# Is Legendre transform related to finding the inverse metrics?

In classical mechanics kinetic part of Hamiltonian is the Legendre transform of kinetic part of Lagrangian.

On the other hand kinetic part of Lagrangian is a metric on the configuration space. At first I though kinetic energy in Hamiltonian setting would be an inverse metric tensor.

However it's not true. In 1D:

$$T_L(\dot x, \dot x) = \frac{1}{2} m \dot x^2$$

$$T_H(p_x, p_x) = \frac{1}{2m} p_x^2$$

where as the inverse of $T_L$ would be

$$T_L^{-1}(p_x, p_x) = \frac{2}{m} p_x^2$$

Is in this case inverse metric tensor related to Legendre transform at all? Can Legendre transform be expressed in coordinate independent way in the case above?

-

For example, the gradient of $\sum_{i,j} a_{ij} x_i x_j$ (defined on Euclidean space) can be thought of as a map $(x_i)\mapsto (2\sum_j a_{ij} x_j)$. This is a linear map with the matrix $2A$. Its inverse has the matrix $A^{-1}/2$. The corresponding quadratic form is $1/4$ of what you would get if you simply inverted pointwise without doing the necessary calculus.
Executive summary: you can get the Legendre transform by inverting the matrix, if you keep the factor of $1/2$ separately - from $m(\dot x^2/2)$ you get $m^{-1}(p_x^2/2)$, etc.