# Can we always locally split the metric as $- N^2dt^2+g_{ij}dx^idx^j$?

In the book by Y. Choquet-Bruhat, General Relativity and the Einstein Equations, the following technical lemma is found on page 9:

A Lorentzian metric can always be written in a small enough neighborhood by a change of coordinates under the form $$-N^2dt^2+g_{ij}dx^idx^j.$$

The proof (I think that's what it's supposed to be) she gives makes little sense:

Indeed, under a change of coordinates $(x'^\alpha)\mapsto (x^\beta)$ with $x^0=x'^0$ we have $$g'_{i0}=\frac{\partial x^j}{\partial x'^i}\left(g_{j0}+g_{jh}\frac{\partial x^h}{\partial x'^0}\right),$$ we make $g_{i0}'=0$ by solving the linear first-order system $g_{j0}+g_{jh}\frac{\partial x^h}{\partial x'^0}=0$ for the functions $x^h(x'^i,x'^0)$.

The reason this is problematic is that we assumed $x^0=x'^0$, so that in fact $x^h$ cannot be a function of $x'^0$, and the linear system falls apart.

Am I misinterpreting what she's saying? Can the proof be salvaged or is this a bad typo? Is the result true?

• I may be overlooking the issue, but I don't see why setting $x'^0 = x^0$ forces $x^h$ not to be a function of $x'^0$ – yoric Jan 7 '17 at 18:57
• @yoric Because $x^h$ is not a function of $x^0$, as these are coordinates, recall $\partial x^\mu/\partial x^\nu=\delta^\mu_\nu$. – Ryan Unger Jan 7 '17 at 19:17
• I still don't get that. Take $(t, \mathbf{x}) \mapsto (t, R(\omega t) \mathbf{x})$ where $R(\theta)$ is some rotation of angle $\theta$. – yoric Jan 7 '17 at 19:23
• @yoric I'm not sure that's a valid coordinate system on the right. It doesn't make sense that the latter coordinates rotate more as you change $t$, even if you're not changing $\mathbf{x}$. – Ryan Unger Jan 7 '17 at 19:31
• Why do you say that it doesn't make sense? You can do whatever you want with your coordinates as long as you have (local) diffeomorphism. And you are changing the $\mathbf{x}$: $\mathbf{x}' = R(\omega t) \mathbf{x}$ – yoric Jan 7 '17 at 19:34

We offer an alternative proof.

Let $(M^{n+1},g)$ be a Lorentzian manifold, and fix $p\in M$. Let $(x^\mu)$ be a coordinate system defined on $U\ni p$ such that $\partial_0$ is a timelike vector field and $\partial_i$ are spacelike vector fields for $i=1,\dotsc,n$. Such a chart is constructed here. In particular, note that $g_{\mu\nu}(p)=\eta_{\mu\nu}$, the Minkowski metric. Thus the inverse metric $g^{-1}$ at $p$ is $\eta^{\mu\nu}$. As the dual of $\{\partial_\mu\}$ is $\{dx^\mu\}$, we have $g^{00}=g^{-1}(dx^0,dx^0)=-1$ at $p$, so $g^{-1}(dx^0,dx^0)<0$ in a neighborhood $U$ of $p$.

Recall the following fact: $g(X,Y)=g^{-1}(X^\flat,Y^\flat)$, where $\flat$ is the lowering operator. To see this, work in a basis, then $g(X,Y)=g_{\mu\nu}X^\mu Y^\nu=g_{\mu\nu}g^{\mu\rho}X_\rho g^{\nu\sigma}Y_\sigma=g^{\rho\sigma}X_\rho Y_\sigma=g^{-1}(X^\flat,Y^\flat).$

Thus we have $g(\text{grad}\,x^0,\text{grad}\,x^0)<0$ in $U$, so $\text{grad}\,x^0$ is timelike there. Consider the hypersurface $\Sigma=[x^0=0]$ in $U$, which contains $p$. It is known that the normal field to $\Sigma$ is $\text{grad}\,x^0$, which is timelike. Thus $\Sigma$ is a spacelike hypersurface containing $p$.

Gaussian coordinates then give the desired form of the metric in some neighborhood of $p$, with $N=1$.

• Yes your proof works, I use the same proof. Gaussian coordinates also proves that locally the choice $N=1$ is always possible as you noticed... – Valter Moretti Jan 7 '17 at 19:23
• @ValterMoretti Incidentally, do you know how to prove that Gaussian normal coordinates are defined for $|t|<\delta$, and that the assignment is a diffeomorphism? I'm having a hard time showing that the geodesics emanating from $\Sigma$ won't cross for small $t$. – Ryan Unger Jan 7 '17 at 21:02
• There is a general proof in O'Neill's textbook, have a look at it. – Valter Moretti Jan 7 '17 at 21:07
• @ValterMoretti I have it next to me, can you give a page please? I didn't see the topic in the index. – Ryan Unger Jan 7 '17 at 21:08
• Found! From page 197 on – Valter Moretti Jan 7 '17 at 21:23

In the comments the OP clarified somehow his argument. Here I show why it fails.

There is nothing that prevent $x^h$ to depend on $x'^0$, even though $x^0 = x'^0$, for example $$u:(t,\mathbf{x}) \mapsto (t',\mathbf{x'}) = (t, R(\omega t)\mathbf{x}),$$ being $R(\theta)$ some rotation of angle $\theta$.

Now, the problem seems coming from the identity $\partial_\nu x^\mu = \delta_\nu^\mu$, that I think to be true. But we have to use if carefully. For instance, in this very example $$\frac{\partial \mathbf x'}{\partial t} = 0$$ but what is nonzero is $$\frac{\partial \mathbf x' \circ u }{\partial t} =\frac{\partial R \mathbf x }{\partial t} = \frac{\partial R}{\partial t} \mathbf x$$ where $u$ is the change of coordinates.

This confusion is a consequence of calling with same letters both the coordinates and the diffeomorhism between them.

• Agreed with the last statement. – Ryan Unger Jan 8 '17 at 1:24