# Can an Einstein metric have the same Levi-Civita connection with a non-Einstein one?

We say that two metrics are affinely equivalent if their Levi-Civita connections coincide. Is it possible that an Einstein (=Ricci tensor is proporional to the metric) is affinely equivalent to a metric which is not Einstein?

Of course, since affinely equivalent metrics have the same Ricci tensor, the question is equivalent to the question whether there exists a metric $g$ such that

(1) its Ricci tensor $Ric$ is nondegenerate (as a bilinear form),

(2) is parallel, $\nabla^g Ric= 0$,

(3) and is not proportional to the metric.

The answer is negative for metrics of Riemannian and Lorentzian signature. In both cases the result follows from the description of affinely equivalent metrics in these signatures and it is hard to generalize it for general signature.

The motivation comes from projective geometry. It is known (Mikes MR0603226 or Kiosak-M arXiv:0806.3169) that if two metrics are projectively, but not affinely equivalent, and one of them is Einstein, then the second is Einstein as well. It is interesting and important to understand whether the assumption that the metrics are not affinely equivalent is necessary.

Edited after the comment of Robert Bryant: Let us assume in addition that the metric is not the direct product of Einstein metrics

• Are you assuming also that the metric is not locally a product? Otherwise, it seems that taking products of irreducible Einstein manifolds with different Einstein constants will give you examples. – Robert Bryant Aug 26 '13 at 13:15
• Actually, I do not know what to answer. The trivial example you gave indeeds answers the question and actually says that the property of the metric to be Einstein in not a projectively invariant property which was my motivation to ask the question. Thank you and if you formulate the answer as an answer accept it. I will edit the question according to your suggestion such that it remains interesting – Vladimir S Matveev Aug 26 '13 at 18:17
• OK. I have done this. – Robert Bryant Aug 26 '13 at 18:34

There are trivial examples that arise just by taking products of irreducible Einstein metrics with different Einstein constants. Whether an irreducible example exists is a much harder question. I do not know the answer to that, but I can think about it. (However, see below, where I answer this question.)

Of course, no such irreducible example can be Riemannian, and I guess, from your statement about the Lorentzian case, it can't be Lorentzian either, though I don't see that immediately.

In light of this, I guess the first case to try would be to see whether or not there could be one in dimension $4$ that is of type $(2,2)$. (See Addition 1.)

Addition 1: Indeed, there is an irreducible example in dimension $4$ of type $(2,2)$. Consider the $6$-dimensional Lie group $G$ with a basis for left-invariant forms satisfying the structure equations \begin{aligned} d\omega^1 &= - \alpha\wedge\omega^1 - \beta \wedge \omega^2 \\ d\omega^2 &= \phantom{-} \beta\wedge\omega^1 - \alpha \wedge \omega^2\\ d\omega^3 &= \phantom{-} \alpha\wedge\omega^3 - \beta \wedge \omega^4 \\ d\omega^4 &= \phantom{-} \beta\wedge\omega^3 + \alpha \wedge \omega^4\\ d\alpha &= c\ \bigl(\omega^3\wedge\omega^1+\omega^4\wedge\omega^2\bigr)\\ d\beta &= c\ \bigl(\omega^4\wedge\omega^1-\omega^3\wedge\omega^2\bigr)\\ \end{aligned} where $c\not=0$ is a constant. Let $H$ be the subgroup defined by $\omega^1=\omega^2=\omega^3=\omega^4=0$ and let $M^4 = G/H$. Then the above structure equations define a torsion-free affine connection $\nabla$ on $M^4$ that satisfies $$\mathrm{Ric}(\nabla) = -4c\ (\omega^1\circ\omega^3+\omega^2\circ\omega^4)$$ while both $h = \omega^1\circ\omega^3+\omega^2\circ\omega^4$ and $g = \omega^1\circ\omega^4-\omega^2\circ\omega^3$ are $\nabla$-parallel. Thus, $g + \lambda h$ is an example of the kind you want for any nonzero constant $\lambda$.

Addition 2: Upon further reflection, I realized that this example points the way to a large number of other examples, all of split type, and most having irreducibly acting holonomy as soon as the (real) dimension gets bigger than $4$.

The reason is that the above example is essentially a holomorphic Riemannian surface of nonzero (but real) constant curvature regarded as a real Riemannian manifold by taking the real part of the holomorphic quadratic form, i.e., $Q = (\omega^1+i\ \omega^2)\circ(\omega^3-i\ \omega^4) = h - i\ g$. (The group $G$ in the above example turns out to just be $\mathrm{SL}(2,\mathbb{C})$ and $H\simeq \mathbb{C}^\times$ is just a Cartan subgroup.)

