# How to calculate scalar curvature in a local chart

I want to find a complete manifold with infinite diameter which has uniformly positive scalar curvature.

And I want to show that $M^n = S^2(r) \times \mathbb{R}^{n-2}$ with $n \geq 3$ is an example which satisfying the properties above.

How do I calculate the scalar curvature of $M$ in a local chart? How do I begin the calculations? Could everyone give me some hints or reference?

Thank you very much!

-

Let $M=S^2(r)\times\mathbb{R}^{n-2}$. I will assume you are using the usual metric both in $S^2(r)$ and in $\mathbb{R}^{n-2}$, wich gives us;

1. The sectional curvature of $S^2(r)$ is everywhere $\frac{1}{r^2}$ and;
2. For each pair of orthonormal vectors $u,v$ in $\mathbb{R}^{n-2}$, the sectional curvature $K(u,v)$ of the plane generated by $u,v$ is $0$.

Let $\phi:U\rightarrow S^2(r)$, with $U\subset\mathbb{R}^2$ some open set, and $I:\mathbb{R}^{n-2}\rightarrow \mathbb{R}^{n-2}$ local charts of the two manifolds considered. Notice that the second one is the identity. Then,

$$\Phi:U\times\mathbb{R}^{n-2}\rightarrow S^2(r)\times\mathbb{R}^{n-2}$$

given by

\begin{eqnarray} \Phi(x,y)&=&(\phi(x),I(y))\\ &=&(\phi(x),y) \end{eqnarray} is a local chart to the manifold $S^2(r)\times\mathbb{R}^{n-2}$. We consider the metric product on this: once

$$T(S^2(r)\times\mathbb{R}^{n-2})=T(S^2(r))\times\mathbb{R}^{n-2}$$

a tangent vector $u\in T(S^2(r)\times\mathbb{R}^{n-2})$ can be decomposed as

$$u=(v,w)$$

with $v\in T(S^2(r))$ and $w\in\mathbb{R}^{n-2}$. Then, if $g$ is the product metric on the product $S^2(r)\times\mathbb{R}^{n-2}$,

$$g(u_1,u_2)=g_{S^2}(v_1,v_2)+\langle w_1,w_2\rangle$$

The important thing here is: with product of two metrics, the sectional curvature behaves in the following way (you need to prove this lemma, but it is straightforward): if we take $u_1=(v_1,w_1)$ and $u_2=(v_2,w_2)$ orthonormal vectors in $T(S^2(r))\times\mathbb{R}^{n-2}$, then

1. $K_M((v_1,0),(v_2,0))=K_{S^2(r)}(v_1,v_2)=\frac{1}{r^r}$
2. $K_M((v_1,0),(0,w_2))=0$
3. $K_M((0,w_1),(0,w_2))=K_{\mathbb{R}^{n-2}}(w_1,w_2)=0$

In another words, the curvature agrees with the decomposition of the tangent bundle. With this in mind, we have:

1. Ricci curvature is constant, because we can take a orthonormal basis of $TM$ consisting of two orthonormal tangent vectors of $S^2(r)$ and the remainning orthonormal tangent vectors from $\mathbb{R}^{n-2}$. It will be an orthonormal basis for $TM$ because we are using the product metric.
2. The scalar curvature is constant, once it is the average of Ricci curvatures.

The formulation in coordinates to the problem is important only to see the decomposition of the tangent bundle and to calculate the sectional curvatures.

-