---- > [!definition] Definition. ([[reducible root system]]) > A [[root system]] $(\Phi, E)$ is said to be **reducible** if there is a partition $\Phi=\Phi_{1} \sqcup \Phi_{2}$, with $\Phi_{1} \neq \emptyset$, such that $\Phi_{1} \perp \Phi_{2}$. > > If $(\Phi,E)$ is not reducible then we say it is **irreducible**. ^definition > [!note] Note. > It would be better to use the terminology 'decomposable' and 'indecomposable', but alas. ^note > [!equivalence] > $(\Phi,E)$ is irreducible if and only if its [[Dynkin diagram]] is [[connected]]. > [!proof] > Suppose $(\Phi, E)$ is *re*ducible. Then its Dynkin diagram is disconnected: if $\Phi=\Phi_{1} \sqcup \Phi_{2}$ with $\Phi_{1} \perp \Phi_{2}$, then each $\Phi_{i}$ is a [[root system]] for its [[submodule generated by a subset]] (see below). Let $\Delta_{i}$ be a [[root basis]] of $\Phi_{i}$. Then $\Delta=\Delta_{1} \cup \Delta_{2}$ is a [[root basis]] of $\Phi$, and $\Delta_{1} \perp \Delta_{2}$ so the Dynkin diagram is disconnected. > > Conversely, suppose the dynkin diagram is *dis*connected. Then $\Delta=\Delta_{1} \sqcup \Delta_{2}$ with $\Delta_{1} \perp \Delta_{2}$. Let $\Phi_{i}=\mathbb{R} \Delta_{i} \cap \Phi$. Then it is clear $\Phi_{1} \perp \Phi_{2}$. Why do we have $\Phi=\Phi_{1} \cup \Phi_{2}$? Let $W$ be the [[Weyl group of a root system|Weyl group]] which is [[the Weyl group and root bases|generated by simple reflections]]. Simple reflections preserve $\Phi_{1}$ and $\Phi_{2}$, hence $W$ preserves $\Phi_{1}$ and $\Phi_{2}$. If $\alpha \in \Phi$ is any root, then it must be of the form $\alpha=w(\alpha_{i})$ for some simple root $\alpha_{i} \in \Delta$, so by the previous sentence $\alpha \in \Phi_{1}$ or $\Phi_{2}$. ^equivalence > [!proposition] > A reducible root system $\Phi=\Phi_{1} \sqcup\Phi_{2}$ gives rise to a [[direct sum of modules|direct sum]] decomposition $E=E_{1} \oplus E_{2}$ where $(E_{1}, \Phi_{1})$, $(E_{2},\Phi_{2})$ are each [[root system|root systems]]. > > Indeed, if $\Phi$ is reducible, $\Phi=\Phi_{1} \sqcup\Phi_{2}$, consider $E_{i}=\mathbb{R}$-span of $\Phi_{i}$, $i=1,2$. Then $E=E_{1} \oplus E_{2}$ because certainly $E=E_{1}+E_{2}$ ($\Phi$ spans) and also $E_{1} \perp E_{2}$ so $E_{2}=E_{1}^{\perp}$ with $\Phi_{i}$ a root system in $E_{i}$. ^proposition ---- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```