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Consider the Lie algebra inclusions $\mathfrak{gl}(n) \subset \mathfrak{so}(2n)$ and $\mathfrak{gl}(n) \subset \mathfrak{sp}(2n)$. Let $\mathfrak{c} \subset \mathfrak{gl}(n)$ denote the centre. Thinking of it as a subspace of $\mathfrak{so}(2n)$ or $\mathfrak{sp}(2n)$, let $\mathfrak{c}^\perp$ denote the orthogonal complement (with respect to the Killing form).

Note that $\mathfrak{c}^\perp$ is not a Lie subalgebra of $\mathfrak{so}(2n)$ or $\mathfrak{sp}(2n)$, but it contains various subalgebras (e.g. $\mathfrak{sl}(n) \subset \mathfrak{c}^\perp$).

What are the (maximal) Lie subalgebras of $\mathfrak{c}^\perp$? Are there any for which the induced $2n$-dimensional representation is simple?

The specific case I care about is when $n=28$, and I look at $\mathfrak{sp}(56)$. Then I want to know if $\mathfrak{c}^\perp$ contains a subalgebra of type $\mathfrak{e}_7$. I suspect it does not, but I cannot prove it.

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    $\begingroup$ @YCor I suspect that the OP means orthocomplement in $\mathfrak{so}$ or $\mathfrak{sp}.$ $\endgroup$ – Vít Tuček Jul 30 '19 at 14:50
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    $\begingroup$ Thanks. In coordinates, the Lie algebra of $\mathfrak{sp}(2n)$ resp. $\mathfrak{so}(2n)$, chosen with respect to the bilinear form $\begin{pmatrix}0 & \pm 1_n\\1_n & 0\end{pmatrix}$, with $\pm=+$ for $\mathfrak{sp}$ and $-$ for $\mathfrak{so}$, consists of those matrices $\begin{pmatrix}A & B \\ C & {-}\,^t\!A \end{pmatrix}$ with $B,C$ symmetric, resp $B,C$ skew-symmetric. In both cases, the copy of $\mathfrak{gl}(n)$ consists of those such matrices with $B=C=0$, and the orthogonal of the center of $\mathfrak{gl}$ consists of those matrices with $A$ being of trace $0$. $\endgroup$ – YCor Jul 30 '19 at 18:13
  • $\begingroup$ @YCor Yes, that's correct. So, for instance, $\{tr(A)=0\text{ and }C=0\}$ defines a Lie subalgebra of $\mathfrak{c}^\perp$. $\endgroup$ – Theo Johnson-Freyd Jul 30 '19 at 20:02
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    $\begingroup$ In discussion with Richard Derryberry, we found the following example. Choose $\lambda \neq 0$. In @YCor's basis, there is a subalgebra cut out by demanding that $A = -^tA$ (so that $A \in \mathfrak{so}(n)$) and $C = \lambda B$. I believe that this algebra is maximal inside of $\mathfrak{c}^\perp$, and that it is isomorphic to $\mathfrak{gl}(n)$. $\endgroup$ – Theo Johnson-Freyd Jul 31 '19 at 0:10
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    $\begingroup$ @TheoJohnson-Freyd about massaging the question, see here. $\endgroup$ – YCor Aug 2 '19 at 14:38
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If you are asking about the inclusions given by embedding of Dynkin diagrams, then I believe that the maximal subalgebras are codimension 1 subalgebras of the parabolic subalgebras whose Levi part is $\mathfrak{gl}(n).$

The Lie algebra $\mathfrak{g} = \mathfrak{sp / so}\; (2n)$ has triangular decomposition $\mathfrak{g}_{-1} \oplus \mathfrak{gl}(n) \oplus \mathfrak{g}_{1}.$ And $\mathfrak{gl}(n)$ acts irreducibly on either $\mathfrak{g}_{-1}$ or $\mathfrak{g}_{1}$. The orthocomplement of the center is in both cases $\mathfrak{g}_{-1} \oplus \mathfrak{sl}(n) \oplus \mathfrak{g}_{1}$ and the center of $\mathfrak{gl}(n)$ can be obtained by bracketing appropriate elements of $\mathfrak{g}_{-1} \oplus \mathfrak{g}_{1}.$ Thus $\mathfrak{sl}(n) \oplus \mathfrak{g}_1$ is a maximal subalgebra. The defining representation of $\mathfrak{g}$ is not simple when restricted to parabolic subalgebra.

As Victor Protsak notes in the comments, similar construction works for other maximal parabolics.

