frak{m}_1}+\cdots+x_s\cdot (-B)|_{\frak{m}_q} $

for some positive real numbers $x_1, x_2, ... x_q$, where $-B$ denotes the negative of the Killing form of $\frak{g}$ (restricted on $\frak{m}$).

Homogeneous Einstein metrics are positive real solutions of the system $\{r_1-r_2=0, ...., r_{q-1}-r_{q}=0\}$ where $r_{i}$ are the components of the invariant Ricci tensor on any isotropy summand. These are given by

\begin{equation}\label{ricc} r_{k}=\frac{1}{2x_{k}}+\frac{1}{4d_{k}}\sum_{i, j}\frac{x_{k}}{x_{i}x_{j}}[ijk]-\frac{1}{2d_{k}}\sum_{i, j}\frac{x_{j}}{x_{k}x_{i}}[kij], \qquad (k=1, \ldots, q). \end{equation} Here $[ijk]$ are the structure constants of $G/K$. Their computation is usually non-trivial. For particular classes of homogeneous spaces (e.g. isotropy irreducible), they are related with the Casimir operator of $\frak{g}$. They are non-zero only for these triples $\frak{m}_i$, $\frak{m}_j$, $\frak{m}_k$, for which

$B([\frak{m}_i, \frak{m}_j], \frak{m}_k)\neq 0$.

Notice that the study of homogeneous Einstein metrics becomes more complicated if some of the isotropy summands are equivalent each other. Finally, it turns out that there are several homogeneous spaces with no invariant Einstein metrics. For example, we know by M. Wang and W. Ziller that the 12-dimensional space $SU(4)/SU(2)$ does not admit any homogeneous Einstein metric.

see: M. Wang and W. Ziller: Existence and non-existence of homogeneous Einstein metrics, Invent.~Math.~84 (1986) 177--194.

This space seems to be the lowest dimensional example of a compact homogeneous manifold with no invariant Einstein metrics.
Other interesting examples of compact homogeneous spaces with no invariant Einstein metrics have been presented for instance by

J-S. Park and Y. Sakane: Invariant Einstein metrics on certain homogeneous spaces, Tokyo J. Math. 20 (1) (1997) 51--61.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.