The Existence of Bases for Root Systems

We’ve defined what a base for a root system is, but we haven’t provided any evidence yet that they even exist. Today we’ll not only see that every root system has a base, but we’ll show how all possible bases arise. This will be sort of a long and dense one.

First of all, we observe that any hyperplane has measure zero, and so any finite collection of them will too. Thus the collection of all the hyperplanes $P_\alpha$ perpendicular to vectors $\alpha\in\Phi$ cannot fill up all of $V$ . We call vectors in one of these hyperplanes “singular”, and vectors in none of them “regular”.

When $\gamma$ is regular, it divides $\Phi$ into two collections. A vector $\alpha$ is in $\Phi^+(\gamma)$ if $\alpha\in\Phi$ and $\langle\alpha,\gamma\rangle>0$ , and we have a similar definition for $\Phi^-(\gamma)$ . It should be clear that $\Phi^-(\gamma)=-\Phi^+(\gamma)$ , and that every vector $\alpha\in\Phi$ is in one or the other; otherwise $\gamma$ would be in $P_\alpha$ . For a regular $\gamma$ , we say that $\alpha\in\Phi^+(\gamma)$ is “decomposable” if $\alpha=\beta_1+\beta_2$ for $\beta_1,\beta_2\in\Phi^+(\gamma)$ . Otherwise, we say that $\alpha$ is “indecomposable”.

Now we can state our existence theorem. Given a regular $\gamma$ , let $\Delta(\gamma)$ be the set of indecomposable roots in $\Phi^+(\gamma)$ . Then $\Delta(\gamma)$ is a base of $\Phi$ , and every base of $\Phi$ arises in this manner. We will prove this in a number of steps.

First off, every vector in $\Phi^+(\gamma)$ is a nonnegative integral linear combination of the vectors in $\Delta(\gamma)$ . Otherwise there is some $\alpha\in\Phi^+(\gamma)$ that can’t be written like that, and we can choose $\alpha$ so that $\langle\gamma,\alpha\rangle$ is as small as possible. $\alpha$ itself can’t be indecomposable, so we must have $\alpha=\beta_1+\beta_2$ for some two vectors $\beta_1,\beta_2\in\Phi^+(\gamma)$ , and so $\langle\gamma,\alpha\rangle=\langle\gamma,\beta_1\rangle+\langle\gamma,\beta_2\rangle$ . Each of these two inner products are strictly positive, so to avoid contradicting the minimality of $\langle\gamma,\alpha\rangle$ we must be able to write each of $\beta_1$ and $\beta_2$ as a nonnegative linear combination of vectors in $\Delta(\gamma)$ . But then we can write $\alpha$ in this form after all! The assertion follows.

Second, if $\alpha$ and $\beta$ are distinct vectors in $\Delta(\gamma)$ then $\langle\alpha,\beta\rangle\leq0$ . Indeed, by our lemma if $\langle\alpha,\beta\rangle>0$ then $\alpha-\beta\in\Phi$ . And so either $\alpha-\beta$ or $\beta-\alpha$ lies in $\Phi^+(\gamma)$ . In the first case, we can write $\alpha=\beta+(\alpha-\beta)$ , so $\alpha$ is decomposable. In the second case, we can similarly show that $\beta$ is decomposable. And thus we have a contradiction and the assertion follows.

Next, $\Delta(\gamma)$ is linearly independent. If we have a linear combination

$\displaystyle\sum\limits_{\alpha\in\Delta(\gamma)}r_\alpha\alpha=0$

then we can separate out the vectors $\alpha$ for which the coefficient $r_\alpha>0$ and those $\beta$ for which $r_\beta<0$ , and write

$\displaystyle\sum\limits_\alpha s_\alpha\alpha=\sum\limits_\beta t_\beta\beta$

with all coefficients positive. Call this common sum $\epsilon$ and calculate

$\displaystyle\langle\epsilon,\epsilon\rangle=\sum\limits_{\alpha,\beta}s_\alpha t_\beta\langle\alpha,\beta\rangle$

Since each $\langle\alpha,\beta\rangle\leq0$ , this whole sum must be nonpositive, which can only happen if $\epsilon=0$ . But then

$\displaystyle0=\langle\gamma,\epsilon\rangle=\sum\limits_\alpha s_\alpha\langle\gamma,\alpha\rangle$

which forces all the $s_\alpha=0$ . Similarly, all the $t_\beta=0$ , and thus the original linear combination must have been trivial. Thus $\Delta(\gamma)$ is linearly independent.

