Another interpretation of the Riemann tensor

One elementary model frame of general relativity is the Minkowski time-space $(\mathbb{R}^3,m)$ where $m$ denotes the Minkowski metric $m=-(dt)^2+(dx)^2+(dy)^2+(dz)^2=m_{ij}dx^idx^j$.

One naive question is thus: can we get a local Minkowski metric in a general time-space? In general, the answer is no. Yet, we can do this at each point. That is to say, at each point, we can choose some coordinate system such that the metric at this point under this coordinate system is the Minkowski metric.

In deed, note that, suppose that $(M,g)$ is a time-space of dimension $n$, i.e. a Lorentzian manifold, and $p\in M$ is any point fixed. Suppose there are two coordinate systems, $(x^{\alpha}),(y^i)$. Under these coordinate systems, the metric writes $g_{\alpha\beta},g_{ij}$, and there is a transformation formula, $g_{\alpha\beta}=g_{ij}\frac{\partial y^i}{\partial x^{\alpha}}\frac{\partial y^j}{\partial x^{\beta}}$. Note that, these functions $\frac{\partial y^i}{\partial x^{\alpha}}$ are independent, or at least we can choose them to be linearly independent. Then, we can choose some coordinate system $(x^{\alpha})$ such that $g_{\alpha\beta}$ is the diagonalized form at point $p$, $g_{\alpha\beta}(p)=diag(-1,1,1,1)$. In $g_{ij}$, there are $n(n+1)/2$ independent entries, whereas we have $n^2$ independent functions $\frac{\partial y^i}{\partial x^{\alpha}}$, so we can indeed diagonalize the metric at one point.

Then we inquire further, whether we can say something about the first derivatives of the metric with respect to the coordinate functions? That is $g_{\alpha\beta}(p+dp)=g_{\alpha\beta}(p)+\frac{\partial g_{\alpha\beta}}{\partial x^{\gamma}}dy^k$. Now in the coordinate system $(y^i)$, we have that $\frac{\partial g_{\alpha\beta}}{\partial x^{\gamma}}=\frac{1}{\partial x^{\gamma}}(g_{ij}\frac{\partial y^i}{\partial x^{\alpha}}\frac{\partial y^j}{\partial x^{\beta}})$. In other words,

$\frac{\partial g_{\alpha\beta}}{\partial x^{\gamma}} =\frac{\partial g_{ij}}{\partial x^{\gamma}} \frac{\partial y^i}{\partial x^{\alpha}} \frac{\partial y^j}{\partial x^{\beta}} +2g_{ij} \frac{\partial^2 y^i}{\partial x^{\gamma} \partial x^{\alpha}} \frac{\partial y^j}{\partial x^{\beta}}$

In this way, we see that the second derivatives of $y^i$ are involved. We can still suppose that these second derivatives $\frac{\partial^2 y^i}{\partial x^{\alpha}\partial x^{\beta}}$ are independent. So, there are $n\times n(n+1)/2$ independent functions. On the other hand, to get that $\frac{\partial g_{\alpha\beta}}{\partial x^{\gamma}}(p)=0$, we should have $n(n+1)/2\times n$ independent functions. These two numbers are equal, showing that we can indeed choose a coordinate system $(x^{\alpha})$ such that $\frac{\partial g_{\alpha\beta}}{\partial x^{\gamma}}(p)=0$. In this coordinate system, we have that the Christoffel symbols are all zero at the point $p$. This, in some sense, ressembles to the normal coordinate system.

One application of this coordinate system is to prove some tensor equalities. Sometimes this can greatly reduce the calculations. For example, under this coordinate system, the Riemann tensor at the point $p$ is a linear function of $g, \partial g, \partial ^2 g$, the quadratic terms all vanish. Using the simplified expression, we can easily deduce the symmetric properties of the Riemann tensor.

Again further, what can we say about the second derivatives of the metric with respect to the coordiante functions? Very similarly, we see that for $\frac{\partial^2 g_{\alpha\gamma}}{\partial x^{\delta} \partial x^{\beta}}$ to vanish at the point $p$, we need $(n(n+1)/2)^2$ independent functions. Yet, this time, the third derivatives $\frac{\partial^3 y^i}{\partial x^{\alpha} \partial x^{\beta} \partial x^{\gamma}}$ are involved. We can suppose that they are independent, thus we have $n\times n(n+1)(n+2)/6$ independent functions. This time, we are not so lucky, because $(n(n+1)/2)^2-n^2(n+1)(n+2)/6=n^2(n^2-1)/12>0$ for $n>1$. So, we can not choose a coordinate system $(x^{\alpha})$ such that the second derivatives of the metric vanish at a point. Note the number $n^2(n^2-1)/12$, very familiar, isn’t it? In fact, we can say that the using the coordinate function $(x^{\alpha})$, we can control $n^2(n+1)(n+2)/6$ components of the second derivatives of the metric. There are $n^2(n^2-1)/12$ components left, which are reflected in the Riemann tensor.

In this way, we can interpret the Riemann tensor as a complement to the coordinate functions, which altogether determine the second derivative of the metric.