The torus $\mathbb{T}^n$, the sphere $\mathbb{S}^n$ and the hyperbolic space $\mathbb{H}^n$ admit metrics of constant (sectional) curvature $0, 1, -1$ respectively. Do they afford metrics of constant curvature in $\{0, 1, -1\}$ for a different value? (e.g. does the torus admit a metric of curvature $-1$, or $1$?)


In the 2-dimensional case, the Gauss-Bonnet theorem answers the question negatively, but what about higher dimensions?



Why don't $\mathbb{T}^n, \mathbb{S}^n, \mathbb{H}^n$ admit other metrics of constant curvature?
The right thing to do is to restrict to compact manifolds of dimension $n\ge 2$. 
Then the answer is: If $M$ admits a metric of (possibly variable) sectional curvature $K<0$, or $K=0$ or $K>0$, then it does not admit a metric of curvature $K'$ with $K'<0, K'=0$ of $K'>0$, unless $KK'>0$ or $K=K'=0$. 


The proof is elementary Riemannian geometry, most of which you can find, say, in do Carmo's book "Riemannian geometry". For instance, it follows from Cartan-Hadamard theorem that a manifold with metric of nonpositive curvature has contractible universal cover. This takes care of $K>0$ (and $K'>0$). 


Consider a manifold $M$ admitting a flat metric. Then $\pi\_1(M)$ contains $Z^2$, since $M$ is covered by the $n$-torus. On the other hand, fundamental groups of compact manifolds of negative curvature cannot contain $Z^2$. Hence, you cannot have a compact manifold which admits both a flat metric and a metric of negative curvature. Actually, a bit more follows from the flat torus theorem: Every metric of nonpositive curvature on the $n$-torus has to be flat. 


