Consider the manifold $P:=S^2 \times \mathbb{R}^2$ equipped with the product metric
$g((x,y),(x',y'))=g_{S^2}(x,x')+g_{\mathbb{R}^2} (y,y')$ where $S^2$ has a constant curvature that is 1. Let $z_1,z_2,z_3,z_4$ be an orthonormal basis of the tangent space such that $z_1 , z_2$ is a basis of the tangent space of $S^2$ and $z_3, z_4$ is one of the tangent space of $\mathbb{R}^2$.
Now what are the surface curvatures of $\sigma_{ij}=< z_i, z_j >_{i\neq j}$ which means what is
$K(z_i,z_j)=\frac{g(R(z_i,z_j)z_j,z_i)}{\|z_i \wedge z_j\|^2}_{i\neq j}$?
Hint: What's the curvature of $\mathbb{R}^2$ ??
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Alright firstly I know that the curvature of $\mathbb{R}^2$ is 0 . Then I suspect that the whole exercise points on using the Gauss-Codazzi equation, right?
So let $\bar{R}$ denote the Riemann curvature of the submanifold $\sigma_{ij}$ and $R$ the Riemann curvature of the Manifold P. Then
$K(\sigma_{ij})=g(\bar{R}(z_i,z_j)z_j,z_i)=g(R(z_i,z_j)z_j,z_i)-g(II(z_i,z_i),II(z_j,z_j))+g(II(z_j,z_i),II(z_i,z_j))=(*)$
Here I don't see how to get on. I guess what comes out at (*) is just
$g(R(z_i,z_j)z_j,z_i)$
But why is
$g(II(z_i,z_i),II(z_j,z_j))=g(II(z_j,z_i),II(z_i,z_j))$ ? How does orthogonality come into play here?
