Root system of a semisimple Lie algebra spans the dual.

Question

Let $L$ be a semisimple Lie algebra and $H$ a Cartan subalgebra. The roots of $L$ with respect to $H$ are the elements of $$ \Phi = \{\alpha \in H^*\setminus 0 : L_\alpha \neq 0\} $$ where $L_\alpha = \{x \in L : [h,x] = \alpha(h)x \text{ for all } h \in H\}$. According to Humphrey's lie algebras book, $\Phi$ spans $H^*$. But the proof somehow uses duality which I cannot seem to understand. This question has been asked before but I also did not understand the accepted answer.

The proof starts like this: if $\Phi$ does not span, then (by duality) there exists a nonzero $h\in H$ such that $\alpha(h) = 0$ for all $\alpha \in \Phi$. This is the part of the proof I don't get.

Answer 1

Let $V$ be a finite-dimensional vector space with dual $V^*$. To make notation consistent (and remove the superficial differences between $V$ and its dual) it is helpful to write down a pairing $\langle -, - \rangle \colon V \times V^* \to \mathbb{C}$ defined by $\langle v, f \rangle = f(v)$.

A nonzero dual vector $f \in V^*$ defines a hyperplane $\langle -, f \rangle = 0$, or equivalently $\ker f$. Similarly, a nonzero vector $v \in V$ defines a hyperplane $\langle v, - \rangle = 0$ (sometimes called the annihilator of $v$). This generalises: for any set $S \subseteq V$ of vectors, we can define $S^\perp \subseteq V^*$ by $$ S^\perp = \{f \in V^* \mid \langle s, f \rangle = 0 \text{ for all } s \in S\} \subseteq V^*.$$ Similarly, we could take a set $T \subseteq V^*$ of dual vectors and define $$ T^\perp = \{v \in V \mid \langle v, t \rangle = 0 \text{ for all } t \in T\} \subseteq V.$$ The key properties of this complement operation are (for our purposes):

1. $S^\perp$ is always a vector subspace of $V^*$.
2. $S^\perp = (\operatorname{span} S)^\perp$.
3. $\dim (\operatorname{span} S) = \operatorname{codim} S^\perp$, where the codimension of a subspace $U \subseteq V$ is $\operatorname{codim} U = \dim V - \dim U$.

These properties can be proven by looking at a basis of $V$ together with its corresponding dual basis in $V^*$.

Now to answer your question: if $\Phi \subseteq V^*$ does not span, then $\operatorname{codim}(\operatorname{span} \Phi) = \dim \Phi^\perp \geq 1$, and so there exists a nonzero vector $v \in \Phi^\perp$ such that $\langle v, \alpha \rangle = 0$ for all $\alpha \in \Phi$.

Answer 2

Of course the other answer is optimal. Let me just add one way to see the result without all the duality machinery (which, however, is a good thing to learn anyway):

Let $V$ be an $n$-dimensional vector space over a field $k$ and $\alpha_1, ... ,\alpha_r$ be linearly independent elements of $V^*$ i.e. nonzero linear maps $V\rightarrow k$.

Then $\mathrm{dim} \left( \cap \mathrm{ker}(\alpha_i) \right) \ge n-r$.

(Proof: Induction over $r$.)

And now use that $V$ and $V^*$ have the same dimension (which is duality in a nutshell).