How are these two definitions of angular momentum related? I've seen angular momentum defined as: 
$$\ L=I \omega\ $$
In dynamics, the notation is different and states: 
$$\ L_o = r  × (mv) \ $$
How are these definitions related, if they are describing the same thing? 
 A: $\mathbf{L} = \mathbf{r} \times (m\mathbf{v}) $ is the general one.
For the special case of a rigid body (i.e. not point mass) rotating freely about an axis`with angular velocity $\omega$, then $ \mathbf{L} = I\mathbf{\omega} $ where $\mathbf{\omega} $ points in the axis of rotation. The moment of intertia $I$ is a scalar.
This follows from $v = \omega /r$ and from $\mathbf{r} \times \mathbf{v}$ pointing in the direction of $\mathbf{\omega}$. $I = mr^2$ for a point mass, and it can be discretised to then be integrated over for a rigid body. 
In the case of a forced rotation (under an external force), then $I$ becomes a tensor and $\mathbf{L}$ and $\omega$ are not parallel anymore.
A: 
diregarding the vectors we have:
$$ L=I  ω $$
$$ω = \frac{(r  v)}{|r^2|}$$
$$I=mr^2$$
So.
$$ Iω= rmv= H_o$$
A: A rigid body can only describe circular motion respect to an axis, because of its definition (rigid ↔ the distances between molecules are always the same; and the distance is invariant under rotations). It can also move linearly, but that's just the CoM.
Consequently, $v$ is that of the circular motion $v=\omega r$, and so
$L=mrv=mr^2\omega$
Then, you call $I\equiv mr^2$, so you have $L=I\omega$
You can also do it with vectors.
A: The angular momentum, with respect to some point $O$, of a particle is defined as
$$\vec L=\vec r\times\vec p,$$
where $\vec r$ is the particle position with respect to $O$.
For any system of particles, the total angular momentum is the sum of the individual ones. For a discrete system
$$\vec L=\sum_i\vec r_i\times\vec p_i,$$
whereas for a continuum distribution
$$\vec L=\int\vec r\times\vec vdm.\tag1$$
For a rigid body rotating with angular velocity $\omega$ about an axis, the velocity of any mass element can be written as $\vec v=\vec\omega\times\vec r$. Plugging this into Eq. (1) we obtain the relation
$$\vec L=\mathbb I\vec\omega,\tag2$$
where $\mathbb I$ is the inertia tensor, a $3\times 3$ matrix with components
$$\mathbb I_{ij}=\int(r^2\delta_{ij}-x_ix_j)dm.$$
As you can see from its construction, Eq. (2) is quite general. For some special cases however, see this answer, the last equation simplifies to 
$$\vec L=I\vec\omega,$$
where $I$ is a scalar. In that case, angular momentum and angular velocity point in the same direction.
