On the uniqueness of the Riemann-Christoffel tensor

Question

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?

Answer 1

UPDATE: The previous version of this answer was completely wrong and was deleted. I have since understood this better and have updated it.

The idea is to use Riemann normal coordinates. This is done as follows. Start with a base point $p\in M$. Choose coordinates $x^\mu(p) = 0$. We can further rotate these coordinates around so that $$ g_{\mu\nu}(p) = \eta_{\mu\nu} $$ The coordinates of a different point in $M$ is described as follows. Let $q \in M$, $q \neq p$. Let $\gamma_q:[0,1]\to M$ be the unique geodesic that starts at $p$ and ends at $q$, i.e. $\gamma_q(0) = p$ and $\gamma_q(1)=q$. Let $v$ be the vector tangent to $\gamma$ at $p$, i.e. $v \in T_pM$ The coordinate for $q$ is then defined as $$ x^\mu(q) = v^\mu~. $$ Let us now work out the Christoffel symbol, etc. in this coordinate system. By construction geodesics passing through $p$ have the form $$ x^\mu(\lambda ) = \lambda v^\mu~. $$ Plugging this into the geodesic equation, we find $$ v^\mu v^\nu \Gamma^\rho_{\mu\nu}(\lambda v) = 0 ~. $$ This is true for any $\lambda$ and any $v^\mu$. Expanding around $\lambda = 0$, we find the relations $$ \partial_{(\mu_1} \cdots \partial_{\mu_n} \Gamma^\rho_{\mu\nu)}(0) = 0 ~, \qquad n \geq 0~. $$ The above condition for $n=0$ and $n=1$ can be manipulated to imply (TRY THIS YOURSELF!) $$ \partial_\mu \partial_\nu g_{\alpha\beta}(p) = \frac{1}{3} \left[ R_{\alpha\mu\nu\beta} (p) + R_{\alpha\nu\mu\beta} (p) \right] ~. $$

Answer 2

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.

Answer 3

Now, if I want to show that the Riemann-Christoffel symbol 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.

Actually, these are not equivalent. Since $\partial^2 g_{\mu\nu} / \partial x^\lambda \partial x^\rho$ is not a tensorial (that is, covariant) object, you can't hope to recover all of the second derivatives of the metric via linear combinations of the Riemann tensor.

To find tensors which can be made of the second derivatives by brute force, you need to plug in the transformations

$$g_{\mu\nu} = \frac{\partial y^a}{\partial x^\mu} \frac{\partial y^b}{\partial x^\nu} g_{ab}, \qquad \frac{\partial}{\partial x^\mu} = \frac{\partial y^a}{\partial x^\mu} \frac{\partial}{\partial y^a},$$

and find the appropriate linear combinations of $\partial_\lambda \partial_\rho g_{\mu\nu}$ (permuting its indices) such that all the inhomogeneous terms cancel. You will find you need to take advantage of the fact that partial derivatives commute, by choosing combinations which are antisymmetric under certain pairs of indices. At the end of the day, you will have the Riemann tensor (with all indices down, of course).

But in any case, since you already believe that the Riemann tensor is the only tensor which is linear in first derivatives of the Christoffel symbols, that should be enough to convince you. As you have written in your question, the first derivatives of the metric are also linear in Christoffel symbols, so the two conditions "linear in 1st derivatives of the Christoffels" and "linear in 2nd derivatives of the metric" are equivalent.