3
$\begingroup$

Let $\Sigma^n \subseteq \mathbb{R}^{n+1}$ be a smooth hypersurface. Let $\lambda > 0$ be a constant and let define $\tilde{\Sigma} := \lambda \Sigma$.

Let $f$ be a smooth function on $\Sigma$. This defines a function $\tilde{f}$ on $\tilde{\Sigma}$ as follows: for every $p \in \tilde{\Sigma}$, $$ \tilde{f}(p) := f(\frac{p}{\lambda}). $$ Let $V$ be a constant vector field in $\mathbb{R}^{n+1}$. I would like to express $$ \langle \nabla^{\Sigma} f, V \rangle_{\mathbb{R}^{n+1}} $$ in terms of $\nabla^{\tilde{\Sigma}} f$.

On a point $p \in \tilde{\Sigma}$ it should hold \begin{equation} \nabla^{\tilde{\Sigma}}\tilde{f}(p) = \frac{1}{\lambda^2} \nabla^{\Sigma}f(\frac{p}{\lambda}). \tag{1} \end{equation} I obtain this formula from expressing the gradient in local coordinates and from the observation that the pull-back metric on $\Sigma$ induced by the metric on $\tilde{\Sigma}$ is basically $\tilde{g}_{ij} = \lambda^2 g_{ij}$, where $g_{ij}$ is the metric on $\Sigma$ induced by the ambient Euclidean metric.

But somehow I feel that there is something fishy. I would expect the gradient to scale as $\frac{1}{\lambda}$ and I would expect a formula of the kind:

$$ \langle \nabla^{\Sigma} f(\frac{p}{\lambda}), V \rangle_{\mathbb{R}^{n+1}} = \langle \nabla^{\tilde{\Sigma}} \tilde{f}(p), \lambda V \rangle_{\mathbb{R}^{n+1}}. \tag{2} $$

Can anyone help me? I'm getting very confused...

EDIT: Consider the case where $f$ is the restriction on $\Sigma$ of a function $F : \mathbb{R}^{n+1} \to \mathbb{R}$. We can always assume that, at least locally. Then it is known that $\nabla^{\Sigma} f = \left( \nabla^{\mathbb{R}^{n+1}} F|_{\Sigma}\right)$. Therefore, given $p \in \tilde{\Sigma}$, and identifying the tangent spaces $T_p \tilde{\Sigma}$ and $T_{\frac{p}{\lambda}} \Sigma$ we have: $$ \nabla^{\tilde{\Sigma}} \tilde{f}(p) = \left( \nabla^{\mathbb{R}^{n+1}} F(\frac{y}{\lambda}) \right)^{\top} = \frac{1}{\lambda} \left(\nabla^{\mathbb{R}^{n+1}} F \right)^{\top}(\frac{p}{\lambda}) = \frac{1}{\lambda} \nabla^{\Sigma}f(\frac{y}{\lambda}). \tag{3} $$ I now believe that $(3)$ is correct and from this $(2)$ follows. Therefore $(1)$ should be wrong. I think that I was mislead by the intrinsic approach showed in the answer by Trevis. I think I got confused from the fact that in the intrinsic approach one thinks about $\Sigma$ and $\tilde{\Sigma}$ as the same manifold but with different Riemannian metrics. The equation of Trevis is of course correct, but then my equation $(1)$ is wrong probably because one should be careful in translating the instrinsic equation back into the extrinsic situation. In fact the same abstract coordinates genereates two different local frames on $\Sigma$ and $\tilde{\Sigma}$ (as hypersurfaces) and one frame is the other one rescaled by a factor of $\lambda$.

$\endgroup$
  • 2
    $\begingroup$ I don't know how to answer. But, as a moral support: these things are confusing. It is just a matter of getting used to them. I also struggled with similar things for quite some time. $\endgroup$ – Giuseppe Negro Apr 9 at 22:20
  • 1
    $\begingroup$ @GiuseppeNegro Thank you for the support! I've been struggling with this all the day and I kind of feel stupid.. $\endgroup$ – Math_tourist Apr 9 at 22:28
  • $\begingroup$ @Math_tourist After $\nabla^\Sigma f$ there comes the Laplacians and all sort of tensors....... $\endgroup$ – Arctic Char Apr 9 at 22:53
4
$\begingroup$

The gradient is intrinsic to (depends only on) the metric structure of the hypersurfaces, so considering $\Bbb R^n$ is perhaps a distraction---and probably a source of confusion.

It's simpler just to consider a fixed abstract Riemannian manifold $(M, g)$ and a metric $\tilde g := \lambda^2 g$, $\lambda > 0$, homothetic to $g$. In particular, this has the advantage that you can think just about smooth functions on $M$, rather than worrying about and comparing gradients of corresponding functions on different surfaces. Now, the respective gradients $\operatorname{grad} f, \widetilde{\operatorname{grad}} f$ of $f$ w.r.t. $g, \tilde g$ are related by $$\boxed{\widetilde{\operatorname{grad}} f = \tilde g^{-1}(df,\,\cdot\,) = (\lambda^2 g)^{-1} (df,\,\cdot\,) = \lambda^{-2} g^{-1} (df,\,\cdot\,) = \lambda^{-2} \operatorname{grad} f} .$$ NB this computation doesn't depend on $\lambda$ being constant, so in fact this formula applies to a conformal rescaling of a metric by a general positive function $\lambda$.

$\endgroup$
  • $\begingroup$ Thank you for your answer. Yes, I did exactly your same "intrinsic" computation in order to get the formula ${\nabla^{\tilde{\Sigma}}} f (p) = \frac{1}{\lambda^2} \nabla^{\Sigma}f.$ The problem is that I am interesting in the "extrinsic" quantity $\langle \nabla^{\Sigma} f, V \rangle$ which I would like to express in terms of ${\nabla^{\tilde{\Sigma}}} f $. $\endgroup$ – Math_tourist Apr 10 at 13:34
  • 1
    $\begingroup$ I'm not sure I understand your question exactly then. But note that writing the scaling map $\phi : \tilde\Sigma \to \Sigma$, $\phi : p \mapsto \lambda^{-1} p$, using the definition of the gradient, the naturality of the exterior derivative (i.e., $d \circ \phi^* = \phi^* \circ d$), and the fact that $T_p \phi \cdot V_p = \lambda^{-1} V_{\phi(p)}$ gives $$\langle (\widetilde{\operatorname{grad}} \tilde f)_p, V_p \rangle_p = \lambda^{-1} \langle (\operatorname{grad} f)_{\phi(p)}, V_{\phi(p)} \rangle_{\phi(p)} ,$$ which seems to agree with (3) in your edit. $\endgroup$ – Travis Apr 10 at 17:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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