Prove that it isn't possible to flatten a Teserract into 2-d space.

Question

Its certainly possible to flatten a Teserract into 3-d space (see: What does a flattened Teserract look like?). But what about 2-d space? It doesn't seem possible without having some faces fall on top of each other. But can we prove this? Alternately, can someone draw out an unfolding with no overlapping faces and prove this conjecture wrong via counter example?

Answer 1

Consider the tesseract in four-dimensional Cartesian space with vertices at $(v_1,v_2,v_3,v_4)$ for all possible combinations of $v_1,v_2,v_3,v_4 \in \{0, 1\}.$ That is, it is a tesseract of side $1$ with edges parallel to the coordinate axes, with one vertex at $(0,0,0,0)$ and the diagonally opposite vertex at $(1,1,1,1).$

The $24$ square faces of this tesseract can be unfolded as shown below.

In this figure each of the four vertices of each square is labeled with the coordinates of the vertex, although for readability of the figure the parentheses around the coordinates and the commas between coordinates have been omitted.

The red lines show where faces that happen to lie adjacent to each other in the unfolding are not connected along that edge in the original tesseract. The non-connection of these faces can also be deduced from their coordinates, but the red lines are meant to be a helpful reminder. In all other places where squares are adjacent along an edge, the squares are connected along that edge in the original tesseract as well.

I derived this unfolding from a fully connected diagram of a tesseract by simple observation, not by any clever algorithm. There are probably much neater, more symmetric unfoldings.

Answer 2

I'm going to attempt an outline of a proof. Imagine two cubes that share a common face.

Claim: the only way to flatten this structure is along the common plane the two cubes share.

The reason is that if we take any other plane, we start with 3 faces making a T-shape (two bottom faces of the two cubes and the common boundary face). And this itself is impossible to flatten without overlap.

This means that if there are three cubes now where one pair share a common face along the x-y plane and the other pair share a common face (say) along the y-z plane, unfolding should become impossible. Which plane would you choose for unfolding onto?

Now, just look at all the unfoldings of the Teserract along 3-d space (see link in the question). In each of them, this configuration of three cubes where the common faces are orthogonal happens somewhere.

Hence, unfolding a Teserract to 2-d space without overlapping faces is impossible.

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EDIT: This proof is incorrect. It actually is possible to unfold along the x-y plane (which contains the two unshaded faces) by moving the shaded face away when you unfold one of the cubes. I'm leaving it in to demonstrate the pitfall of this line of reasoning. And also because there is some hope if can be saved.