# Is the chromatic number of the real plane invariant under the norm?

Recall that chromatic number of $\mathbb{R}^2$ is the least $n$ such that there exists a function $f$ from $\mathbb{R}^2$ into a set of colors ${C_1,\ldots,C_n}$ with $f(x)\neq f(y)$ for $||x-y||_2=1$.

As far as I know, the problem which number this exactly is is still open. I was wondering whether this number is invariant under the norm $||\cdot||$ that is chosen.

-
Could you kindly remind us what are the known bounds on the chromatic number? –  Joel David Hamkins Jan 3 '11 at 14:38
–  Qiaochu Yuan Jan 3 '11 at 14:41
In any case, I don't see any good reason for this to be true. The corresponding graphs are certainly not isomorphic: the clique number of the graph with the Euclidean norm is 3 but the clique number of the graph with the sup norm is at least 4. –  Qiaochu Yuan Jan 3 '11 at 15:51

For the $L^\infty$ norm (or equivalently the $L^1$ norm rotating by 45 degrees) the chromatic number is easy to calculate: 4. Just colour translates of $[0,1)\times[0,1)$ by $\mathbb Z^2$ in a $2\times 2$ repeating pattern. It must be at least 4 because ${0,1}\times{0,1}$ are all distance 1 apart.
There's certainly no simple periodic colouring with 4 colours that will work for the $L^2$ norm.
This is a good response, but it should be remarked that as far as we know the chromatic number in the $L^2$ norm could be $4$. I think it is also a worthwhile question if on more general spaces are there any monotonicity/convexity properties of the chromatic number in the $L^p$ norm (as $p$ varies). –  GH from MO Jan 3 '11 at 21:06
Yes, its not quite an answer because it could still be equal. But since a proof of the equality would prove that the chromatic number of $\mathbb{R}$ equals 4, I assume that nothing of that kind can be said yet. Thus, I accept it as an answer, thank you. –  Christoph-Simon Senjak Jan 4 '11 at 21:37