2
$\begingroup$

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

Is the Leibniz rule $\nabla_X(fZ)=X(f)\cdot Z+f\nabla_XZ$ already holds? If I set $X=X_1+X_2$ and $Z=Z_1+Z_2$ then:

$\nabla_X(fZ) = \nabla^1_{X_1}(fZ_1)+ \nabla^2_{X_2}(fZ_2)$

$= (X_1(f)\cdot Z_1 + f\nabla^1_{X_1}Z_1) + (X_2(f)\cdot Z_2 + f\nabla^2_{X_2}Z_2)=$

$= f\nabla_XZ + (X_1(f)Z_1+X_2(f)Z_2)$.

Is it correct?

$\endgroup$

1 Answer 1

0
$\begingroup$

To satisfy the Leibniz rule we need that: $$\nabla_{X_1 + X_2}(f(Z_1 + Z_2)) = (X_1 + X_2)(f)\cdot (Z_1 + Z_2) + f\nabla_{X_1 + X_2}(Z_1 + Z_2)$$

Clearly your expression satisfies the second term, so we look to the first. On the product manifold $M_1\times M_2$, we can certainly say locally that $\Gamma(T(U_1 \times U_2)) = \Gamma(TM_1) \oplus \Gamma(TM_2)$. So it makes sense to rewrite the first term as: \begin{align*} (X_1 + X_2)(f)\cdot (Z_1 + Z_2) &= \pi_1\big((X_1 + X_2)(f)\cdot (Z_1 + Z_2)\big) \oplus \pi_2\big((X_1 + X_2)(f)\cdot (Z_1 + Z_2)\big)\\ &= \pi_1(X_1 + X_2)(\pi_1f)\cdot \pi_1(Z_1 + Z_2) \oplus \pi_2(X_1 + X_2)(\pi_2 f)\cdot \pi_2(Z_1 + Z_2)\\ &= X_1 (\pi_1 f)\cdot Z_1 \oplus X_2 (\pi_2 f)\cdot Z_2 \end{align*}

We can identify $X_1 \oplus X_2$ with $X_1 + X_2$ in the product manifold. If we write $f = \pi_1f \oplus \pi_2 f$ it is clear that $X_i(f) = X_i(\pi_i f)$ since the $X_i$ act over smooth functions defined on the individual $M_i$. Therefore this is equal to the expression you derived, satisfying the Leibniz rule.

I think another way to say this, in more or less the same way, is that locally you can identify $X_1 \in \Gamma(TM_1)$ with $(X_1,0)\in \Gamma(T(M_1\times M_2))$ and $X_2 \in \Gamma(TM_2)$ with $(0,X_2)\in \Gamma(T(M_1\times M_2))$. Then you can directly calculate the first term and see it is equal to your result.

$\endgroup$

You must log in to answer this question.