2
$\begingroup$

I am following Humphreys' Lie algebra, and solving an exercise. First definition according to book: it is exactly as in the book mentioned.

Root system $B_l$: Let $E=\mathbb{R}^l$ Euclidean space ($l\geq 2$), and $I=\mathbb{Z}e_1+\cdots + \mathbb{Z}e_l$. Define $$\Phi(B_l)=\{\alpha\in I : (\alpha,\alpha)=1\mbox{ or } 2\}. $$

Root system $C_l$: $C_l$ ($l\geq 3$) may be viewed most conventionally as the root system dual to $B_l$, i.e. $$\Phi(C_l)=\Big{\{} \alpha^* : \alpha^* = \frac{2\alpha}{(\alpha,\alpha)}, \alpha\in\Phi(B_l) \Big{\}}.$$ Exercise In constructing (root system) $C_l$, would it be correct to characterize $\Phi(C_l)$ as the set of all vectors in $I$ of squared length $2$ or $4$? Explain.

My answer: Since $(\alpha^*,\alpha^*)=2.2\frac{(\alpha,\alpha)}{(\alpha,\alpha)(\alpha,\alpha)}=\frac{4}{(\alpha,\alpha)}$, so $$(\alpha,\alpha)=1 \mbox{ or } 2 \Longleftrightarrow (\alpha^*,\alpha^*)=4 \mbox{ or } 2. $$ So answer is YES, it is correct to characterize $\Phi(C_l)$ as said in exercise.

I just want to convince whether my answer is right?

1 Answers 1

2

You forgot to check the detail that $\alpha^*$ should also be an element of $I$. Let me show by an example what goes wrong if you don't do that.

If $\ell\ge4$ the set $I$ contains vectors such as $\beta=\pm e_1\pm e_2\pm e_3\pm e_4$ that satisfy $(\beta,\beta)=4$ (you can also use a set of subscripts other than $1,2,3,4$ when $\ell>4$). Yet such vectors should not be included in the root system.

The reason is that for such vectors $$ \beta^*=\frac{2\beta}{(\beta,\beta)}=\pm\frac12e_1\pm\frac12e_2\pm\frac12e_3\pm\frac12e_4\notin I. $$

On the other hand, if $\alpha\in I$ and $(\alpha,\alpha)=2$ then it follows that $\alpha^*\in I$ also.

A correct alternative description would be that $C_\ell$ consists of those vectors $\alpha\in I$ such that $(\alpha,\alpha)$ is $2$ or $4$ AND $\alpha^*\in I$.