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.

  • 1
    $\begingroup$ @YCor I suspect that the OP means orthocomplement in $\mathfrak{so}$ or $\mathfrak{sp}.$ $\endgroup$ – Vít Tuček Jul 30 '19 at 14:50
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ @TheoJohnson-Freyd about massaging the question, see here. $\endgroup$ – 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.

  • $\begingroup$ 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

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 $\mathbf{32}_+$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.