I'm new here, so if I need to correct something, let me know. Calculating some things I have noticed the following: If $f:(M,g_M)\to (N,g_N)$ is a differentiable mapping, where $g_N$ is a metric of $N$ and $g_M$ is a metric of $M$. Suppose that $M$ has constant curvature $K_M$ and if the pullback satisfies that $f^*g_N=\alpha\cdot g_M$, with $\alpha\in\mathbb R$, $\alpha \neq 0$ then $f(M)\subset N$ has constant curvature $K_N=K_M/\alpha$.
That's right? The calculations I have done in $\mathbb R^3$ tell me yes, but I have some questions, for example:
- What would the demonstration look like in general for Riemannian manifolds? Since in $\mathbb R^3$ I just went to play a little with the Gaussian curvature formulas.
- Is the $K_N=K_M/\alpha$ curvature of the $f(M)$ image a local curvature or is the curvature constant throughout the image?
I show you the example that I have in mind, I can take $f:\mathbb H_+^2=\{(u,v):u>0\}\subset\mathbb H^2\to\mathbb R^3$ such that $f(u,v)=(x_1,x_2,x_3)$ where $x_1=\displaystyle\frac{\cos(\alpha v)}{u}$, $x_2=\displaystyle\frac{\sin(\alpha v)}{u}$ and $x_3=\displaystyle\int_{1/\alpha}^{u}\frac{\sqrt{\alpha^2u^2-1}}{u^2}du$. Doing the calculations I get to $$dx_1^2+dx_2^2+dx_3^2=\alpha^2\left(\displaystyle\frac{du^2+dv^2}{u^2}\right).$$
I need to calculate the sectional curvature of the image $f(\mathbb H_+^2)\subset\mathbb R^3$, in this case it is simple because the sectional curvature coincides with the gaussian curvature, but considering the same application $f(u,v)=(x_1,x_2,x_3,x_4,x_5)$ defined analogously to above. The same would not happen anymore, and the calculation does not come out of the Riemann tensor or the Cristofell symbols, so I conjectured what I put at the beginning.
The answer to your conjecture is negative. Let $(X, g)$ be a manifold with constant sectional curvature. Let $\emptyset \ne U \subset X$ be open. Let $\varphi : X \to (0, \infty)$ be a smooth function such that $\varphi = 1$ on $\overline U$. Let $(M, g_M) = (U, g | _U)$ and $(N, g_N) = (X, \varphi g)$. Clearly, if $i : M = U \hookrightarrow X = N$ is the natural inclusion then $i^* g_N = g_M$ trivially. The curvature of $M = U$ is constant, but you can always play with $\varphi$ in order to make the metric $g_N$ not have constant curvature.