## Testing *My* Intuition (34): Tiling High Dimension with an Arbitrary Low-Dimensional Tile.

Test your intuition 34 asked the following:

A tile $T$ is a finite subset of $\mathbb Z^d$. We can ask if $\mathbb Z^d$ can or cannot be partitioned into copies of $T$. If  $\mathbb Z^d$ can be partitioned into copies of $T$ we say that $T$ tiles $\mathbb Z^d$.

Here is a simpe example. Let $T$ consists of 24 points of the 5 by 5 planar grid minus the center point. $T$ cannot tile $\mathbb Z^2$.

Test your intuition: Does $T$ tiles $\mathbb Z^d$ for some $d>2$?

We had a poll and 58% of voters said YES. The answer is

YES!

As a matter of fact Adam Chalcraft have made the beautiful conjecture that every tile $T$ in $\mathbb Z^d$ tiles $\mathbb Z^n$ for some large $n$.  This conjecture was proved by Vytautas Gruslys, Imre Leader,  and Ta Sheng Tan in their remarkable paper Tiling with arbitrary tiles.

Theorem (Vytautas Gruslys, Imre Leader,  and Ta Sheng Tan): Let $T \subset$  $\mathbb Z^d$ be a tile. Then $T$ tiles $\mathbb Z^n$ for some $n \ge d$.

But wait, what about our tile T? After seeing the abstract of Imre Leader’s lecture, looking briefly at the paper which contained the 5 by 5 minus the middle example, I posted the question on my blog.  But then driving to Jerusalem I suddenly was sure that there is no way in the world the hole in T can be filled up by another tile of the same shape.  T is simply too fat, I thought. I must have missed something –  some extra condition or subtelty that I overlooked. It turned out that my intuition was wrong already after I saw the right answer. (This does happen from time to time.)

When I had a chance I looked again at the paper, and saw a beautiful picture explaining how the hole can be filled in four dimension. (BTW, I don’t know what is the minimum dimension that T can tile.)

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