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Holonomy of a Ricci-flat affine connection

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There is some link between Ricci-flatness and reduction of holonomy. For example a Kahler manifold is Ricci-flat if and only if it has at most $SU(n)$ holonomy rather than $U(n)$, and it's apparently open to construct a closed, simply-connected Ricci-flat manifold with full $SO(n)$ holonomy.

On a manifold with an arbitrary affine connection, the Ricci curvature still makes sense. By any chance, does being Ricci-flat imply a reduction of the holonomy? In particular, does it imply that the connection is compatible with some (pseudo-Riemannian) metric (i.e. that the holonomy is contained in some $O(p,q)$)? I'm happy to assume the connection is torsion-free. And I'm really most interested in the local holonomy, so assume the manifold is simply-connected if that makes a difference.

(The motivation for this question comes from thinking very naively about general relativity as a pure gauge theory, which I think makes sense in a vacuum at least. Maybe in this case the metric constraint comes for free!)

This post imported from StackExchange MathOverflow at 2016-04-24 15:42 (UTC), posted by SE-user Tim Campion
asked Feb 19, 2016 in Theoretical Physics by Tim Campion (20 points) [ no revision ]
retagged Apr 24, 2016

2 Answers

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The answer depends on the dimension. When $n=2$, Ricci-flatness of a connection implies that it is flat, so, in that case, yes, you get holonomy reduction locally. However, when $n>2$, Ricci-flatness of a torsion-free connection only implies that the (local) holonomy lies in $\mathrm{SL}(n,\mathbb{R})$. You do not generally get any further reduction than that.

Remark 1: While writing down the general torsion-free connection with vanishing Ricci tensor and holonomy $\mathrm{SL}(n,\mathbb{R})$ (for $n>2$) does not appear to be easy, it is not hard to construct specific examples: Let $M=\mathbb{R}^n$ have its standard coordinates $x^1,\ldots,x^n$ and, for notational simplicity, assume that the indices are taken modulo $n$, i.e., we have $x^{n+1} = x^1,\ x^{-1} = x^{n-1}$, etc.. Let $E_i$ be the standard coordinate vector fields on $\mathbb{R}^n$ and consider the connection $\nabla$ defined by setting $$ \nabla_{E_i} E_i = E_{i-1}\qquad\text{while}\qquad \nabla_{E_i} E_j = 0\qquad \text{when $i\not\equiv j\mod n$}. $$ Then $\nabla$ is torsion-free and its curvature tensor is $$ R^\nabla = \sum_{i=1}^n\ E_{i-1}\otimes \mathrm{d}x^{i+1}\otimes \bigl(\mathrm{d}x^{i}\wedge\mathrm{d}x^{i+1}\bigr). $$ Since $n>2$, one has $\mathrm{Ric}(\nabla) = 0$. The image of the curvature operator in $TM\otimes T^*M$ is spanned by the (nilpotent) linear transformations $$ N_i = E_{i-1}\otimes \mathrm{d}x^{i+1}, $$ and these span an $n$-dimensional subbundle of $\mathrm{End}(TM)$ whose iterated commutators span the entire Lie algebra ${\frak{sl}}(n,\mathbb{R})$ at every point. Thus, by the Ambrose-Singer Holonomy Theorem, the holonomy of $\nabla$ is $\mathrm{SL}(n,\mathbb{R})$.

Note that $\nabla$ is translation-invariant, so it is well-defined on the quotient $\mathbb{T}^n = \mathbb{R}^n/\mathbb{Z}^n$, a compact torus. Thus, compactness does not obstruct the existence of a Ricci-flat torsion-free connection with full holonomy $\mathrm{SL}(n,\mathbb{R})$.

This post imported from StackExchange MathOverflow at 2016-04-24 15:42 (UTC), posted by SE-user Robert Bryant
answered Feb 19, 2016 by Robert Bryant (40 points) [ no revision ]
This answers my question. But I imagine it's still a bit of work to construct Ricci-flat connections with $SL(n)$ holonomy?

This post imported from StackExchange MathOverflow at 2016-04-24 15:42 (UTC), posted by SE-user Tim Campion
It's easy to prove that, when $n>2$, they exist (locally), as they are the solutions of an underdetermined elliptic equation, and the generic solution will have full holonomy $\mathrm{SL}(n,\mathbb{R})$. I don't know how hard it is to construct explicit examples, but I wouldn't expect it to be very hard.

This post imported from StackExchange MathOverflow at 2016-04-24 15:42 (UTC), posted by SE-user Robert Bryant
OK, that makes sense. Thanks so much!

This post imported from StackExchange MathOverflow at 2016-04-24 15:42 (UTC), posted by SE-user Tim Campion
@TimCampion: You're welcome. I had some time on a flight this morning, so I wrote down a specific example, in case you are interested. I have appended it to my answer as a Remark.

This post imported from StackExchange MathOverflow at 2016-04-24 15:42 (UTC), posted by SE-user Robert Bryant
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By a Theorem of Hano-Ozeki (see here http://projecteuclid.org/download/pdf_1/euclid.nmj/1118799772) any connected Lie subgroup $G$ of $GL(n,\mathbb{R})$ can be realized as the holonomy group of an affine connection in an $n$-dimensional manifold ($n>1$). Take $G = SL(n,\mathbb{R})$ and apply Hano-Ozeki to get a connection whose holonomy is $SL(n,\mathbb{R})$. Now by the Holonomy theorem (e.g. Theorem 9.1., page 151 Kobayashi-Nomizu Vol I) the curvature tensor $R$ belongs to the Lie algebra of $SL(n,\mathbb{R})$ hence $trace(R) = 0$ (i.e. the connection is Ricci-flat). Finally it is not difficult to see that $SL(n,\mathbb{R})$ is not contained in any $O(p,q)$, $(p+q = n)$ hence there are no compatible pseudo-Riemannian metric.

This post imported from StackExchange MathOverflow at 2016-04-24 15:42 (UTC), posted by SE-user Holonomia
answered Feb 19, 2016 by Holonomia (0 points) [ no revision ]
Actually, having holonomy in $\mathrm{SL}(n,\mathbb{R})$ only implies that the Ricci tensor is symmetric, not that it vanishes. Also, this does not answer the question when you assume that the connection is torsion-free, in which case, Hano-Ozeki does not apply.

This post imported from StackExchange MathOverflow at 2016-04-24 15:42 (UTC), posted by SE-user Robert Bryant
I think I have a wrong definition of the Ricci tensor of a connection $\nabla$... For me the Ricci tensor is $Ric(X,Y) = trace(R(X,Y))$. By the holonomy theorem, for fixed $X,Y$, $R(X,Y)$ belongs to the holonomy algebra hence $Ric(X,Y) = 0$ whenever the Lie algebra is contained in $sl(n,\mathbb{R})$.

This post imported from StackExchange MathOverflow at 2016-04-24 15:42 (UTC), posted by SE-user Holonomia
@Robert Bryant: Indeed, it is wrong to think that $Ric(X,Y) = trace(R(X,Y))$ otherwise all Riemannian manifolds would be Ricci flat since $R(X,X) = 0$. I see I made a deep mistake.

This post imported from StackExchange MathOverflow at 2016-04-24 15:42 (UTC), posted by SE-user Holonomia

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