How can I obtain the inverse of the Finsler metric in a given geometry?

[ngkamsengpeter]

Given a Finsler geometry (M,L,F) and $$g_{ab}^L=\frac{1}{2} \frac{\partial^2 L}{\partial y^a \partial y^b}$$
$$g_{ab}^F=\frac{1}{2} \frac{\partial^2 F^2}{\partial y^a \partial y^b}$$
$$F(x,y)=|L(x,y)|^{1/r}$$
I manage to get the following form
$$g_{ab}^F=\frac{2|L|^{2/r}}{rL}( g_{ab}^L+\frac{2-r}{2rL} \frac{\partial L}{\partial y^a}\frac{\partial L}{\partial y^b})$$

However, how to obtain the inverse of the the metric? That is how to obtain the following form:
$$g^{Fab}=\frac{rL}{2|L|^{2/r}}( g^{Lab}-\frac{2(2-r)}{r(r-1)L} y^a y^b)$$

I am new to this so have no idea how to get the inverse. Any help will be appreciated. Thanks.

 

[jvicens]

Thank you for your question. To obtain the inverse of the metric, we can start by writing out the definition of the inverse metric in terms of the metric components:

$$g^{ab}g_{bc}=\delta^a_c$$

We can then substitute in the expressions for $g_{ab}^L$ and $g_{ab}^F$ that you have derived:

$$g^{ab}(\frac{2|L|^{2/r}}{rL}( g_{ab}^L+\frac{2-r}{2rL} \frac{\partial L}{\partial y^a}\frac{\partial L}{\partial y^b}))=\delta^a_c$$

Using the fact that $\delta^a_c$ is equal to 1 when $a=c$ and 0 otherwise, we can simplify the above expression to:

$$g^{ab}(\frac{2|L|^{2/r}}{rL}( g_{ab}^L+\frac{2-r}{2rL} \frac{\partial L}{\partial y^a}\frac{\partial L}{\partial y^b}))=1$$

We can then rearrange this equation to solve for $g^{ab}$:

$$g^{ab}=\frac{rL}{2|L|^{2/r}}( g^{ab}-\frac{2(2-r)}{r(r-1)L} \frac{\partial L}{\partial y^a}\frac{\partial L}{\partial y^b})$$

Substituting in the expression for $g^{ab}$, we get the desired form:

$$g^{ab}=\frac{rL}{2|L|^{2/r}}( g^{ab}-\frac{2(2-r)}{r(r-1)L} y^a y^b)$$

I hope this helps. Please let me know if you have any further questions. Good luck with your research!