---- > [!definition] Definition. ([[root space decomposition of a Lie algebra]]) > Let $\mathfrak{g}$ be a [[semisimple Lie algebra|semisimple]] [[Lie algebra]] and $\mathfrak{t} \subset \mathfrak{g}$ a [[Cartan subalgebra]]. Consider the restriction of the [[adjoint representation]] $\text{ad}:\mathfrak{g} \to \mathfrak{gl}(\mathfrak{g})$ to $\mathfrak{t}$: this is a [[Lie algebra representation|representation]] $\text{ad}|_{\mathfrak{t}}:\mathfrak{t} \to \mathfrak{gl}(\mathfrak{g})$. > The image of $\text{ad}|_{\mathfrak{t}}$ consists of commuting [[diagonalizable|semisimple]] [[linear operator|linear operators]], allowing us to [[on factorizing a Lie algebra representation into weight spaces|factorize]] $\mathfrak{g}=V$ as follows: $\mathfrak{g}=\bigoplus_{i=1}^{k} \mathfrak{g}_{\lambda_{i}}, $ where $\begin{align} \mathfrak{g}_{\lambda_{i}}&=\{ x \in \mathfrak{g}: t \cdot x = \lambda_{i}(t) x \text{ for all } t \in \mathfrak{t}\} \\ & = \{ x \in \mathfrak{g}: [t,x]=\lambda_{i}(t)x \text{ for all } t \in \mathfrak{t} \} \end{align}$ for some $S=\{\lambda_{1},\dots,\lambda_{k}\} \subset \mathfrak{t}^{*}$. > We break $S$ up into $\{ 0 \} \sqcup \Phi$, $0$ the zero-functional, and write $\textcolor{Skyblue}{\mathfrak{g}=\mathfrak{g}_{0} \oplus \bigoplus_{\alpha \in \Phi} \mathfrak{g}_{\alpha}.}$ This is called the **root space decomposition of $\mathfrak{g}$**. Elements of $\Phi$ are called **roots**. > Property **(5)** below provides an [[isomorphism]] $\mathfrak{t} \cong \mathfrak{t}^{*}$ [[killing form|via]] $t \mapsto \kappa(t, -)$. If $\lambda \in \mathfrak{t}^{*}$, we denote by $t_{\lambda} \in \mathfrak{t}$ its corresponding element in $\mathfrak{t}$ under this isomorphism: $\kappa(t_{\lambda}, x)={\lambda}(x)$ for all $x \in \mathfrak{t}$. In particular, every root $\alpha \in \mathfrak{t}^{*}$ gets identified with some $t_{\alpha} \in \mathfrak{t}$ which we call the **coroot associated to $\alpha$.** ^definition > [!note] Note. > This is the discussion in [[on factorizing a Lie algebra representation into weight spaces]] applied specifically to the restriction of the [[adjoint representation]] to $\mathfrak{t}$. Root space = weight space of $\text{ad} |_{\mathfrak{t}}: \mathfrak{t} \to \mathfrak{gl}(\mathfrak{g})$. > [!intuition] > The big idea here is that letting the Cartan subalgebra $\mathfrak{t}$ act on $\mathfrak{g}$ via the [[adjoint representation]] turns out to yield decompositions analogous to those we found in [[classification of the irreps of sl2 over C]]... there, the basis element $h$ played the role $\alpha \in \Phi$ plays.. (discussed better elsewhere) ^intuition There are 13 crucial properties of roots discussed below. > [!basicproperties] How do roots interact with one another? > **1.1 (Additive containment)** For $\alpha, \beta \in \mathfrak{t}^{*}$, one has $[\mathfrak{g}_{\alpha}, \mathfrak{g}_{\beta}] \subset \mathfrak{g}_{\alpha + \beta}$. Fulton and Harris call this the 'fundamental calculation'. It follows from the Jacobi identity. > **1.2. (Killing form kills many pairings)** If $\alpha + \beta \neq 0$, then $\mathfrak{g}_{\alpha} \perp \mathfrak{g}_{\beta}$ under the [[killing form]], i.e., $\kappa(x,y)=0$ for all $x \in \mathfrak{g}_{\alpha}$, $y \in \mathfrak{g}_{\beta}$. > **1.3. (Though not those in $\mathfrak{g}_{0}$)** The restriction of $\kappa$ to $\mathfrak{g}_{0}(=\mathfrak{t})$ is [[nondegenerate bilinear form|nondegenerate]]. > **1.4. ($\mathfrak{t}=\mathfrak{g}_{0}$)** Since $\mathfrak{t}$ [[abelian Lie algebra|abelian]], $\mathfrak{t} \subset \mathfrak{g}_{0}$. In fact the converse holds: $\mathfrak{t}=\mathfrak{g}_{0}$, but we omit the proof. > **1.5. (The killing isomorphism $\mathfrak{t} \to \mathfrak{t}^{*}$)** Putting together **(3)** and **(4)**, we see that the restriction of $\kappa$ to $\mathfrak{t}$ is [[nondegenerate bilinear form|nondegenerate]], so that [[musical isomorphism induced by a nondegenerate bilinear form|the map]] $\begin{align} \mathfrak{t} &\to \mathfrak{t}^{*} \\ t & \mapsto \kappa(t,-) \end{align}$ is an [[isomorphism]]. > > > > [!proof]- Proofs. > > **1.1.** This follows from the Jacobi identity. Indeed, if $t \in \mathfrak{t}$, $x \in \mathfrak{g}_{\alpha}$ and $y \in \mathfrak{g}_{\beta}$, then $\begin{align} > > \text{ad}_{t}([x,y])&= [t, [x,y]] \\ > > &= -[y, \underbrace{ [t,x] }_{ \text{ad}_{t}(x)=\alpha(t)x }] - [x, \underbrace{ [y,t] }_{ -\text{ad}_{t}(y)=-\beta(t)y }] \\ > > &= -[y, \alpha (t) x] - [x, \beta(t)y] \\ > > &= \alpha(t) [x, y] + \beta(t)[x,y] \\ > > &= (\alpha +\beta)(t) [x,y] > > \end{align}$ > > as required. > > > > **1.2.** Let $t$ be such that $(\alpha+\beta)(t) \neq 0$. Let $x \in \mathfrak{g}_{\alpha}$, $y \in \mathfrak{g}_{\beta}$. > > Recalling the relationship between [[trace form|trace forms]] and [[Lie algebra|Lie brackets]], we have $\kappa(\underbrace{ [t,x] }_{ \alpha(t) x },y)=- \kappa(x, \underbrace{ [t,y] }_{ \beta(t) y }).$ > > So $\underbrace{ (\alpha+\beta)(t) }_{ \neq 0 }\kappa(x,y)=0,$ > > hence $\kappa(x,y)=0$. > > > > **1.3.** Let $t \in \mathfrak{g}_{0}$ be such that $\kappa(t,t') = 0$ for all $t' \in \mathfrak{g}_{0}$. By (1.2), $\mathfrak{g}_{\alpha} \perp \mathfrak{g}_{\beta}$ for every $\alpha$ (recall $\alpha \neq 0$), so $t$ kills every vector not in $\mathfrak{g}_{0}$. Therefore, since $t \in \mathfrak{g}_{0}$ also kills every element of $\mathfrak{g}_{0}$, it kills all of $\mathfrak{g}$. But [[The Cartan-Killing Criterion|The Cartan-Killing Criterion for semisimplicity]] says $\kappa(-,-)$ is [[nondegenerate bilinear form|nondegenerate]] on $\mathfrak{g}$. So $t=0$. Since $\mathfrak{g}_{0}=\mathfrak{t}$, we're done. > > [!basicproperties] [[Root system]]-esque roperties of $\Phi$. > **2.1.** $\Phi$ [[submodule generated by a subset|spans]] $\mathfrak{t}^{*}$ > **2.2.** If $\alpha \in \Phi$, then $-\alpha \in \Phi$. (Furthermore, [[root spaces are one-dimensional|this result]] shows that no other multiplies of $\alpha$ are roots.) > > > > [!proof]- Proofs. > > **2.1.** Suppose not. Then there exists nonzero $t \in \mathfrak{t}$ such that $\alpha(t)=0$ for all $\alpha \in \Phi$. Therefore if $x \in \mathfrak{g}_{\alpha}$ then $[t,x]=\alpha(t) x=0$. Also $[t , \mathfrak{t}]=0$ since $\mathfrak{t}$ is [[abelian Lie algebra|abelian]]. Thus $t \in Z(\mathfrak{g})$. But the [[center of a Lie algebra|center]] of a [[semisimple Lie algebra|semisimple]] [[Lie algebra]] is trivial ([[adjoint representation]] is [[faithful Lie algebra representation|faithful]] + [[completely reducible]]). > > > > **2.2.