The Unapologetic Mathematician

Mathematics for the interested outsider

Bases for Root Systems

We don’t always want to deal with a whole root system \Phi\subseteq V. Indeed, that’s sort of like using a whole group when all the information is contained in some much smaller generating set. For a vector space we call such a small generating set a basis. For a root system, we call it a base. Specifically, a subset \Delta\subseteq\Phi is called a base if first of all \Delta is a basis for V, and if each vector \beta\in\Phi can be written as a linear combination


where the coefficients k_\alpha are either all nonnegative integers or all nonpositive integers.

Some observations are immediate. Because \Delta is a basis, it contains exactly n=\dim(V) vectors of \Phi. It also tells us that the decomposition of each \beta is unique. In fact, as for any basis, every vector in V can be written uniquely as a linear combination of the vectors in \Delta. What we’re emphasizing here is that for vectors in \Phi, the coefficients are all integers, and they’re either all nonnegative or all nonpositive.

Another thing a choice of base gives us is a partial order \preceq on the root system \Phi. We say that \beta is a “positive root” with respect to \Delta (and write \beta\succeq0) if all of its coefficients are nonnegative integers. Similarly, we say that \beta is a “negative root” with respect to \Delta (and write \beta\preceq0) if all of its coefficients are nonpositive integers. We extend this to a partial order by defining \beta\preceq\alpha if \beta-\alpha\preceq0.

Every root is either positive or negative. We write \Phi^+ for the collection of positive roots with respect to a base \Delta and \Phi^- for the collection of negative roots. It should be clear that \Delta\subseteq\Phi^+, and also that \Phi^-=-\Phi^+ — the negative roots are exactly the negatives of the positive roots.

We can also define a kind of size of a vector \beta\in\Phi. Given the above (unique) decomposition, we define the “height” of \beta relative to \Delta as


This will be useful when it comes to proving statements about all vectors in \Phi^+ by induction on their heights.

If \alpha\neq\beta are two vectors in a base \Delta\subseteq\Phi, then we know that \langle\alpha,\beta\rangle\leq0 and \alpha-\beta\notin\Phi. Indeed, our lemma tells us that if \langle\alpha,\beta\rangle>0 then \alpha-\beta would be in \Phi. But this is impossible, because every vector in \Phi can only be written as a linear combination of vectors in \Delta in one way, and that way cannot have some positive signs and some negative signs like \alpha-\beta does.

What this tells us (among other things) is that \beta must be one end of the \alpha root string through \beta. The other end must be \sigma_\alpha(\beta), and the root string must be unbroken between these two ends. Every vector \beta+k\alpha with 0\leq k\leq-\beta\rtimes\alpha must be in \Phi^+.


February 1, 2010 - Posted by | Geometry, Root Systems


  1. Does every root system have a base (I’m guessing yes)? Is the answer to that question obvious?

    Comment by Chad | February 1, 2010 | Reply

  2. Good eye, Chad. It’s not obvious, and that’s exactly the question I’m set to take up tomorrow.

    Comment by John Armstrong | February 1, 2010 | Reply

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