crumbled paper is Eucleadian or non Euclidean? Greeting.
I was reading about relativity theory and started thinking about the following.
if I have an A4 flat paper and draw a right triangle on it. It is flat and satisfy the Euclidean geometry. If I make the paper wiggle or curvy shape, still have my triangle the same so I would say the Eucleadian metric still in effect, and the space still flat Eucleadian.
Now if I crumbled or crushed the paper in ball like shape, my tringle still in there, but
0- did my triangle changed? does it still have same area and side lengths?.
1- did I change the metric?
2-does the curvature tensor equal zero?
or what is the resulting space? curved or flat?
3- would the metric stay the same, or would it be something different? is it smooth? if not smooth, can we smooth it out? if yes, what's the name of the technique.?
please correct me if I am wrong at anything.
and sorry for the tons of questions
thanks
 A: There are different mathematical models of crumbled paper, you can find one in these slides of Robert Kohn. Below is a model which is much less sophisticated. 
A flat piece of paper is a region $R\subset E^2$ bounded by a simple polygonal loop. A crumbled piece of paper is a 1-1 continuous map $f: R\to E^3$ such that there is a finite piecewise-geodesic subgraph $G\subset R$ such that  $f$ restricts to a distance-preserving map on each component of $R- G$. 
In particular, $f$ preserves lengths of paths in $R$: If $c: [0,1]\to R$ is a path then the length of $c$ equals the length of $f\circ c$. In this sense the metric (understood as the "interior" or the "path-metric") on the crumbled piece of paper did not change. In other words, the intrinsic geometry of $R$ is left unchanged by $f$. Accordingly, the area did not change either: $Area(R)=Area(f(R))$. 
Of course, the extrinsic geometry of $R$ did change a lot. For instance, typically, for points $p, q$ in distinct components of $R-G$ we will have
$$
||f(p)- f(q)||< ||p-q||,
$$
where $||v||$ is the Euclidean norm of a vector $v$. The quantity 
$||f(p)- f(q)||$ is the extrinsic metric on $f(R)$. 
Now, to your questions:
0- did my triangle changed? YES and NO: The intrinsic geometry did not change but the extrinsic geometry did. 
does it still have same area and side lengths? YES
1- did I change the metric? Intrinsic - NO, Extrinsic -YES. 
2-does the curvature tensor equal zero?  MEANINGLESS: The surface $f(R)$ is not smooth, so you cannot talk about its curvature tensor.
or what is the resulting space? curved or flat? It is intrinsically flat. 
3- would the metric stay the same, or would it be something different? 
See above. 
is it smooth? Meaningless. There is no Riemannian metric tensor to speak of, so I do not see what would "smooth metric" even mean in this context.
if not smooth, can we smooth it out? if yes, what's the name of the technique.?
Yes, we can. For instance, you can take some small $\epsilon>0$ and replace $f(R)$ with its $\epsilon$-neighborhood. This will be a $C^1$-smooth surface, so you can talk about its metric tensor  and even its curvature (in the distributional sense). There are other ways to approximate Lipschitz maps such as $f$ by smooth maps. Feel free to explore. Some mathematicians spend their lives doing exactly this. 
Caveat: Geometers use the word "metric" to denote different things, for instance, a (pseudo)Riemannian metric is the metric tensor of suitable signature on a smooth manifold. On the other hand, in metric geometry the word "metric" means a distance function on a set satisfying some axioms (see here).  You should think what "metric" you are actually interested in. 
