MathOverflow is a question and answer site for professional mathematicians. 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

I am probably making some very annoying mistake but for whatever it is worth let me ask it here.

Consider a hypersurface $S$ in $\mathbb{R}^{n+1}$. Let $\nu$ be the unit outward normal. Let $f:\mathbb{R}^{n+1}\rightarrow\mathbb{R}$ be defined as $f(p)=r^2$, where $r$ is the radial distance of $p$ to the origin. Let $z=f|_S$, $\sigma$ be the second fundamental tensor, and for an orthonormal local framefield $\{E_i\}$ let $h=\frac{1}{n}\sum\sigma(E_i,E_i)$. Then we have the following -

$\Delta z = 2n(1+\langle h,\vec{r}\rangle)$, This comes from the (possibly preprint) paper of Oscar Garay titled "application of reilly's formula"

Where $\Delta$ is defined as $\Delta g=-\textrm{Trace}(X\mapsto\nabla_X\textrm{grad}g)$

My problem is I am getting the wrong sign always ... I get $\Delta z = 2n(-1+\langle h,\vec{r}\rangle)$

share|cite|improve this question
What is the convention for the sign of the second fundamental form? Once you know that you can just do a simple sanity check by looking at the unit sphere $\mathbb{S}^n$. If the sign convention is that for a sphere the mean curvature vector is outward pointing, then you are right (up to an overall minus sign outside the parentheses). (Since $z \equiv 1$ on $S$...) – Willie Wong Mar 1 '13 at 1:38
The convention is $\bar{\nabla} = \nabla + \sigma$ where $\bar{\nabla} $ is the connection on $\mathbb{R}^{n+1}$, $\nabla$ is the connection on $S$ and $\sigma$ is the second fundamental tensor. The laplacian sign convention I have already described. I dont think it is a problem of sign convention. Any other opinions ? – magguu Mar 1 '13 at 6:05

Your Answer


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

Browse other questions tagged or ask your own question.