18 votes
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English translation of paper: Sur les variétés riemanniennes à groupe d'holonomie G2 ou Spin(7)

My library has provided a copy of the article of Bonan. Here is a summary of its contents: Section 1: The author introduces the inner product algebra of octonions (algèbra des octaves de Cayley), ...
Robert Bryant's user avatar
18 votes
Accepted

Does the curvature locally determine the connection?

The answer is 'not always'. Here is a simple case where you cannot recover the connection up to gauge transformation from the curvature: Let $n=2$, let the rank of $E$ be $m$, and, since $E$ is ...
Robert Bryant's user avatar
12 votes
Accepted

Are there mistakes in Kovalev's "Twisted connected sums and special Riemannian holonomy"?

The error in Kovalev's paper is described in arXiv:1206.227 (see the discussion following theorem 2.6). An alternative proof is in arXiv:1212.6929. Building on the previous work of Tian–Yau, Kovalev ...
Carlo Beenakker's user avatar
10 votes
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How does one complexify a real $n$-dimensional Riemannian manifold $(M,g)$?

I believe the following is meant: Every smooth (real) manifold $M$ has a (unique) real-analytic structure compatible with the smooth structure. So, cover $M$ with real-analytic charts, i.e. whose ...
M.G.'s user avatar
  • 6,730
10 votes

Manifolds with special holonomy especially $G_2$

Joyce's book Riemannian Holonomy Groups and Calibrated Geometry is an extended version of the research monograph you are reading, with more details and background material, aimed at providing a ...
Ben McKay's user avatar
  • 25.6k
9 votes
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A consequence of Ambrose-Singer theorem on holonomy

Your first question is a bit ambiguous. Are you asking whether, for each $p\in U$, the matrices $S_k(p)$ span the Lie algebra of $\mathrm{Hol}^0_p(\nabla)$ or are you asking whether, after taking the ...
Robert Bryant's user avatar
9 votes

Alternative (easier) Proof of Ambrose Singer Holonomy theorem

I always find it easier to work with the vector bundle induced by a linear representation of the structure group. I believe this theorem is a consequence of the following loop formula (a terse proof ...
Deane Yang's user avatar
8 votes
Accepted

Holonomy of a Ricci-flat affine connection

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-...
Robert Bryant's user avatar
8 votes
Accepted

Compact quaternionic Kahler manifolds of negative curvature: examples

Any (Riemannian) symmetric space admits a cocompact lattice. This is due to A. Borel, Compact Clifford-Klein forms of symmetric spaces, Topology 2, 1963, pp.111-122. The quaternionic hyperbolic space ...
Igor Belegradek's user avatar
7 votes
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Riemannian holonomy of generic manifolds

Here are proofs for the Riemannian and Kähler case which rely on the fact that the curvature can be seen as parallel transport around infinitesimal loops. It uses explicit deformations which are hard ...
Thomas Richard's user avatar
7 votes
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Holonomy bounded in terms of area and the curvature

There are in fact more precise versions, expressing the parallel translation around a loop as the identity map plus a curvature integral over a homotopy. References: Section 3.1 of Werner Ballmann's ...
user127309's user avatar
6 votes
Accepted

Killing vector fields on a compact $G_2$ manifold

A parallel vector field implies a reduction of holonomy. Any form of $G_2$ does not preserve any nonzero vector when acting in its nontrivial 7-dimensional representation. So the holonomy must be the ...
Ben McKay's user avatar
  • 25.6k
6 votes
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Holonomy groups of compact Riemannian symmetric spaces

At the request of the OP I put my comment as an answer: in general, the holonomy group and the isotropy group have the same identity component (this is a theorem of E. Cartan). So if you assume that $...
abx's user avatar
  • 37.3k
5 votes

Alternative (easier) Proof of Ambrose Singer Holonomy theorem

I think of this result in terms of Sussmann's orbit theorem. I wrote a paper about his theorem (https://arxiv.org/pdf/math/0508121.pdf), in which I show that a map of manifolds $P \to M$ which takes ...
Ben McKay's user avatar
  • 25.6k
5 votes

Holonomy as integration of curvature for principal $G$-bundles?

