Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $\Phi$ be a root system in a real inner product space $E$. Define $\alpha^\vee = \frac{2\alpha}{(\alpha, \alpha)}$. Then the set $\Phi^\vee = \{\alpha^\vee: \alpha \in \Phi \}$ is also a root system.

Let $B$ be a base for the root system $\Phi$, ie. $B$ is a basis for $E$ and each $\alpha \in \Phi$ can be written as $\alpha = \sum_{\beta \in B} {k_\beta} \beta$ such that $k_\beta$ are integers of same sign.

Question: How to prove that $B^\vee = \{ \alpha^\vee: \alpha \in B\}$ is a base for the root system $\Phi^\vee$?

It is easy to see that $B^\vee$ is a basis for $E$. For the other property, here is what I have so far. Let $\alpha = \sum_{\beta \in B} {k_\beta} \beta$ be a root. Then

$$\alpha^\vee = \sum_{\beta \in B} \frac{k_\beta (\beta, \beta)}{(\alpha, \alpha)} \beta^\vee$$

so to prove the claim, it is necessary to prove that $\frac{k_\beta (\beta, \beta)}{(\alpha, \alpha)}$ is an integer. I don't see how that follows. In the case where $(\alpha, \beta) = 0$ the proportion $\frac{(\beta, \beta)}{(\alpha, \alpha)}$ could be anything, so I am a bit confused.

share|cite|improve this question

3 Answers 3

up vote 3 down vote accepted

This is an exercise in Humphreys, namely Exercise 1 of Chapter 10.

Let $\Phi^\vee$ be the dual to $\Phi$ and $\Delta^\vee = \{\alpha^\vee : \alpha\in \Delta\}$. By property of the Weyl group $W$, every $\alpha\in \Phi$ can be written as $ \alpha = \sigma(\beta)$ for some $\beta\in \Delta$ and $\sigma\in W$. Then $$\alpha^\vee = \sigma(\beta)^\vee = \frac{2\sigma(\beta)}{(\sigma(\beta),\sigma(\beta))} = \frac{2}{(\beta,\beta)}\sigma(\beta) = \sigma(\beta^\vee)$$ Let $\Delta = \{\alpha_1,\cdots, \alpha_l\}$, claim for any $\sigma\in W$, $\sigma(\alpha_i^\vee)$ is a linear combination of $\alpha_i^\vee,\cdots, \alpha_l^\vee$ with integer coefficients. Note that we may write $\sigma = \sigma_{\alpha_{i_1}}\cdots \sigma_{\alpha_{i_l}}$, it is enough to show that $\sigma_{\alpha_j}(\alpha_i^\vee)$ is a linear combination of $\alpha_i^\vee,\cdots, \alpha_l^\vee$ with integer coefficients. $$\sigma_{\alpha_j}(\alpha_i^\vee) = \alpha_i^\vee - \langle \alpha_i^\vee, \alpha_j\rangle \alpha_j = \alpha_i^\vee - \frac{2(\alpha_i^\vee,\alpha_j)}{(\alpha_j,\alpha_j)\alpha_j}$$ $$ = \alpha_i^\vee - \frac{4(\alpha_i,\alpha_j)}{(\alpha_i,\alpha_i)(\alpha_j,\alpha_i)}\alpha_j = \alpha_i^\vee - \frac{2(\alpha_i,\alpha_j)}{(\alpha_i,\alpha_i)}\alpha_j^\vee = \alpha_i^\vee - \langle \alpha_j,\alpha_i\rangle \alpha_j^\vee$$ We are done since $\langle \alpha_j,\alpha_i\rangle$ is an integer. Hence for $\alpha\in \Phi$ $$\alpha^\vee = \sum_{i = 1}^l k_i \alpha_i^\vee,\;\;k_i\in\mathbb{Z}$$ Now we show $k_i$ are all nonnegative or nonpositive. Since we can write $\alpha = \sum_{i = 1}^l k_i' \alpha_i$ with all nonnegative or nonpositive coefficients, then $$\alpha^\vee = \frac{2\alpha}{(\alpha,\alpha)} = \frac{2}{(\alpha,\alpha)}\sum_{i = 1}^lk_i'\alpha_i= \sum_{i = 1}^l \frac{(\alpha_i,\alpha_i)}{(\alpha,\alpha)}k_i'\alpha_i^\vee$$ Note that $\{\alpha_i^\vee,\cdots,\alpha_l^\vee\}$ is linearly independent, whence $$ k_i =\frac{(\alpha_i,\alpha_i)}{(\alpha,\alpha)} k_i'$$ and since $\frac{(\alpha_i,\alpha_i)}{(\alpha,\alpha)}>0$ for all $i = 1,\cdots, l$ and $\alpha\in\Phi$, we conclude that $k_i$ and $k_i'$ have the same sign.

share|cite|improve this answer
Thanks. Why do we need the part in the beginning? To me this works even if we start from "By property of the Weyl group, every.." I mean, we know that any root $\beta$ is equal to $\sigma(\alpha)$ for some $\sigma$ in the weyl group, $\alpha$ from the basis. –  spin Apr 2 '14 at 9:08
@spin I think you are right. –  mez Apr 2 '14 at 9:34
In Humphreys book the hint "Compare Weyl Chambers of $\Phi$ and $\Phi^\vee$." is given. I wonder what is the intended solution? –  spin Apr 2 '14 at 14:33
@spin Here is possibly what he means, which is essentially the same proof. Look at the two theorems on page 48. Take $\gamma$ from the fundamental chamber of $\Delta$, then split $(\Phi^\vee)^+ = \{\beta\in \Phi^\vee: (\gamma,\beta)>0\}$. Then you show that $\Delta^\vee$ is exactly the set of indecomposable roots in $(\Phi^\vee)^+$, which is a base of $\Phi^\vee$ by theorem. –  mez Apr 2 '14 at 19:00
Thanks, I will take a look. –  spin Apr 2 '14 at 19:08

This is covered in Chapter 11 (p. 86) of the course Introduction to Lie Algebras.

share|cite|improve this answer

Here is a different solution, which is perhaps the one hinted at in the book by Humphreys. See the comments in the answer by mezhang. The notation and terminology is that used by Humphreys in Chapter III of his book.

Now $\Delta = \Delta(\gamma)$ for some regular element $\gamma$ in the Euclidean space $E$ where the root system $\Phi$ resides. Since $\alpha$ and $\alpha^\vee$ determine the same hyperplanes, $\gamma$ is regular also with respect to $\Phi^\vee$. Thus $\Delta^\vee(\gamma)$ forms a base for $\Phi^\vee$. Here $\Delta^\vee(\gamma)$ is the set of roots of $\Phi^\vee$ that are indecomposable with respect to $\gamma$.

We prove that $\Delta^\vee \subseteq \Delta^\vee(\gamma)$, which gives equality since both sets are bases for $E$. Let $\alpha^\vee \in \Delta^\vee$. If $\alpha^\vee = \beta_1^\vee + \beta_2^\vee$ for $\beta_i \in (\Phi)^+(\gamma)$, then $\alpha$ is a linear combination of two distinct positive roots. But this is not possible since $\Delta$ is a base. Thus $\alpha^\vee$ is indecomposable, ie. $\alpha^\vee \in \Delta^\vee(\gamma)$.

share|cite|improve this answer
I think this is correct. –  mez May 11 '14 at 17:50
@mezhang: great, thanks. –  spin May 11 '14 at 19:26

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.