According to Section 6.2, Gravitation and Cosmology by Weinberg, the Riemann-Christoffel tensor is the only tensor that can be constructed out of the second (or lower) order derivatives of the metric tensor and is linear in the second order derivatives. The reasoning behind the same goes like this:
In a class of frames where $\Gamma^{\lambda}_{\mu\nu}
=0$, the transformation rule for $\dfrac{\partial \Gamma^{\lambda}_{\mu\nu}}{\partial x^{\kappa}}$ involves an inhomogeneous term which is symmetric in $\mu$, $\nu$, and $\kappa$. Thus, if one is to construct a tensor which is a linear combination of the first order derivatives of the Christoffel symbol then the only way to do so is by eliminating the inhomogeneous part of the transformation and this could be done only by making the combination explicitly antisymmetric in $\mu$ and $\kappa$. Since in these frames, $R^{\lambda}_{\mu\nu\rho}$ $ = \dfrac{\partial \Gamma^{\lambda}_{\mu\nu}}{\partial x^{\rho}} – \dfrac{\partial \Gamma^{\lambda}_{\mu\rho}}{\partial x^{\kappa}} $, $R^{\lambda}_{\mu\nu\rho}$ is the only tensor that can be formulated using the second (or lower) order derivatives of the metric tensor and is linear in the second order derivatives.
I think the presented argument can only suffice to prove that $R^{\lambda}_{\mu\nu\rho}$ is the only tensor that can be formulated from the first derivatives of the Christoffel symbol and is linear in them. I can't figure out why this suffices to assert that $R^{\lambda}_{\mu\nu\rho}$ is the only tensor that can be constructed out of the second (or lower) order derivatives of the metric tensor and is linear in the second order derivatives.
Edit: As the metric is covariantly constant, $\dfrac{\partial g_{{\mu}{\nu}}}{\partial x^{\rho}} = \Gamma^{\kappa}_{{\mu}{\rho}}g_{{\kappa}{\nu}}+\Gamma^{\kappa}_{{\nu}{\rho}}g_{{\kappa}{\mu}}$.
Therefore,
$\dfrac{\partial^2 g_{{\mu}{\nu}}}{\partial x^{\xi}\partial x^{\rho}} = \bigg(\dfrac{\partial \Gamma^{\sigma}_{{\mu}{\rho}}}{\partial x^{\xi}} + \Gamma^{\sigma_1}_{{\mu}{\rho}}\Gamma^{\sigma}_{{\sigma_1}{\xi}}\bigg)g_{{\sigma}{\nu}}+\bigg(\dfrac{\partial \Gamma^{\sigma}_{{\nu}{\rho}}}{\partial x^{\xi}} + \Gamma^{\sigma_1}_{{\nu}{\rho}}\Gamma^{\sigma}_{{\sigma_1}{\xi}}\bigg)g_{{\sigma}{\mu}} + \bigg(\Gamma^{\sigma}_{{\mu}{\rho}}\Gamma^{\sigma_1}_{{\nu}{\xi}}+\Gamma^{\sigma}_{{\nu}{\rho}}\Gamma^{\sigma_1}_{{\mu}{\xi}}\bigg)g_{{\sigma}{\sigma_1}} $
Now, if I want to show that the Riemann-Christoffel tensor is the only (non-trivial and independent) tensor that can be formulated out of the linear combinations of the second order derivatives of the metric tensor then I can equivalently show that the above expansion can be expressed linearly in terms of the Riemann-Christoffel tensor. But I am stuck over how to do it. Also, I am a little bit unclear as to what a linear combination means in this context. The coefficient of the second order term can be only a scalar constant or can it be the metric or can it be the metric that is getting summed over any of the indices as well?
Best Answer
Note that we're working in a normal coordinate system, so your equation for the second derivative of the metric simplifies to $$\tag{$1$}\frac{\partial^2 g_{\mu\nu}}{\partial x^\xi \partial x^\rho}=\frac{\partial\Gamma^\sigma{}_{\mu\rho}}{\partial x^\xi}g_{\sigma\nu}+\frac{\partial\Gamma^\sigma{}_{\nu\rho}}{\partial x^\xi}g_{\sigma\mu}.$$ So from this it should be clear that if we try to build a tensor out of second derivatives of the metric, we will be building it out of first derivatives of the Christoffel symbols. But let's be more explicit.
