The symplectic lie algebra defined by $sp\left(n\right)=\left\{ X\in gl_{2n}\,|\, X^{t}J+JX=0\right\}$ when $J=\begin{pmatrix}0 & I\\ -I & 0\end{pmatrix}$. So $X\in sp\left(n\right)$ is of the sort $X=\begin{pmatrix}A & B\\ C & -A^{t}\end{pmatrix}$ when $B=B^{t},\, C=C^{t}$.

This far I was able to get, but how can I prove that it is simple?


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    $\begingroup$ Can you prove it is semisimple? $\endgroup$ – Mariano Suárez-Álvarez Feb 1 '12 at 19:38
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    $\begingroup$ I can't. Although I'm pretty sure you'll have to show that the killing form is non-degenerate. Can't see exactly how. $\endgroup$ – IBS Feb 1 '12 at 21:39

Let ${\mathfrak g}=sp(n)$.

Short answer, from an advanced viewpoint : $\mathfrak g$ is semisimple (because $K(x,y)=tr(xy)$ defines a nondegenerate invariant bilinear form on $\mathfrak g$), so it will be simple iff its Dynkin diagram is connected. But we know that the diagram is $C_n$ and this is clearly connected.

Longer answer, from a more pedestrian viewpoint : a standard basis $\cal B$ for $\mathfrak g$ is $\lbrace h_i,a_i,b_i\rbrace_{1 \leq i \leq n} \cup \lbrace c_{i,j},d_{i,j},f_{i,j},g_{i,j}\rbrace_{1 \leq i<j \leq n}$ where

$$ h_i=e_{ii}-e_{n+i,n+i}, \ \ \ a_i=e_{i,i+n}, \ \ \ b_i=e_{i+n,i}, $$

$$ c_{i,j}=e_{ij}-e_{i+n,j+n}, \ \ \ d_{i,j}=e_{ji}-e_{j+n,i+n}, $$

$$ f_{i,j}=e_{i,j+n}-e_{j,i+n}, \ \ \ g_{i,j}=e_{i+n,j}-e_{j+n,i} $$

(and $e_{ij}$ is the matrix all of whose coefficients are zero except for the $(i,j)$-th which is $1$).

Now let $\mathfrak i$ be an ideal of $\mathfrak g$. Then $\mathfrak i$ is stable by $ad(x)$ for any $x\in {\mathfrak g}$. In particular, if $H$ is the vector space spanned by $h_1,h_2, \ldots ,h_n$, then $\mathfrak i$ is stable by all the $ad(h)$, for $h \in H$. Now those $ad(h)$ are simultaneously diagonalized by the basis we have just described above. We deduce that $\mathfrak i=H' \oplus {\sf Vect}(F)$, where $H'$ is a linear subspace of $H$ and $F$ is a subset of ${\cal B}'={\cal B} \setminus \lbrace h_1,h_2, \ldots ,h_n \rbrace$. Let us now consider the root spaces $$ AB_i={\sf Vect}(h_i,a_i,b_i), \ CD_{ij}={\sf Vect}(h_i-h_j,c_{ij},d_{ij}), \ FG_{ij}={\sf Vect}(h_i+h_j,f_{ij},g_{ij}) $$ Each root space $R$ is a Lie subalgebra of $\mathfrak g$ isomorphic to ${\mathfrak sl}_2$. It easily follows that either ${\mathfrak i} \cap R=\lbrace 0 \rbrace$ or $R \subseteq {\mathfrak i}$ for any $R$. Combining this with the decomposition of $\mathfrak i$ above, we see that $\mathfrak i$ is the direct sum of some root spaces, plus some subspace of $H$.

Suppose $F$ contains $a_i$ for some $i$. Then $\mathfrak i$ contains the whole root space ${\sf Vect}(h_i,a_i,b_i)$. For $j>i$, $\mathfrak i$ also contains $[a_i,g_{ij}]=c_{ij}$, $h_i-[c_{ij},d_{ij}]=h_j, [h_j,\frac{a_j}{2}]=a_j$. Similarly, for $k<i$, $\mathfrak i$ also contains $[a_i,-g_{ki}]=d_{ki}$, $h_i+[c_{ki},d_{ki}]=h_k, [h_k,\frac{a_k}{2}]=a_k$. So $\mathfrak i$ contains all the $a_i$, and we easily deduce that $\mathfrak i=A$.

Suppose $F$ contains $c_{ij}$ for some $i<j$. Then $\mathfrak i$ contains the whole root space ${\sf Vect}(h_i-h_j,c_{ij},d_{ij})$. And $\mathfrak i$ also contains $h_i+h_j=[c_{ij},f_{ij}]$. So $\mathfrak i$ contains both $h_i$ and $h_j$, it also contains the root space ${\sf Vect}(h_i,a_i,b_i)$, so $\mathfrak i=A$ by the preceding paragraph.

Suppose $F$ contains $f_{ij}$ for some $i<j$. Then $\mathfrak i$ contains the whole root space ${\sf Vect}(h_i+h_j,f_{ij},g_{ij})$. And $\mathfrak i$ also contains $c_{ij}=[f_{ij},b_{j}]$, so $\mathfrak i=A$ by the preceding paragraph.

Finally we see that $\mathfrak i=A$ unless $F$ is empty. In that case, $\mathfrak i$ is a linear subspace of $H$. For any $i$, $ad (a_i)$ must be zero on $\mathfrak i$, otherwise $\mathfrak i$ would contain $a_i$. It is easily deduced that ${\mathfrak i}=\lbrace 0 \rbrace$, so $\mathfrak g$ is indeed simple.

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    $\begingroup$ To show semisimplicty one could also show that the obvious module is simple and faithful, so that the algebra is reductive, and then check that the center is trivial. This is simpler than checking that the Killing form is non-degenerate, I think.. $\endgroup$ – Mariano Suárez-Álvarez Apr 2 '12 at 7:13
  • $\begingroup$ In fact there is no need to compute the Killing form. All we need to do is exhibit one nondegenerate invariant bilinear form (which is tr(xy)). Then, we shall learn a posteriori that our initial form was a multiple of the Killing form. $\endgroup$ – Ewan Delanoy Apr 2 '12 at 9:56
  • $\begingroup$ To check that tr(xy) is nondegenerate, it suffices to note that $\mathfrak g$ behaves nicely with respect to the usual decomposition symmetric+antisymmetric, and that tr(xy) is positive definite on symmetric matrices, and negative definite on antisymmetric matrices. $\endgroup$ – Ewan Delanoy Apr 2 '12 at 9:59
  • $\begingroup$ The basis for the symplectic Lie algebra has a typo in it: it should be $e_{i,j} -e_{j+n, i+n}$, and not $e_{i,j} - e_{i+n, j+n}$, etc. $\endgroup$ – Raj Jul 28 '17 at 16:17

$sp(n)$ is a Lie subalgebra of the more familiar lie algebra $sl(n)$ because $Sp(n)$ is a Lie subgroup of $SL(n)$ (determinant $= 1$ condition). Since $sl(n)$ is semisimple, and therefore has non-degenerate Killing form, $tr(xy)$ would also be non-degenerate on its lie subalgebras.

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    $\begingroup$ This is false. For example, pick the one-dimensional subalgebra generated by $\left[\begin{smallmatrix}0 & 1 \\ 0 & 0\end{smallmatrix}\right]$ in $\mathfrak{sl}_2\mathbb{R}$. $\endgroup$ – Robin Goodfellow Nov 24 '17 at 18:13

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