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Let $R$ and $R'$ be irreducible root systems in the real inner product spaces $E$ and $E'$. Prove that $R$ and $R'$ are isomorphic iff there exists a scalar $\lambda \in \mathbb{R}$ and a vector space isomorphism $\varphi: $ $E \to E'$ such that $\varphi(R)=R'$ and $(\varphi(u),\varphi(v))=\lambda(u,v)\text{ for all }u,v \in E$. (Introduction to Lie algebra Erdmann Karin- Mark Wildon, Exercise 11.15, page 124.)

I just can prove the "if" part and I get stuck with the "only if part". I highly appreciate who can give me some ideas.

Thank in advance

An Khuong Doan
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    Hard to say for sure for I don't know how they define an isomorphism of roots systems. Presumably in terms of $\langle\alpha,\beta^{\vee}\rangle$? Anyway, an isomorphism of root systems surely takes a set of simple roots to another. Those form vector space bases, so that mapping as a unique extension to a linear isomorphism. The length data can be deduced from the root-coroot data, so long (resp. short) roots get mapped appropriately with the angles and ratios of lengths being preserved. Should follow from that. – Jyrki Lahtonen Jul 24 '15 at 21:58

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Two root systems $R$ and $R’$ are said to be isomorphic if there is a vector space isomorphism $\phi : E \to E’$ such that:

(1) $\phi(R) = R’$

(2) for any two roots $\alpha$ and $\beta \in R$, $\langle \alpha, \beta \rangle = \langle \phi(\alpha), \phi(\beta) \rangle$.

Now, for the only if direction, you only need to prove that $\lambda(u,v) = (\phi(u), \phi(v))$. Notice that if $u,v \in R$ this follows easily from condition (2), by definition of $\langle u,v \rangle$ and $\langle \phi(u), \phi(v) \rangle$.

cip
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