Levi Civita connection of the product manifold

Question

I am stuck on the same question as this - the Levi-Civita connection on a product of Riemannian manifolds. For convenience, I am putting the question here -

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

I am wondering whether the result is even true.

Consider the case of $\mathbb{R}^2$ viewed as the product $\mathbb{R}\times\mathbb{R}$. Say we have vector fields $X = f_1(x,y) \frac{\partial}{\partial x} + f_2(x,y) \frac{\partial}{\partial y}$ and $Y = g_1(x,y) \frac{\partial}{\partial x} + g_2(x,y) \frac{\partial}{\partial y}$. The decompositon will be $X_1 = f_1(x,y) \frac{\partial}{\partial x}$, $X_2 = f_2(x,y) \frac{\partial}{\partial y}$ and $Y_1 = g_1(x,y) \frac{\partial}{\partial x}$, $Y_2 = g_2(x,y) \frac{\partial}{\partial y}$, which gives us $\nabla_{Y_1} X_1 = g_1 \frac{\partial f_1}{\partial x}\frac{\partial}{\partial x}$ and $\nabla_{Y_2} X_2 = g_2 \frac{\partial f_2}{\partial y}\frac{\partial}{\partial y}$

This means $\nabla_{Y_1} X_1 + \nabla_{Y_2} X_2 = g_1 \frac{\partial f_1}{\partial x}\frac{\partial}{\partial x} + g_2 \frac{\partial f_2}{\partial y}\frac{\partial}{\partial y}$, but we know that $\nabla_{Y} X = \big( g_1 \frac{\partial f_1}{\partial x} + g_2 \frac{\partial f_1}{\partial y}\big)\frac{\partial}{\partial x} + \big( g_1 \frac{\partial f_2}{\partial x} + g_2 \frac{\partial f_2}{\partial y}\big) \frac{\partial}{\partial y}$

Where am I going wrong?

Answer 1

I dont know if my answer is of use, but hope it helps if somebody else stumbles into this question.

Considering $g = g_1 \oplus g_2$ the product metric in $M_1 \times M_2$ we know that there is an unique riemannian conection induced by the metric, so we can proceed by proving that the definition given in the problem is a compatible and symmetric conection. This is, to proof the two statements (using the notation of the question)

• For $X_1,Y_1 \in \mathcal{X}(M_1)$ and $X_2,Y_2 \in \mathcal{X}(M_2)$ we have that \begin{equation} [X_1 + X_2,Y_1 + Y_2] = \nabla_{X_1 + X_2}(Y_1 + Y_2) - \nabla_{Y_1 + Y_2}(X_1 + X_2) \end{equation}

• For $Z = Z_1 + Z_2 \in \mathcal{X}(M_1\times M_2)$ \begin{equation} (Z_1 + Z_2)g(X_1+X_2,Y_1+Y_2) = g(\nabla_{Z_1 + Z_2}(X_1+X_2),Y_1+Y_2) + g(X_1 + X_2, \nabla_{Z_1 + Z_2}(Y_1 + Y_2) \end{equation}

Which look like a straightforward computation, and then conclude by the unicity I mentioned above.