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Let $\{a,b,c\}$ be a set of simple roots of the Lie algebra $B_3$ and suppose $|a|=|c|$ and $\langle b,c\rangle=0$. I want to find the corresponding Cartan matrix.

I know that it's a $3\times3$ matrix with diagonal elements equal to $2$. Also since $\langle b,c\rangle=0$ then entry $(2,3)$ in the Cartan matrix is $\langle b^\vee,c\rangle=\frac{2}{\langle b,b\rangle}\langle b,c\rangle=0$ and so entry $(3,2)$ is also zero.

So far the Cartan matrix is $\begin{pmatrix} 2 & a_{12} & a_{13} \\ a_{21} & 2 & 0 \\ a_{31} & 0 & 2 \end{pmatrix}$. The determinant is $-2a_{21}a_{12}-2a_{31}a_{13}+8$, which should be strictly positive.

Since $|a|=|c|$ then $a_{13}=a_{31}$ by symmetry and so we must have that $a_{13}=a_{31}\in\{0,-1\}$. If $a_{31}=a_{13}=0$ then we don't have much more info about $a_{12}$ and $a_{21}$. If $a_{31}=a_{13}=-1$, then we know that none of $a_{12}$ or $a_{21}$ cannot be $-3$, or else it would violate that the determinant should be strictly positve.

So this is where I'm stuck. I haven't really managed to get any further.

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    $\begingroup$ If we look at the Dynkin diagram, vertex $2$ and $3$ are not connected. But since the graph must be connected, vertex $3$ should be connected to vertex $1$. Is that right? In that case the $a_{31}\neq 0$. $\endgroup$
    – nobody
    Mar 16, 2020 at 12:59
  • $\begingroup$ Maybe you should write down what $B_3$ is to you, so we know what more information we can use. $\endgroup$ Mar 16, 2020 at 14:44

1 Answer 1

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Here $b$ is the small simple root, $a,c$ the long simple roots. Hence the Cartan matrix will be $\begin{pmatrix} 2 &-2 &-1 \\ -1& 2& 0 \\ -1& 0& 2 \end{pmatrix}$

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    $\begingroup$ Every irreducible root system has atmost $2$ possible root lengths, the roots having the shorter root length is called the short root and the roots having the longer root length called the long roots. Now every root system has a base whose elements are called the simple roots, which is a basis of the underlying vector space and any root is either a non-negative or a non-positive integer linear combination of simple roots (so no mixed signs involved). Given a base, we write the Cartan matrix, and associate to it a Dynkin diagram. Now we know the Dynkin diagram of $B3$ root system... $\endgroup$
    – nobody
    Mar 16, 2020 at 13:16
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    $\begingroup$ ...and I concluded it from there. But I have in a way used too much theory maybe there is an elementary way of doing it. $\endgroup$
    – nobody
    Mar 16, 2020 at 13:18
  • $\begingroup$ (Just so I don't look like a fool, the OP asked me what long and short roots are but somehow deleted the question) $\endgroup$
    – nobody
    Mar 16, 2020 at 13:19
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    $\begingroup$ Thank you for the explanation. How do we know that $b$ i small and $a,c$ is long and not the other way around? $\endgroup$
    – user760071
    Mar 16, 2020 at 13:20
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    $\begingroup$ Because in type $B$ there is one short simple root and all other simple roots are long. Here $b$ and $c$ have the same length so they better not be short. $\endgroup$
    – nobody
    Mar 16, 2020 at 13:21

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