Now we can show that $\Delta(\gamma)$ is a base. Every vector in $\Phi^+(\gamma)$ is indeed a nonnegative integral linear combination of the vectors in $\Delta(\gamma)$ . Since $\Phi^-(\gamma)=-\Phi^+(\gamma)$ , every vector in this set is a nonpositive integral linear combination of the vectors in $\Delta(\gamma)$ . And every vector in $\Phi$ is in one or the other of these sets. Also, since $\Phi$ spans $V$ we find that $\Delta(\gamma)$ spans $V$ as well. But since it’s linearly independent, it must be a basis. And so it satisfies both of the criteria to be a base.

Finally, every base $\Delta$ is of the form $\Delta(\gamma)$ for some regular $\gamma$ . Indeed, we just have to find some $\gamma$ for which $\langle\gamma,\alpha\rangle>0$ for each $\alpha\in\Delta$ . Then since any $\beta\in\Phi$ is an integral linear combination of $\alpha\in\Delta$ we can verify that $\langle\gamma,\beta\rangle\neq0$ for all $\beta\in\Phi$ , proving that $\gamma$ is regular. and $\Phi^+=\Phi^+(\gamma)$ . Then the vectors $\alpha\in\Delta$ are clearly indecomposable, showing that $\Delta\subseteq\Delta(\gamma)$ . But these sets contain the same number of elements since they’re both bases of $V$ , and so $\Delta=\Delta(\gamma)$ .

The only loose end is showing that such a $\gamma$ exists. I’ll actually go one better and show that for any basis $\{\eta_i\}_{i=1}^{\dim(V)}$ the intersection of the “half-spaces” $\{\gamma\vert\langle\gamma,\eta_i\rangle\}$ is nonempty. To see this, define

$\displaystyle\delta_i=\eta_i-\sum\limits_{\substack{1\leq j\leq\dim(V)\\j\neq i}}\frac{\langle\eta_i,\eta_j\rangle}{\langle\eta_j,\eta_j\rangle}\eta_j$

This is what’s left of the basis vector $\eta_i$ after subtracting off its projection onto each of the other basis vectors $\eta_j$ , leaving its projection onto the line perpendicular to all of them. Then consider the vector $\gamma=r^i\delta_i$ where each $r^i>0$ . It’s a straightforward computation to show that $\langle\gamma,\eta_k\rangle=r^i\langle\delta_k,\eta_k\rangle>0$ , and so $\gamma$ is just such a vector as we’re claiming exists.

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

5 Comments »

Your final summation doesn’t actually give you a vector orthogonal to the remaining basis vectors, unless everything is already orthogonal. I think you want a gram schmidt process here, applied independently to each basis vector.

Also, out of curiosity, do you have any plans to draw out any of the polytopal or crystollographic connections to root systems?

Comment by Gilbert Bernstein | February 3, 2010 | Reply
You’re right, Gilbert. But in the end the resulting vector $\gamma$ still has the properties we want.

As for applications, I’m just looking at classification for now. I may return to applications at some future point.

Comment by John Armstrong | February 3, 2010 | Reply
[…] A very useful concept in our study of root systems will be that of a Weyl chamber. As we showed at the beginning of last time, the hyperplanes for cannot fill up all of . What’s left over […]

Pingback by Weyl Chambers « The Unapologetic Mathematician | February 3, 2010 | Reply
[…] then is a (positive) root for some simple . If for all , then the same argument we used when we showed is linearly independent would show that is linearly independent. But this is impossible because […]

Pingback by Some Lemmas on Simple Roots « The Unapologetic Mathematician | February 4, 2010 | Reply
[…] is any regular vector, then there is some so that for all . That is, sends the Weyl chamber to the fundamental […]

Pingback by The Action of the Weyl Group on Weyl Chambers « The Unapologetic Mathematician | February 5, 2010 | Reply

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