# Levi-Civita connection for euclidian spaces and surfaces

One approach to construct the Levi-Civita connection for a surface $$S \subseteq \mathbb{R}^3$$ is to construct the connection first for $$\mathbb{R}^3$$ and then apply a suitable orthogonal projection to get the connection for $$S$$. This approach is for example discussed in the textbooks "Notes on Differential Geometry" by Hicks and "Manifolds and Differential Geometry" by Lee. One step in this construction consists of applying the Levi-Civita connection for $$\mathbb{R}^3$$ to vector fields of the surface $$S$$. My question is why this is possible.

I'll try to explain this in some detail:

Let $$X,Y$$ be vector fields on $$\mathbb{R}^3$$, i.e. smooth mappings $$X,Y:\mathbb{R}^3 \to \mathbb{R}^3$$. Write $$Y=(Y_1,Y_2,Y_3)$$ where $$Y_i:\mathbb{R}^3 \to \mathbb{R}$$. Define the vector field $$D_X Y$$ by $$(D_X Y)(p)=\bigl((D_{X_p} Y_1)(p),(D_{X_p} Y_2)(p),(D_{X_p} Y_3)(p)\bigr),\phantom{aaa} p \in \mathbb{R}^3,$$ where $$(D_{X_p} Y_i)(p)$$ is the derivation of $$Y_i$$ in direction of the vector $$X_p$$ at point $$p$$, i.e. $$(D_{X_p} Y_i)(p)=X_p \cdot \text{grad }Y_i(p)$$ The Levi-Civita connection for $$\mathbb{R}^3$$ is then denoted by $$D$$ and defined via $$D: \ (X,Y) \mapsto D_X Y,\phantom{aaa}X,Y\text{ vector fiels on }\mathbb{R}^3.$$

My problem arises when this is used to define the Levi-Civita connection for a surface $$S \subseteq \mathbb{R}^3$$. Denote this connection by $$\nabla$$. For two vector fields $$X,Y$$ on $$S$$, the basic idea seems to be to define the vector field $$\nabla_X Y$$ by $$(\nabla_X Y)(p)=\text{ orthogonal projection of }(D_X Y)(p)\text{ to }T_p S,\phantom{aaa}p \in S$$ Question 1: In the above definition of $$\nabla_X Y$$, how can we apply $$D_X$$ to $$Y$$? My problem is that $$Y$$ is defined on $$S$$ and since $$S$$ is in general not open in $$\mathbb{R}^3$$, we can not differentiate $$Y$$ in order to get the directional derivatives. To put it otherwise: $$D$$ is only defined for vector fields on $$\mathbb{R}^3$$ but not for vector fields on $$S$$.

The authors of the textbooks I mentioned seem to suggest that this is somehow possible because $$D_X Y$$ can also be expressed via curves in the following way: If $$X,Y$$ are vector fields on $$\mathbb{R}^3$$ and if $$\gamma:(-\epsilon,\epsilon) \to \mathbb{R }^3$$ is a smooth curve with $$\gamma(0)=p$$ and $$\gamma'(0)=X_p$$, then $$(D_X Y)(p)=(Y\circ \gamma)'(0)$$ I understand this part but I don't know how this could help in applying $$D$$ to vector fields of $$S$$. My only idea so far would be that we use the above equation as a definition for $$D$$ applied to vector fields on $$S$$ but then I am not able to show that this is independent of the curve $$\gamma$$. I can also make a second question out this idea:

Question 2: Let $$S \subseteq \mathbb{R}^3$$ be surface and $$Y$$ a vector field on $$S$$. Let $$\gamma$$ and $$\tau$$ be curves in $$S$$ with $$\gamma(0)=\tau(0)$$ and $$\gamma'(0)=\tau'(0)$$. Is it true that $$(Y\circ \gamma)'(0)=(Y \circ \tau)'(0)?$$

