Let $M$ be a manifold, $TM$ its tangent bundle, and $N:TM\to TM$ a vector bundle morphism. It is possible to find a torsionless linear connection $\nabla$ on $TM$ such that $\nabla N=0$?
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$\begingroup$ The canonical followup question is, of course, what endomorphisms are flat for some connection. $\endgroup$– Mariano Suárez-ÁlvarezCommented Sep 1, 2013 at 9:32
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$\begingroup$ Actually, I'm interested in a fixed endomorphism $N$, because it is the anchor of a certain Lie algebroid. Rather, a less demanding question would be if there exists a symmetric connection such that $(\nabla_X N)(Y)=(\nabla_Y N)X$ (which is enough for my purposes), but I'm afraid that some integrability conditions will be also required in this case, so I'll try another approach. Anyway, the discussion has been very interesting and helpful! $\endgroup$– GrimolattoCommented Sep 1, 2013 at 16:53
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2 Answers
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My answer is almost contained in the answers/comments of Mariano Suárez-Alvarez, Ben McKay and Robert Bryant; I summarize the answers and give a reference.
There exists a torsion-free connection such that a given endomorphism $A$ is parallel if and only if the following two conditions are fulfilled: (1) The Jordan type of the endomorphism is the same at all points (2) The Nijenhujs torsion of the endomorphism vanishes. (3) For every k and every eigenvalue $c$ the distribution $Kern(A-c Id)^k $ is integrable.
A reference is Thompson, Gerard The integrability of a field of endomorphisms. Math. Bohem. 127 (2002), no. 4, 605–611 and it is downloadable from https://eudml.org/doc/249030 (actually, this result was reproved and republished many times, first time possible by Shirokov in the 50th who unfortunately published most his works in obscure places and possibly the last time by Boubel arXiv:1003.0979)
answered Sep 1, 2013 at 10:54
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1$\begingroup$ If I understand the proof correctly, Thompson's result is local. Let $f(z_1,z_2)=(2z_1,4z_2+z_1^2)$. Let $M=(\mathbb{C}^2-0)/(z \sim f(z))$. A holomorphic affine connection on $M$ is an $f$-invariant holomorphic connection on $\mathbb{C}^2-0$. By Hartog's, such a connection would extend to $\mathbb{C}^2$ which you can check is impossible by taking a Taylor expansion. A torsion-free connection under which the standard complex structure $J$ on $M$ is parallel is holomorphic: $J$ is not parallel under any torsion-free connection, even though it is locally. $\endgroup$ Commented Sep 1, 2013 at 12:25
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$\begingroup$ Thompson has more than one result in his paper and most probably some results are local. The answer on the question of Grimolatto I have summarized above is in my understanding global. The conditions (1,2,3) are geometric and if they are fulfilled locally in a neighborhood of every point than they are fulfilled everywhere, so they are global. Now, if there exists, in a neighborhood of every point, a local endomorphism-preserving connection, one can glue it into a global connection with the help of the partition of unity (to see this, check that $f \Gamma+ (1-f) \Gamma'$ is a connection) $\endgroup$ Commented Sep 1, 2013 at 16:43
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$\begingroup$ OK, my global counterexample is wrong; there is torsion-free connection for which $J$ is parallel, but it is not holomorphic. $\endgroup$ Commented Sep 1, 2013 at 17:18
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No. To be parallel under a connection, $N$ would have to have the same type at each point. In other words, for any two points, there must be a linear map identifying tangent spaces which identifies $N$ at those two tangent spaces. If $N$ vanishes somewhere, it would have to vanish everywhere, for example.
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$\begingroup$ Suppose then the Jordan type is constant :-) $\endgroup$ Commented Sep 1, 2013 at 9:26
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$\begingroup$ That's not enough either. For example, if $N$ has exactly two eigenvalues $1$ and $-1$ and they are of constant multiplicity, then there will exist a torsion-free connection making $N$ parallel if and only if the two eigenbundles in $TM$ are Frobenius. $\endgroup$ Commented Sep 1, 2013 at 9:27
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1$\begingroup$ But then the generalized eigenspaces are also parallel: $(N-\lambda I)^k=0$. By torsion-freedom, that should make them Frobenius. $\endgroup$ Commented Sep 1, 2013 at 9:39
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1$\begingroup$ The Nijenhuis tensor of $N$ also has to vanish. (So the notation $N$ is perhaps not ideal here.) $\endgroup$ Commented Sep 1, 2013 at 9:50