---- Motivation: take the discussion in [[orthogonal Lie algebra]], but now we care about [[alternating multilinear map|skew-symmetric]] bilinear forms instead of symmetric ones. > [!definition] Definition. ([[symplectic Lie algebra]]) > Let $k$ be a [[field|field]] ($\text{char }k \neq 2$) and $V$ a $k$-[[vector space]] endowed with an [[alternating multilinear map|alternating]], [[nondegenerate bilinear form|nondegenerate]] [[bilinear map|bilinear form]] $\langle -, - \rangle: V \times V \to k$. Then with $\mathfrak{gl}(V)$ denoting the [[general linear Lie algebra|general linear]] [[Lie algebra]] on $V$ define $\mathfrak{sp}(V)=\{ f \in \mathfrak{gl}(V) : \langle f(v), w \rangle + \langle v, f(w) \rangle =0 \text{ for all } v,w \in V \}.$ This is a [[Lie subalgebra]] of $\mathfrak{gl}(V)$ called the **symplectic Lie algebra**. The matrix version is denoted $\mathfrak{sp}_{n}$. > If we choose an [[linear isomorphism|isomorphism]] $V \cong \mathbb{F}^{\oplus n}$ (i.e., choose a [[basis]] of $V$), then $\langle -,- \rangle$ [[matrix of a bilinear form|corresponds to a (nonsingular)]] [[alternating multilinear map|skew-symmetric]] [[matrix]] $A \in \text{Mat}_{n}(k)$, $\langle v,w \rangle=v^{\top}Aw$ and we have [^1] $\mathfrak{sp}(V, \langle -,- \rangle ) \cong \{ X \in \mathfrak{gl}_{n}(\mathbb{F}): X^{\top}A + AX=0 \}.$ > Over $\mathbb{C}$, [[the isomorphism classes of orthogonal and symplectic Lie algebras do not depend on choice of bilinear form|It turns out]] that any two choices of [[bilinear map|bilinear form]] $\langle -,- \rangle$, $\langle -,- \rangle'$ yield [[Lie algebra homomorphism|isomorphic]] symplectic Lie algebras. In coordinates, we usually take the form $\langle -,- \rangle$ to be the [[skew-symmetric matrix|skew-symmetric]] form given by $A=\begin{bmatrix} 0 & I_{\ell} \\ -I_{\ell} & 0 \end{bmatrix}$where $n=2\ell$. > Explicitly, have $\mathfrak{sp}_{2\ell}=\{ \begin{bmatrix} P & Q \\ R & -P^{\top} \end{bmatrix} : P,Q,R \in \mathfrak{gl}_{\ell} , Q=Q^{\top}, R=R^{\top} \}.$ Mnemonic: For 'symplectic, $Q, R$ are symmetric.' ^7a882e > [!justification] Basis and dimension for $\mathfrak{sp}_{2 \ell}$. > For $n=2\ell$, one has $\begin{align} > \mathfrak{sp}(V, \langle -,- \rangle_{\text{preferred}} ) & \cong \{ X \in \mathfrak{gl}_{n}: X^{\top} A + AX = 0 \} \\ > &= \{ \begin{bmatrix}P &Q \\ R & S\end{bmatrix} \in \mathfrak{gl}_{n} : \begin{bmatrix}P^{\top} & R^{\top} \\ Q^{\top} & S^{\top}\end{bmatrix} \begin{bmatrix}0 & I_{\ell} \\-I_{\ell} & 0 > \end{bmatrix} + \begin{bmatrix}P & Q \\ R & S\end{bmatrix}\begin{bmatrix}0 & I_{\ell} \\-I_{\ell} & 0 > \end{bmatrix} = 0 \}. > \end{align}$ > We see that > $ > \begin{align} > \begin{bmatrix}P^{\top} & R^{\top}\\ Q^{\top} & S^{\top}\end{bmatrix} \begin{bmatrix}0 & I_{\ell} \\-I_{\ell} & 0 > \end{bmatrix} + \begin{bmatrix}0 & I_{\ell} \\-I_{\ell} & 0 > \end{bmatrix} \begin{bmatrix}P & Q \\ R & S\end{bmatrix} & = \begin{bmatrix} > -R^{\top} & P^{\top} \\ -S^{\top} & Q^{\top} > \end{bmatrix} + \begin{bmatrix} > R & S \\ -P & -Q > \end{bmatrix} > \end{align} > $ > and so the condition $X^{\top}A+AX=0$ in this case amounts to asserting > $R^{\top} = R, P^{\top} = -S, S^{\top} = -P, Q^{\top} = Q,$ > in other words, we have that the matrices comprising $\mathfrak{sp}_{2 \ell}$ have the form $\begin{bmatrix} > P & Q \\ > R & -P^{\top} > \end{bmatrix}$ > where $Q,R$ are symmetric. Then we get a $\frac{\ell(\ell + 1)}{2}$ basis elements contributed from $Q \text{ and }R$ each and $\ell^{2}$ basis elements contributed from $P$ for a total of $\ell(\ell+1)+\ell^{2}=2 \ell^{2}+\ell$ basis elements in total. > > For example, for $\mathfrak{sp}_4(\mathbb{C})$ a full basis is: > $\begin{align} > E_{13} \\ > E_{14} + E_{23} \\ > E_{24} \\ > \ \\ > E_{31} \\ > E_{32} + E_{41} \\ > E_{42} \\ > \ \\ >\textcolor{Skyblue}{ E_{11}-E_{33} }\\ > \textcolor{Skyblue}{E_{22} - E_{44} }\\ > E_{12} - E_{43} \\ > E_{21} - E_{34} > \end{align}$ > > > > One reason for choosing the form $A$ like we did was that it makes the [[root space decomposition of a Lie algebra|root space decomposition]] of $\mathfrak{sp}_{2 \ell}$ easier to compute: the basis vectors in white form a [[simultaneously diagonalizable|simultaneous eigenbasis]] for the [[adjoint representation|adjoint actions]] of those in $\textcolor{Skyblue}{\text{blue}}$. This is related to the [[root space decomposition of a Lie algebra|root space decomposition]] below. Root space decomposition. $\mathfrak{sp}_{4}$ has [[Cartan subalgebra|CSA]] $\mathfrak{t}:=\langle E_{11}-E_{33}, E_{22}- E_{4 4} \rangle=\mathfrak{g}_{0}$. Denote by $e^{1}, e^{2}$ the [[dual basis]]. Since $\text{dim }\mathfrak{sp}_{4}=10$, there are $8$ other [[root system|roots]] ($4$ up to sign). The form $A$ was chosen so that the [[basis]] described above [[simultaneously diagonalizable|simultaneously diagonalizes]] the elements in $\mathfrak{t}$. In other words, we already know the root spaces are $\langle E_{13} \rangle, \langle E_{14 }+E_{23} \rangle, \dots$. We just need to find the corresponding roots, which is a matter of computation. [[vector space of m-by-n matrices|Recall]] the formula $E_{ij}E_{k \ell}=\delta_{jk}E_{i \ell}$. - For $E_{13}$, compute $\begin{align} [\textcolor{Skyblue}{E_{11}-E_{33}}, E_{13}] = [E_{11}, E_{13} ] - [E_{3 3}, E_{13}] = 2E_{13} \\ [\textcolor{Skyblue}{E_{22}-E_{44}}, E_{13}]=0. \end{align}$ so the corresponding [[root system|root]] is $2e^{1} \in \mathfrak{t}^{*}$. $E_{31}$ clearly has the negated root $-2e^{1}$. Similarly, the roots for $E_{24}$ and $E_{42}$ are $2e^{2}$ and $-2e^{2}$ respectively. - For $E_{14}+E_{23}$, can show the root is $e_{1}+e_{2}$... - others are all same reasoning, no time right now In all, the roots are $\Phi= \{ e_{i}-e_{j} : 1 \leq i \neq j \leq 2 \} \sqcup \{ \pm (e_{i}+e_{j}):1 \leq i \leq j \leq 2 \}=\{ \pm(e_{1}-e_{2}), \pm(e_{1}+e_{2}), \pm 2e_{1}, \pm 2e_{2} \}.$ Drawing picture, this is the $C_{2}$ [[root system]]. Note that the $C_{2}$ [[root system]] is [[root system isomorphism|isomorphic]] to $B_{2}$, and $B_{2}$ is more straightforward to draw, so many times we'll analyze $\mathfrak{sp}_{4}$ using $B_{2}$. [[classification of complex semisimple Lie algebras|In general]], [[the classical Lie algebras under the Cartan-Killing classification|though]], $\Phi(\mathfrak{sp}_{2n} )\cong C_{n} \not \cong B_{n}$. ![[CleanShot 2025-06-02 at [email protected]]] $\mathfrak{t}$ has basis $E_{11}-E_{4 4}, E_{2 2}-E_{5 5}, E_{3 3}-E_{6 6}$; denote the dual basis $e_{1},e_{2},e_{3}$. Then: $\Phi=\{ \pm e_{i} \pm e_{j} : 1 \leq i \neq j \leq 3 \} \cup \{ \pm 2e_{i} \}.$ A root basis is $\Delta =\{ e_{1}-e_{2}, e_{2}-e_{3}, 2e_{3} \}$. [[Dynkin diagram]]: ![[Pasted image 20250602102301.png|500]] The Weyl group is $S_{3}\ltimes C_{2}^{3}$. ![[Pasted image 20250602104144.png]] ---- #### [^1]: For the same reasoning as is provided in [[orthogonal Lie algebra]]. ---- #### 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 > ```