# Confusion about projection of connection in manifold

If $$M$$ is a submanifold of $$S$$ and if there is for each point $$p$$ a mapping $$\pi_p$$ from $$T_{p}(S)$$ to $$T_{p}(M)$$, then we may use the definition of connection on $$M$$. Assume that $$\pi_{p}:T_{p}(S)\to T_{p}(M)$$ is a linear mapping and that $$\pi_{p}(D)=D$$ for every $$S\in T_{p}(M)$$,and that the relation $$p\mapsto \pi_{p}$$ is $$C^{\infty}$$. Now suppose, for each $$X,Y$$ vector fields in $$M$$, we define a vector field $$\nabla_{X}^{(\pi)}$$ on $$M$$ in the following way: $$(\nabla_{X}^{(\pi)}Y)_{p}=\pi_{p}((\nabla_{X}Y)_{p}),\quad \forall p\in M \tag{1}$$ Then $$\nabla^{(\pi)}$$ is a connection on $$M$$. In particular, if a Riemannian metric $$g=\langle \bullet ,\bullet\rangle$$ is given on $$S$$, we may take as $$\pi_{p}$$ the orthogonal projection with respect to $$g$$. This is defined such that, for all $$D\in T_{p}(S)$$ and all $$D'\in T_{p}(M)$$ $$\langle \pi_{p}(D),D'\rangle_p =\langle D,D'\rangle_{p} \tag{2}.$$ We call such $$\nabla^{(\pi)}$$ the projection of $$\nabla$$ onto $$M$$ with respect to $$g.$$

I am trying to understand the above definition but find it difficult to visualize it or maybe I am missing something. Let me explain what I have understood (Please correct me if I am wrong.)

We have a submanifold $$M$$ of a manifold $$S$$, if we take $$X,Y$$ two vector fields on $$M$$ then we know that $$\nabla_{X}Y$$ need not be a vector field in $$M$$. So to define the connection on the submanifold $$M$$, we have defined a linear map $$\pi_{p}$$ and then we have defined the connection on $$M$$ by $$\nabla^{(\pi)}$$ by $$(1)$$. This is my understanding for $$(1)$$ and $$(2)$$, I am not getting how we got that. If someone explain this that will be great help for me, Thanks.

• Which definition are you trying to understand? Commented Nov 8, 2022 at 18:03
• @Didier Actually I want to know that the way I understood equation $(1)$ is correct or not and I want to know how we got equation $(2)$. Commented Nov 8, 2022 at 18:07
• I mean what is the role of orthogonal projection in defining $(2)$ Commented Nov 8, 2022 at 18:13
• (2) is the definition of the orthogonal projection Commented Nov 8, 2022 at 18:38

First of all, forget that we are on a manifold. Let $$(E,\langle\cdot,\cdot\rangle)$$ be a Euclidean vector space. Let $$F\subset E$$ a linear subspace. Then the orthogonal projection on $$F$$ $$\pi\colon E\to E$$ is the unique linear endomorphism of $$E$$ such that $$\forall v \in E,\quad \pi(v)\in F \quad \text{and} \quad v -\pi(v) \perp F.$$ It is also uniquely determined by the following: $$\forall v\in E,\forall w \in F,\quad \langle v,w\rangle = \langle \pi(v),w\rangle.$$ (this is your equation (2)). This is a particular case of general projections: the extra assumption is that they satisfy $$\mathrm{Im}(\pi)\perp \ker (\pi)$$ for the inner product $$\langle \cdot,\cdot\rangle$$. As a projection, it takes values onto the subspace on which it projects. Hence, the orthogonal projection on $$F$$ takes values in $$F$$.
Now, go back to a Riemannian manifold $$(S,g)$$ and consider a submanifold $$M\subset S$$. If $$x\in M$$, then $$(T_xS,g_x)$$ is an Euclidean vector space, with a distinguished linear subspace given by $$T_xM\subset T_xS$$. You can then consider the orthogonal projection on $$T_xM$$, call it $$\pi_x\colon T_xS\to T_xS$$. It takes values in $$T_xM$$.
If $$\nabla$$ is the Levi-Civita connection of $$(S,g)$$, then for two vector fields $$X$$ and $$Y$$ on $$S$$ and for a point $$x\in M$$, you can consider $$\pi_x\left(\nabla_XY(x)\right)\in T_xM.$$ This gives you a map $$x\in M \mapsto \pi_x\left(\nabla_XY(x)\right).$$ It turns out that $$\pi \colon x\mapsto \pi_x$$ is a smooth section of $$\mathrm{End}(TS|_M)$$, from which deduce that in fact $$x \mapsto \pi_x\left(\nabla_XY(x)\right)$$ defines a vector field on $$M$$. This then gives you a map $$\nabla^{(\pi)}= \pi\circ \nabla \colon \Gamma(TM)\times \Gamma(TM)\times \Gamma(M)$$. We can show that is in fact an affine connection, and using the fact that $$\nabla$$ is torsion-free and $$g$$ is $$\nabla$$-parallel, we can deduce that it is actually the Levi-Civita connection of $$g|_M$$ (see here for instance).