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:

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

crystallographicusually 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.) $\endgroup$ – Jim Humphreys Feb 24 '17 at 17:462more comments