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I'm having a hard time proving the following:

If $M$ is an $n$-dimensional indefinite Riemannian manifold whose metric $g$ has index $s$, then the metric of Sasaki $g^{D}$ is an indefinite metric on $TM$ whose index is $2s$. Let $J'$ be the natural almost complex structure on TM, then $TM(J',g^{D})$ is an indefinite almost Kähler manifold (that means the fundamental 2-form is closed). Moreover $(TM,g^{D} )$ is Kähler if and only if $M$ is locally flat.

This assertion is from the article Indefinite Kähler manifolds by Manuel Barros and Alfonso Romero.

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  • $\begingroup$ By $D$, you mean the Levi-Civita connection, right? Otherwise, for an arbitrary affine connection, the requirement for the Sasaki metric to be Kahler is for $D$ to be flat and to be Codazzi-coupled to the metric $g$. $\endgroup$
    – Gabe K
    Commented May 16, 2020 at 22:06
  • $\begingroup$ actully $D$ represents covariant differentiation: $Dv^{i}=dv^{i}+Γ_{jk}^{i}v^{j}dx^{k}$ $\endgroup$
    – MOULI Kawther
    Commented May 16, 2020 at 22:13
  • $\begingroup$ Right. Choosing a connection is equivalent to choosing the connection coefficients $\Gamma_{jk}^i$. The point is that the theorem is only true when the connection used is the Levi-Civita connection. $\endgroup$
    – Gabe K
    Commented May 16, 2020 at 22:37
  • $\begingroup$ assuming the connection used is the Levi-Civita connection connection, can you please help me with the proof? $\endgroup$
    – MOULI Kawther
    Commented May 16, 2020 at 23:02

1 Answer 1

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For a proof of this result (and a more general version), there's a paper by Satoh [1] which has a lot of detail. The main idea is that for $TM$ to be Hermitian, the Nijenhuis tensor of $J^\prime$ needs to vanish. However, when you calculate the Nijenhuis tensor, you find that it vaishes if and only if the torsion and curvature of $D$ vanish (i.e. D is flat). See the equation on the bottom of page 8 of [1] for the exact formula.

For a Riemannian manifold, $g$ is flat iff the Levi-Civita connection is a flat connection, so I presume this is what the authors are using. For a more general connection $D$ (i.e. not the Levi-Civita connection), $(TM, J^\prime,g^D)$ is Kahler iff $(M,g,D)$ is a so-called Hessian manifold, which means that $D$ is flat and satisfies $$(D_X g)(Y,Z)=(D_Y g)(X,Z)$$ for all vector fields $X,Y$ and $Z$. This relationship between the metric and connection is also known as Codazzi-coupling. The proof of this is also in that reference.

[1] Satoh, Hiroyasu, Almost Hermitian structures on tangent bundles, Suh, Young Jin (ed.) et al., Proceedings of the 11th international workshop on differential geometry, Taegu, Korea, November 9–11, 2006. Taegu: Kyungpook National University. 105-118 (2007). ZBL1125.53022.

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  • $\begingroup$ thanks a lot, can we use the same argument if $TM$ is associated with $J^h$ instead where h is the horisental lifting? $\endgroup$
    – MOULI Kawther
    Commented May 17, 2020 at 3:18

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