Hyperbolic structure on torus Is there an quick elementary argument of why the standard torus can not be equipped with a hyperbolic structure?
 A: If you do not like the answer given by Mike Miller in the comments, here is a "scissor and hyperbolic paper" explanation (maybe a bit sloppy and not the best, but maybe you can get the idea)
Assume you have a hyperbolic structure on the torus. Pick a point on the torus and cut along a simple loop passing through this point. As a result you get a cylinder, whose boundary consists of two copies of the loop and there two copies of the point on these two boundary loops. Next, cut along a simple arc connecting the two copies of the point. You obtain a quadrilateral (possibly with curvilinear edges) with hyperbolic structure on it. This means you can lay it "flat" (hm, sound like an oxymoron... maybe it is...) on the hyperbolic plane. The point initially fixed on the torus now turns into the four vertices of the quadrilateral in the hyperbolic plane. Let's call them $A, B, C, D$ ordered in a cyclic order, so that $AB, BC, CD$ and $DA$ are the pairs of vertices connected by (curved, non-geodesic) edges of the quadrilateral. Next, draw the unique hyperbolic geodesic $l_{AB}$ between $A$ and $B$, geodesic $l_{BC}$ between $B$ and $C$, geodesic $l_{CD}$ between $C$ and $D$ and geodesic $l_{CA}$ between $C$ and $A$. Now, as it turns out the two geodesics $l_{AB}$ and $l_{CD}$ have the same length, and the other two geodesics $l_{BC}$ and $l_{DA}$ also have the same length. Gluing back the quadrilateral now first along $l_{AB}$ and $l_{CD}$, and then along $l_{BC}$ and $l_{DA}$ leads to the same torus with the same hyperbolic structure. As a result of this gluing, the four points  $A, B, C, D$ come together to the initially chosen point on the torus and as the hyperbolic structure is "smooth" around each point, it is smooth around the fixed point in particular, so the four angles at the vertices of the geodesic version of the  quadrilateral $ABCD$ come together to form angle $2\pi$. But the geodesic quadrilateral $ABCD$ is in the hyperbolic plane and according to hyperbolic geometry, the sum of it's angles should be strictly less than $2\pi$, which is a contradiction.       
