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I have the following Cartan matrix and I have to compute all the roots. I know there are 18. \begin{equation} \begin{bmatrix} 2 & -1 & 0\\ -1 & 2 & -2\\ 0 & -1 & 2 \end{bmatrix} \end{equation}

So if name my basis $\Delta=\{a,b,c\}$ I can get from the Cartan matrix :

$s_b(a)=a+b \to$ new root
$s_c(a)=a$
$s_a(b)=a+b$
$s_c(b)=b+2c \to$ new root
$s_a(c)=c$
$s_b(c)=c+b$

so in this step I got the new roots $\{a+b, b+c, b+2c\}$

So I start again :

$s_a(a+b)=b$
$s_c(a+b)=a+b+2c \to$ new root
$s_a(b+2c)=a+b+2c$
$s_b(b+2c)=b+2c$
$s_a(b+c)=c+b+a \to$ new root
$s_c(c+b)=b$

So I got the new roots $\{a+b+2c, c+b+a\}$

and I continue. But Am I using all the tricks I could use here?

Ben Grossmann
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roi_saumon
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1 Answers1

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You are right in principle. Little correction: $s_c(b+c)=b+c$. Just for completeness, in a next step you would get the new positive root $s_b(a+b+2c)=a+2b+2c$, and then we have all 9 positive roots. By the way, this is the root system $B_3$.

You can make the procedure a bit more efficient by using the methods of this answer. Then e.g. in the first step, the entry $-2$ at position $a_{23}$ immediately tells you that the $c$-string through $b$ is $b, b+c, b+2c$, and you get two new roots for the price of one.

Unfortunately, for further steps, that answer tacitly assumes that we have a $2 \times 2$-Cartan matrix. But the idea is easily generalised.

Let $(n_{ij})_{ij}$ bet the Cartan matrix. Take a simple root $\alpha_k$ and a positive root $\lambda =\sum \lambda_i \alpha_i$ ($\lambda_i \in \Bbb Z^+$). If $r:= -\sum_i \lambda_i n_{ik}$ is positive, then the $\alpha_k$-string through $\lambda$ is $\lambda, \lambda+\alpha_k, ..., \lambda + r\alpha_k$.

Here, in your second step, you can spare yourself computing $a$- or $b$-strings ($s_a$ or $s_b$) of $a+b$ because you already got it both through $a$- and $b$-strings; you don't have to look for further $c$-strings from $b+c$ or $b+2c$ either. On the other hand, instead of computing only $s_c(a+b)$, you easily get that the $c$-string through $a+b$ has length $-(a_{13}+a_{23}) =2$, so again you get two new roots $a+b+c$ and $a+b+2c$ at once.

And finally in that third step, one would see the length of a $b$-string through $a+b+2c$ is $-(a_{12}+a_{22}+2a_{32}) =1$, which again gives the last positive root $a+2b+2c$.

Torsten Schoeneberg
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  • In the definition of $r$ do we multiply the coefficients of $\lambda$ with the $k^{th}$ row or $k^{th}$ column? According to this answer it is the $k^{th}$ column but according to the answer by David Hill to which you linked it is $k^{th}$ row. I might be misunderstanding something, as I am just a beginner. – Sanjana Mar 24 '24 at 12:05
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    @Sanjana: That's the disagreement between David Hill and myself discussed in the comments to that linked answer. It depends on convention of root labelling AND that for some people, the Cartan matrix is the transpose of other people's Cartan matrix. – Torsten Schoeneberg Mar 24 '24 at 16:16
  • So is it true that using these more well-known conventions, in the first step too, the rule is modified i.e. for $i \ne j$, the $\alpha_i$ string through $\beta_j$ is $\beta_j,\beta_j+\alpha_i,...,\beta_j-n_{ji} \alpha_i$ instead of $\beta_j,\beta_j+\alpha_i,...,\beta_j-n_{ij} \alpha_i$ as you probably have used in the 2nd paragraph of your answer? – Sanjana Mar 24 '24 at 17:23
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    I do think that (up to $\beta_j-n_{ji}\alpha_i$) is indeed the convention I am working with, and so does e.g. Bourbaki as shown in the images in my comments to the other answers. – Torsten Schoeneberg Mar 26 '24 at 03:53