# Harmonic/conformal map composition between manifolds in either order?

Suppose $\mathcal{M}$, $\mathcal{N}$, and $\mathcal{P}$ are Riemannian manifolds (compact and of dimension 2, if it matters). It seems well-known that if $\phi:\mathcal{M}\rightarrow\mathcal{N}$ is (weakly) conformal and $f:\mathcal{N}\rightarrow\mathcal{P}$ is harmonic, then $f\circ\phi:\mathcal{M}\rightarrow\mathcal{P}$ is harmonic. (See this book for a reference)

But, I can't find any references that reverse this relationship. Namely, if $f:\mathcal{M}\rightarrow\mathcal{N}$ is harmonic and $\phi:\mathcal{N}\rightarrow\mathcal{P}$ is (weakly) conformal, then is $\phi\circ f:\mathcal{M}\rightarrow\mathcal{P}$ harmonic?

-
@Justin: Maybe you should be asking a slightly different question: What property of $\phi:\mathcal{N}\to\mathcal{P}$ has the property that $\phi\circ f$ is harmonic whenever $f:\mathcal{M}\to\mathcal{N}$ is harmonic? I seem to remember that the condition is that $\phi$ has to preserve constant-speed geodesics, i.e., whenever $\gamma:(a,b)\to\mathcal{N}$ is a constant-speed geodesic, the composition $\phi\circ\gamma:(a,b)\to\mathcal{P}$ must be a constant speed geodesic. This is certainly necessary, but I seem to remember that it is sufficient, too. I just don't have time to check it now. –  Robert Bryant Jul 24 '12 at 20:17
@Justin: There's an asymmetry here that you should think about. To determine the harmonicity of a map $f:\mathcal{M}\to\mathcal{N}$ you need a metric on $\mathcal{M}$ (although, when $\mathrm{dim}\mathcal{M}=2$, you only really need the conformal structure) and you need an affine structure (i.e., a connection) on $\mathcal{N}$. That's why the two compositions behave so differently. For $\phi:\mathcal{N}\to\mathcal{P}$ to preserve harmonicity in compositions $\phi\circ f$, you need $\phi$ to 'respect' affine structures on $\mathcal{N}$ and $\mathcal{P}$, not necessarily metric structures. –  Robert Bryant Jul 25 '12 at 0:10
Ugh, MathOverflow didn't email me that you responded to my questions! Thanks for the additional clarifications -- I'm still thinking about related questions, and these are good pointers into the literature. –  Justin Oct 4 '12 at 4:07

Because it is not true. For example, suppose that $f: D^n \to D^n$ (the unit disk in $R^n$ with Euclidean metric) is harmonic -- i.e. the components are bounded harmonic functions in the ordinary sense satisfying $f_1^2 + \ldots + f_n^2 < 1$. Let $\phi: D^n \to D^n$ be the identity map, where the domain has the Euclidean and the range has the hyperbolic metric. Then $\phi \circ f$ is almost never harmonic.