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 !


You need to calculate the Levi-Civita connection with respect to the product metric. You have not done it yet.

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 $\Gamma_{is}^t =0$.


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