# do Carmo Riemannian Geometry Exercise 2.3: definition of $\nabla$ for an immersion

The following is Exercise 3 of Chapter 2 of my Brazilian edition of do Carmo's Riemannian Geometry:

Let $$f: M^n \to \overline M^{n + k}$$ be an immersion from a differentiable manifold $$M$$ to a Riemannian manifold $$\overline M$$. Assume in $$M$$ the Riemannian metric induced by $$f$$: $$\langle u, v \rangle_p = \langle df_p(u), df_p(v) \rangle_{f(p)}.$$ Let $$p \in M$$ and $$U \subset M$$ be a neighborhood of $$p$$ such that $$f(U) \subset \overline M$$ be a submanifold of $$\overline M$$. Let $$X, Y$$ be vector fields on $$f(U)$$ and extend then to vector fields $$\overline X, \overline Y$$ on an open subset of $$\overline M$$. Define $$(\nabla_X Y)(p) = \text{tangential component of } \overline \nabla_{\overline X} \overline Y(p),$$ where $$\overline \nabla$$ is the Riemannian connection of $$\overline M$$. Prove that $$\nabla$$ is the Riemannain connection of $$M$$.

My questions are:

• $$X$$ and $$Y$$ are vector fields on $$f(U) \subset M$$. Then $$\nabla_X Y(p)$$ does not make sense. Shouldn't it be $$\nabla_X Y(f(p))$$? Also, $$\nabla$$ is not a connection on $$M$$, but on $$f(U)$$, isn't it? So what does the problem want us to prove?
• What does "tangential component" mean?
• Yes. It's sloppy writing by identify $U$ as $f(U)$. Since $\overline{M}$ is a Riemannian manifold, each tangent space at $f(U)$ split into tangent space of the submanifold and its orthogonal complement. Tangential component obviously the projection onto tangent space of submanifold. Jul 31 '20 at 20:18

There is a bit of abusing notation here. The goal is to define $$\nabla$$, which is a connection on the tangent bundle of $$M$$. Thus the goal IS to define $$\nabla _X Y(p)$$, where $$p\in M$$ and $$X, Y$$ are local vector fields of $$M$$ around $$p$$. (I think it is very important to know that they are not defining something at $$f(p)$$: there might be $$p\neq q$$ so that $$f(p) = f(q)$$)

The way to do so is

(1) push-forward the local vector fields $$X, Y$$ to $$\mathrm df(X), \mathrm df(Y)$$ respectively, if $$X, Y$$ are on $$U$$, then $$\mathrm df(X), \mathrm df(Y)$$ are on $$f(U)$$ (they abuse notations here, identifying $$X, Y$$ with $$\mathrm df(X), \mathrm df(Y)$$)

(2) extend $$\mathrm df(X), \mathrm df(Y)$$ to a local vector fields $$\overline X, \overline Y$$ respectively on $$\overline M$$ around $$f(p)$$, and

(3) define $$\nabla_X Y(p) := \text{tangential component of }\overline\nabla _{\overline X} \overline Y (f(p))$$ (Note that it is $$f(p)$$ on the right hand side. I guess it is a typo) As suggested in the comment, the tangent space at $$T_{f(p)} \overline M$$ split into $$df (T_pM)$$ and $$(df (T_pM))^\perp$$, the orthogonal complement. The tangential component are taken with respect to this decomposition. Thus the more precise definition should be $$\nabla_X Y(p) :=(\mathrm df)^{-1} \bigg( \text{tangential component of }\overline\nabla _{\overline X} \overline Y (f(p))\bigg)$$

I suppose they went on to show that $$\nabla$$ is well defined, independent of the extension $$\overline X, \overline Y$$. Indeed $$\nabla$$ is the Levi-Civita connection on $$M$$ with respect to the pullback metric $$\langle u, v \rangle_p := \langle df_p(u), df_p(v) \rangle_{f(p)}.$$

• I guess the choice of $U$ would make $f|_U$ embedding no ? Therefore $U$ and $f(U)$ diffeomorphic. Jul 31 '20 at 20:57
• Yes (I think I don't understand your comment, can you be more direct/specific?) @SiKucing Jul 31 '20 at 20:59
• It is said in the question that $U$ is a nbd of $p$ such that $f(U)$ is a submanifold of $\overline{M}$, which i interpret as embedded submanifold. With this choice, $f|U$ is embedding and therefore we can't have $f(p)=f(q)$ for $p\neq q$ isn't it ? Jul 31 '20 at 21:04
• Ar I see. My point is that there could be $q\notin U$ so that $f(q) = f(p)$. Think of the immersion $f: \mathbb R \to \mathbb R^2$, $f(t) = (\cos t, \sin t)$. It is important to know that $\nabla$ is defined on $\mathbb R$, but not on the image $\mathbb S^1$. @SiKucing Jul 31 '20 at 21:07
• Did the answer here help? @DaniloGregorin Aug 6 '20 at 17:56