Let $\mathbb{E} \cong \mathbb{R}^n$ be a vector space with an inner product $\langle\cdot,\cdot\rangle$. A subset $R \subset \mathbb{E} \setminus \{0\}$ is called a root system, if $R$ has the following properties
- $R$ is finite and spans $\mathbb{E}$
- if $\alpha \in R$ then $-\alpha \in R$ and $\pm \alpha$ are the only multiples of $\alpha$ in $R$
- R is invariant under the reflection in the hyperplane orthogonal to any $\alpha \in R$, i.e, $\forall_{\alpha, \beta} \in R, \; s_\alpha(\beta) = \beta - 2\operatorname{proj}_\alpha(\beta) \in R$
- $\forall \alpha, \beta \in R, \; 2 \cdot \frac{\langle\beta, \alpha\rangle}{\langle\alpha,\alpha\rangle} \in \mathbb{Z}$
I'm having trouble coming up with examples of such vector subspaces. I tried $R = \{(1,0), (-1,0), (0,1),(0,-1)\}$ where $(a,b)$ is vector (I don't know what other notation to use for a vector since the usual notation is being used by the inner product) but this subspace didn't satisfy the invariant condition.
I actually see that for $n=1$, $R = \{\alpha, - \alpha\}$ works. I think that is it for $n=1$ though.
Can anyone come up with some examples so i can see what is going on?
Your $R$ should have worked, you probably messed up computations. In general, let $\mathcal{B} = (e_1,\ldots,e_n)$ be an orthonormal basis of $\Bbb E$ and put $$R = (e_1,-e_1,\ldots,e_n,-e_n).$$Conditions 1 and 2 are automatic. For 3, we reason as follows: fix $e_i$. We have $$s_{e_i}(\pm e_j) = \pm e_j - 2\frac{\langle \pm e_j,e_i\rangle}{\langle e_i,e_i\rangle}e_i = \pm e_j \mp 2\delta_{ij}e_i = \begin{cases} \pm e_j \mbox{ if }j\neq i, \\ \mp e_i \mbox{ if }j=i. \end{cases}$$Similarly$$s_{-e_i}(\pm e_j) = \pm e_j - 2\frac{\langle \pm e_j,-e_i\rangle}{\langle -e_i,-e_i\rangle}(-e_i) = \pm e_j \mp 2\delta_{ij}e_i = \begin{cases} \pm e_j \mbox{ if }j\neq i, \\ \mp e_i \mbox{ if }j=i. \end{cases}$$Condition 4 also follows from the above.