Degrees of freedom of Riemann curvature tensor

I know the argument that uses the symmetries

1. $$R_{a b c d} = -R_{b a c d} = R_{c d a b}$$
2. $$R_{a b c d} + R_{b c a d} + R_{c a b d} = 0$$

of the Riemann curvature tensor $$R$$ of an $$n$$-dimensional Riemannian manifold to show that it (the curvature tensor) has at most $$\frac1{12} n^2 (n^2 - 1)$$ degrees of freedom (see for example here, section 7) at a given point.

As far as I know $$\frac1{12} n^2 (n^2 - 1)$$ is the actual number of degrees of freedom, so my question is how does one get the lower bound?

EDIT: To clarify: The argument I linked to shows that using the mentioned symmetries $$\frac1{12}n^2 (n^2 - 1)$$ of the $$n^4$$ entries of $$R$$ already determine it. However there could be other constraints like symmetries or inequalities that make the actual number of degrees of freedom even lower. As far as I know, this is not the case and my but I would like a proof of this.

(The only wat to prove this I can think of would be constructing a family of manifolds parametrized by $$\mathbb{R}^{n^2 (n^2 - 1)/12}$$ such that for a given $$p \in \mathbb{R}^{n^2 (n^2 - 1)/12}$$ the corresponding manifold has the curvature tensor corresponding to $$p$$ at some point.)

• lower bound on what ? – Amr Dec 30 '18 at 2:11
• @Amr on the degrees of freedom. – 0x539 Dec 30 '18 at 2:25
• Wouldn't $\frac{1}{12}n^2(n^2-1)$ count as a lower bound already ? – Amr Dec 30 '18 at 3:08
• @Amr Yes it is, but i want to know why. – 0x539 Dec 30 '18 at 3:19
• Duplicate of math.stackexchange.com/questions/2235253/… (no vote to close due to open bounty) – Dap Jan 16 at 11:12

In this answer I will work out the degrees of freedom of the Riemann tensor in the most general way, and the answer will be exactly equal to $$\frac{1}{12}n^2(n^2-1)$$.

The components of the Riemann tensor characterizes the genuine curvature of the space-time. Since this is the genuine curvature, the number cannot be further reduced at any region of the space-time.

Consider a general coordinate transformation from $$x^i$$ to $$x'^i$$ in the neighborhood of the origin of the coordinate system $$x^i$$. Then, $$x'^i$$ can be expanded in Taylor series as follows $$x'^i=B^i_kx^k+C^i_{jk}x^jx^k+D^i_{jkl}x^jx^kx^l+...$$ where the coefficients $$B^i_k,C^i_{jk},D^i_{jkl}$$ are symmetric in the lower indices.

Our aim is to make the metric as close to Cartesian as possible near the origin i.e., we must have $$g_{ab}=\eta_{ab}$$ at the origin and make as many (first order, second order, etc) derivatives of the metric tensor vanish at the origin as possible. The requirements are as follows:

1. The choice of $$B^i_k$$ such that the transformed metric is $$g'_{ik}=\eta_{ik}$$:

The number of independent components of $$B^i_k$$ is $$n^2$$.The number of conditions corresponding to $$g'_{ik}=\eta_{ik}$$ is $$\frac{1}{2}n(n+1)$$, for the symmetric pair (i,k). So, the $$B^i_k$$ can be chosen so as to impose the condition $$g'_{ik}=\eta_{ik}$$ with $$[n^2-\frac{1}{2}n(n+1)]=\frac{1}{2}n(n-1)$$ parameters to spare. This is precisely the number of free parameters required to specify Lorentz transformations and spatial rotations in n dimensions.

2. The choice of $$C^i_{jk}$$ such that the first derivatives of the metric vanishes:

The number of independent components of $$C^i_{jk}$$ is $$\frac{1}{2}n^2(n+1)$$. Now the condition of the vanishing of the first derivatives of $$g'_{ik}$$, i.e., the condition $$\frac{\partial g'_{ik}}{\partial x'^a}=0$$ requires $$n\times\frac{1}{2}n(n+1)=\frac{1}{2}n^2(n+1)$$ conditions, for the symmetric pair (i,k) corresponding to each index a. Notice that the number of conditions for the vanishing first derivatives and the number of independent components of $$C^i_{jk}$$ are both equal. This means that all the first derivatives of the metric tensor vanishes at the origin.

3. The choice of $$D^i_{jkl}$$ such that the second derivatives of the metric vanishes:

The number of independent components of $$D^i_{jkl}$$ is $$\frac{1}{6}n^2(n+1)(n+2)$$. Now the condition of the vanishing of the second derivatives of $$g'_{ik}$$, i.e., the condition $$\frac{\partial^2 g'_{ik}}{\partial x'^a \partial x'^b}=0$$ requires $$\frac{1}{2}n(n+1)\times\frac{1}{2}n(n+1)=\frac{1}{4}n^2(n+1)^2$$ conditions, for the symmetric pairs (i,k) and (a,b). Notice that this value for the independent components of the second derivatives of the metric tensor is greater than the number of components of $$D^i_{jkl}$$. This clearly means that all the second derivatives of the metric tensor can never vanish in any region of the space-time.

The number of non-vanishing second derivatives of the metric tensor can be calculated as $$\frac{1}{4}n^2(n+1)^2-\frac{1}{6}n^2(n+1)(n+2)=\frac{1}{12}n^2(n^2-1)$$ This is precisely equal to the degrees of freedom of the Riemann tensor and no further symmetries can reduce the degrees of freedom below this value.