Is the definition of angular momentum $\vec{L} = \vec{r} \times \vec{p}$ rigorous?

Question

Let $\vec{r}$ be the position vector in $3\mathrm{D}$ Euclidean space $\mathbb{R}^3$, and $\vec{p}$ the linear momentum of point mass at $P\in\mathbb{R}^3$. The angular momentum $\vec{L}$ of that point mass with respect to the origin is defined to be $$\vec{L} = \vec{r}\times\vec{p} = m\vec{r}\times\vec{v}.$$

This feels weird to me, as $\vec{r}$ is a vector in $\mathbb{R}^3$ and $\vec{v}$ is a vector in $T_P\mathbb{R}^3$... aren't they? If so, we shouldn't be able to take their cross product, so how can this definition be fixed for the operations to make sense?

Answer 1

Building on @Vercassivelaunos 's comment, I think you need to rephrase your question in terms of more general manifolds. You will then naturally build the notion of moment map.

Good luck on this long journey.

Answer 2

$\vec{r}$ is a vector field. It has a value in the tangent space of $P$ given by,

$$ \vec{r} = x(P) \partial_x + y(P) \partial_y + z(P) \partial_z $$