# Holonomy of a Ricci-flat affine connection

+ 4 like - 0 dislike
454 views

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
retagged Apr 24, 2016

+ 4 like - 0 dislike

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 (40 points)
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
+ 0 like - 0 dislike

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 (0 points)
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

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ysicsOverf$\varnothing$owThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.