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Let $M$ and $\tilde{M}$ be smooth manifolds, each with a metric $g_{ab}$ and $\tilde{g}_{ab}$, assumed here to be smooth symmetric invertible tensor fields, which are non-degenerate but not necessarily positive-definite. Let $\nabla_a$ be the derivative operator that is 'compatible' with $g_{ab}$ in that $\nabla_a g_{bc}=\mathbf{0}$, and similarly let $\tilde{\nabla}_a \tilde{g}_{bc}=\mathbf{0}$. (These derivative operators are sometimes called 'connections' and are unique.) Let $\varphi:M\rightarrow\tilde{M}$ be a diffeomorphism with pushforward $\varphi_*$.

If $\varphi$ is an isometry ($\varphi_*g_{ab} = \tilde{g}_{ab})$, is it true that for an arbitrary tensor field like $\lambda^{bc}_d$,

$\varphi_*\left(\nabla_a \lambda^{bc}_d\right) = \tilde{\nabla}_a \varphi_*\left(\lambda^{bc}_d\right)$?

It is clearly true in some special cases, but I'm interested in understanding this general context. A reference would be very helpful. I'm curious also if there are any non-isometries that also have this property. Thanks! -- Bryan

soulphysics
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1 Answers1

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I figured it out myself: the answer is 'always'. Consider the mapping $\hat{\nabla}$ defined by,

$\hat{\nabla}:\lambda^{bc}_d\mapsto \varphi^*(\tilde{\nabla}_a \varphi_* \lambda^{bc}_d).$

The first step is to show that this mapping is a derivative operator. It obviously commutes with addition, index substitution and contraction because all three maps do ($\varphi_*$, $\varphi^*$ and $\tilde{\nabla}_a$). It's also easy to show that it satisfies the Leibniz rule, since the pull-backs and push-forwards commute with multiplication and $\tilde{\nabla}_a$ satisfies the Leibniz rule. Finally, for all vectors $\xi^n$ and all scalar fields $\alpha$, it satisfies the condition that $\xi^n \hat{\nabla}_n\alpha = \xi(\alpha)$. These are the conditions for being a derivative operator. No assumption about $\varphi$ being an isometry is needed for this, but only that it is a diffeomorphism.

The second step is to show that $\hat{\nabla}$ is compatible with the metric, in the sense that $\hat{\nabla}_a g_{bc}=\mathbf{0}$. That's almost immediate when $\varphi$ is a (bijective) isometry.

Finally, I just use the known fact that there is a unique compatible derivative operator associated with a given metric. So, $\hat{\nabla}$ and $\nabla$ are the same. Therefore,

$\nabla_a \lambda^{bc}_d = \hat{\nabla}_a\lambda^{bc}_d = \varphi^*(\tilde{\nabla}_a \varphi_* \lambda^{bc}_d)$,

or, pushing-forward the left-most and right-most sides with $\varphi_*$,

$\varphi_* (\nabla_a \lambda^{bc}_d) = \tilde{\nabla}_a \varphi_* \lambda^{bc}_d,$

which proves the claim. It's still not obvious to me whether or not there's a non-isometry $\varphi$ that has this same property, but I suppose it's clear how to figure it out now.

EDIT: I recognize that nobody seems to care about this question, but for completeness: there are some non-isometries that have this property. What you can show (using little more than the argument I sketched above) is that a diffeomorphism $\varphi$ satisfies the equality above for all tensors if and only if $\tilde{\nabla}_a\varphi_* g_{bc}=\mathbf{0}$. In particular it holds of isometries. But, an example of a non-isometry that satisfies the equality is a conformal transformation for which the conformal factor $\Omega$ is a constant scalar field, $\tilde{\nabla}_a\Omega=0$. These are in fact the only conformal transformations for which the equality holds. --- Thanks Bryan! --- You're welcome, Bryan!

soulphysics
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