If we go to this Wiki page, we find the following formula for the Riemann tensor in normal coordines (Weinberg calls them inertial coordinates) $$\tag{$2$}R_{\mu\nu\rho\sigma}=\frac{1}{2}\left(\frac{\partial g_{\mu\sigma}}{\partial x^\nu\partial x^\rho}+\frac{\partial g_{\nu\rho}}{\partial x^\mu\partial x^\sigma}-\frac{\partial g_{\mu\rho}}{\partial x^\nu\partial x^\sigma}-\frac{\partial g_{\nu\sigma}}{\partial x^\mu\partial x^\rho}\right).$$
From Weinberg, we have $$g_{\mu\nu}'=g_{\rho\sigma}\frac{\partial x^\rho}{\partial x'^\mu}\frac{\partial x^\sigma}{\partial x'^\nu},$$ $$\frac{\partial \Gamma'^\tau{}_{\rho\sigma}}{\partial x'^\eta}=\frac{\partial x^\mu}{\partial x'^\rho}\frac{\partial x^\nu}{\partial x'^\sigma}\frac{\partial x^\kappa}{\partial x'^\eta}\left(\frac{\partial x'^\tau}{\partial x^\lambda}\frac{\partial \Gamma^\lambda{}_{\mu\nu}}{\partial x^\kappa}-\frac{\partial^3 x'^\tau}{\partial x^\kappa\partial x^\mu\partial x^\nu}\right).$$ Inserting these into $(1)$, we have \begin{align*}\frac{\partial^2 g'_{\mu\nu}}{\partial x'^\xi \partial x'^\rho}&=\frac{\partial\Gamma'^\sigma{}_{\mu\rho}}{\partial x'^\xi}g_{\sigma\nu}'+\frac{\partial\Gamma'^\sigma{}_{\nu\rho}}{\partial x'^\xi}g_{\sigma\mu}'\\ &=\frac{\partial x^\omega}{\partial x'^\mu}\frac{\partial x^\zeta}{\partial x'^\rho}\frac{\partial x^\kappa}{\partial x'^\xi}\frac{\partial x^\alpha}{\partial x'^\sigma}\frac{\partial x^\beta}{\partial x'^\nu}\left(\frac{\partial x'^\sigma}{\partial x^\lambda}\frac{\partial \Gamma^\lambda{}_{\omega\zeta}}{\partial x^\kappa}-\frac{\partial^3 x'^\sigma}{\partial x^\kappa\partial x^\omega\partial x^\zeta}\right)g_{\alpha\beta} \\&\quad+\frac{\partial x^\omega}{\partial x'^\nu}\frac{\partial x^\zeta}{\partial x'^\rho}\frac{\partial x^\kappa}{\partial x'^\xi}\frac{\partial x^\alpha}{\partial x'^\sigma}\frac{\partial x^\beta}{\partial x'^\mu}\left(\frac{\partial x'^\sigma}{\partial x^\lambda}\frac{\partial \Gamma^\lambda{}_{\omega\zeta}}{\partial x^\kappa}-\frac{\partial^3 x'^\sigma}{\partial x^\kappa\partial x^\omega\partial x^\zeta}\right)g_{\alpha\beta}.\end{align*} This is equal to \begin{align*} \frac{\partial^2 g'_{\mu\nu}}{\partial x'^\xi \partial x'^\rho}&=\frac{\partial x^\omega}{\partial x'^\mu}\frac{\partial x^\zeta}{\partial x'^\rho}\frac{\partial x^\kappa}{\partial x'^\xi}\frac{\partial x^\beta}{\partial x'^\nu}\frac{\partial \Gamma^\alpha{}_{\omega\zeta}}{\partial x^\kappa}g_{\alpha\beta}+\frac{\partial x^\omega}{\partial x'^\nu}\frac{\partial x^\zeta}{\partial x'^\rho}\frac{\partial x^\kappa}{\partial x'^\xi}\frac{\partial x^\beta}{\partial x'^\mu}\frac{\partial \Gamma^\alpha{}_{\omega\zeta}}{\partial x^\kappa}g_{\alpha\beta}\\ &\quad -\frac{\partial x^\omega}{\partial x'^\mu}\frac{\partial x^\zeta}{\partial x'^\rho}\frac{\partial x^\kappa}{\partial x'^\xi}\frac{\partial x^\alpha}{\partial x'^\sigma}\frac{\partial x^\beta}{\partial x'^\nu}\frac{\partial^3 x'^\sigma}{\partial