the Levi-Civita connection on a product of Riemannian manifolds

Question

I'm working on exercise 1(a) of chapter 6 in do Carmo's Riemannian Geometry:

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

Of course the first thing is to show that $\nabla$ is a connection at all, and this is turning out to be more subtle than I had originally thought. First and foremost, it's not even immediately clear that the given formula uniquely determines $\nabla$, since $\Gamma(T(M_1\times M_2))\supsetneq \Gamma(TM_1)\oplus \Gamma(TM_2)$.

I'm having particular trouble showing that the Leibniz rule $\nabla_X(fZ)=X(f)\cdot Z+f\nabla_XZ$ holds. My original thought was to write $X=X_1+X_2$ and $Z=Z_1+Z_2$ and then calculate \begin{equation*} \nabla_X(fZ) = \nabla^1_{X_1}(fZ_1)+ \nabla^2_{X_2}(fZ_2) \end{equation*} \begin{equation*} = (X_1(f)\cdot Z_1 + f\nabla^1_{X_1}Z_1) + (X_2(f)\cdot Z_2 + f\nabla^2_{X_2}Z_2) = f\nabla_XZ + (X_1(f)Z_1+X_2(f)Z_2). \end{equation*} But this is definitely not looking like what I want. This is right iff $X(f)Z = X_1(f)Z_1+X_2(f)Z_2$, which is certainly not going to hold in general. Of course this shouldn't be right, because it's not like $Z=Z_1+Z_2 \in \Gamma(T(M_1\times M_2))$ is going have $Z_i$ be pulled back via the projections.

So my next guess was instead to integrate $X$ by a curve $\alpha:(-\epsilon,\epsilon)\rightarrow M_1\times M_2$, which I can even assume is a geodesic (meaning it projects to a geodesic in both factors). Then, along $\alpha$ I can hope to decompose $Z=Z_1+Z_2$, where $Z_i\in \Gamma(\alpha^* TM_i)$ (where I'm considering $TM_i \rightarrow M_1\times M_2$ as a subbundle of the tangent bundle $T(M_1\times M_2)$). In other words, I'm hoping to turn $Z|_\alpha$ into a sum of pullbacks. But whether or not I can even do this (which I can't in general if $\alpha$ is constant in one or the other factor), this gives me the same equations as above, which just as above is a problem.

In the above paragraph, I think I'm actually modifying $f$ to be a pullback too, but I think this should be alright since ultimately the only thing that matters is the value of $fZ$ along $\alpha$.

So, questions: (1) Is $\nabla$ uniquely determined by the given formula? (2) What am I doing wrong?

Answer 1

The problem of verifying the Leibniz rule is closely related to your question as to what extent the formula given actually gives a well-defined map.

The point is that any vector field on $M_1\times M_2$ can locally be written as a linear combination of vector fields on $M_1$ and $M_2$, with the coeffients being functions on the product $M_1\times M_2$. The formula given should then be extended to all vector fields by assuming the Leibniz rule. (In short, the candidate connection on the product will satisfy the Leibniz rule by definition.)

You then have to check that what you have is well-defined, torsion-free, and compatible with the product metric. These should all be straightforward exercises, though.

Answer 2

This equality is obviously not true in general. To see it, take $M=N=(0,+\infty)$. On the hand, $$\nabla_{(xe_1 + ye_2)}(-ye_1+xe_2)=-ye_1+xe_2,$$ on the other hand, $$\nabla_{xe_1}(-ye_1) = \nabla_{ye_2}(xe_2) = 0.$$ It happens because $-ye_1+xe_2$ is not a "decomposable" vector fields. Let's say that $X$ is decomposable if $X(p,q) = X_1(p) + X_2(q)$, $p \in M, q \in N,$ for $X_1 \in \mathfrak{X}(M), X_2 \in \mathfrak{X}(N).$

To be more precise, embed $M \hookrightarrow{} M \times \{q\} \hookrightarrow M \times N$ for every $q \in N$. The collection of pushforwards of $X$, still denoted by $X$, is a smooth vector field on $M\times N$.

What is true is that, if X and Y are decomposable, then $$\nabla_{Y}X = \nabla^1_{Y_1}X_1 + \nabla^2_{Y_2}X_2,$$ which is equivalent to prove that $$\nabla^1_{Y_1}X_2=\nabla^1_{Y_2}X_1=\nabla^1_{Y_2}X_2 = 0.$$ But $X_1$ does not vary on $N$ direction, so $\nabla_{Y_2}X_1 = 0. $The same for $\nabla_{Y_1}X_2$. It remains to show the last equality. Take local coordinates $(x_1, \dots, x_m)$ on $M^m$ and $(y_1, \dots, y_n)$ on $N^n$. Using Koszul formula, we have

