I'm working on exercise 1(a) of chapter 6 in do Carmo's Riemannian Geometry:
Let $M_1$ and $M_2$ be Riemannian manifolds, and consider the product $M_1\times M_2$, with the product metric. Let $\nabla^1$ be the Riemannian connection of $M_1$ and let $\nabla^2$ be the Riemannian connection of $M_2$. Part (a): Show that the Riemannian connection $\nabla$ of $M_1\times M_2$ is given by $\nabla_{Y_1+Y_2}(X_1+X_2) = \nabla_{Y_1}^1 X_1 + \nabla_{Y_2}^2 X_2$, where $X_i,Y_i\in \Gamma(TM_i)$.
Of course the first thing is to show that $\nabla$ is a connection at all, and this is turning out to be more subtle than I had originally thought. First and foremost, it's not even immediately clear that the given formula uniquely determines $\nabla$, since $\Gamma(T(M_1\times M_2))\supsetneq \Gamma(TM_1)\oplus \Gamma(TM_2)$.
I'm having particular trouble showing that the Leibniz rule $\nabla_X(fZ)=X(f)\cdot Z+f\nabla_XZ$ holds. My original thought was to write $X=X_1+X_2$ and $Z=Z_1+Z_2$ and then calculate
\begin{equation*}
\nabla_X(fZ) = \nabla^1_{X_1}(fZ_1)+ \nabla^2_{X_2}(fZ_2)
\end{equation*}
\begin{equation*}
= (X_1(f)\cdot Z_1 + f\nabla^1_{X_1}Z_1) + (X_2(f)\cdot Z_2 + f\nabla^2_{X_2}Z_2) = f\nabla_XZ + (X_1(f)Z_1+X_2(f)Z_2).
\end{equation*}
But this is definitely not looking like what I want. This is right iff $X(f)Z = X_1(f)Z_1+X_2(f)Z_2$, which is certainly not going to hold in general. Of course this shouldn't be right, because it's not like $Z=Z_1+Z_2 \in \Gamma(T(M_1\times M_2))$ is going have $Z_i$ be pulled back via the projections.
So my next guess was instead to integrate $X$ by a curve $\alpha:(-\epsilon,\epsilon)\rightarrow M_1\times M_2$, which I can even assume is a geodesic (meaning it projects to a geodesic in both factors). Then, along $\alpha$ I can hope to decompose $Z=Z_1+Z_2$, where $Z_i\in \Gamma(\alpha^* TM_i)$ (where I'm considering $TM_i \rightarrow M_1\times M_2$ as a subbundle of the tangent bundle $T(M_1\times M_2)$). In other words, I'm hoping to turn $Z|_\alpha$ into a sum of pullbacks. But whether or not I can even do this (which I can't in general if $\alpha$ is constant in one or the other factor), this gives me the same equations as above, which just as above is a problem.
In the above paragraph, I think I'm actually modifying $f$ to be a pullback too, but I think this should be alright since ultimately the only thing that matters is the value of $fZ$ along $\alpha$.
So, questions: (1) Is $\nabla$ uniquely determined by the given formula? (2) What am I doing wrong?
Best Answer
The problem of verifying the Leibniz rule is closely related to your question as to what extent the formula given actually gives a well-defined map.
The point is that any vector field on $M_1\times M_2$ can locally be written as a linear combination of vector fields on $M_1$ and $M_2$, with the coeffients being functions on the product $M_1\times M_2$. The formula given should then be extended to all vector fields by assuming the Leibniz rule. (In short, the candidate connection on the product will satisfy the Leibniz rule by definition.)
You then have to check that what you have is well-defined, torsion-free, and compatible with the product metric. These should all be straightforward exercises, though.