Ricci curvature on Product manifold Let $(M,g)$ and $(N,g)$ be two Riemann manifolds and let $M \times N$ be a Riemannian product manifold.
Let $Ric_1(X_1,Y_1)$ the Ricci curvature on $M$, with $X_1$ and $Y_1$ vectors fields on $M$.
Let $Ric_2(X_2,Y_2)$ the Ricci curvature on $N$, with $X_2$ and $Y_2$ vectors fields on $N$.
Which of the following writing is correct?

*

*$Ric(X_1+X_2,Y_1+Y_2)=Ric_1(X_1,Y_1)+Ric_2(X_2,Y_2)$,


*$Ric(X_1+X_2,Y_1+Y_2)=Ric_1(X_1,Y_1) \oplus Ric_2(X_2,Y_2)$,
(with $Ric$ the Ricci curvature on $M \times N$).
 A: $\DeclareMathOperator{\Ric}{Ric}$For uniformity, let's denote our Riemannian factors as $(M_{i}, g_{i})$ for $i = 1$, $2$, and denote by $\pi_{i}:M_{1} \times M_{2} \to M_{i}$ the projections. The product metric is
$$
g = \pi_{1}^{*}g_{1} + \pi_{2}^{*}g_{2},
$$
which we might denote $g_{1} \oplus g_{2}$ or $g_{1} \times g_{2}$ (depending whether we view $TM$ as $\pi_{1}^{*}TM_{1} \oplus \pi_{2}^{*}TM_{2}$ or as $TM_{1} \times TM_{2}$), and similarly for the Ricci tensor.
If $X_{1}$ is a vector field on $M_{1}$, there is a unique vector field $\widehat{X}_{1}$ on $M$ satisfying $(\pi_{1})_{*}\widehat{X}^{1} = X_{1}$ and $(\pi_{2})_{*}\widehat{X}_{1} = 0$. Literally correct articulations include
\begin{align*}
(\Ric_{1} \oplus \Ric_{2})(\widehat{X}_{1} + \widehat{X}_{2}, \widehat{Y}_{1} + \widehat{Y}_{2})
&= \Ric(\widehat{X}_{1} + \widehat{X}_{2}, \widehat{Y}_{1} + \widehat{Y}_{2}) \\
&= \Ric_{1}(X_{1}, Y_{1}) + \Ric(X_{2}, Y_{2}).
\end{align*}
It's not entirely unreasonable to drop the hats (which gives your first alternative), but doing so is strictly an abuse of notation. Ultimately, you'll have to be guided (in a reader-dependent way) by the contrary constraints of (i) what gets the point across (ii) without ambiguity.
