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 $E_6$?

  • $\begingroup$ 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... $\endgroup$ – dan_fulea Sep 15 '18 at 3:16
  • $\begingroup$ @dan_fulea Yes you are right, otherwise there are trivial counter-examples. I added some conditions. $\endgroup$ – Akatsuki Sep 15 '18 at 3:55
  • $\begingroup$ Why the restriction to types ADE? Do you have counterexamples for the other cases -- maybe they are worth a look. $\endgroup$ – Torsten Schoeneberg Oct 9 '18 at 19:39
  • 1
    $\begingroup$ @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. $\endgroup$ – 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 $W$.

  • $\begingroup$ What is $v_i$? Is it the same as $v_i$ in the question setting? $\endgroup$ – Akatsuki Oct 29 '18 at 9:02
  • $\begingroup$ Yes, the $v_i$'s are in $\Phi$ and span $\Lambda$ as a $\mathbb{Z}$-module - as in the question. $\endgroup$ – user213008 Oct 29 '18 at 10:58
  • $\begingroup$ Then how do you pick $n$ of them? $\endgroup$ – Akatsuki Oct 29 '18 at 11:49
  • $\begingroup$ $n$ 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. $\endgroup$ – user213008 Oct 29 '18 at 12:02
  • $\begingroup$ Sorry I still cannot follow. Why $\{v_i\}$ forms a basis (I suppose you mean base, i.e. simple system) up to sign? $\endgroup$ – Akatsuki Oct 29 '18 at 13:05

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