The Unapologetic Mathematician

Mathematics for the interested outsider

From Cartan Matrix to Root System

Yesterday, we showed that a Cartan matrix determines its root system up to isomorphism. That is, in principle if we have a collection of simple roots and the data about how each projects onto each other, that is enough to determine the root system itself. Today, we will show how to carry this procedure out.

But first, we should point out what we don’t know: which Cartan matrices actually come from root systems! We know some information, though. First off, the diagonal entries must all be {2}. Why? Well, it’s a simple calculation to see that for any vector \alpha

\displaystyle\alpha\rtimes\alpha=2\frac{\langle\alpha,\alpha\rangle}{\langle\alpha,\alpha\rangle}=2

The off-diagonal entries, on the other hand, must all be negative or zero. Indeed, our simple roots must be part of a base \Delta, and any two vectors \alpha,\beta\in\Delta must satisfy \langle\alpha,\beta\rangle\leq0. Even better, we have a lot of information about pairs of roots. If one off-diagonal \alpha_i\rtimes\alpha_j element is zero, so must the corresponding one \alpha_j\rtimes\alpha_i on the other side of the diagonal be zero. And if they’re nonzero, we have a very tightly controlled number of options. One must be -1, and the other must be -1, -2, or -3.

But beyond that, we don’t know which Cartan matrices actually arise, and that’s the whole point of our project. For now, though, we will assume that our matrix does in fact arise from a real root system, and see how to use it to construct a root system whose Cartan matrix is the given one. And our method will hinge on considering root strings.

What we really need is to build up all the positive roots \Phi^+, and then the negative roots \Phi^- will just be a reflected copy of \Phi^+. We also know that since there are only finitely many roots, there can be only finitely many heights, and so there is some largest height. And we know that we can get to any positive root of any height by adding more and more simple roots. So we will proceed by building up all the roots of height {1}, then height {2}, and so on until we cannot find any higher roots, at which point we will be done.

So let’s start with roots of height {1}. These are exactly the simple roots, and we are just given those to begin with. We know all of them, and we know that there is nothing at all below them (among positive roots, at least).

Next we come to the roots of height {2}. Every one of these will be a root of height {1} plus another simple root. The problem is that we can’t add just any simple root to a root of height {1} to get another root of height {2}. If we step in the wrong direction we’ll fall right off the root system! We need to know which directions are safe, and that’s where root strings come to the rescue. We start with a root \beta with \mathrm{ht}(\beta)=1, and a simple root \alpha. We know that the length of the \alpha string through \beta must be \beta\rtimes\alpha. But we also know that we can’t step backwards, because \beta-\alpha would be (in this case) a linear combination of simple roots with both positive and negative coefficients! If \beta\rtimes\alpha=0 then we can’t step forwards either, because we’ve already got the whole root string. But if \beta\rtimes\alpha>0 then we have room to take a step in the \alpha direction from \beta, giving a root \beta+\alpha with height \mathrm{ht}(\beta+\alpha)=2. As we repeat this over all roots \beta of height {1} and all simple roots \alpha, we must eventually cover all of the roots of height {2}.

Next are the roots of height {3}. Every one of these will be a root of height {2} plus another simple root. The problem is that we can’t add just any simple root to a root of height {2} to get another root of height {3}. If we step in the wrong direction we’ll fall right off the root system! We need to know which directions are safe, and that’s where root strings come to the rescue… again. We start with a root \beta with \mathrm{ht}(\beta)=2, and a simple root \alpha. We know that the length of the \alpha string through \beta must again be \beta\rtimes\alpha. But now we may be able to take a step backwards! That is, it may turn out that \beta-\alpha is a root, and that complicates matters. But this is okay, because if \beta-\alpha is a root, then it must be of height {1}, and we know that we already know all of these! So, look up \beta-\alpha in our list of roots of height {1} and see if it shows up. If it doesn’t, then the \alpha string through beta starts at \beta, just like before. If it does show up, then the root string must start at \beta-\alpha. Indeed, if we took another step backwards, we’ve have a root of height {0}, which doesn’t exist. Thus we know where the root string starts. We can also tell how long it is, because we can calculate \beta\rtimes\alpha by adding up the Cartan integers \gamma\rtimes\alpha for each of the simple roots \gamma we’ve used to build \beta. And so we can tell whether or not it’s safe to take another step in the direction of \alpha from \beta, and in this way we can build up each and every root of height {3}.

And so on at each level we start with the roots of height n and look from each one \beta in the direction of each simple root \alpha. In each case, we can carefully step backwards to determine where the \alpha string through \beta begins, and we can calculate the length \beta\rtimes\alpha of the string, and so we can tell whether or not it’s safe to take another step in the direction of \alpha from \beta, and we can build up each and every root of height n+1. Of course, it may just happen (and eventually it must happen) that we find no roots of height n+1. At this point, there can be no roots of any larger height either, and we’re done. We’ve built up all the positive roots, and the negative roots are just their negatives.

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February 17, 2010 - Posted by | Geometry, Root Systems

6 Comments »

  1. [...] root system we can construct a connected Dynkin diagram, which determines a Cartan matrix, which determines the root system itself, up to isomorphism. So what he have to find now is a list of Dynkin diagrams which can possibly [...]

    Pingback by The Classification of (Possible) Root Systems « The Unapologetic Mathematician | February 19, 2010 | Reply

  2. [...] classify these, we defined Cartan matrices and verified that we can use it to reconstruct a root system. Then we turned Cartan matrices into Dynkin [...]

    Pingback by Root Systems Recap « The Unapologetic Mathematician | March 12, 2010 | Reply

  3. Thanks for spelling this algorithm out; I found Humphrey’s description (p.56) a bit condensed. One thing I found a bit confusing though, is that you say that the [latex]\alpha[/latex] string [latex]\beta[/latex] has ‘length’ [latex]\beta \rtimes\alpha[/latex]. I think you mean that r-q (in your and Humphreys’ notation) equals this number; the length, i.e. the number of roots in the string, would be r+q+1.

    Comment by Landau | May 2, 2011 | Reply

  4. Damn you, Latex! I should have used dollar signs. (Here I read that both dollar signs and [ latex ] [/ latex ] would work:
    http://wordpress.org/extend/plugins/wp-latex/faq/ )

    Thanks for spelling this algorithm out; I found Humphrey’s description (p.56) a bit condensed. One thing I found a bit confusing though, is that you say that the \alpha string through \beta has ‘length’ \beta \rtimes\alpha. I think you mean that r-q (in your and Humphreys’ notation) equals this number; the length, i.e. the number of roots in the string, would be r+q+1.

    Comment by Landau | May 2, 2011 | Reply

  5. But how do I find the scalar products where α are the simple roots from the cartan matrix? I know 2/, from this how do I find out ? I basically want to construct the root diagram for the algebra.

    Comment by Ramanujan | March 2, 2012 | Reply

  6. Who said anything about an algebra? I’m just talking about a collection of vectors and associated transformations satisfying certain symmetry properties on the one hand, and an integer-entry matrix on the other hand. No algebras here!

    Comment by John Armstrong | March 2, 2012 | Reply


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