# Non-reduced, non-crystallographic root systems

Let $V$ be a $n$-dimensional real vector space with standard inner product $(\cdot,\cdot)$. For any $\alpha \neq 0 \in V$, set $\alpha^\vee := \frac{2}{(\alpha,\alpha)}\alpha$. For $\alpha \neq 0,\beta \in V$ set $n_{\alpha}(\beta) := (\beta,\alpha^\vee)$ and $s_\alpha(\beta) := \beta - n_{\alpha}(\beta)\cdot\alpha$.

A root system in $V$ is a finite set $\Phi$ of non-zero vectors in $V$ satisfying:

1. $\Phi$ spans $V$;
2. $s_{\alpha}(\Phi) = \Phi$ for any $\alpha \in \Phi$;
3. $\mathrm{Span}\{\alpha\} \cap \Phi = \{\alpha,-\alpha\}$ for any $\alpha \in \Phi$;
4. $n_{\alpha}(\beta) \in \mathbb{Z}$ for any $\alpha,\beta \in \Phi$.

This is by now a standard definition and there is a very satisfying classification theory for root systems based on Dynkin diagrams (the so-called Cartan-Killing classification).

Nevertheless, sometimes small modifications to this definition are considered. For instance, sometimes condition 4 above is omitted and root systems satisfying condition 4 are called crystallographic. However, considering non-crystallographic root systems doesn't change much: there are only a few more families of irreducible non-crystallographic root systems.

Similarly, sometimes condition 3 is omitted and root systems satisfying condition 3 are called reduced. Once again this does not change the structure theory so much: from 4 it follows that for any $\alpha \in \Phi$ we have $\mathrm{Span}\{\alpha\} \cap \Phi \subseteq \{2\alpha,\alpha,-\alpha\,-2\alpha\}$, and I think then it is not hard to show that any irreducible non-reduced root system is of the form $A \cup B \cup 2A$ where $A\cup B$ and $2A \cup B$ are irreducible reduced root systems (see Proposition 13, Section 1.4, Chapter VI of Bourbaki's "Lie Groups and Lie Algebras").

I wonder if anyone has ever considered what happens when we eliminate both 3 and 4 from the above. Now things get a bit worse: even in rank one (i.e. $n=1$) there are infinitely many different root systems- any symmetric set of finite vectors in $\mathbb{R}^1$ is a root system by this definition. From what I can gather from some quick searches on the internet, nobody ever tries to remove both 3 and 4 from the definition of root systems, and maybe that's because the resulting theory is horrible. However, I wonder if this is really the case: is there a nice structural classification of non-reduced, non-crystallographic root systems or not?

• The formulation is confusing to me. For instance, what does the term crystallographic usually mean? (Also, note that many authors including Bourbaki do not require in advance that $V$ has a given inner product, instead defining $\alpha^\vee$ to be a suitable element of the dual space $V^*$. My own identification of $V$ with $V^*$ arose from the limited context of traditional Lie algebra theory.) – Jim Humphreys Feb 24 '17 at 17:46
• Crystallographic usually means what I said- the inner product of any root with any coroot is an integer. (It's true you can avoid identifying V with its dual but anyways I hope this more naive definition still gets across what I am attempting to ask.) – Sam Hopkins Feb 24 '17 at 17:55
• For example, Wikipedia makes the same remarks about "crystallographic" and "reduced" that I have (subject to a renumbering of the properties): en.wikipedia.org/wiki/Root_system#Definition – Sam Hopkins Feb 24 '17 at 18:07
• I think a non-crystallographic root system should be thought of as a collection of rays (or points of the corresponding projective space). It's only for crystallographic root systems that we can start really caring about the length of the roots, so that the reduced/nonreduced distinction makes sense. In any case, the Weyl group should always be a Coxeter group, and these are classified, so you can't expect anything really new. – Gro-Tsen Feb 24 '17 at 18:18
• @Sam Hopkins: I wouldn't rely too much on Wikipedia entries for advanced mathematics (especially the anonymous ones). As far as I know, the integrality condition is usually imposed, leading to the traditional matrix of Cartan integers. (But there are by now varying notions of "root system" in the literature.) The notion of "crystallographic" root system depends on context, but usually one requires that the product $s_\alpha s_\beta$ of any two reflections has order 2, 3, 4, or 6. – Jim Humphreys Feb 24 '17 at 19:28

Let $\Phi_0$ be the set of normalized roots $\frac{\alpha}{||\alpha||}$, $\alpha\in\Phi$. This is a root system satisfying 1, 2 and 3. Thus it is in the list of not necessarily crystallographic root systems. To reconstruct $\Phi$ we need for every $\alpha_0\in\Phi_0$ the length spectrum $L(\alpha_0)=\{||\alpha||\mid\alpha\in\mathbb{R}_{>0}\alpha_0\}\subseteq\mathbb{R}_{>0}$. These spectra can be chosen arbitrarily, subject to the condition that $$L(w\alpha_0)=L(\alpha_0)\text{ for all w in the Weyl group W of \Phi_0}$$ Thus root systems with 1 and 2 are classified by a non-crystallographic root system and a finite set of positive numbers for each $W$-conjugacy class of roots.
This can be made more concrete. For this, let $S$ be a set of simple roots of $\Phi_0$. Attached to it is a labeled graph where the label $n_{\alpha\beta}$ above the edge between $\alpha$ and $\beta$ indicates the order of $s_\alpha s_\beta$. Each root is conjugate to a simple root. So it suffices to know $L(\alpha)$ for $\alpha\in S$. If the $n_{\alpha\beta}$ is odd then $\alpha$ is conjugate to $\beta$ (easy exercise). Thus we need $$\alpha,\beta\in S\text{ with }n_{\alpha\beta}\text{ odd}\Rightarrow L(\alpha)=L(\beta).$$ A not so easy theorem on Coxeter groups (somewhere in Bourbaki) implies that this condition above is also sufficient.
• Hmm... I now realize that I am not sure what the proper notion of isomorphism is for non-crystallographic root systems (and thus in what sense they are classified). For crystallographic root systems, one normally requires an isomorphism of root systems to preserve the numbers $n_{\alpha}(\beta)$. But as Gro-Tsen points out in another comment, it makes sense to think of a non-crystallographic root system just as a collection of rays, in which case we might require only that the isomorphism preserve angles between vectors- but with this notion of isomorphism $B_n$ and $C_n$ would be isomorphic. – Sam Hopkins Feb 25 '17 at 23:04
• Yes, $B_n$ and $C_n$ are isomorphic as non-cristallographic root systems. If you look at a list you'll see that $C_n$ is missing or there is an entry $BC_n$ instead of $B_n$ and $C_n$. – Friedrich Knop Feb 26 '17 at 16:52