Infinite root systems

Question

The definition of root system on Wikipedia specifies that a root system must be a finite set of vectors. If we relax this restriction (while still in finite dimensions, i.e. $\Bbb R^n$), do we get any new root systems? Or to rephrase without referencing the definition of "root system":

Does there exist a set $\Phi\subseteq\Bbb R^n$ such that:

1. $\Phi$ is a spanning set
2. If $x\in\Phi$ then $ax\in\Phi\to a=\pm1$
3. If $x,y\in\Phi$ then $\langle y,x\rangle :=2\dfrac{(x,y)}{(x,x)}$ is an integer, and $y-\langle y,x\rangle x\in\Phi$
4. $\Phi$ is infinite?

Obviously the answer is no in dimensions $0,1$, but even in dimension $2$ I'm not seeing any clear argument for or against. The strict constraints on possible finite solutions suggests to me that no infinite solutions exist, but I'm not clear on the usual argumentation here and I can't see any nontrivial reductions of the problem.

Answer 1

No. It's still true that there are only a finite set of possible angles between two roots in a root system. Given infinitely many vectors in $\mathbb{R}^n$ (which aren't scalar multiples of each other), the angles between them must get arbitrarily small (basically because the unit sphere $S^{n-1}$ is compact).

Answer 2

For a positive definite bilinear form, the classified finite root systems are the only ones that can exist (up to isomorphism). If we allow arbitrary, possibly degenerate, bilinear forms we can get infinite root systems that satisfy these axioms. These correspond to the root systems of infinite dimensional Kac-Moody Lie algebras. If we then drop the condition that $\langle x,y\rangle$ be an integer, we essentially obtain the concept of the root system of an arbitrary finitely generated Coxeter group, which is usually infinite.