# Finding inverse metric tensor when there are off-diagonal terms

1. Jan 25, 2014

### Nabigh R

How do you find the inverse of metric tensor when there are off-diagonals?
More specifivally, given the (Kerr) metric,
$$d \tau^2 = g_{tt} dt^2 + 2g_{t \phi} dt d\phi +g_{rr} dr^2 + g_{\theta \theta} d \theta^2 + g_{\phi \phi} d \phi^2 +$$
we have the metric tensor;
$$g_{\mu \nu} = \begin{pmatrix} g_{tt} & 0 & 0 & g_{t \phi} \\ 0 & g_{rr} & 0 & 0 \\ 0 & 0 & g_{\theta \theta} & 0 \\ g_{\phi t} & 0 & 0 & g_{\phi \phi} \\ \end{pmatrix}$$
In "A First Course in General Relativity", Schutz say that since the only off-diagonal element involves $t$ and $\phi$, the contravariant components $g^{rr}$ and $g^{\theta \theta}$ are given by $g^{rr} = \frac{1}{g_{rr}}$ and $g^{\theta \theta} = \frac{1}{g_{\theta \theta}}$. And then invert the matrix
\begin{pmatrix}
g_{tt} & g_{t \phi} \\
g_{\phi t} & g_{\phi \phi} \\
\end{pmatrix}
to find $g^{tt}$, $g^{\phi t}$ and $g^{\phi \phi}$. I don't get why we can do that. Is it some kind on generalised version for the inverse of a diagonal matrix.
If $A = \begin{pmatrix} a_{11} & 0 & 0 & 0 \\ 0 & a_{22} & 0 & 0 \\ 0 & 0 & a_{33} & 0 \\ 0 & 0 & 0 & a_{44} \\ \end{pmatrix}$
then $A^{-1} = \begin{pmatrix} \frac{1}{a_{11}} & 0 & 0 & 0 \\ 0 & \frac{1}{a_{22}} & 0 & 0 \\ 0 & 0 & \frac{1}{a_{33}} & 0 \\ 0 & 0 & 0 & \frac{1}{a_{44}} \\ \end{pmatrix}$

2. Jan 25, 2014

### Mentz114

3. Jan 25, 2014

### Nabigh R

Thanks Mentz :-D I know I can get $g^{\mu \nu}$ by inverting the matrix representation of $g_{\mu \nu}$. But what I want to know is reasoning Schutz used to simplify the problem of finding the inverse of a $4 \times 4$ matrix to that of finding the inverse of a $2 \times 2$ matrix :-)
If there is a general theorem or something that allows it, then it sure can save me a lot of work.

4. Jan 25, 2014

### DrGreg

It's perhaps easier to see by writing the coordinates in a different order:
$$g_{\mu \nu} = \begin{pmatrix} g_{tt} & g_{t \phi} & 0 & 0 \\ g_{\phi t} & g_{\phi \phi} & 0 & 0 \\ 0 & 0 & g_{\theta \theta} & 0 \\ 0 & 0 & 0 & g_{rr} \\ \end{pmatrix}$$
and the observation that if you can decompose a matrix into smaller sub-matrices
$$\textbf{A} = \left( \begin{array}{c|c} \textbf{P} & \textbf{0} \\ \hline \textbf{0} & \textbf{Q} \end{array} \right)$$
then it inverts as
$$\textbf{A}^{-1} = \left( \begin{array}{c|c} \textbf{P}^{-1} & \textbf{0} \\ \hline \textbf{0} & \textbf{Q}^{-1} \end{array} \right)$$
(then finally put the coordinates back into the original order).

5. Jan 25, 2014

### Nabigh R

Thanks a lot Greg. That's just what I was looking for. Just saw blockwise inversion theorem of matrices on Wikipedia. Since it didn't occur me to change the order of coordinates, I didn't make the connection. Thanks again