# Reflections along root vectors in a basis generate the Weyl group?

Let $$\Phi$$ be a root system of type ADE, $$\Lambda$$ be the lattice in the Euclidean space spanned by $$\Phi$$. If $$\Lambda$$ is spanned by $$\{v_i\}\subset \Phi$$ (as a $$\mathbb Z$$-module), and let $$\sigma_i$$ be the reflection along hyperplane associated to $$v_i$$. Is it true that $$\{\sigma_i\}$$ generate the Weyl group $$W(\Phi)$$?

My approach

I know there is a proposition says reflections along vectors in a base (of a root system, or in some book called simple system) generate the full $$W(\Phi)$$. So I tried to show $$\{v_i\}$$ contains a base, but have no idea how to do this. More precisely, I don't know how to use the condition $$\{v_i\}$$ generate $$\Lambda$$.

If this too general, what about the case $$\Phi=E_6$$? If $$6$$ vectors $$v_1,\ldots,v_6$$ in the root system generate the lattice, is it true that after possibally replacing $$v_i$$ by $$-v_i$$ for some $$i$$, they will form a base of $$E_6$$?

• A root system should appear somehow, "spanned by $\{v_i\}$..." is too general, this comment tries to be shorter than the question, not so simple... – dan_fulea Sep 15 '18 at 3:16
• @dan_fulea Yes you are right, otherwise there are trivial counter-examples. I added some conditions. – Akatsuki Sep 15 '18 at 3:55
• Why the restriction to types ADE? Do you have counterexamples for the other cases -- maybe they are worth a look. – Torsten Schoeneberg Oct 9 '18 at 19:39
• @TorstenSchoeneberg Because otherwise there will be trivil counterexample. For example in $B_2$ the two minimal vectors form a basis of the lattice but did not generate the Weyl group. – Akatsuki Oct 10 '18 at 6:53

Let $$\Phi'$$ be the root subsystem of $$\Phi$$ generated by $$v_1,\ldots,v_n$$ where $$n$$ is the rank of $$\Phi$$. Then, $$\Phi'$$ itself also has rank $$n$$ (because the $$v_1,\ldots,v_n$$ are linearly independent). Because $$\Phi$$ is simply laced, it follows that $$\Phi=\Phi'$$. Indeed, the Dynkin diagram of $$\Phi'$$ is a subdiagram of the one of $$\Phi$$ (each time with respect to some base) and both have $$n$$ vertices. Since there are only simple edges, we conclude that the diagrams are equal. It follows that $$v_1,\ldots,v_n$$ form a base up to sign. By replacing $$v_i$$ with $$-v_i$$ if necessary, which does not change the reflection along $$v_i$$, we may assume that $$v_1,\ldots,v_n$$ form a base of $$\Phi$$. Hence, $$s_{v_1},\ldots,s_{v_n}$$ generate $$W$$.
• What is $v_i$? Is it the same as $v_i$ in the question setting? – Akatsuki Oct 29 '18 at 9:02
• Yes, the $v_i$'s are in $\Phi$ and span $\Lambda$ as a $\mathbb{Z}$-module - as in the question. – user213008 Oct 29 '18 at 10:58
• Then how do you pick $n$ of them? – Akatsuki Oct 29 '18 at 11:49
• $n$ is the rank of the lattice $\Lambda$. If $v_i$ span the lattice, there must be at least $n$ of them (invariant basis number for commutative rings). If there are more, you can pick $n$ which form a basis. – user213008 Oct 29 '18 at 12:02
• Sorry I still cannot follow. Why $\{v_i\}$ forms a basis (I suppose you mean base, i.e. simple system) up to sign? – Akatsuki Oct 29 '18 at 13:05