# What subalgebras of $\mathfrak{so}(2n)$ or $\mathfrak{sp}(2n)$ are orthogonal to the centre of $\mathfrak{gl}(n)$?

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.

• @YCor I suspect that the OP means orthocomplement in $\mathfrak{so}$ or $\mathfrak{sp}.$ – Vít Tuček Jul 30 '19 at 14:50
• 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 $0$. – YCor Jul 30 '19 at 18:13
• @YCor Yes, that's correct. So, for instance, $\{tr(A)=0\text{ and }C=0\}$ defines a Lie subalgebra of $\mathfrak{c}^\perp$. – Theo Johnson-Freyd Jul 30 '19 at 20:02
• 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 $\mathfrak{gl}(n)$. – Theo Johnson-Freyd Jul 31 '19 at 0:10
• @TheoJohnson-Freyd about massaging the question, see here. – YCor Aug 2 '19 at 14:38

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.

• 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. – Victor Protsak Jul 30 '19 at 18:41
• Should "whose Levi part is $\mathfrak{gl}(n)$" instead read "whose Levi part is $\mathfrak{sl}(n)$"? – Theo Johnson-Freyd Jul 30 '19 at 20:04
• 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. – Victor Protsak Jul 30 '19 at 20:35
• You are correct, I have edited my answer. – Vít Tuček Jul 30 '19 at 20:57
• @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? – Vít Tuček Jul 30 '19 at 21:01

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 $$\mathbf{32}_+$$.