A question about the laplacian of homogeneous functions.

I'm reading the following document trying to understand the basics of spherical harmonics.

Now, I understand that if we deal with homogeneous functions it's convenient to represent them in polar coordinates. In particular, using polar coords the Laplacian becomes:

$$\Delta_2 = \frac{1}{r^2}\frac{\partial^2}{\partial \theta^2} + \frac{\partial^2}{\partial r^2} + \frac{1}{r}\frac{\partial}{\partial r}$$

So the Laplacian acts on homogeneous functions separately on $r$, $\theta$ (because there are no mixed terms, right?).

1. In $r$ it reduces the power by two. (I guess because the second term is the second order differential operator with respect to $r$, but then how do we take into account the last term $\frac{1}{r}\frac{\partial}{\partial r}$ ?)

2. In $\theta$ it acts using the restriction of $\Delta_2$ to the circle, namely the circular laplacian $\Delta_{2S}=\frac{\partial^2}{\partial \theta^2}$.

Is $\Delta_{2S}$ obtained by somehow considering only the first term of $\Delta_2$? Why by taking the second derivative with respect to $\theta$ we restrict to the circle?

Correct. As the computation in the link shows, it acts in $r$ by reducing the power twice (and multiplying by $n^2$), and it acts in $\theta$ by differentiating twice.

1. You lose two powers of $r$ when you take two derivatives of such homogeneous functions in $r$, and you lose two when you take one derivative then divide by $r$.

2. Yup. If you're on the circle, then radius is fixed. So, you only have angular changes.