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Let $(M,g)$ be a Riemannian manifold with Levi Civita connection $\nabla$. Then $\nabla$ satisfies a compatibility condition:
$(\nabla_ZX,Y)+(X,\nabla_ZY)=Z((X,Y))$
where $(\cdot,-)$ is a Hermitian pairing. In general, if we have a connection $\nabla$ on bundle $E$ (in our case $E=TM$) one can define the dual connection with the formula
$\nabla'_Z(\alpha)=Z(\alpha(\cdot))-\alpha(\nabla_Z(\cdot))$
where $\alpha$ is a one form. My question is the following:

Does the dual connection satisfy compatibility condition?

I did some computation and get that compatibility is equivalent to the condition: $Z(g^{ij})\alpha_i\beta_j=-g^{ij}Z^p\Gamma_{pi}^q\alpha_q\beta_j-g^{ij}Z^p\Gamma_{pj}^q\alpha_i\beta_q$
where $\Gamma_{ij}^k$ are defined by $\nabla_{\partial_i}\partial_j=\Gamma_{ij}^k\partial_k$, $g^{ij}$ are components of the inverse matrix to the$(g_{ij})_{i,j}$ where $g_{i,j}=g(\partial_i,\partial_j)$ (I used the Einstein summation convention).

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It's a bit too late in the evening for me to start the calculation, but in general the Levi Civita connection extends to the higher order tensor bundles (which all inherit a scalar product) and satisfies there the corresponding compatibility conditions. Without having checked I think your $\nabla^\prime$ is just that.

This is discussed in detail in the second volume of Michael Spivak's "Comprehensive Introduction to Differential Geometry", where you will also find all the formulae both in coordinate free as well as in coordinate dependent formulation.

What you are after specifically with your question is something which is called 'Ricci's Lemma' (also often referred to by saying that 'the metric is parallel wrt to it's induced (Levi Civita) connection'), which you will find in the index formulation as Lemma 3 in Chapter 5 in that book.

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  • $\begingroup$ Thank you! What was exactly needed: it was the formula $(**)$ in the first volume of Spivak's book (see page 331). It seems that the proof uses the exact formula for Christoffels symbols for Levi Civita connection, not only the fact that it is compatible with the metric: nevertheless it solves my problem so once again-thank you! $\endgroup$ – truebaran Sep 8 '15 at 22:52

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