The Unapologetic Mathematician

Mathematics for the interested outsider

Construction of D-Series Root Systems

Starting from our setup, we construct root systems corresponding to the D_n Dynkin diagrams (for n\geq4).

The construction is similar to that of the A_n series, but instead of starting with a hyperplane in n+1-dimensional space, we just start with n-dimensional space itself with the lattice I of integer-coefficient vectors. We again take \Phi to be the collection of vectors \alpha\in I of squared-length 2: \langle\alpha,\alpha\rangle=2. Explicitly, this is the collection of vectors \pm(\epsilon_i\pm\epsilon_j) for i\neq j, where we can choose the two signs independently.

Similarly to the A_n case, we define \alpha_i=\epsilon_i-\epsilon_{i+1} for 1\leq i\leq n-1, but these can only give vectors whose coefficients sum to {0}. To get other vectors, we throw in \alpha_n=\epsilon_{n-1}+\epsilon_n, which is independent of the others. The linearly independent collection \Delta=\{\alpha_i\} has n vectors, and so must be a basis of the n-dimensional space.

As before, any vector in \phi of the form \epsilon_i-\epsilon_j for i<j can be written as

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

while vectors of the form \epsilon_i+\epsilon_j are a little more complicated. We can start with

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

and from this we can always build 2\epsilon_j=(\epsilon_j-\epsilon_n)+(\epsilon_j+\epsilon_n) for 1\leq j\leq n-1. Then if i<j\leq n-1 we can write \epsilon_i+\epsilon_j=(\epsilon_i-\epsilon_j)+2\epsilon_j. This proves that \Delta is a base for \Phi.

Again, we calculate the Cartan integers. The calculation for i and j both less than n is exactly as before, showing that these vectors form a simple chain in the Dynkin diagram of length n-1. However, when we involve \alpha_n we find

\displaystyle\frac{2\langle\epsilon_i-\epsilon_{i+1},\epsilon_{n-1}+\epsilon_n\rangle}{\langle\epsilon_i-\epsilon_{i+1},\epsilon_i-\epsilon_{i+1}\rangle}=\langle\epsilon_i-\epsilon_{i+1},\epsilon_{n-1}+\epsilon_n\rangle

For i<n-2, this is automatically {0}; for i=n-2, we get the value -1; and for i=n-1 we again get {0}. This shows that the Dynkin diagram of \Delta is D_n.

Finally, we consider the reflections with respect to the \alpha_i. As in the A_n case, we find that \sigma_{\alpha_i} swaps the coefficients of \epsilon_i and \epsilon_{i+1} for 1\leq i\leq n-1. But what about \alpha_n?

\displaystyle\begin{aligned}\sigma_{\alpha_n}(v)&=v-\frac{2\langle v,\alpha_n\rangle}{\langle\alpha_n,\alpha_n\rangle}\alpha_n\\&=v-\langle v,\alpha_n\rangle\alpha_n\\&=v-(v^{n-1}+v^n)(\epsilon_{n-1}+\epsilon_n)\\&=v-\langle v,\alpha_n\rangle\alpha_n\\&=v-(v^{n-1}\epsilon_{n-1}+v^n\epsilon_n)-(v^n\epsilon_{n-1}+v^{n-1}\epsilon_n)\end{aligned}

This swaps the last two coefficients of v and flips their sign. Clearly, this sends the lattice I back to itself, showing that \Phi is indeed a root system.

Now we can use \sigma_{\alpha_n}\sigma_{\alpha_{n-1}} to flip the signs of coefficients of v, two at a time. We use whatever of the \sigma_{\alpha_i} we need to get the two coefficients we want into the last two slots, hit it with \sigma_{\alpha_n}\sigma_{\alpha_{n-1}} to flip them, and then invert the first permutation to move everything back where it started from. In fact, this is a lot like what we saw way back with the Rubik’s cube, when dealing with the edge group. We can effect whatever permutation we want on the coefficients, and we can flip any even number of them.

The Weyl group of D_n is then the subgroup of the wreath product S_n\wr\mathbb{Z}_2 consisting of those transformations with an even number of flips coming from the \mathbb{Z}_2 components. Explicitly, we can write \mathbb{Z}_2^{n-1} as the subgroup of \mathbb{Z}_2^n with sum zero. Then we can let S_n act on \mathbb{Z}_2^n by permuting the components, and use this to give an action of S_n on \mathbb{Z}_2^{n-1}, and thus form the semidirect product S_n\ltimes\mathbb{Z}_2^{n-1}.

March 3, 2010 Posted by | Geometry, Root Systems | 2 Comments

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}.

March 2, 2010 Posted by | Geometry, Root Systems | 3 Comments

Construction of Root Systems (setup)

Now that we’ve proven the classification theorem, we know all about root systems, right? No! All we know is which Dynkin diagrams could possibly arise from root systems. We don’t know whether there actually exists a root system for any given one of them. The situation is sort of like what we found way back when we solved Rubik’s magic cube: first we established some restrictions on allowable moves, and then we showed that everything else actually happened.

And so we must construct some actual root systems. For this task, we let E stand for a finite-dimensional real vector space \mathbb{R}^n for various n, equipped with its usual inner product. We pick an orthonormal basis \{\epsilon_1,\dots,\epsilon_n\} and let the integral linear combinations of these basis vectors form the lattice I. Here, I do not mean “lattice” in the order-theory sense. I mean that this is a discrete collection of points in the vector space that is closed under addition.

In every case we’re going to take either the lattice I, or a slightly modified lattice J. We’ll define our root system \Phi to be the collection of vectors in the lattice of either one or two specified lengths (since there can be at most two root lengths). That is, we’re considering the intersection of a discrete collection of points with one or two spheres. These spheres are closed and bounded, and thus compact. The collection \Phi must be finite or else it would have an accumulation point by Bolzano-Weierstrass, and thus wouldn’t be discrete!

Any one of our constructed collections will span E, and in fact an explicit basis will be shown in each case, in case it’s not clear. It should also be clear that none of them can contain the vector {0}, and so the first condition of being a root system will hold. Our choice of lengths will make it clear that there are no possible scalar multiples of a root besides itself and its negative. On the other hand, it should be clear that if \alpha is in a lattice I and on a sphere \lVert\alpha\rVert^2=r^2, then -\alpha is also in both, and thus the second condition holds.

The reflection \sigma_\alpha preserves lengths, and so it sends the spheres back to themselves. We’ll have to check in each case that \sigma_\alpha sends every vector in our collection back into the lattice, which will establish the third condition.

As to the fourth condition, the inner product \langle\alpha,\beta\rangle is automatically going to be in \mathbb{Z} when we pick \alpha and \beta from a lattice, and so picking the squared radii of our spheres to divide 2 should be enough to guarantee that \frac{2\langle\beta,\alpha\rangle}{\langle\alpha,\alpha\rangle}\in\mathbb{Z}.

Tomorrow we start in constructing our root systems, towards the theorem: For each Dynkin diagam allowed by the classification theorem, there exists an irreducible root system having that diagram.

March 1, 2010 Posted by | Geometry, Root Systems | 7 Comments

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