Now, the same phenomenon occurs in all dimensions: Let $(M^{2n},Q)$ be a holomorphic Einstein manifold with a nonzero real Einstein constant and write $Q = h - i\ g$ where $h$ and $g$ are real quadratic forms. Then $(M,h)$ will be an Einstein manifold of type $(n,n)$ (with a nonzero Einstein constant) and $g$ will be parallel with respect to the Levi-Civita connection of $h$. Thus, all of the split metrics $g+\lambda\ h$ for $\lambda$ real will have the same Levi-Civita connection as $h$ and none of them will be Einstein. If the (holomorphic) holonomy of $Q$ is $\mathrm{SO}(n,\mathbb{C})$, then the holonomy of $h$ will be $\mathrm{SO}(n,\mathbb{C})\subset\mathrm{SO}(n,n)$, which acts $\mathbb{R}$-irreducibly on $\mathbb{R}^{2n}=\mathbb{C}^n$ when $n\ge3$.

(Constructing examples of non-split type might be interesting$\ldots$)

Addition 3: By examining the Berger classification (suitably corrected by later work), one can see that, if $M$ is simply connected and if $h$ is a non-symmetric pseudo-Riemannian metric on $M$ with irreducibly acting holonomy whose space of $\nabla$-parallel symmetric $2$-tensors has dimension greater than $1$, then the dimension of $M$ must be even, say, $2n$, and the holonomy of the metric $h$ must lie in $\mathrm{SO}(n,\mathbb{C})$. Of the possible irreducible holonomies in this case, only the subgroups $\mathrm{SO}(n,\mathbb{C})$ and $\mathrm{Sp}(m,\mathbb{C})\cdot\mathrm{SL}(2,\mathbb{C})$ (when $n=2m$) can occur if the metric is to be Einstein with a nonzero Einstein constant. Both of these cases do occur, and, in each case, the space of $\nabla$-parallel symmetric $2$-tensors has dimension exactly $2$. Thus, the construction outlined in Addition 2 gives all of the examples of desired pairs $(h,g)$ for which the holonomy is irreducible and that are not locally symmetric.

To make sure that we get the full list of examples with irreducible holonomy, we'd have to examine Berger's list of irreducible pseudo-Riemannian symmetric spaces for other possible candidates. (I suspect that, even there, the examples will turn out to be holomorphic metrics in disguise, but I have not yet checked Berger's list to be sure.)

The case in which the metric is irreducible but the holonomy is not remains, and it may not be easy to resolve with known technology.

• The nonexistence of an irreducible counterexample in the lorentzian case is due to the following `'almost linear algebra'' observation: existence of a parallel symmetric tensor nonproportial to the metric implies either the reducibility of the metric or the existence of a parallel vector field. The parallel vector field lies in the nullity of the ricci tensor so the ricci tensor can not be nondegenerate. – Vladimir S Matveev Aug 26 '13 at 18:56
• Hmmm. I claim that there are no irreducible Riemannian examples. The reason is that both $g$ and $\mathrm{Ric}(g)$ would be $\nabla$-parallel and hence the holonomy would have to preserve the eigenspaces of $\mathrm{Ric}(g)$ with respect to $g$ and hence (since they are not multiples of one another by hypothesis) the metric would have to be a local product by deRham (actually by Cartan, but everyone seems to think the local decomposition is due to deRham even though deRham only proved the global version after Cartan gave the local version). – Robert Bryant Aug 26 '13 at 18:59
• That's an interesting fact about Cartan which I did not know and I shall credit him from now on! – Paul Reynolds Aug 26 '13 at 21:48
• I am really impressed by your examples in the additions. Thank you very much indeed. – Vladimir S Matveev Sep 3 '13 at 9:00
• @Vladimir: You're welcome; it is an interesting question. By the way, I believe that the 4-dimensional example listed in Addition 1 is the only one in that dimension that is not a product. Its holonomy acts reducibly, though. The 6-dimensional symmetric example $\mathrm{SO}(4,\mathbb{C})/\mathrm{SO}(3,\mathbb{C})$ appears to be the lowest dimensional example for which the holonomy acts irreducibly. – Robert Bryant Sep 4 '13 at 9:36