I am studying the book Introduction to Lie algebras. In page 122 there is something I don't understand and I am looking for some help.

In the beginning of that page the authors give the definition of what it means two root systems to be isomorphic:

Definition 1:
Two root systems $R_1$ and $R_2$ in the real inner product spaces $E_1$ and $E_2$ respectively are isomorphic if there is a vector space isomorphism $\phi:E_1\to E_2$ such that:
(a) $\phi(R_1)=R_2$,
(b) $\color{brown}{ \text{ for any two roots }\alpha,\beta\in R_1, \ \langle\alpha,\beta\rangle=\langle\phi(\alpha),\phi(\beta)\rangle}$.

In the end of that page the authors prove a proposition saying that if the Dynkin diagrams of two root systems $R_1$ and $R_2$ are the same then the root systems are isomorphic.The proof begins as follows:

We may choose bases $B_1 = \{\alpha_1,\ldots , \alpha_\ell\}$ in $R_1$ and $B_2 = \{\alpha'_1,\ldots , \alpha'_\ell\}$ in $R_2$ so that for all $i, j$ one has $\langle \alpha_i,\alpha_j\rangle =\langle \alpha'_i,\alpha'_j\rangle$. Let $\phi: E \to E$ be the linear map which maps $\alpha_i$ to $\alpha'_i$. By definition, this is a vector space isomorphism satisfying $\color{brown}{\text{condition } 1(b)}$.

It is clear to me that for any two roots $\alpha,\beta\in B_1, \ \langle\alpha,\beta\rangle=\langle\phi(\alpha),\phi(\beta)\rangle$ but not clear at all that $\color{red}{\text{ for any two roots } \alpha,\beta\in R_1, \ \langle\alpha,\beta\rangle=\langle\phi(\alpha),\phi(\beta)\rangle}$. So can someone explain to me why this is true?

Note that here $\langle\alpha,\beta\rangle=\dfrac{2(\alpha,\beta)}{(\beta,\beta)}$ where $(,)$ is the inner product.


Do you mean $2\frac{\langle\alpha,\beta\rangle}{\langle\beta,\beta\rangle}$ instead of $\langle\alpha,\beta\rangle$? Isomorphic root systems need not preserve the Killing form.

The proof that I'm familiar with works because the Weyl groups $W=\langle S_{\alpha_i}\rangle$ and $W'=\langle S_{\alpha_i'}\rangle$ are isomorphic, $\phi$ as you've defined is an isomorphism of $W$-modules, and $R_1=WB_1=\lbrace w(\alpha_i):w\in W\rbrace$ and $R_2=WB_2=\lbrace w(\alpha'_i):w\in W\rbrace$.

EDIT. Sorry, I hadn't seen that you'd clarified that you meant $2\frac{\langle\alpha,\beta\rangle}{\langle\beta,\beta\rangle}$.

EDIT 2. Perhaps this is a better way. If $\alpha\in B_1,\beta\in R_1$, then $\langle\alpha,\beta\rangle=\langle\phi(\alpha),\phi(\beta)\rangle$ by the linearity of $\phi$ and $\langle\cdot,\cdot\rangle$ in the second variable. If $\alpha\in R_1$, then $\alpha=w(\alpha_i)$, and $$\phi(\alpha)=(\phi\circ w)(\alpha_i)=(\phi\circ\phi^{-1}\circ w'\circ\phi)(\alpha_i)=w'(\phi(\alpha_i))=w'(\alpha'_i)$$ and hence: \begin{align*} \langle\phi(\alpha),\phi(\beta)\rangle&=\langle w'(\alpha'_i),\phi(\beta)\rangle=\langle\alpha'_i,(w')^{-1}\phi(\beta)\rangle=\langle\alpha'_i,(\phi\circ w\circ\phi^{-1})^{-1}\phi(\beta)\rangle \\ &=\langle\alpha'_i,(\phi\circ w'\circ\phi^{-1}\circ\phi)(\beta)\rangle=\langle\phi^{-1}(\alpha'_i),w^{-1}(\beta)\rangle=\langle\alpha_i,w^{-1}(\beta)\rangle=\langle w(\alpha_i),\beta\rangle \\ &=\langle\alpha,\beta\rangle \end{align*}

  • $\begingroup$ Sorry but I still don't understand. If $\alpha,\beta\in R_1$ then $\alpha=w_1(\alpha_i)$ and $\beta=w_2(\alpha_j)$ for some $w_1,w_2\in W$. Then $\langle\alpha,\beta\rangle=\langle w_1(\alpha_i),w_2(\alpha_j)\rangle=?$ and $\langle\phi(\alpha),\phi(\beta)\rangle=\langle \phi(w_1(\alpha_i)),\phi(w_2(\alpha_j))\rangle=\langle w_1(\phi(\alpha_i)),w_2(\phi(\alpha_j))\rangle=\langle w_1(\alpha'_i),w_2(\alpha'_j)\rangle=?$ $\endgroup$ – George Apr 18 '13 at 20:06
  • $\begingroup$ Since $S_\beta(\alpha)=\alpha-\langle\alpha,\beta\rangle\beta$, $\langle\alpha,\beta\rangle$ is determined by the reflection $S_\beta$ acting on $\alpha$. The corresponding formula for $R_2$, together with the fact that $S_{\phi(\beta)}=\phi\circ S_\beta\circ\phi^{-1}$ ($w\mapsto\phi\circ w\circ\phi^{-1}$ is an isomorphism of $W$ onto $W'$) is what implies that $\langle\alpha,\beta\rangle=\langle\phi(\alpha),\phi(\beta)\rangle$. I'll try to think of a better way to explain if that hasn't helped! $\endgroup$ – Warren Moore Apr 18 '13 at 20:39
  • $\begingroup$ @George I've edited my answer to explicitly show why $\langle\alpha,\beta\rangle=\langle\phi(\alpha),\phi(\beta)\rangle$. I hope this helps! $\endgroup$ – Warren Moore Apr 18 '13 at 23:13

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