----- > [!proposition] Proposition. ([[root spaces are one-dimensional]]) > Using the [[classification of the irreps of sl2 over C|representation theory]] [[any representation of sl2(C) is completely reducible|of]] $\mathfrak{sl}_{2}(\mathbb{C})$, we may deduce the following. > > If $\alpha \in \Phi$, then $\text{dim }\mathfrak{g}_{\alpha}=1$. Moreover, if $c \in \mathbb{C}$, then $c \alpha \in \Phi \iff c=\pm 1$. ^proposition > [!proposition] Corollary. > If $\alpha \in \Phi$, then $\mathfrak{m}_{\alpha}=\mathfrak{g}_{\alpha} \oplus [\mathfrak{g}_{\alpha}, \mathfrak{g}_{-\alpha}] \oplus \mathfrak{g}_{-\alpha}$. ^proposition > [!proof]- Proof. ([[root spaces are one-dimensional]]) > Summary: > 1. Invoke [[finding sl2-triples]] to get a copy $\mathfrak{m}_{\alpha}$ of $\mathfrak{sl}_{2}$ inside $\mathfrak{g}$. > 2. Define $V=\mathfrak{t} \oplus \bigoplus_{c \in \mathbb{C}}\mathfrak{g}_{c \alpha}$, an adjoint representation of $\mathfrak{m}_{\alpha}$ > 3. Show in fact $V=U:=\mathfrak{t}+\mathfrak{m}_{\alpha}=\mathfrak{t} \oplus \mathfrak{g}_{\alpha} \oplus \mathfrak{g}_{-\alpha}$ (from which the result follows). Do so by examining a prospective complement $W$ (exists by complete reducibility) and showing $W=0$. > 1. To show $W=0$, look at its weights as an $\mathfrak{sl}_{2}$-rep... neither 0 nor 1 will be a weight, so nothing can be! > > Obtain using [[finding sl2-triples]] a triple $(e_{\alpha}, h_{\alpha}, f_{\alpha})$ satisfying the $\mathfrak{sl}_{2}(\mathbb{C})$-relations. Denote by $\mathfrak{m}_{\alpha}$ the [[Lie subalgebra|subalgebra]] it [[spans]]. > > Define $V:=\mathfrak{t} \oplus \bigoplus_{c \in \mathbb{C}}\mathfrak{g}_{c \alpha}.$ > We will show that $V=U:=\mathfrak{t}+\mathfrak{m}_{\alpha}=\mathfrak{t} \oplus \mathfrak{g}_{\alpha} \oplus \mathfrak{g}_{-\alpha}$, from which the result will follow. > > > By [[root space decomposition of a Lie algebra|additive containment]], $V$ is stable under the [[adjoint representation|adjoint action]] of $\mathfrak{m}_{\alpha}$, and so defines a (adjoint) [[Lie algebra representation|representation]] of $\mathfrak{m}_{\alpha} \cong \mathfrak{sl}_{2}$. Define $U:=\mathfrak{t}+\mathfrak{m}_{\alpha}$. Since $[\mathfrak{t}, \mathfrak{m}_{\alpha}] \subset \mathfrak{m}_{\alpha}$ and $\mathfrak{m}_{\alpha}$ is a subalgebra, $U$ is an $\mathfrak{m}_{\alpha}$-subrepresentation of $V$. By [[completely reducible|complete reducibility]] of $\mathfrak{sl}_{2}$ there then exists a [[complement of a linear subspace|complementary subrepresentation]] $W \subset V$ such that $V=U \oplus W$. We are done if we can show $W=0$. > > To do so, we show its weights are trivial. Let $S=\{ c \in \mathbb{C}: c \alpha \in \Phi \}$. First note the [[weight space for sl2(C)|weights]] of $V$ are the [[eigenvalue|eigenvalues]] of $h_{\alpha}$ on $V$, in other words $0$ (corresponding to $\mathfrak{t}$) and $c \alpha(h_{\alpha})=2c$ for $c \in S$.