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 $$\lambda$$.

• 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. – Giuseppe Negro Apr 9 at 22:20
• @GiuseppeNegro Thank you for the support! I've been struggling with this all the day and I kind of feel stupid.. – Math_tourist Apr 9 at 22:28
• @Math_tourist After $\nabla^\Sigma f$ there comes the Laplacians and all sort of tensors....... – Arctic Char Apr 9 at 22:53

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 $$\lambda$$.
• 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 ${\nabla^{\tilde{\Sigma}}} f$. – Math_tourist Apr 10 at 13:34
• 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. – Travis Apr 10 at 17:41