It can be done, and I think that is known to many physicists. The main difficulty is to deal with a non-commutative structure group $G$. Consider a loop $\gamma\subset M$ bounding a disk $D\subset M$, ...
Roberto Ladu's user avatar
5 votes
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Learning roadmap for holonomy theory

Dominic Joyce has two relevant books: Compact Manifolds with Special Holonomy and Riemannian Holonomy Groups and Calibrated Geometry for the study of holonomy groups of Riemannian manifolds, the ...
Ben McKay's user avatar
  • 25.6k
4 votes
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Representation variety vs. space of flat connections

The answer to your question is yes, the bijection between the Betti moduli space and the de Rham moduli space is a homeomorphism. See here for a nice exposition on this topic with references for ...
Sean Lawton's user avatar
  • 8,394
4 votes

Manifolds with special holonomy especially $G_2$

Salamon's Riemannian geometry and holonomy groups is another nice place to start.
Fran Burstall's user avatar
4 votes
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Flat connections, curvature and holonomy

The Stokes theorem must be modified first to deal with the nonabelian case. See http://arxiv.org/abs/0802.0663, Section 3.2, Theorem 3.4 and the displayed formula on top of page 48 for an appropriate ...
Dmitri Pavlov's user avatar
3 votes

Holonomy groups of Hermitian, and hyper-Hermitian, manifolds

This is really a comment, but it's far too long to go into a comment window. Your questions need to be made more precise. First, if $(M,J)$ is complex and $g$ is a metric on $M$ that is $J$-...
Robert Bryant's user avatar
3 votes

Examples of Calabi-Yau manifolds with $\mathbb{T}^2$ symmetry

As far as I understand, such metrics can be obtained by the so-called "Calabi-anszats": http://www.numdam.org/article/ASENS_1979_4_12_2_269_0.pdf Namely, one can construct a Calabi-Yau metric on the ...
aglearner's user avatar
  • 14k
3 votes

Holonomy of a Warped Product Metric

In general, just knowing the holonomy of $M$ will not tell you much about the holonomy of $\tilde M$. For example, if $M$ is isometric to the sphere of radius $r>0$ (and the dimension of $M$ is at ...
Robert Bryant's user avatar
3 votes

Nowhere vanishing section implies reduction of structure group

Informally, say a type of structure $S$ which can form "bundles" over a manifold is a "symmetry structure" if all fibers are isomorphic (in an appropriate, say categorical, sense). Let $G(S)$ be the ...
Dmitry Vaintrob's user avatar
3 votes

holonomy of connection on gerbes

Take a closed surface, $\Sigma$, and map it into your manifold. If $\Sigma \hookrightarrow U_{i}$, then you can simply integrate $F_{i}$ over $\Sigma$ and obtain the holonomy in $S^1$, evaluated on $\...
cheyne's user avatar
  • 1,396
3 votes

holonomy of connection on gerbes

$A_{\alpha\beta}=df_{\alpha\beta}+B_\beta-B_\alpha$ implies that $iA_{\alpha\beta}+iA_{\beta\gamma}+iA_{\gamma\alpha}=$ $i(df_{\alpha\beta}+B_\beta-B_\alpha+df_{\beta\gamma}+B_\gamma-B_\beta+df_{\...
Tsemo Aristide's user avatar
3 votes
Accepted

Analytic conjugacy of vanishing holonomy groups implies analytic conjugacy of foliations

No, in general the holonomy group is not solvable. Yet, this is how I'd tackle the exercise. (By the way, as it is posed the exercise cannot be solved: you need to assume that each eigenratio of both ...
Loïc Teyssier's user avatar
2 votes

Flatness in a neighborhood of a point condition

There must be a glitch in the formulation because the answer to the first question is an obvious "no": take a round sphere and flatten it around the North Pole. Also note that every sufficiently ...
alvarezpaiva's user avatar
  • 13.2k
2 votes

Alternative (easier) Proof of Ambrose Singer Holonomy theorem

The best overview on holonomy is perhaps in Besse's Einstein Manifolds. Regarding "Ambrose-Singer-Nijenhuis" they (pl) have a remark 10.59 about Nijenhuis' "more intuitive" approach with some "double ...
Martin Gisser's user avatar
2 votes

Alternative (easier) Proof of Ambrose Singer Holonomy theorem

There is another approach in JP. Magnot; Structure Groups and Holonomy in Infinite Dimensions, Bull. Sci. Math. 128 (2004) 513–529 based on the very nice book from Lichnerowicz Théorie globale des ...
user251364's user avatar

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