[^1] Since the weights of an $\mathfrak{sl}_{2}$-rep [[weights characterize any representation of sl2(C)|are known to be integers]], $S \subset \frac{1}{2}\mathbb{Z}$. > > Now look at $W$. Since $0$ is not a weight of $W$ (the 0-weight space is already taken by $\mathfrak{t} \subset U$) $W$ can't have any even weights ($\mathfrak{sl}_{2}$ theory - subtracting off from highest weight can't hit zero). In particular $2 \alpha \notin \Phi$, since $2\alpha(h_{\alpha})=4$ would be a weight of $W$. We have shown at this point $2 \alpha \notin \Phi$ for every $\alpha \in \Phi$. But this implies that $\frac{1}{2} \alpha \notin \Phi$ for every $\alpha \in \Phi$: indeed, if $\beta=\frac{1}{2}\alpha$ were a root, then $2 \beta=\alpha$ cannot be a root, contradiction. So $\frac{1}{2} \alpha \notin \Phi$, hence $1=\frac{1}{2}\alpha(h_{\alpha})$ is not a weight of $W$. > > But if neither $0$ nor $1$ are weights of $W$... nothing is. So $W=\{ 0 \}$ and we see that $\mathfrak{t}+\mathfrak{m}_{\alpha}=U=V$. ----- #### - [[special linear Lie subalgebra]] [^1]: This is by definition of $\mathfrak{g}_{c \alpha}$ as [[simultaneously diagonalizable|simultaneous eigenspaces]], don't think too hard. [^2]: For then $V=\mathfrak{t} \oplus \bigoplus_{c \in \mathbb{C}^{\times}}\mathfrak{g}_{c \alpha}=\mathfrak{t}+\mathfrak{m}_{\alpha}=\mathfrak{t} \oplus \langle\overbrace{ e_{\alpha} }^{ \subset \mathfrak{g}_{\alpha} } \rangle \oplus \langle \overbrace{ f_{\alpha} }^{ \subset \mathfrak{g}_{-\alpha} } \rangle$ ... think. ---- Consider the subspace $U=\mathfrak{t}+\mathfrak{m}_{\alpha} \subset V$. Since $[\mathfrak{t}, \mathfrak{m}_{\alpha}] \subset \mathfrak{m}_{\alpha}$ and $\mathfrak{m}_{\alpha}$ is a subalgebra, $U$ is an $\mathfrak{m}_{\alpha}$-subrepresentation of $V$. By [[completely reducible|complete reducibility]] there then exists a [[complement of a linear subspace|complementary subrepresentation]] $W \subset V$ such that $V=U \oplus W$. *Claim: $W=0$.* Note that we are done once we show this.[^2] To show $W=0$, we'll look at its weights. Since $0$ is not a weight of $W$ (the 0-weight space is already taken by $\mathfrak{t} \subset U$) $W$ can't have any even weights ($\mathfrak{sl}_{2}$ theory - subtracting off from highest weight can't hit zero). In particular $2 \alpha \notin \Phi$, since $2\alpha(h_{\alpha})=4$ would be a weight of $W$. We have shown at this point $2 \alpha \notin \Phi$ for every $\alpha \in \Phi$. But this implies that $\frac{1}{2} \alpha \notin \Phi$ for every $\alpha \in \Phi$: indeed, if $\beta=\frac{1}{2}\alpha$ were a root, then $2 \beta=\alpha$ cannot be a root, contradiction. So $\frac{1}{2} \alpha \notin \Phi$, hence $1=\frac{1}{2}\alpha(h_{\alpha})$ is not a weight of $W$. But if neither $0$ nor $1$ are weights of $W$... nothing is. So $W=\{ 0 \}$ and we see that $\mathfrak{t}+\mathfrak{m}_{\alpha}=U=V$. #### 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 > ```