### David Roberts

Shared publicly -Take a rectangular sheet of paper, and glue two opposite sides of the rectangle to each other to make a cylinder. Now glue the other two opposite sides to each other — you'll probably have to crumple or fold the paper to do it, because a round cylinder won't bend around smoothly to let you do this step. The result is a surface with the topology of a torus, but (using distances measured in terms of paths within the surface, not shortcutting through the space around it) the geometry of a flat Euclidean plane everywhere — the points in the middle of the rectangle already started with this Euclidean geometry, but the gluing process made it so that the points that were originally on the boundary of the rectangle look just the same, so they also have flat Euclidean neighborhoods in the torus. You can do the same gluing process with a square, a parallelogram, or a regular hexagon, and in each case get a flat torus.

Flat tori have nice smooth embeddings in four-dimensional space (take the Cartesian product of two perpendicular circles) but some crumpling seems to be required in 3d. Nevertheless, they can be embedded into 3d, and this video shows one way to do it using fractal crumpling rather than folds.

Via https://plus.google.com/+KevinClift/posts/N9Qx8JYRfnZ and https://plus.maths.org/content/abel-prize-2015-all-wrapped which have more on how this sort of embedding problem relates to the recent award of the Abel Prize to John Nash and Louis Nirenberg.

Flat tori have nice smooth embeddings in four-dimensional space (take the Cartesian product of two perpendicular circles) but some crumpling seems to be required in 3d. Nevertheless, they can be embedded into 3d, and this video shows one way to do it using fractal crumpling rather than folds.

Via https://plus.google.com/+KevinClift/posts/N9Qx8JYRfnZ and https://plus.maths.org/content/abel-prize-2015-all-wrapped which have more on how this sort of embedding problem relates to the recent award of the Abel Prize to John Nash and Louis Nirenberg.

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