Energy functional of smooth map $S^2\to M$ Let $f:S^2\to M$ be a smooth map where $(M,g)$ is a Riemannian manifold. Typically the energy functional of $f$ is written as an integral over $S^2$, so it exists by compactness. However, in Ricci Flow and the Sphere Theorem by Brendle, he identifies $S^2$ with $\Bbb R^2\cup\{\infty\}$ via stereographic projection, then defines 
$$\mathscr{E}(f):=\frac{1}{2}\int \left(\left|\frac{\partial f}{\partial x}\right|^2+\left|\frac{\partial f}{\partial y}\right|^2\right)\,\mathrm dx\,\mathrm dy,$$
where $(x,y)$ are the standard coordinates on the plane. In general, this functional seems to diverge, even for quite reasonable functions $S^2\to M$. Does the stereographic projection force the integral to be finite somehow? If it doesn't, do we have to restrict attention to a restricted class of functions to vary this functional and find its critical points? 
 A: This energy is in fact exactly equal to the standard Dirichlet energy $\frac12 \int_{S^2} |Df|^2 dA$. This is a special case of the fact that the Dirichlet energy of maps from surfaces is invariant under conformal maps; i.e. $E(f \circ \phi) = E(f)$ for all conformal maps $\phi$. To prove this, remember that in the stereographic coordinates we have $$dA = \psi\ dx\ dy$$ and $$g = \psi\ (dx^2 + dy^2)$$ where $\psi=4(1+x^2+y^2)^{-2}$ is the conformal factor. Thus we get $$|Df|^2 dA = g^{ij} \partial_i f \cdot \partial_j f\ dA= \psi^{-1}(|\partial_x f|^2+|\partial_y f|^2)\psi\ dx\ dy,$$ and the $\psi$ factors cancel.
Intuitively, the fact that $f$ is continuous at the pole of projection means that the corresponding map from the plane converges to a constant at infinity; so "reasonable functions" $S^2 \to M$ are more restricted than "reasonable functions" $\mathbb R^2 \to M$. In particular they have derivatives that decay to zero at infinity sufficiently fast to guarantee the integral converges.
A: The two dimensional energy functional is conformally invariant, and the stereographic projection is a conformal map from the sphere minus a point to the Euclidean plane with it's standard coordinates. So the integral is just the same as the usual Energy on the sphere (or the Energy of $f$ taken with respect to the pull back metric of inverse stereographic projection).
To see conformal invariance just look at the general form of the energy functional with respect to a Riemannian metric $g_{ij}$:
$$2E(f)=\int \sum_{i,j}  g^{ij}f_{x^i} f_{x^j} \sqrt{g}dxî dx^j$$
where $g$ denotes the determinant of $g_{ij}$. If you now replace $g_{ij}$ by $\lambda g_{ij}$ the inverse becomes $\frac{1}{\lambda} g^{ij} $ while you get a $\lambda^2$ in the determinant.