x^\kappa\partial x^\omega\partial x^\zeta}g_{\alpha\beta}\\ &\quad-\frac{\partial x^\omega}{\partial x'^\nu}\frac{\partial x^\zeta}{\partial x'^\rho}\frac{\partial x^\kappa}{\partial x'^\xi}\frac{\partial x^\alpha}{\partial x'^\sigma}\frac{\partial x^\beta}{\partial x'^\mu}\frac{\partial^3 x'^\sigma}{\partial x^\kappa\partial x^\omega\partial x^\zeta}g_{\alpha\beta}. \end{align*} We need to kill the last two terms by taking linear combinations (in the linear algebra sense) of this object with indices permuted. For our purposes we can drop the Christoffel terms, because they do have the correct transformation behavior. So $$\tag{$3$}\left.\frac{\partial^2 g'_{\mu\nu}}{\partial x'^\xi \partial x'^\rho}\right|_{\text{bad}}=-\left(\frac{\partial x^\omega}{\partial x'^\mu}\frac{\partial x^\zeta}{\partial x'^\rho}\frac{\partial x^\kappa}{\partial x'^\xi}\frac{\partial x^\alpha}{\partial x'^\sigma}\frac{\partial x^\beta}{\partial x'^\nu}+\frac{\partial x^\omega}{\partial x'^\nu}\frac{\partial x^\zeta}{\partial x'^\rho}\frac{\partial x^\kappa}{\partial x'^\xi}\frac{\partial x^\alpha}{\partial x'^\sigma}\frac{\partial x^\beta}{\partial x'^\mu}\right)\frac{\partial^3 x'^\sigma}{\partial x^\kappa\partial x^\omega\partial x^\zeta}g_{\alpha\beta}.$$ We write this as $$G_{\mu\nu\xi\rho}=\left(D^{\omega\zeta\alpha\kappa\beta}_{\mu\rho\xi\sigma\nu}+D^{\omega\zeta\alpha\kappa\beta}_{\nu\rho\xi\sigma\mu}\right)T^\sigma_{\kappa\omega\zeta}g_{\alpha\beta}$$ for clarity. We want to write a linear combination of these terms so that it vanishes. The key observation is that the interchanges $\kappa\leftrightarrow\omega\leftrightarrow\zeta$ don't do anything because partial derivatives commute. If we do $\nu\leftrightarrow \rho$ and subtract, we kill the second term but end up with $$G_{\mu\nu\xi\rho}-G_{\mu\rho\zeta\nu}=\left(D^{\omega\zeta\kappa\alpha\beta}_{\mu\rho\xi\sigma\nu}-D^{\omega\zeta\kappa\alpha\beta}_{\mu\nu\xi\sigma\rho}\right)T^\sigma_{\kappa\omega\zeta}g_{\alpha\beta}.$$ If we do $\mu\leftrightarrow \zeta$ and subtract, we kill the first term and end up with $$G_{\mu\nu\xi\rho}-G_{\mu\rho\zeta\nu}-G_{\xi\nu\mu\rho}=(-D^{\omega\zeta\kappa\alpha\beta}_{\mu\nu\xi\sigma\rho}-D^{\omega\zeta\kappa\alpha\beta}_{\nu\rho\mu\sigma\zeta})T^\sigma_{\kappa\omega\zeta}g_{\alpha\beta}.$$ The only move left is to do $\mu\leftrightarrow \zeta$ and $\rho\leftrightarrow\nu$. Explicitly, this is $$G_{\xi\rho\mu\nu}=\left(D^{\omega\zeta\kappa\alpha\beta}_{\xi\nu\mu\sigma\rho}+D^{\omega\zeta\kappa\alpha\beta}_{\rho\nu\mu\sigma\xi}\right)T^\sigma_{\kappa\omega\zeta}g_{\alpha\beta}.$$ Using invariance one more, we see that this equals the negative of the previous equation, so $$G_{\mu\nu\xi\rho}-G_{\mu\rho\zeta\nu}-G_{\xi\nu\mu\rho}+G_{\xi\rho\mu\nu}=0.$$ So we're done. This is the simplest linear combination that can kill the bad terms. Compare with (2) above to verify that this the Riemann tensor.