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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?

fgraderboy
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  • 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 $y$. – Andreas Blass May 08 '19 at 04:15
  • 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. – Andreas Blass May 08 '19 at 04:19
  • @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. – fgraderboy May 08 '19 at 04:39
  • @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 $M_2$. – fgraderboy May 08 '19 at 04:50
  • 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 $M_1\times M_2$. – Andreas Blass May 08 '19 at 12:17
  • I think I see what you mean. The connection $\nabla_{Y} X$ depends on the behaviour of $X$ along the integral curve of $Y$, and this behaviour need not necessarily decompose into two vector fields on $M_1$ and $M_2$. Assuming this reasoning is correct, how do I then proceed to solve the question? – fgraderboy May 08 '19 at 16:00
  • As far as I can tell (perhaps the Riemannian geometers on this site will have a better idea), the problem with the passage you quoted is the words "given by". If they were replaced by "satisfies" then I think it would be OK. In other words, the Riemannian connection on $M_1\times M_2$ satisfies the claimed equation for those vector fields that happen to be of the specified $X_1+X_2$ form, but I don't see that this equation suffices to determine $\nabla$ completely (maybe the experts will see it). – Andreas Blass May 08 '19 at 17:50

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