$$2\langle \nabla_{Y_2}X_2, \frac{\partial}{\partial x^i}\rangle = X_2 \underbrace{\langle \frac{\partial}{\partial x^i},Y_2\rangle}_{=0} +Y_2 \underbrace{\langle X_2,\frac{\partial}{\partial x^i} \rangle}_{=0} -\frac{\partial}{\partial x^i}\underbrace{\langle X_2,Y_2\rangle}_{\rm{don't \ vary \ on \ M}} \\ -\langle[X_2,\frac{\partial}{\partial x^i}],Y_2 \rangle -\langle[Y_2,\frac{\partial}{\partial x^i}],X_2 \rangle- \langle\underbrace{[X_2,Y_2]}_{\in \mathfrak{X}(N)},\frac{\partial}{\partial x^i} \rangle$$ Writting $X_2(x,y) = X_2(y) = \phi^j(y)\frac{\partial}{\partial y^j}$, $$ [X_2,\frac{\partial}{\partial x^i}] = \phi^j[\frac{\partial}{\partial y^j}, \frac{\partial}{\partial x^i}] - \frac{\partial \phi^j}{\partial x^i}\frac{\partial}{\partial y^j} =0, $$ which proves the desired formula.

Observe that this is enough to prove the remaining itens of the exercise.

Answer 3

I think that this can be solved by using the fact that the canonical maps from M1 and M2 into their product are immersions, and then applying exercise 3 of chapter 6.

Let $p\in M_1, q\in M_2.$ Observe that the canonical maps $M_1\rightarrow M_1\times \{q\}$ and $M_2\rightarrow \{p\}\times M_2$ are immersions of $M_1$ and $M_2$ into $M_1\times M_2.$ Next, notice that metrics induced on these immersed submanifolds are the same as their original metrics. Now, let $X_1,Y_1\in \chi (M_1\times\{q\})$ and $X_2,Y_2\in \chi(\{p\}\times M_2)$

Now, let $\overline\nabla$ be the Levi-Civita connection on $M_1\times M_2,$ by exercise 3 of chapter 6 we get,

$$\nabla^{1}_{X_1}Y_1=\big(\overline\nabla_{\overline{X}}\overline{Y}\big)^{Tan_{M_1}}$$ where $\overline{X}$ and $\overline{Y}$ are extensions of $X_1$ and $Y_1.$

We get a similar formula for $\nabla^{2}_{X_2}Y_2.$

Now let us define $\overline{X}$ and $\overline{Y}$ by;

$\overline{X}(a,b)=X_1(a,q)+X_2(p,b)$

$\overline{Y}(a,b)=Y_1(a,q)+Y_2(p,b)\quad \forall (a,b)\in M_1\times M_2.$

This gives us \begin{align*} \nabla^{1}_{X_1}Y_1+ \nabla^{2}_{X_2}Y_2&=\big(\overline\nabla_{\overline{X}}\overline{Y}\big)^{Tan_{M_1}}+\big(\overline\nabla_{\overline{X}}\overline{Y}\big)^{Tan_{M_2}} \\ &=\overline{\nabla}_{\overline{X}}\overline{Y}\end{align*} as required.

Answer 4

As pointed out by Rafael Farias, the formula proposed in Do Carmo's exercise applies only for vector fields $V$ which are 'decomposable', in the sense that $V(x_1, x_2) = (V_1(x_1), V_2(x_2))$, whereas in general $V_1$ and $V_2$ would depend on both $x_1$ and $x_2$.

Here is a more general formula.

Let $M_1$ and $M_2$ be two manifolds, respectively equipped with connections $\nabla^{(1)}$ and $\nabla^{(2)}$. We let $M = M_1 \times M_2$ be the product manifold. Then, the map $\nabla$ defined as follows is a connection on $M$ (let's call it the product connection): $$ \nabla_{(u_1, u_2)} (V_1, V_2) = \Big( \nabla^{(1)}_{u_1} V_1(\cdot, x_2) + \mathrm{D} V_1(x_1, \cdot)(x_2)[u_2], \nabla^{(2)}_{u_2} V_2(x_1, \cdot) + \mathrm{D} V_2(\cdot, x_2)(x_1)[u_1] \Big) $$ for all $(u_1, u_2)$ tangent to $M$ at $(x_1, x_2)$. Notation such as $V_1(\cdot, x_2)$ denotes the map obtained from $V_1 \colon M_1 \times M_2 \to TM_1$ by fixing the second input to $x_2$. In particular, $V_1(\cdot, x_2)$ is a vector field on $M_1$, while $V_1(x_1, \cdot)$ is a map from $M_2$ to the fixed linear space $T_{x_1}M_1$ (hence we can differentiate it in the usual way, as denoted by $\mathrm{D}$). If $V$ is separable, then $V_1(x_1, \cdot)$ is constant hence its differential is zero and we recover do Carmo's formula.

As one would expect, if $\nabla^{(1)}$ and $\nabla^{(2)}$ are the Riemannian connections on $M_1, M_2$ and if $M$ is equipped with the product metric, then $\nabla$ as above is the Riemannian connection on $M$.

These statements appear as exercises in my book here, specifically in Section 5.3 and 5.4. As already pointed out in the discussions here, the step that requires the most work is establishing the Leibniz rule. This can be done by first establishing a general formula for the differential of maps on product manifolds: a "total derivative" of sorts (Exercise in Section 3.4).