# The Unapologetic Mathematician

## Orthogonal and Symplectic Lie Algebras

For the next three families of linear Lie algebras we equip our vector space $V$ with a bilinear form $B$. We’re going to consider the endomorphisms $f\in\mathfrak{gl}(V)$ such that

$\displaystyle B(f(x),y)=-B(x,f(y))$

If we pick a basis $\{e_i\}$ of $V$, then we have a matrix for the bilinear form

$\displaystyle B_{ij}=B(e_i,e_j)$

and one for the endomorphism

$\displaystyle f(e_i)=\sum\limits_jf_i^je_j$

So the condition in terms of matrices in $\mathfrak{gl}(n,\mathbb{F})$ comes down to

$\displaystyle\sum\limits_kB_{kj}f_i^k=-\sum_kf_j^kB_{ik}$

or, more abstractly, $Bf=-f^TB$.

So do these form a subalgebra of $\mathfrak{gl}(V)$? Linearity is easy; we must check that this condition is closed under the bracket. That is, if $f$ and $g$ both satisfy this condition, what about their commutator $[f,g]$?

$\displaystyle\begin{aligned}B([f,g](x),y)&=B(f(g(x))-g(f(x)),y)\\&=B(f(g(x)),y)-B(g(f(x)),y)\\&=-B(g(x),f(y))+B(f(x),g(y))\\&=B(x,g(f(y)))-B(x,f(g(y)))\\&=-B(x,f(g(y))-g(f(y)))\\&=-B(x,[f,g](y))\end{aligned}$

So this condition will always give us a linear Lie algebra.

We have three different families of these algebras. First, we consider the case where $\mathrm{dim}(V)=2l+1$ is odd, and we let $B$ be the symmetric, nondegenerate bilinear form with matrix

$\displaystyle\begin{pmatrix}1&0&0\\ 0&0&I_l\\ 0&I_l&0\end{pmatrix}$

where $I_l$ is the $l\times l$ identity matrix. If we write the matrix of our endomorphism in a similar form

$\displaystyle\begin{pmatrix}a&b_1&b_2\\c_1&m&n\\c_2&p&q\end{pmatrix}$

our matrix conditions turn into

$\displaystyle\begin{aligned}a&=0\\c_1&=-b_2^T\\c_2&=-b_1^T\\q&=-m^T\\n&=-n^T\\p&=-p^T\end{aligned}$

From here it’s straightforward to count out $2l$ basis elements that satisfy the conditions on the first row and column, $\frac{1}{2}(l^2-l)$ that satisfy the antisymmetry for $p$, another $\frac{1}{2}(l^2-1)$ that satisfy the antisymmetry for $n$, and $l^2$ that satisfy the condition between $m$ and $q$, for a total of $2l^2+l$ basis elements. We call this Lie algebra the orthogonal algebra of $V$, and write $\mathfrak{o}(V)$ or $\mathfrak{o}(2l+1,\mathbb{F})$. Sometimes we refer to the isomorphism class of this algebra as $B_l$.

Next up, in the case where $\mathrm{dim}(V)=2l$ is even we let the matrix of $B$ look like

$\displaystyle\begin{pmatrix}0&I_l\\I_l&0\end{pmatrix}$

A similar approach to that above gives a basis with $2l^2-l$ elements. We also call this the orthogonal algebra of $V$, and write $\mathfrak{o}(V)$ or $\mathfrak{o}(2l,\mathbb{F})$. Sometimes we refer to the isomorphism class of this algebra as $D_l$.

Finally, we again take an even-dimensional $V$, but this time we use the skew-symmetric form

$\displaystyle\begin{pmatrix}0&I_l\\-I_l&0\end{pmatrix}$

This time we get a basis with $2l+2$ elements. We call this the symplectic algebra of $V$, and write $\mathfrak{sp}(V)$ or $\mathfrak{sp}(2l,\mathbb{F})$. Sometimes we refer to the isomorphism class of this algebra as $C_l$.

Along with the special linear Lie algebras, these form the “classical” Lie algebras. It’s a tedious but straightforward exercise to check that for any classical Lie algebra $L$, each basis element $e$ of $L$ can be written as a bracket of two other elements of $L$. That is, we have $[L,L]=L$. Since $L\subseteq\mathfrak{gl}(V)$ for some $V$, and since we know that $[\mathfrak{gl}(V),\mathfrak{gl}(V)]=\mathfrak{sl}(V)$, this establishes that $L\subseteq\mathfrak{sl}(V)$ for all classical $L$.

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Posted by John Armstrong | Algebra, Lie Algebras

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