Killing fields on product metrics

Question

Let $(M_i,g_i)$ be Riemannian manifolds, $i=1,2$. (Save Euclidiean factors) Is it true that a Killing field $Z$ on $(M_1\times M_2,g_1\times g_2)$ will split as a sum of Killing fields $Z=X+Y$, where $X$ is Killing on $M_1$ and $Y$ on $M_2$?

The converse is obviously true: if $X$ and $Y$ are Killing, so it is $Z$; and it is obviously false for a product of Euclidean spaces: $(\mathbb R^2,dx^2+dy^2)=(\mathbb R, dt^2)\times(\mathbb R, dt^2)$ and the isometry group of $\mathbb R^2$ is pretty bigger then the product of the groups of $\mathbb R$.

The question arises from a question on foliations: does Riemannian foliations with (reducible) totally geodesic leaves (locally) splits as products of orthogonal Riemannian foliations?

Answer 1

Use the Corollary of this paper:

If a compact Riemannian manifold $M$ splits as $M = M_1 \times M_2$, then the identity component of the isometry group splits as $I_0 (M) = I_0(M_1) \times I_0(M_2)$.

(This is authors' stated Corollary, but from their main theorem it looks like we can weaken compactness to your requirement that $M$ has no Euclidean factor.)

Let $\zeta_t : \mathbb R \to I_0(M)$ be the one-parameter group of isometries generated by $Z$. The above fact means we can write $\zeta_t = (\xi_t, \upsilon_t)$ with $\xi_t$ acting on $M_1$, $\upsilon_t$ acting on $M_2$. Since $\zeta_t$ is a one-parameter subgroup, both $\xi_t$ and $\upsilon_t$ must be as well, and their generators $X,Y$ will satisfy $X+Y=Z$.

Answer 2

Let $L$ denote the Lie derivative. If $X$ is tangent to the product $X=(X_1,X_2)$ with $X_i$ tangent to $M_i$, and the metric splits as $g=(g_1,g_2)$. Then $L_{(X_1,X_2)}(g_1 ,g_2)= (L_{X_1}g_1, L_{X_2}g_2)$, so $L_Xg=0 <=> L_{X_1}g_1 = 0 $and $ L_{X_2}g_2=0$