Short version: Let $V$ be a 7-dimensional linear space of (real) square matrices. Suppose further that $[V,V]$ (the linear space spanned $[X,Y]$, $X,Y\in V$) is actually a subalgebra isomorphic to $\mathfrak{so}(7)$. Can one prove that $$V\cap [V,V]=\{0\}?$$

(Maybe there is nothing special about 7 here, just in the long version.)

Long version: This is really about Cartan-Killing fields on spheres, and related to totally geodesic submanifolds on Lie groups. Given a $n\times n$ skew-symmetric matrix $\xi$, it defines a Killing vector field $\chi_\xi:S^n\to TS^n$ via $x\mapsto \xi x$. This Killing field is called Clifford-Wolf if it has constant length.

A Cartan-Killing space is a linear subspace of Clifford-Wolf Killing fields. It is known that the only non-euclidean symmetric spaces that admit non-trivial Cartan-Killing spaces are compact Lie groups, even dimensional spheres and SU(2n)/Sp(n), $n\geq 3$ (these results are part of a nice paper that settles the subject: http://projecteuclid.org/euclid.jdg/1251122544).

I find myself in the following situation: let $CK$ be the space of Clifford-Wolf fields defined by octonionic multiplication (or any other seven-dimensional space of constant-length Killing fields on $S^7$, which we think as a subspace of $\mathfrak{so}(8)$ through $\chi$). There is a vector space $\mathcal{H}\subset \mathfrak{so}(8)$ and an injective linear map $A:CK\to End(\mathcal{H})$, such that $ad_\xi ad_\eta$ preserves $\mathcal{H}$ and $A^\xi A^\eta=ad_\xi ad_\eta$ for all $\xi,\eta\in CK$. It is known that $[CK,CK]$ (in the sense of the brackets in the short answer) is isomorphic to $\mathfrak{so}(7)$ and that $CK\cap [CK,CK]=\{0\}$.

A stronger question would be: is (in some sense) $A^\xi=ad_\xi$? But I would be happy(ier) with just a weak one: does $A(CK)\cap [A(CK),A(CK)]=\{0\}$?


1) $ad_\xi (X)=[\xi,X]$ (the Lie brackets).

2) We denote $A^\xi=A(\xi)$ to simplify notation.

Thanks in advance

  • $\begingroup$ "a linear space is isomorphic to $\mathfrak{so}(7)$": what do you mean? to say that a linear space is isomorphic to $\mathfrak{so}(7)$ just means that it has the same dimension (21). Maybe you mean it is stable under bracket and isomorphic to $\mathfrak{so}(7)$ as a Lie algebra? $\endgroup$ – YCor Jun 28 '16 at 23:14
  • $\begingroup$ @YCor sorry for that. It is isomorphic as a Lie algebra. $\endgroup$ – Llohann Jun 28 '16 at 23:16
  • 1
    $\begingroup$ It is probably well to specify (assuming you do) in the body of the question that you mean that $[V, V]$ is a sub-Lie algebra of $\mathfrak{gl}_n$, rather than just that it is an abstract Lie algebra. $\endgroup$ – LSpice Jun 29 '16 at 3:32
  • $\begingroup$ @LSpice Like that? $\endgroup$ – Llohann Jun 29 '16 at 10:53
  • 1
    $\begingroup$ I think there is something at least a little special about 7 in the short version, as it is not true for 3. On the other hand I don't know what happens for 5 so perhaps the special property of 7 is just 'being bigger than 3'. $\endgroup$ – Vincent Jun 29 '16 at 12:11

Your Answer

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

Browse other questions tagged or ask your own question.