# Levi-Civita Connection, Riemannian Product Manifold and Leibniz product rule

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 $$X_i,Y_i\in \Gamma(TM_i)$$.

Is the Leibniz rule $$\nabla_X(fZ)=X(f)\cdot Z+f\nabla_XZ$$ already holds? If I set $$X=X_1+X_2$$ and $$Z=Z_1+Z_2$$ then:

$$\nabla_X(fZ) = \nabla^1_{X_1}(fZ_1)+ \nabla^2_{X_2}(fZ_2)$$

$$= (X_1(f)\cdot Z_1 + f\nabla^1_{X_1}Z_1) + (X_2(f)\cdot Z_2 + f\nabla^2_{X_2}Z_2)=$$

$$= f\nabla_XZ + (X_1(f)Z_1+X_2(f)Z_2)$$.

Is it correct?

To satisfy the Leibniz rule we need that: $$\nabla_{X_1 + X_2}(f(Z_1 + Z_2)) = (X_1 + X_2)(f)\cdot (Z_1 + Z_2) + f\nabla_{X_1 + X_2}(Z_1 + Z_2)$$
Clearly your expression satisfies the second term, so we look to the first. On the product manifold $$M_1\times M_2$$, we can certainly say locally that $$\Gamma(T(U_1 \times U_2)) = \Gamma(TM_1) \oplus \Gamma(TM_2)$$. So it makes sense to rewrite the first term as: \begin{align*} (X_1 + X_2)(f)\cdot (Z_1 + Z_2) &= \pi_1\big((X_1 + X_2)(f)\cdot (Z_1 + Z_2)\big) \oplus \pi_2\big((X_1 + X_2)(f)\cdot (Z_1 + Z_2)\big)\\ &= \pi_1(X_1 + X_2)(\pi_1f)\cdot \pi_1(Z_1 + Z_2) \oplus \pi_2(X_1 + X_2)(\pi_2 f)\cdot \pi_2(Z_1 + Z_2)\\ &= X_1 (\pi_1 f)\cdot Z_1 \oplus X_2 (\pi_2 f)\cdot Z_2 \end{align*}
We can identify $$X_1 \oplus X_2$$ with $$X_1 + X_2$$ in the product manifold. If we write $$f = \pi_1f \oplus \pi_2 f$$ it is clear that $$X_i(f) = X_i(\pi_i f)$$ since the $$X_i$$ act over smooth functions defined on the individual $$M_i$$. Therefore this is equal to the expression you derived, satisfying the Leibniz rule.
I think another way to say this, in more or less the same way, is that locally you can identify $$X_1 \in \Gamma(TM_1)$$ with $$(X_1,0)\in \Gamma(T(M_1\times M_2))$$ and $$X_2 \in \Gamma(TM_2)$$ with $$(0,X_2)\in \Gamma(T(M_1\times M_2))$$. Then you can directly calculate the first term and see it is equal to your result.