# Euler's equation for a rotating frame when the inertia tensor is non-diagonal

Wikipedia's entry for Euler's equation states:

In classical mechanics, Euler's rotation equations are a vectorial quasilinear first-order ordinary differential equation describing the rotation of a rigid body, using a rotating reference frame with its axes fixed to the body and parallel to the body's principal axes of inertia. Their general form is:

$$I\dot{\omega}+\omega\times (I \omega) = M.$$

Is the bolded part really necessary? That would mean that the equation only works when $$I$$ is diagonal, and that doesn't seem right to me.

Also, just to be certain, both $$\omega$$ and $$M$$ in the equation are expressed in the fixed body frame, right?

$$\mathrm{I}\dot{\boldsymbol{\omega}} + \boldsymbol{\omega}\times\left(\mathrm{I} \boldsymbol \omega\right) = \boldsymbol M$$
Also, just to be certain, both $\boldsymbol \omega$ and $\boldsymbol M$ in the equation are expressed in the fixed body frame, right?
Correct. Being pedantically correct, $\boldsymbol \omega$ is the angular velocity of the object in question with respect to an inertial frame but expressed in body frame coordinates, and $\boldsymbol M$ is the external non-fictitious torque on the body, once again expressed in body frame coordinates. One way of looking at the $\boldsymbol{\omega}\times\left(\mathrm{I} \boldsymbol \omega\right)$ term is that it is a fictitious torque, the torque analogy of the fictitious centrifugal, Coriolis, and Euler forces.