• Question 2: yes! This is fundamental to many concepts of differential geometry. And it does allow you to define the connection on $S$ without having to extend a vector field on $S$ to an open neighborhood of it. Commented Jun 18, 2021 at 4:14
• @Deane, any hints of how to prove it? My problem is that I cannot apply the chain rule since $Y$ is not $C^1$ (it is not defined on an open set but only on $S$). Commented Jun 18, 2021 at 12:01
• You have to use the definition of a surface to do any of this. In the books you cite, $S$ is a surface if, for any $p \in S$, there exists an open set $O \subset \mathbb{R}^3$ containing $p$, an open $D \subset\mathbb{R}^2$, and a smooth (but $C^1$ is good enough) embedding $u: D\rightarrow O$ such that $u(D) O\cap S$. Any map $F: O\cap S \rightarrow \mathbb{R}^3$ is smooth if and only if $F\circ u$ is smooth. You use that to define smooth vector fields on $S$. Commented Jun 18, 2021 at 15:23
• Thanks for your comment. I know the definitions I guess. To make my problem more precise: $Y$ is defined on $S$ which is in general not an open subset of $\mathbb{R}^3$. So, we cannot differentiate $Y$ in the usual way, i.e. like a function defined on an open subset of $\mathbb{R}^3$. Therefore, I cannot apply the chain rule to compute $(Y \circ γ)'$. (If that rule could be applied, it would easily solve the problem.) I already tried to consider $Y$ on a chart of $S$ or equivalently, use an embedding as you described but this didn't help my so far. Commented Jun 18, 2021 at 23:51
• What type of proof do you have in mind? Commented Jun 18, 2021 at 23:53

Briefly, what happens is that if $$S \subseteq \Bbb R^3$$ is the surface, and $$X,Y \in \mathfrak{X}(S)$$, then for each $$p \in S$$, you choose extensions $$\widetilde{X}$$ and $$\widetilde{Y}$$ for $$X$$ and $$Y$$ near $$p$$, and set $$(\nabla_XY)_p= \mbox{orthogonal projection of }(D_{\widetilde{X}}\widetilde{Y})_p\mbox{ to }T_pS.$$Here's the magic: this is independent of the choice of extensions $$\widetilde{X}$$ and $$\widetilde{Y}$$. Check the chapter on submanifold theory on do Carmo's Riemannian Geometry for more details.

• With the extensions, do you mean that they are defined on some set $U$ with $p \in U$ and $U$ open in $\mathbb{R}^3$? Commented Jun 18, 2021 at 2:40
• Yes. ${}{}{}{}{}$ Commented Jun 18, 2021 at 3:02
• Thanks for your answer. I see how this solves my problems. It can even be used to give an answer and proof to question 2! Therefore accepted. A last question: Could you be more precise on where to find the relevant part in do Carmo's book? Commented Jun 19, 2021 at 13:14

$$\newcommand{\R}{\mathbb{R}}$$ I'm pretty sure this is all discussed carefully in the books you cite, but here is a quick summary:

The idea is to use the fact that $$S$$ is an abstract manifold and transfer everything to an open $$D \subset \R^2$$, where you can use classical calculus.

$$S\subset\R^3$$ is a $$C^1$$ surface if, for any $$p \in S$$, there exists an open set $$O \subset \R^3$$, and open set $$D \subset \R^2$$, and a $$C^1$$ embedding $$u: D \rightarrow \R^3$$ such that $$u(D) = O\cap S$$ and $$p \in u(D)$$. It is straightforward, using the implicit function theorem, to show that $$S$$ is an abstract $$2$$-manifold.

The partial derivatives of the coordinate map $$u$$ define, for each $$x \in D$$, a linear isomorphism $$u_*(x): \R^2 \rightarrow T_{u(x)}S$$. The tangent space $$T_{u(x)}S$$ is a subspace of $$\R^3$$, but we will avoid using that here.

If $$S$$ is viewed as an abstract $$2$$-manifold, then a curve $$c: I \rightarrow S$$, where $$c(0) = p$$ is defined to be $$C^1$$ if $$u^{-1}\circ c: I \rightarrow D$$ is $$C^1$$. The velocity of a $$C^1$$ curve $$c: I \rightarrow u(D)\subset S$$ at $$c(0)$$ is defined to be $$u_*((u^{-1}\circ c)'(0))$$.

