Construction of A-Series Root Systems

Starting from our setup, we construct root systems corresponding to the $A_n$ Dynkin diagrams.

We start with the $n+1$ -dimensional space $\mathbb{R}^{n+1}$ with orthonormal basis $\{\epsilon_0,\dots,\epsilon_n\}$ , and cut out the $n$ -dimensional subspace $E$ orthogonal to the vector $\epsilon_0+\dots+\epsilon_n$ . This consists of those vectors $v=\sum_{k=0}^nv^k\epsilon_k$ for which the coefficients sum to zero: $\sum_{k=0}^nv^k=0$ . We let $J=I\cap E$ , consisting of the lattice vectors whose (integer) coefficients sum to zero. Finally, we define our root system $\Phi$ to consist of those vectors $\alpha\in J$ such that $\langle\alpha,\alpha\rangle=2$ .

From this construction it should be clear that $\Phi$ consists of the vectors $\{\epsilon_i-\epsilon_j\vert i\neq j\}$ . The $n$ vectors $\Delta=\{\alpha_i=\epsilon_{i-1}-\epsilon_i\}$ are independent, and thus form a basis of the $n$ -dimensional space $E$ . This establishes that $\Phi$ spans $E$ . In particular, if $i<j$ we can write

$\displaystyle(\epsilon_i-\epsilon_j)=(\epsilon_i-\epsilon_{i+1})+(\epsilon_{i+1}-\epsilon_{i+2})+\dots+(\epsilon_{j-1}-\epsilon_j)$

showing that $\Delta$ forms a base for $\Phi$ .

We calculate the Cartan integers for this base

$\displaystyle\frac{2\langle\epsilon_{j-1}-\epsilon_j,\epsilon_{i-1}-\epsilon_i\rangle}{\langle\epsilon_{i-1}-\epsilon_i,\epsilon_{i-1}-\epsilon_i\rangle}=\langle\epsilon_{j-1}-\epsilon_j,\epsilon_{i-1}-\epsilon_i\rangle$

For $i=j$ we get the value ${2}$ ; for $j+1$ or $j-1$ we get the value $-1$ ; otherwise we get the value ${0}$ . This clearly gives us the Dynkin diagram $A_n$ .

Finally, the reflections with respect to the $\alpha_i$ should generate the entire Weyl group. We must verify that these leave the lattice $J$ invariant to be sure that we have a root system. We calculate

$\displaystyle\begin{aligned}\sigma_{\alpha_i}(v)&=v-\frac{2\langle v,\alpha_i\rangle}{\langle\alpha_i,\alpha_i\rangle}\alpha_i\\&=v-\langle v,\alpha_i\rangle\alpha_i\\&=v-(v^{i-1}-v^i)(\epsilon_{i-1}-\epsilon_i)\\&=v-(v^{i-1}\epsilon_{i-1}+v^i\epsilon_i)+(v^i\epsilon_{i-1}+v^{i-1}\epsilon_i)\end{aligned}$

That is, it swaps the coefficients of $\epsilon_{i-1}$ and $\epsilon_i$ , and thus sends the lattice $J$ back to itself, as we need.

We can also see from this effect that any combination of the $\sigma_{\alpha_i}$ serves to permute the $n+1$ coefficients of a given vector. That is, the Weyl group of the $A_n$ system is naturally isomorphic to the symmetric group $S_{n+1}$ .

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March 2, 2010 - Posted by John Armstrong | Geometry, Root Systems

4 Comments »

[...] construction is similar to that of the series, but instead of starting with a hyperplane in -dimensional space, we just start with -dimensional [...]

Pingback by Construction of D-Series Root Systems « The Unapologetic Mathematician | March 3, 2010 | Reply
[...] to the series, and exactly as in the series, we define for . This time, though, to get vectors whose [...]

Pingback by Construction of B- and C-Series Root Systems « The Unapologetic Mathematician | March 4, 2010 | Reply
[...] root system is, as we can see by looking at it, closely related to the root system. And so we start again with the -dimensional subspace of consisting of vectors with coefficients [...]

Pingback by Construction of the G2 Root System « The Unapologetic Mathematician | March 8, 2010 | Reply
There is a small type in the line “From this construction it should be clear that $\phi$ consists of the vectors..you missed a – sign there, and a pipe further down the line.

Comment by Sangeeta Bhatia | July 28, 2012 | Reply

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