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If $\mathcal{M}$ is a Riemannian manifold of constant curvature, is the manifold $\mathcal{M}^n$ with the product metric, of constant curvature? (and why?)

Thank you

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  • $\begingroup$ I think that thw scalar curvature of a product manifold is the sum of the scalar curvatures of its factors. But you should double check on some textbook on Riemannian geometry. $\endgroup$ – Giuseppe Negro Dec 30 '15 at 21:47
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For your question, the answer is

Proposition 1: Let M be a Riemannian manifold with constant curvature, then $M^n$ has constant curvature if and only if $M$ has constant curvature zero.

For example, the $S^2\times S^2$ is not a constant curvature space as the 2 dimensional plain spanned by two vectors which come from the tangent spaces of two $S^2$ has sectional curvature zero.

Let's prove a more general result here.

Proposition 2: Let $M=M_1\times M_2$ be the product of two riemannian manifolds, and R be its curvature tensor, $R_1, R_2$ be curvature tensor for $M_1$ and $M_2$ respectively, then one can relate $R, R_1$ and $R_2$ by

$$R(X_1+X_2,Y_1+Y_2,Z_1+Z_2,W_1+W_2)=R_1(X_1,Y_1,Z_1,W_1)+R_2(X_2,Y_2,Z_2,W_2)$$

where $X_i, Y_i, Z_i, W_i\in TM_i$.

To show this, you should use:

(1)$\langle X_1+X_2,Y_1+Y_2 \rangle_{M}=\langle X_1,Y_1 \rangle_{M_1}+\langle X_2,Y_2 \rangle_{M_2}$;

(2)$[X_1+X_2,Y_1+Y_2]_{M}=[X_1,Y_1]_{M_1}+[X_2,Y_2]_{M_2}$;

(3)$\nabla_{X_1+X_2}^M(Y_1+Y_2)=\nabla_{X_1}^{M1}(Y_1)+\nabla_{Y_1}^{M2}(Y_2)$.

(1) is simply by definition of product Riemannian manifold, (2) can be shown in local coordinates, and (3) can be shown by (1) and (2) and along with Koszul formula. Also, you may find this post useful.

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