A vector field on $$S$$ is a map $$V: S \rightarrow \R^3$$ such that $$V(p) \in T_pS$$, for each $$p \in S$$. $$V$$ is $$C^1$$, if, for any coordinate map $$u: D \rightarrow S$$, $$(u_*)^{-1}\circ V\circ u: D \rightarrow \R^2$$ is $$C^1$$. This implies that for any $$X \in T_pS$$ and $$C^1$$ vector field $$Y$$ on $$S$$, you can define $$$$(1)\ \ \ D_XY(p) = \left.\frac{d}{dt}\right|_{t=0}(Y\circ u)\circ(u^{-1}\circ c)(t) = \sum_{i=1}^2\partial_i(Y\circ u)(u^{-1}\circ c)_i'(0).$$$$ where $$c(0) = p$$ and $$c'(0) = X$$, and $$Y$$ is treated as a $$\R^3$$-valued function.

Given any two curves $$c_1$$ and $$c_2$$, such that $$c_1'(0) = c'(0) = X$$, the definition of $$c'$$ above implies $$(u^{-1}\circ c_1)'(0) = (u^{-1}\circ c_2)'(0) = (u_*)^{-1}(X).$$ Substituting this into (1) shows that the definition of $$D_XY$$ is independent of the curve used.

• How can we apply the chain rule to $(Y \circ c)'(0)$? In order to do this, don't we need the partial derivatives of $Y$ in the sense of "classical" calculus, i.e. $\partial_i Y_j(x):=\lim_{h\to 0} \frac{1}{h}(Y_j(x_1,...,x_i+h,...,x_3)−Y_j(x))$? In general, the points $(x_1,..,x_i+h,...,x_3)$ are not in $S$ and therefore, $Y_j$ is not defined for those points. Commented Jun 19, 2021 at 12:45
• @russoo, you're right, and my explanation is inadequate. Thanks for patiently pointing it out again. Here's the trick: You apply the chain rule not to $Y$ and $c$ but to $c\circ u^{-1}: I \rightarrow D$ and $Y\circ u: D \rightarrow \mathbb{R}^3$. Both of these maps are $C^1$. Since $D \subset \mathbb{R}^2$ is open, you can now use classical calculus (I like that term). Commented Jun 19, 2021 at 14:32
• I guess you mean $u^{-1} \circ c: I \to D$ instead of $c \circ u^{-1}$? And then, is your idea to write $(Y \circ c)'=(Y \circ u \circ u^{-1} \circ c)'$ and applying the chain rule to the right-hand side? Commented Jun 19, 2021 at 14:37
• Yes, that's right. Commented Jun 19, 2021 at 14:42
• I still don't understand your last claim about the equality of the $(u^{-1} \circ c_i)'$. I decided to ask a new question about this particular problem since it is quite specific. Anyway, thanks for your help! :-) Commented Jun 20, 2021 at 1:32

$$\newcommand{\R}{\mathbb{R}}$$ Here's another way to define the Levi-Civita connection:

Given any $$C^2$$ coordinate map $$u=(u^1,u^2,u^3): D \rightarrow S \subset \R^3$$, the partial derivatives $$\partial_1u, \partial_2u$$ are a basis of $$T_{u(x)}S$$. Therefore, a vector field $$Y=(Y^1,Y^2,Y^3)$$ on $$S$$ can be written as $$Y(u(x)) = Y^i(u(x))\partial_iu(x).$$ $$Y$$ is $$C^1$$ iff $$Y\circ u$$ is $$C^1$$. Differentiating, we get $$\partial_j(Y\circ u)(x) =\partial_j(Y^i\circ u)\partial_iu(x) + (Y^i\circ u)(x)\partial^2_{ij}u(x)$$ If $$X = X^i\partial_iu(x)$$, then we can define $$D_XY(u(x)) = X^j\partial_j(Y^i\circ u)(x)\partial_iu(x). + X^j(Y^i\circ u(x))\pi_{u(x)}(\partial^2_{ij}u(x)),$$ where $$\pi_{p}: \R^3 \rightarrow T_pS$$ is orthogonal projection.