# Understanding a local definition of the gradient

In this topic, the tangential gradient on a submanifold $$M \subset \mathbb{R}^d$$, embedded by a map $$F: M\rightarrow \mathbb{R}^d$$, is defined as $$\nabla^M f = \nabla^i f \frac{\partial F}{\partial x_i} = g^{ij} \frac{\partial f}{\partial x_j} \frac{\partial F}{\partial x_i},$$ where $$g_{ij}=\langle\frac{\partial F}{\partial x_i}, \frac{\partial F}{\partial x_j}\rangle$$.

I'm stack in understanding the embedding $$F$$ and how to calculate $$\frac{\partial F}{\partial x_i}$$ even in the basic example of a sphere ($$d=3$$). Any hint would really be helpful.

Note that for $$F\colon M \to\mathbb{R}^d$$, we have the differential map $$dF_p\colon T_pM\to T_{F(p)}\mathbb{R}^d$$ for all $$p\in M$$, i.e. $$dF_p$$ transports tangential vectors in $$T_pM$$ to tangential vectors in $$T_{F(p)}\mathbb{R}^d$$. The tangential vector they transport here via the differential map is the gradient of a function $$f$$ on $$M$$. So the given notation means $$(\nabla^M f)_p := dF_p((\nabla f)_p)$$ for all points $$p\in M$$, where they denote $$dF_p((\nabla f)_p)$$ with respect to local coordinates.
Looking at an example like the embedding of a 2-sphere into $$\mathbb{R}^3$$ is certainly a good way to get a better understanding of the definition. Using standard coordinates on $$\mathbb{S}^2$$, an embedding can (almost everywhere) be given by $$F\colon [0,\pi]\times[0,2\pi)\to\mathbb{R}^3,\quad (\theta,\phi)\mapsto\left(\begin{array}{c}\sin\theta\cos\phi \\ \sin\theta\sin\phi\\ \cos\theta\end{array}\right),$$ which yields $$dF = \left(\frac{\partial F}{\partial x_i}\right)_{x_1 = \theta, x_2 = \phi} = \left(\begin{array}{cc}\cos\theta\cos\phi & -\sin\theta\sin\phi \\ \cos\theta\sin\phi & \sin\theta\cos\phi\\ -\sin\theta & 0\end{array}\right).$$ Then, with respect to these coordinates, you can compute the gradient vector $$\nabla^{\mathbb{S}^2}f$$ in $$\mathbb{R}^3$$, which is tangential to the sphere embedded into $$\mathbb{R}^3$$, from the gradient vector $$\nabla f$$ on $$\mathbb{S}^2$$ by multiplication (for any differentiable function $$f$$): $$\nabla^{\mathbb{S}^2}f = DF\cdot\nabla f = \left(\begin{array}{cc}\cos\theta\cos\phi & -\sin\theta\sin\phi \\ \cos\theta\sin\phi & \sin\theta\cos\phi\\ -\sin\theta & 0\end{array}\right)\cdot \nabla f.$$ Simple example: $$f=\phi$$ gives you $$\nabla^i \phi = g^{ij}(df)_j$$, hence $$\nabla\phi = \left(\begin{array}{cc} 1 & 0 \\ 0 & \frac{1}{\sin^2\theta}\end{array}\right)\left(\begin{array}{c} 0 \\ 1 \end{array}\right) = \left(\begin{array}{c} 0 \\ \frac{1}{\sin^2\theta}\end{array}\right)$$ and thus $$\nabla^{\mathbb{S}^2}\phi = \left(\begin{array}{c} -\frac{\sin\phi}{\sin\theta} \\ \frac{\cos\phi}{\sin\theta}\\0\end{array}\right)$$ tangential to $$\mathbb{S}^2$$ in $$\mathbb{R}^3$$.
• Thanks. $F$ is defined on $M$ but in the example you defined it on $[0,\pi]\times [0,2\pi)$. Also $\nabla f$ usually denotes the gradient on $\mathbb R^d$ not $S^2$. Is there another notion of gradient that you mean (say $\nabla_{S^2}$) other than tangential gradient you define above? Commented Mar 28, 2021 at 19:53
• 1) $[0,\pi]\times[0,2\pi)$ technically denotes (part of) a chart on the 2-sphere. For some concrete computation you can only do it wrt a chart. 2) In the case of the example $M=\mathbb{S}^2$. 3) I would not say that $\nabla f$ usually denotes the gradient on the ambient space. That depends on how the author defines the notation. 4) There are different notions for the gradient: One intrinsic to $M$ (elements of $TM$) and one vectors from the ambient space tangential to the embedding. Commented Mar 29, 2021 at 17:14
• I know that $M=S^2$, so I'm still not understanding why you replace $M$ by a chart? Is this what is meant by $F: M\to R^d$? it then should be $F: U\to R^d$ for a chart $U$. My question is motivated by understanding the tangential gradient, but the problems is there are many inconsistent definitions, see e.g math.stackexchange.com/q/4024360/787068 Commented Mar 29, 2021 at 22:31
• Yes, certainly the chart only covers a subset $U$ of the manifold in general. But whenever you would like to do some concrete computations and gain some understand of the geometric situation, using a chart is quit helpful. The definition using the orthogonal projection mentioned here math.stackexchange.com/questions/4024360/… is tricky as it requires you to extend the definition of $f$ to a neighborhood of $M$. But if you do that, it will yield the same result. Commented Mar 31, 2021 at 6:54