---- > [!definition] Definition. ([[Heisenberg Lie algebra]]) > > Let $\mathfrak{h}$ be a 3-dimensional [[vector space]] over $\mathbb{C}$ with [[basis]] $\{ x,y,z \}$. Define a [[bilinear map]] $[-,-]:\mathfrak{h \times h} \to \mathfrak{h}$ by setting $[x,x]=[y,y]=[z,z]=0$ > and $[x,z]=[z,x]=[y,z]=[z,y]=0$ > and $[x,y]=-[y,z]=z$ > and extending [[bilinear map|bilinearly]]. This is a [[Lie subalgebra]] of the [[general linear Lie algebra]] $\mathfrak{gl}_{3}$ known as the **Heisenberg algebra**. It consists of the $3 \times 3$ [[strictly triangular matrix|strictly upper triangular]] matrices over $\mathbb{C}$. > ---- #### > [!justification] > The claim is that the [[matrix|matrices]] $X=\begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}, Y=\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}, Z = \begin{bmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$ form a [[basis]] for $\mathfrak{h}$ realizing it as a [[Lie subalgebra]] of $\mathfrak{gl}_{2}$. Clearly they span a 3-dimensional subspace of $\mathfrak{gl}_{2}$; we just have to check that the [[commutator]] operation $[-,-]$ on $\mathfrak{gl}_{2}$ preserves the prescribed bracket under the embedding $x \mapsto X, y \mapsto Y, z \mapsto Z$ of $\mathfrak{h}$ into $\mathfrak{gl}_{2}$. The conditions $[x,x]=[y,y]=[z,z]=0$ are satisfied by definition. The conditions $[x,z]=[z,x]=0$ and $[y,z]=[z,y]=0$ are satisfied because $X$ and $Z$ commute, as do $Y$ and $Z$; indeed, $XZ=ZX=0$ and $YZ=ZY=0$. Finally, we have $[x,y]=-[y,x]=z$ because $XY-YX=\begin{bmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} - \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}=Z$ as required. Note that through these computations we have also computed $[X,Y],[Y,X]$ and $[X,Z]$, thus showing stability of or embedding $\mathfrak{h} \hookrightarrow \mathfrak{gl}_{3}$ under the commutator. ^justification > [!basicexample] > The [[adjoint representation]] of $\mathfrak{h}$ is the [[Lie algebra homomorphism]] $\begin{align} \rho: \mathfrak{h} &\to \mathfrak{gl}_{3}(\mathfrak{h}) \\ h & \mapsto [h, -], \end{align}$where $[h,-]=\text{ad}_{h}$ is the map $\mathfrak{h} \to \mathfrak{h}$ given by $\text{ad}_{h}(\ell)=[h,\ell]$. In other words, an element $h \in \mathfrak{h}$ acts on another element $h' \in \mathfrak{h}$ as $h \cdot h'=[h,h']$. > We will find a [[Lie algebra subrepresentation|subrepresentation]] of $(\rho, \mathfrak{h})$ with no $\mathfrak{h}$-invariant [[complement of a linear subspace|complement]] and will then be done by the result from October 18th's lecture. Define $W=\text{span}(z)$. Note that this [[linear subspace]] is indeed a subrepresentation, since $[x,z]=[y,z]=[z,z]=0$ and so $W$ is stable under the action. However, $W$ does not admit a $\mathfrak{h}$-invariant complement in $\mathfrak{h}$. Indeed, let $W'$ be such that $W \cap W'=(0)$ and suppose $W'$ is stable under the action. The general element of $W'$ is a vector of the form $v=ax+by+cz$. $[y,v]=[y, ax+by+cz]=a[y,x]+b[y,y]+c[y,z]=a[y,x]=-az$ and so $a=0$. Similarly, $[x,v]=[x, ax+by+cz]=[x, by]=bz$ implying $b=0$. But if $v=cz$ then since $v \notin W$ we have $c=0$. It follows that $v=0$, and thus that $W'=(0)$. ^dc6d47 - [ ] need that result in obsidian ---- #### 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 > ```