# Riemannian connection on product manifolds vanishes

Let $$X$$ and $$Y$$ be Riemannian manifolds with metric $$g$$ and $$h$$, respectively. Let $$(x_1, ..., x_n)$$ and $$(y_1, ..., y_m)$$ be local coordinates on $$X$$ and $$Y$$.

Then in local coordinates form $$g = \sum_{i,j = 1}^n g_{(i,j)}(x)dx_i \otimes dx_j \ \mbox{and} \ h = \sum_{s,t=1}^m h_{(s,t)}(y) dy_s \otimes dy_t.$$

Forming a Riemannian manifold $$X \times Y$$ with the metric $$G := g \times h = \sum_{i,j = 1}^n g_{(i,j)}(x)dx_i \otimes dx_j + \sum_{s,t=1}^m h_{(s,t)}(y) dy_s \otimes dy_t.$$

Let $$F_1 = \sum_{i=1}^n a_i(x)\frac{\partial}{\partial x_i}$$ and $$F_2 = \sum_{j=1}^m b_j(y)\frac{\partial}{\partial y_j}$$ be two vector fields on $$X \times Y$$.

Then I want to show that $$\nabla_{F_1}F_2 = 0, \nabla :=$$ the Riemannian connection on $$X \times Y.$$

Attempt Write $$\nabla_{F_1}F_2$$ in local coordinate form as $$\nabla_{F_1}F_2 = \sum_{i=1}^n\sum_{j=1}^ma_ib_j\nabla_{\frac{\partial}{\partial x_i}}\frac{\partial}{\partial y_j} + \sum_{i=1}^n\sum_{j=1}^ma_i\frac{\partial}{\partial x_i}b_j\frac{\partial}{\partial y_j}.$$

Since $$b_j = b_j(y) = b_j(y_1, y_2, ..., y_m)$$, $$\frac{\partial}{\partial x_i}b_j(y) = 0.$$

so what left is the first sum, which I am not sure why it is zero ? I guess it might due to symmetry of $$\nabla$$, but the terms $$\nabla_{\frac{\partial}{\partial y_j}}\frac{\partial}{\partial x_i}$$ do not present in the sum. I am not sure how to proceed !

Since $$\nabla_{\partial x_i} \partial_ {y_s} = \Gamma^{k}_{is} \partial _{x_k} + \Gamma_{is}^t \partial_{y_t},$$
we need to show $$\Gamma^{k}_{is} = \Gamma_{is}^t = 0.$$ This is easy: let $$G$$ be the product metric, then
\begin{align} \Gamma^{k}_{is} &= \sum_{l} G^{kl} (-\partial_l G_{is} + \partial_i G_{sl} + \partial_s G_{li}) + \sum_{t} G^{kt} (-\partial_t G_{is} + \partial_i G_{st} + \partial_s G_{ti}) \\ &= \sum_{l} G^{kl} (-\partial_l G_{is} + \partial_i G_{sl} + \partial_s G_{li}) \qquad (\text{since } G^{kt} = 0) \\ &= 0 \end{align} since $$G_{is} = G_{sl} = 0$$ and $$G_{li}= g_{li}$$ is independent of $$y$$. Similarly you can show that $$\Gamma_{is}^t =0$$.