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By definition, if $\Phi$ and $\Phi'$ are root systems of the Euclidean spaces $E$ and $E'$, respectively, then an isomorphism $\Phi\to\Phi'$ is one that is induced by an isomorphism $E\to E'$ which preserves the inner product on pairs of roots, but is not necessarily an isometry.

I read that since the axioms of a root system are unchanged by scaling the inner products be a positive real number, we can assume that the isomorphism is induced by an isometry.

How is this possible? It seems like this suggests that given an isomorphism $\psi\colon E\to E'$, it's possible to scale the inner products on $E$ and $E'$ so that $s(\psi(x),\psi(y))=r(x,y)$ for all $x,y\in E$ and some fixed $s,r>0$. But this doesn't seem like it'd be true at all.

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  • $\begingroup$ But if $v_1, ..., v_n$ is any generating set of $E$, and if $(\psi(v_i), \psi(v_j))' = (v_i, v_j)$ for all $1\le i,j\le n$ (including $i=j$), then $\psi$ is already an isometry because $(\psi(\sum a_i v_i), \psi(\sum b_j v_j))' = \sum_{i,j}a_ib_j(\psi(v_i), \psi(v_j))' = \sum_{i,j}a_ib_j(v_i, v_j) = (\sum a_i v_i, \sum b_j v_j).$ So when you say "preserves the inner product on pairs of roots", do you (or your source) maybe mean $(\psi(\alpha), \psi(\beta))' = (\alpha, \beta)$ just for all $\alpha, \beta \in \Phi$ with $\color{red}{\alpha \neq \pm\beta}$? $\endgroup$ – Torsten Schoeneberg Dec 29 '17 at 23:11
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I only know the case where $\Phi,\Phi'$ are assumed to be irreducible root systems.

Recall that for irreducible systems, roots of the same length are conjugate under the Weyl group $\mathcal{W}$.

(These can be found in Humphreys' Introduction to Lie Algebras and Representation Theory, Section 10.4)

Since and isomorphism of roots system comes from an isomorphism of vector space preserving the $<,>$ operation, so in order to make the map $\pi:E\to E'$ an isometry, we only need to show that

$$(\alpha,\alpha)_E=(\pi\alpha,\pi\alpha)_{E'}$$

for each $\alpha\in E$. Since for each $\sigma\in\mathcal{W}$ that

$$(\sigma\alpha,\sigma\alpha)_E=(\alpha,\alpha)_E$$

(similarly for $\mathcal{W}',E'$), we see that for $\alpha,\beta\in\Phi$, we have

$$(\alpha,\alpha)_E=(\beta,\beta)_E\quad\Rightarrow\quad(\pi\alpha,\pi\alpha)_{E'}=(\pi\beta,\pi\beta)_{E'}$$

For roots of distinct lengths, notice

$$\frac{(\pi\alpha,\pi\alpha)_{E'}}{(\pi\beta,\pi\beta)_{E'}}=\frac{<\pi\alpha,\pi\beta>_{E'}}{<\pi\beta,\pi\alpha>_{E'}}=\frac{<\alpha,\beta>_{E}}{<\beta,\alpha>_{E}}=\frac{(\alpha,\alpha)_E}{(\beta,\beta)_E}$$

(Warning : for this equation to make sense, we need to assume that $(\alpha,\beta)\neq 0$)

These together shows the assertion.

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  • $\begingroup$ After rereading this, I found out that there are still some possible issues with this : in the last equation, we need to assume that $(\alpha,\beta)\neq 0$ so that the quotient makes sense, but this defect can be remedied by the fact that in irreducible root systems, at most two root length occur and that roots of same length are conjugates, so using the irreducibility of the root system, we can find roots of distinct length such that there inner product isn't 0. The rest should be clear. $\endgroup$ – Huang Samuel May 31 at 3:27
  • $\begingroup$ As for the case when the root systems are no irreducible, the restrictions to the irreducible components are injective, so we can "regard" the map as a combination of isomorphisms between irreducible systems. Notice that the decomposition of root system into its irreducible parts, induces a decomposition of its underlying euclidean spaces $E,E'$, so that applying the case where $\Phi,\Phi'$ are irreducibles, we have found a way to justify the assumption that $\pi$ comes from an isometry; the only warning is that the "scaling" might be different on which different direct summands of $E$. $\endgroup$ – Huang Samuel May 31 at 3:36

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