# Levi Civita connection of the product manifold

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?

• The definition you quoted says that the formula for $\nabla_{Y_1+Y_2}(X_1+X_2)$ is for $X_i,Y_i$ sections of the tangent bundle of $M_i$. In particular, the coefficient $f_1$ in $X_1$ should be a function of only $x$, and $f_2$ should be a function of only $y$. Commented May 8, 2019 at 4:15
• I hope, of course, that the source of this definition includes an explanation of how vector fields $X_1$ on $M_1$ and $X_2$ on $M_2$ are to be regarded as vector fields on $M_!\times M_2$, which is where $X_1+X_2$ is supposed to live. Commented May 8, 2019 at 4:19
• @AndreasBlass We evaluate the connection at a point, say $(p,q)$. My understanding is that $\nabla_{Y_1}X_1$ is to be calculated as $\nabla_{g_1(x,q)\frac{\partial}{\partial x}}(f_1(x,q)\frac{\partial}{\partial x})$, and similarly for $\nabla_{Y_2}X_2$. So, we view the coefficients as being dependent on one variable only by fixing the other variable. Commented May 8, 2019 at 4:39
• @AndreasBlass We can view vectors at each point on the product as the sum of two vectors - one from the tangent space of $M_1$ and the other from $M_2$. So, at $(p,q)$, we assign the vector $X_1(p) + X_2(q)$. In this way, we obtain a vector field on the product from vector fields on the individual manifolds. The converse is not true, the information in a vector field on the product manifold is not necessarily contained in one vector field on $M_1$ and one on $M_2$. Commented May 8, 2019 at 4:50
• Both of your comments are true, but they are only tangentially relevant to what you asked, why your calculation gave unexpected results. I still think the reason for that is that the formula you quoted is, as stated at the end of the quote, only about sums of a vector field $X_1$ on $M_1$ and a vector field $X_2$ on $M_2$ (both viewed as vector fields on $M_1\times M_2$ as in your first comment), not about arbitrary vector fields on $M_1\times M_2$. Commented May 8, 2019 at 12:17

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