Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
3  
@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
1  
@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
add comment

1 Answer 1

up vote 8 down vote accepted

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.

share|improve this answer
    
This makes sense! Thanks! –  Justin Jul 24 '12 at 17:55
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.