# Equivalent definitions of a root system.

For studying root systems many authors start from a vector space $V$ over $\mathbb{R}$ with a positive definite scalar product $(\cdot,\cdot)$, in which a reflection $\sigma_\alpha$ is a linear application that fixes the hyperplane $H_\alpha$ and send $\alpha$ to its opposite. In formulas \begin{gather} \sigma_\alpha(\beta)=\beta- <\beta,\alpha>\alpha \end{gather} where $<\beta,\alpha>=$$2(\alpha,\beta)\over (\alpha,\alpha) . Then a root system is defined as a subset R of V such that • \langle R \rangle =V, R finite and 0 \notin R • \mathbb{R}\{ \alpha\} \cap R=\{\pm \alpha \} if \alpha \in R • for every \alpha, \beta \in R, R is invariant under \sigma_\alpha and <\beta,\alpha> is an integer. This is a quite strong structure, but all the properties envolved arise naturally in the form of the weights \mu \in H^*, where H is the Cartan subalgebra of a complex Lie algebra L (we are considering the adjoint representation). Then we can consider on H^* the dual of the Killing form, that is again symmetric and positive definite. In this environment, we can restrict to V_\mathbb{Q}, the \mathbb{Q}-span of the non zero weights in H^*, (indeed one can prove that the dual of the killing form take rational values, see the chapter "Integrality Properties" of "Introduction of Lie Algebras and Representation theory", Humphreys) then consider V=V_\mathbb{Q} \otimes_\mathbb{Q} \mathbb{R} and check that these are a root system in V. For proving the characterization of semisimple lie algebras, this can be a start point. Nevethless, some authors need to weak slightly the definition of a root system, starting from a vector space V on \mathbb{R} (without scalar product) and define a root system R as a subset of V such that • R is finite, generates V and 0 \notin R • \alpha \in R \implies$$ \alpha \over 2$$\notin R$
• for each $\alpha, \beta \in R$ there is $\alpha^\vee \in V^*$ with $\alpha^\vee (\alpha)=2$ and $\alpha^\vee(\beta) \in \mathbb{Z}$ and $s_{\alpha^\vee,\alpha}(R) \subseteq R$

Where for $\lambda \in V^*, w \in V$ $s_{\lambda,v}(w)=w-\lambda(w)v$. Then, considerd $G$ the finite subgroup of $GL(V)$ that preserves $R$ and $(\cdot,\cdot)$ a generic positive definite scalar product, we define \begin{gather} (\alpha,\beta)'=\sum_{g\in G}(g\alpha,g\beta) \end{gather} After realizing that $(\cdot,\cdot)'$ is a positive definite scalar product by which $G$ acts by isometries, we can prove that

1. $\alpha^\vee$ is uniquely determined by $\alpha$
2. $\alpha^\vee(\lambda)=2\frac{(\alpha,\lambda)'}{(\alpha,\alpha)'}$

Here the identifications: $\alpha^\vee$ is the element $h_\alpha$ such that $x_\alpha,y_\alpha,[x_\alpha,y_\alpha]=h_\alpha$ are the usual generators of a copy of $sl_2$ and $\alpha^\vee(\lambda)=\lambda(h_\alpha)$ after $H^{**}=H$.

My first question: what is the relation between $(\cdot,\cdot)'$ and the dual of the Killing form? My guess is that they differ from a scalar, but I cant prove it directly.

My second question: the second approach has the property that make easier to prove that the non zero weights of a Lie algebra via the adjoint representation are a root system, indeed we can avoid the use of the "orthogonality relations", but we need more work to return to the stronger situation of the first definition. So what is the real advantage of the second way?

• Ok thank you David, I correct! Commented Mar 4, 2016 at 16:22
• Now that you have recently bumped this anyway, you may want to correct the axioms in the first description. The second bullet is wrong. Commented Mar 10, 2016 at 13:31
• Without proving why the only scalar multiples of a root are the root iself and it's opposite, seems that also on Wikipedia there is this assumption. Commented Mar 10, 2016 at 13:40
• en.wikipedia.org/wiki/Root_system?wprov=sfla1 Commented Mar 10, 2016 at 13:40
• Ah ok sorry, I saw, I correct now! Commented Mar 10, 2016 at 13:41

For whom that may concern, some days ago I found a solution for my first question, I write only now since I was busy before. One can suppose that there exists $c \in \mathbb{R}$ such that $(\alpha,\alpha)=c(\alpha,\alpha)'$. Then necessary we have that \begin{gather} 2\frac{(\alpha,\beta)}{(\alpha,\alpha)}=<\beta,\alpha>=2\frac{(\alpha,\beta)'}{(\alpha,\alpha)'} \end{gather} But this implies that $(\alpha,\beta)=c(\alpha,\beta)'$. Then using the same way we can prove that $(\beta,\beta)=c(\beta,\beta)'$ and again go further with a root $\gamma$ and prove that $(\beta,\gamma)=c(\beta,\gamma)'$. Proceding in this way we can expand the proportionality along one connected component of the Dynkin diagram. So we deduce that for an irreducible root system all the bilinear forms that we can replace in the first definition differ from a scalar. This mainly because the "information" of a root system is only a matter of ratios between vectors.
I gladly noticed that we can use this argument for proving that in one simple Lie algebra $L$ there is at most one invariant non degerate bilinear form, up to constants. First notice that we can replace the Killing form with a generic non degenerate symmetric form, invariant under the Lie bracket. Indeed for arriving to the root space decompositions, all the proofs involve only this property and don't rely on the mere definition of the Killing form (or at least in the book of Humphreys "Introduction to Lie Algebras and Representation Theory" every proof works fine for every such a form). Then dualizing any two such a forms in $L$ we get two scalar product on the ambient space of root system associated. For what I wrote before these two scalar product must be multiple, then going backward the two forms on $L$ that we started with, they must be multiple.