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  • $\begingroup$ The $n$th max parabolic ${\frak p}_n={\frak gl}(n)\oplus {\frak g}_1$ is not orthogonal to ${\frak c}$: as you yourself have noted, ${\frak c}^{\perp}$ intersected with ${\frak gl}(n)$ is ${\frak sl}(n)$, so one needs to consider the codimension 1 subalgebra ${\frak sl}(n)\oplus{\frak g}_1\subset {\frak p}_n$. A similar construction works for other maximal parabolics ${\frak p}_k$: replace the Levi component ${\frak l}$ with ${\frak l}^{\prime}={\frak l}\cap {\frak sl}(n)$, then ${\frak l}^{\prime}\oplus{\frak g}_1\subset{\frak c}^{\perp}$ and is a maximal Lie subalgebra with this property. $\endgroup$ – Victor Protsak Jul 30 '19 at 18:41
  • $\begingroup$ Should "whose Levi part is $\mathfrak{gl}(n)$" instead read "whose Levi part is $\mathfrak{sl}(n)$"? $\endgroup$ – Theo Johnson-Freyd Jul 30 '19 at 20:04
  • $\begingroup$ Correction The second sentence in my comment should read: "A similar construction works for other maximal parabolics ${\frak p}_k={\frak l}\oplus{\frak n}$: replace the Levi component ${\frak l}$ with ${\frak l}^{\prime}={\frak l}\cap {\frak sl}(n)$, then ${\frak l}^{\prime}\oplus{\frak n}\subset{\frak c}^{\perp}$ and is a maximal Lie subalgebra with this property." @Theo: The Levi component ${\frak l}$ is uniquely determined by the parabolic ${\frak p}$, and cannot be ${\frak sl}(n)$, e.g. because the center of ${\frak l}$ is non-trivial and has dimension 1 for max parabolics. $\endgroup$ – Victor Protsak Jul 30 '19 at 20:35
  • $\begingroup$ You are correct, I have edited my answer. $\endgroup$ – Vít Tuček Jul 30 '19 at 20:57
  • $\begingroup$ @VictorProtsak I will have to think more about the other maximal parabolics that you suggest. Perhaps this even works for maximal subgroups subalgebras coming from Borel de Siebenthal theory? $\endgroup$ – Vít Tuček Jul 30 '19 at 21:01
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Here is an example of a Lie subalgebra of $\mathfrak{c}^\perp$ for which the $2n$-dimensional remains simple, with $n=16$.

Consider the Lie group $\mathrm{Spin}(12)$, and its vector representation, which I will call $\mathbf{12}$, and its two half-spin representations, each of dimension $32$, which I will call $\mathbf{32}_\pm$. So for instance the adjoint representation is $\mathfrak{so}(12) = \operatorname{Alt}^2(\mathbf{12})$. Each half-spin representation supports a $\mathfrak{so}(12)$-invariant symplectic form.

Mathieu's group $\mathrm{M}_{12}$ has no 12-dimensional irreps, but its double cover $2\mathrm{M}_{12}$ has one (up to isomorphism) 12-dimensional irrep. It supports a symmetric invariant form, so it defines a (conjugacy class of) map(s) $2\mathrm{M}_{12} \to \mathrm{O}(12)$, and there is only one conjugacy class of irreducible maps like this. However, there are two conjugacy classes of irreducible maps $2\mathrm{M}_{12} \to \mathrm{SO}(12)$, exchanged by the outer automorphism thereof, because any copy of the 12-dimensional $2\mathrm{M}_{12}$-irrep is chiral.

Choose one of these maps $2\mathrm{M}_{12} \to \mathrm{SO}(12)$. The choice breaks the symmetry between the half-spin representations $\mathbf{32}_\pm$. Namely, one of them, which I will arbitrarily call $\mathbf{32}_+$, descends from $2\mathrm{M}_{12}$ to $\mathrm{M}_{12}$ and splits as the sum $\mathbf{16} \oplus \overline{\mathbf{16}}$ of a 16-dimensional complex irrep and its dual, and the other, $\mathbf{32}_-$, is nontrivially charged under the centre of $2\mathrm{M}_{12}$ and remains simple upon restriction.

Thus we have a commutative square: $$ \begin{matrix} 2\mathrm{M}_{12} & \hookrightarrow & \mathrm{Spin}(12) \\ \downarrow & & \downarrow \\ \mathrm{GL}(16) & \hookrightarrow & \mathrm{Sp}(32) \end{matrix} $$ where the map $\mathrm{Spin}(12) \to \mathrm{Sp}(32)$ is via the representation $\mathbf{32}_+$. (Both horizontal arrows are inclusions, and both downward arrows have kernel of order $2$.)

Now look at the adjoint representation $\mathfrak{sp}(32)$, on which $2\mathrm{M}_{12}$ acts orthogonally, and the image therein $\mathfrak{c}$ of the centre of $\mathfrak{gl}(16)$. On the one hand, this centre is fixed by $2\mathrm{M}_{12}$, and in fact is the only fixed subspace. On the other hand, $\mathfrak{so}(12) \subset \mathfrak{sp}(32)$ remains simple when restricted to $2\mathrm{M}_{12}$. It follows that $\mathfrak{so}(12)$ and $\mathfrak{c}$ are orthogonal.

Thus $\mathfrak{g} = \mathfrak{so}(12) \subset \mathfrak{c}^\perp$. But by construction the $32$-dimensional defining represenation of $\mathfrak{sp}(32)$ restricts to $\mathfrak{g}$ as the irrep $\mathbf{32}_+$.

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