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Let $\mathfrak{g}$ be a complex semisimple Lie algebra. It is well known that there is a universal enveloping algebra $U(\mathfrak{g})$ over $\mathbb{C}$ generated by generators $e_1, \dotsc, e_n, f_1, \dotsc, f_n, h_1, \dotsc, h_n$ with relations

$$ [e_i, f_j]=\delta_{ij}h_i, \quad [h_i,h_j]=0 $$ $$ [h_i,e_j]=a_{ij}e_j, \quad [h_i,f_j]=-a_{ij}f_j $$ $$ \sum_{\ell=0}^{1-a_{ij}} (-1)^\ell\begin{pmatrix}1-a_{ij}-\ell\\\ell \end{pmatrix} e_i^{1-a_{ij}-\ell} e_j e_i^\ell=0$$ $$ \sum_{\ell=0}^{1-a_{ij}} (-1)^\ell\begin{pmatrix}1-a_{ij}-\ell\\\ell \end{pmatrix} f_i^{1-a_{ij}-\ell} f_j f_i^\ell=0$$

where $a=(a_{ij})$ is the Cartan matrix corresponding to $\mathfrak{g}$, and $n$ is the rank of $\mathfrak{g}$.

$\mathfrak{g}_1=\mathrm{sl}_4$ has rank 3 and Cartan matrix $a=\begin{pmatrix} 2 & -1 & 0\\ -1 & 2 & -1\\ 0& -1&2 \end{pmatrix}$, so $U(sl_4)=\langle e_i, f_i,h_i : i=1,2,3\rangle$ with the relations above.

$\mathfrak{g}_2=\mathrm{sp}_2$ (other authors may denote it $\mathrm{sp}_4$) has rank 2 and Cartan matrix $a=\begin{pmatrix} 2 & -1 \\-2 & 2 \end{pmatrix}$, so $U(\mathrm{sp}_2)=\langle\tilde{e_i}, \tilde{f_i}, \tilde{h_i} : i=1,2\rangle$ (I use tilde just to make difference between generators of $U(\mathrm{sp}_2)$ and $U(\mathrm{sl}_4)$).

Is there a subalgebra of $U(\mathrm{sl}_4)$ isomorphic to $U(\mathrm{sp}_2)$? More generally, is there a subalgebra of $U(\mathrm{sl}_{2n})$ isomorphic to $U(\mathrm{sp}_n)$?

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  • $\begingroup$ I am confused about the sense of "rank" here... E.g., $sl_4$ has rank 3, and $sp_2$ (or 4) also has rank $2$. Can you clarify? $\endgroup$ May 9, 2018 at 23:43
  • $\begingroup$ sorry, you are right. It was just my mistake. I will edit it $\endgroup$
    – R. Kasyfil
    May 9, 2018 at 23:51
  • $\begingroup$ I have edited to clean up a little bit. Don't use the relation symbols $<>$ as delimiters; use $\langle\rangle$ (\langle\rangle) instead. $\endgroup$
    – LSpice
    May 10, 2018 at 0:53
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    $\begingroup$ If $\mathfrak h \subset \mathfrak g$ is an embedding of Lie algebras, then $U(\mathfrak h) \subset U(\mathfrak g)$ is an inclusion of algebras. $\endgroup$ May 10, 2018 at 12:27

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