MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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

Let $\Gamma$ be a $C^k$ $(n-1)$-dimensional hypersurface embedded in $\mathbb{R}^n$. Let $H=L^2(\Gamma)$ and $V=H^1(\Gamma)$.

Suppose that $\{v_j\}$ is a basis for $H$ and $V$ (not necessarily orthogonal).

Let $V_m = \text{span}(v_1, ..., v_m)$.

Define a projection operator $P_m:H \to V_m$ satisfying $$(P_m h - h, v_m) = 0 \qquad\text{for all $v_m \in V_m$}.$$

Is it true that $$P_m (\nabla_{\Gamma}h) = \nabla_{\Gamma}(P_mh)$$ (on the LHS, the projection operator is applied to each element of the gradient vector) and $$P_m (\Delta_{\Gamma}h) = \Delta_{\Gamma}(P_mh)?$$ Where $\nabla_{\Gamma}$ is the surface gradient (defined by $\nabla_\Gamma f= \nabla f - \nabla f \cdot \nu \nu$ with $\nu$ the unit normal) and $\Delta_\Gamma$ is the Laplace-Beltrami operator. I.e. does the projection operator and the surface gradient operator commute??

share|cite|improve this question
The surface gradient is a vector field on $\Gamma$, so how do you make sense of it being in $V_m$, a space of functions? Do you use the embedding in some higher dimensional manifold? I guess you want $\Gamma$ to be a hypersurface in $\mathbb{R}^n$, but maybe you could make that clearer. – Ben McKay Feb 14 '14 at 13:25
Thanks for the comments. I edited my post. Yes $\Gamma$ is $(n-1)$-dimensional hypersurface in $\mathbb{R}^n$. So I think everything makes sense. – weasd Feb 14 '14 at 13:41

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.