Is angular momentum conservation Galilean invariant? Suppose I have a system of particles with constant total angular momentum $\mathbf{L} = \sum_a m_a \mathbf{r_a \times v_a}$ in frame K. If frame K' moves with velocity $V$ with respect to K and their origins coincide at t=0, then $\mathbf{r_a} = \mathbf{r'_a} + \mathbf{V}t$ and $\mathbf{v_a} = \mathbf{v'_a} + \mathbf{V}$. Substituting these into the equation for $\mathbf{L}$ I get
$$\mathbf{L} = \mathbf{L'} + \sum_a m_a (\mathbf{r'_a} - \mathbf{v'_a}t) \times \mathbf{V}$$
Now in general, the second term in RHS should vary with time so that even if $\mathbf{L}$ does not vary with time, $\mathbf{L'}$ must. But this doesn't make sense to me because the rotational invariance of a system doesn't seem to change when I change to a new inertial reference frame. Did I make a mistake in my analysis, and if so, what is it?
 A: The answer is no.
To begin with, a Galilean transformation in three dimensional Euclidean space(time) consists of

*

*space-time translation: $(t,\vec{x})\rightarrow(t+s,\vec{x}+\vec{a})$

*spatial rotation: $(t,\vec{x})\rightarrow(t,R\vec{x})$, where $R\in O(3)$

*Galilean boost: $(t,\vec{x})\rightarrow(t,\vec{x}+\vec{v}t)$
One can easily check that generic Galilean transformations take the form $$\mathrm{Gal}(3)\cdot(t,\vec{x})=(t+s,R\vec{x}+\vec{v}t+\vec{a}),$$
and they form a group, which is also a Lie group. One can explicitly work out its matrix representation $\rho:\mathrm{Gal}(3)\rightarrow\mathrm{GL}(5,\mathbb{R})$, which is given by $$(s,\vec{a},\vec{v},R)\rightarrow\begin{pmatrix}
R & \vec{v} & \vec{a} \\
0 & 1 & s \\
0 & 0 & 1
\end{pmatrix}$$
To study the effects of any infinitesimal Galilean transformations, one has to compute the commutation relations of its Lie algebra. In matrix representation, a generic infinitesimal Galilean transformation takes the form $$\mathfrak{gal}(3)=\begin{pmatrix}
X & \vec{u} & \vec{b} \\
0 & 0 & \epsilon \\
0 & 0 & 0
\end{pmatrix},$$
where $X\in\mathfrak{o}(3)$, $\epsilon\in\mathbb{R}$, and $\vec{u}$, $\vec{b}\in\mathbb{R}^{3}$. It has 10 generators, in which $H$ generates the time translation, $\vec{P}$ generates spatial translation, $\vec{K}$ generates Galilean boosts, and $\vec{J}$ generates rotation. Their commutation relations are listed below:
$$[H,J^{i}]=0\quad [H,P^{i}]=0\quad [H,K^{i}]=-P^{i}$$
$$[J^{i},J^{j}]=\epsilon^{ij}_{\,\,\,\,k}J^{k}\quad [J_{i},K_{j}]=\epsilon^{ij}_{\,\,\,\,k}K^{k}\quad[J^{i},P^{j}]=\epsilon^{ij}_{\,\,\,\,k}P^{k}$$
$$[K^{i},P^{j}]=0\quad[K^{i},K^{j}]=0$$
$$[P^{i},P^{j}]=0$$
from which one finds that the angular momenta $\vec{J}$ and Galilean boosts $\vec{K}$ do not commute. In other words, performing a Galilean boost first, followed by a rotation is not the same as performing a rotation first and then a Galilean boost. One concludes that the angular momentum is not conserved by a Galilean boost.
Specifically, in Newtonian mechanics, one considers a Galilean invariant free particle whose action is given by $$S=\int dt L(\vec{x},\dot{\vec{x}},t)=\frac{m}{2}\int dt|\dot{\vec{x}}(t)|^{2}.$$
Here, the Lagrangian is strickly invariant under the action of the Euclidean group, i.e rotation and space-time translation, but is invariant up to a total time derivative under the Galilean boost. This has to do with the fact that the Galilean group has non-trivial second cohomology group, which forbids its projective representation being lifted to a linear representation. This was explained in details here.
But since the Lagrangian only changes by a total time-derivative under a Galilean boost, one can still apply the Noether's theorem for the Galilean symmetry to the above action, and obtain the following conserved charges: $$\vec{p}=\frac{\partial L}{\partial\dot{\vec{x}}}=m\dot{\vec{x}}$$
$$E=\vec{p}\cdot\dot{\vec{x}}-L=\frac{m}{2}|\dot{\vec{x}}|^{2}$$
$$\vec{l}=\vec{p}\times\vec{x}=m\dot{\vec{x}}\times\vec{x}$$
$$\vec{g}=m\vec{x}-\vec{p}t=m(\vec{x}-\dot{\vec{x}}t)$$
In the above charges, the energy $E$ is generated by $H\in\mathfrak{gal}(3)$, the momentum $\vec{p}$ is generated by $\vec{P}\in\mathfrak{gal}(3)$, the angular momentum $\vec{l}$ is generated by $\vec{J}\in\mathfrak{gal}(3)$, and the Galilean momentum $\vec{g}$ is generated by $\vec{K}\in\mathfrak{gal}(3)$.
In the Hamiltonian formalism, one is interested in the Poisson bracket of the above conserved charges. Naively, one expects that the Poisson bracket would preserve the commutation relations of $\mathfrak{gal}(3)$. However, this is not the case. As mentioned previously, the Galilean group has a non-trivial second cohomology group, which does not allow its projective representation to be lifted as a linear representation.
Now with the Poisson bracket $$\left\{x^{i},p_{j}\right\}=\delta^{i}_{j}$$
in mind, a careful calculation shows that there appears a classical anommaly $$\left\{g^{i},p_{j}\right\}=m\delta^{i}_{j}.$$
In contrast, the original Lie bracket was $[K^{i},P^{j}]=0$. We recognize that the Poisson bracket realization of the conserved charges associated with Galilean transformations of a free particle is in fact its central extension parameterized by the mass.
The essence of the above discussion is that, even though the Galilean group acts on the Lagrangian of a free particle as a symmetry. The Poisson brackets of the associated Noether charges span the central extension of the Galilei group.
A: 
But this doesn't make sense to me because the rotational invariance of a system doesn't seem to change when I change to a new inertial reference frame.

It can change. Torque depends on origin. As an example...

If C is origin, angular momentum will remain zero. If O is origin, the system will gain clockwise angular momentum
