I know from many sources, including Samelson's Notes on Lie Algebras, that the Lie algebras of even size skew-symmetric matrices, $\mathfrak{so}_{2n}$, are isomorphic to the Lie algebras of type $D_n$ and the odd size ones are isomorphic to the Lie algebras of type $B_n$. I cannot find an explicit isomorphism at the basis level anywhere. To clarify, in the case of $\mathfrak{so}_{2n}$ for example, I'm hoping for an explicit isomorphism that sends the basis elements $\{e_{i,j}-e_{j,i}\mid 1\leq i<j\leq 2n\}$ for $\mathfrak{so}_{2n}$ to a $2n\times 2n$ matrix of the form $\left(\begin{matrix}A&B\\C&-A^T\end{matrix}\right)$ where $A,B,C\in\mathfrak{gl}_n$, $B=-B^T$, and $C=-C^T$, which is the form of the elements of $D_n$ given in Humphreys Introduction to Lie Algebras and Representation Theory. I would also love one for the odd size matrices to the basis for $B_n$ given in the same source. Could someone please provide these isomorphisms?
An Explicit isomorphism between the orthogonal Lie algebras $\mathfrak{so}_n$ and the Lie algebras of type $B_n$ or $D_n$.
abstract-algebralie-algebras
Related Solutions
I found myself working on this same problem (for homework), and I think I've written a fairly detailed solution. So I will post it here, in case it is helpful to anyone else.
Let $\mathfrak{g}$ be a 1-dimensional Lie algebra, and let $\{E_1\}$ be a basis for $\mathfrak{g}$. Then for any two vector fields $X,Y\in\mathfrak{g}$, we have $X=aE_1$ and $Y=bE_1$, for some $a,b\in\mathbb{R}$. Thus, $$[X,Y]=[aE_1,bE_1]=ab[E_1,E_1]=0$$ for all $X,Y\in\mathfrak{g}$. Therefore, the only 1-dimensional Lie algebra is the trivial one. The map $$\varphi:\mathfrak{g}\rightarrow\mathfrak{gl}(2,\mathbb{R})$$ $$\varphi:aE_1\mapsto \left(\begin{array}{ll} a&0\\ 0&0 \end{array}\right)$$ is a Lie algebra homomorphism, since $$\varphi([aE_1,bE_1])=\varphi(0)=\left(\begin{array}{ll} 0&0\\ 0&0 \end{array}\right)\mbox{, and}$$ $$[\varphi(aE_1),\varphi(bE_1)]=\left(\begin{array}{ll} a&0\\ 0&0 \end{array}\right)\left(\begin{array}{ll} b&0\\ 0&0 \end{array}\right)-\left(\begin{array}{ll} b&0\\ 0&0 \end{array}\right)\left(\begin{array}{ll} a&0\\ 0&0 \end{array}\right)$$ $$=\left(\begin{array}{ll} 0&0\\ 0&0 \end{array}\right).$$ Thus, $\mathfrak{g}$ is isomorphic to the (abelian) Lie subalgebra $$\varphi(\mathfrak{g})=\left\{\left(\begin{array}{ll} a&0\\ 0&0 \end{array}\right)\in\mathfrak{gl}(2,\mathbb{R}):a\in\mathbb{R}\right\}\subset\mathfrak{gl}(2,\mathbb{R}).$$
Now let $\mathfrak{h}$ be a 2-dimensional Lie algebra, and let $\{E_1,E_2\}$ be a basis for $\mathfrak{h}$. Then for any two vector fields $X,Y\in\mathfrak{h}$, we have $X=aE_1+bE_2$ and $Y=cE_1+dE_2$, for some $a,b,c,d\in\mathbb{R}$. Thus, $$\begin{array}{ll} [X,Y]&=[aE_1+bE_2,cE_1+dE_2]\\ &=a[E_1,cE_1+dE_2]+b[E_2,cE_1+dE_2]\\ &=ac[E_1,E_1]+ad[E_1,E_2]+bc[E_2,E_1]+bd[E_2,E_2]\\ &=(ad-bc)[E_1,E_2]. \end{array}$$
If $[E_1,E_2]=0$, then we have the trivial 2-dimensional Lie algebra. The map $$\varphi:\mathfrak{h}\rightarrow\mathfrak{gl}(2,\mathbb{R})$$ $$\varphi:aE_1+bE_2\mapsto \left(\begin{array}{ll} a&0\\ 0&b \end{array}\right)$$ is a Lie algebra homomorphism, since $$\varphi([aE_1+bE_2,cE_1+dE_2])=\varphi(0)=\left(\begin{array}{ll} 0&0\\ 0&0 \end{array}\right)\mbox{, and}$$