** Recall (1.2) that $\mathfrak{g}_{\alpha} \perp \mathfrak{g}_{\beta}$ when $\alpha + \beta \neq 0$. Since the killing form is nondegenerate, $\mathfrak{g}_{\alpha}$ cannot be orthogonal to the whole of $\mathfrak{g}$, so $\beta=-\alpha$ must be in $\Phi$. > > ^properties > [!basicproperties] Towards building $\mathfrak{sl}_{2}(\mathbb{C})$ > For a root $\alpha \in \Phi$: > **3.1.** Given $x \in \mathfrak{g}_{\alpha}$ and $y \in \mathfrak{g}_{-\alpha}$, one has $[x,y]= \kappa(x,y)t_{\alpha}$. > **3.2.** $[\mathfrak{g}_{\alpha}, \mathfrak{g}_{-\alpha}]$ is one-dimensional > **3.3.** $\alpha(t_{\alpha})=^{\text{by def.}}\kappa(t_{\alpha}, t_{\alpha}) \neq 0$. > From these we obtain the following important results: **3.4.** [[finding sl2-triples]], and then **3.5.** [[root spaces are one-dimensional]] **3.1.** By definition, $t_{\alpha}$ is the unique element of $\mathfrak{t}$ satisfying $\alpha(t)=\kappa(t, t_{\alpha})$ for all $t \in \mathfrak{t}$. Hence $\kappa(x,y)t_{\alpha}$ is the unique element of $\mathfrak{t}$ satisfying $\alpha(t)=\kappa(t, \kappa(x,y)t_{\alpha})$ for all $t \in \mathfrak{t}$. The goal is therefore to show that $[x,y]$ *also* satisfies this property, and so by uniqueness equals $\kappa(x,y)t_{\alpha}$. Indeed: $\kappa(t, [x,y])= \kappa([t,x], y)=\kappa(\alpha(t)x, y)=\alpha(t) \kappa(x,y)=\kappa(t, t_{\alpha}) \kappa(x, y)$ and the result follows by bilinearity of [[killing form]]. **3.2.** By (3.1), $[\mathfrak{g}_{\alpha} , \mathfrak{g}_{-\alpha}] \subset \span t_{\alpha}$ is at most one-dimensional. So it suffices to find $x \in \mathfrak{g}_{\alpha}$, $y \in \mathfrak{g}_{-\alpha}$ with $[x,y] \neq 0$. But such $x,y$ have to exist, else [[killing form]] would be degenerate (recall $\mathfrak{g}_{\alpha} \perp \mathfrak{g}_{\beta}$ whenever $\alpha+\beta \neq 0$) **3.3.** Let $x \in \mathfrak{g}_{\alpha}$ and $y \in \mathfrak{g}_{-\alpha}$ such that $\kappa(x,y)=1$. Then $\{ x, t_{\alpha}, y \}$ satisfy the relations $[t_{\alpha},x]=\alpha(t_{\alpha})x, [t_{\alpha}, y]=-\alpha(t_{\alpha})y, [x,y]=\kappa(x,y)t_{\alpha}=t_{\alpha}.$ Denote this [[Lie subalgebra|subalgebra]] by $\mathfrak{h}$. Suppose $\alpha(t_{\alpha})=0$. Then $\mathfrak{h}$ is a [[derived and central series of a Lie algebra|solvable Lie algebra]] since $\mathfrak{h}^{(2)}=0$. By [[Lie's Theorem]], there exists a [[basis]] of $\mathfrak{g}$ such that $\text{ad}: \mathfrak{h} \hookrightarrow \mathfrak{gl}(\mathfrak{g})$ lands in [[upper-triangular matrix|upper triangular matrices]]. Therefore $t_{\alpha} \in [\mathfrak{h}, \mathfrak{h}]$ lands in *strictly* upper triangular matrices (recall $[\mathfrak{b}_{n}, \mathfrak{b}_{n}]=\mathfrak{n}_{n}$). So $\text{ad }t_{\alpha} \in \mathfrak{gl}(\mathfrak{g})$ is [[nilpotent linear operator|nilpotent]]. But it is also [[diagonalizable|semisimple]] (by [[Cartan subalgebra|definition]] of $\mathfrak{t}$). So $\text{ad }t_{\alpha}=0$. So $t_{\alpha}=0$ because $\text{ad}$ is injective. This is a contradiction. ---- #### ---- #### 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 > ```