Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In Hamilton's 1997 paper on four-manifolds with positive isotropic curvature, he considers a local diffeomorphism of Riemannian $n$-manifolds $$ P: (N,\bar g) \to (M, g). $$ Such a map is harmonic if

$$ \operatorname{tr}_{\bar g} \nabla d P^\alpha = \bar g^{jk} \left( \frac{\partial P^\alpha}{\partial x^j \partial x^k} + \Gamma_{\mu\nu}^\alpha \frac{\partial P^\mu}{\partial x^j} \frac{\partial P^\nu}{\partial x^k} - \bar \Gamma^l_{jk} \frac{\partial P^\alpha}{\partial x^l} \right) = 0 $$

where $\Gamma$ is the Levi-Civita connection of the pullback metric $ P^* g $. He then claims that when $(N, \bar g)$ is the round sphere $S^n$, this is equivalent to $$ \bar g^{jk} ( \bar\Gamma_{jk}^i - \Gamma_{jk}^i ) =0.$$

There must be some property of the connection on the round sphere that makes this work, but I'm not seeing it - any pointers?

Also, in the first formula for the Laplacian I gave above, $\Gamma$ is usually interpreted as the pullback connection on $P^* TM$ (which I have given Greek indices) - this works in more general cases than when $P$ is a local diffeomorphism. Am I correct that in this case both interpretations give the same result?

share|cite|improve this question
up vote 2 down vote accepted

In the second equation you quoted, Hamilton is in effect redefining $g$ to be the pullback metric $P^*g$ on $S^n$, and $P$ to be the identity map of $S^n$. So in this case the Jacobian of $P$ is the identity matrix, and its second partial derivatives all vanish. It has nothing to do with whether the manifold is the round sphere or not; it's just what the harmonic map equation reduces to when you apply it to the identity map between two metrics on the same manifold.

share|cite|improve this answer
Thankyou very much, it seems obvious in retrospect. I guess the wording "or equivalently on $S^n$" led me too far down that path without stopping to think it through properly... – Anthony Carapetis Aug 22 '13 at 23:32

Your Answer


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.