How to prove that $B^\vee$ is a base for coroots?

Question

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.

Answer 1

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, with positive coefficients. But this is not possible since $\Delta$ is a base. Thus $\alpha^\vee$ is indecomposable, ie. $\alpha^\vee \in \Delta^\vee(\gamma)$.

Answer 2

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

Solution
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.

Answer 3

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