$$[\varphi(aE_1+bE_2),\varphi(cE_1+dE_2)]=\left(\begin{array}{ll} a&0\\ 0&b \end{array}\right)\left(\begin{array}{ll} c&0\\ 0&d \end{array}\right)-\left(\begin{array}{ll} c&0\\ 0&d \end{array}\right)\left(\begin{array}{ll} a&0\\ 0&b \end{array}\right)$$ $$=\left(\begin{array}{ll} 0&0\\ 0&0 \end{array}\right).$$Furthermore, this map is faithful (injective). Thus, $\mathfrak{h}$ is isomorphic to the (abelian) Lie subalgebra $$\varphi(\mathfrak{h})=\left\{\left(\begin{array}{ll} a&0\\ 0&b \end{array}\right)\in\mathfrak{gl}(2,\mathbb{R}):a,b\in\mathbb{R}\right\}\subset\mathfrak{gl}(2,\mathbb{R}).$$
If $[E_1,E_2]\neq0$, then set $E_3=[E_1,E_2]$. Then for all $X,Y\in\mathfrak{h}$ we have $[X,Y]=\lambda E_3$ for some $\lambda\in\mathbb{R}$. In particular, for any $E_4\in\mathfrak{g}$ such that $E_4$ and $E_3$ are linearly independent, we have $[E_4,E_3]=\lambda_0 E_3$. Replacing $E_4$ with $1/\lambda_0 E_4$, we now have a basis $\{E_4, E_3\}$ for $\mathfrak{g}$ such that $[E_4, E_3]=E_3$. The map $$\varphi:\mathfrak{h}\rightarrow\mathfrak{gl}(2,\mathbb{R})$$ $$\varphi:aE_4+bE_3\mapsto \left(\begin{array}{ll} a&b\\ 0&0 \end{array}\right)$$ is a Lie algebra homomorphism, since $$\varphi([aE_4+bE_3,cE_4+dE_3])=\varphi((ad-bc)E_3)=\left(\begin{array}{ll} 0&ad-bc\\ 0&0 \end{array}\right)\mbox{, and}$$
$$[\varphi(aE_4+bE_3),\varphi(cE_4+dE_3)]=\left(\begin{array}{ll} a&b\\ 0&0 \end{array}\right)\left(\begin{array}{ll} c&d\\ 0&0 \end{array}\right)-\left(\begin{array}{ll} c&d\\ 0&0 \end{array}\right)\left(\begin{array}{ll} a&b\\ 0&0 \end{array}\right)$$ $$=\left(\begin{array}{ll} 0&ad-bc\\ 0&0 \end{array}\right).$$Furthermore, this map is faithful (injective). Thus, $\mathfrak{h}$ is isomorphic to the (non-abelian) Lie subalgebra $$\varphi(\mathfrak{h})=\left\{\left(\begin{array}{ll} a&b\\ 0&0 \end{array}\right)\in\mathfrak{gl}(2,\mathbb{R}):a,b\in\mathbb{R}\right\}\subset\mathfrak{gl}(2,\mathbb{R}).$$
Your approach works without problems, if you write the condition $[Ax,Ay]=A[x,y]$ for all $x,y$ in terms of the $9$ coefficients of the matrix $A$. The polynomial equations in these $9$ unknowns over $\mathbb{R}$ quickly yield $\det(A)=0$, a contradiction.
Another elementary argument is the following. $\mathfrak{sl}(2,\mathbb{R})$ has a $2$-dimensional subalgebra, e.g., $\mathfrak{a}=\langle f_1,f_2\rangle$, but $\mathfrak{su}(2)$ has no $2$-dimensional subalgebra. Hence they cannot be isomorphic.
Best Answer
This works with exactly the kind of calculations that I did in step 2 of my answer to your previous question. Concretely, what we are doing is a base change from the symmetric bilinear form given by the identity matrix
$$S_1=\pmatrix{1 & 0 &\dots & 0\\ 0&1& &0\\ && \ddots &\\ 0&0&&1}$$ (giving the skew-symmetric matrices) to the one with, depending on the parity of $n$, $$S_2=\begin{pmatrix} 1&0&0\\ 0&O&I_l\\ 0&I_l&O \end{pmatrix} \hspace{.5in}\text{or}\hspace{.5in} S_2=\begin{pmatrix} O&I_l\\ I_l&O \end{pmatrix} $$
(with $l=\lfloor n/2\rfloor$) or some scalar multiple thereof, giving the split forms as described by Humphreys, where the diagonal matrices form a nice CSA and the roots are visible.
Note that in the other answer, $S_2$ was a bit different again, but that $S_2$ and Humphreys' $S_2$ are easily seen to have an easy base change between them which works over almost any field. Meaning that like in the other answer, this entire base change should work, i.e. you can find a matrix $P$ with $^tPS_1P=S_2$, if and only if the field over which we are working contains a square root of $-1$, which will come up in the base change matrix and in the explicit isomorphism. In particular, the described Lie algebras are indeed isomorphic e.g. over $\mathbb C$, but not over $\mathbb R$. (Cf. Two Definitions of the Special Orthogonal Lie Algebra, note my comment to the accepted answer.)