3
$\begingroup$

If $\mathbb{K},\mathbb{L} \in \{\mathbb{R},\mathbb{C},\mathbb{H},\mathbb{O}\}$ then the Rosenfeld projective ("elliptic"?) plane $\mathbb{P}^2(\mathbb{K}\otimes\mathbb{L})$ is "the" compact Riemannian symmetric space associated with the non-compact Lie algebra $\tilde L_3(\mathbb{K},\mathbb{L})$ given by the following version of the Tits-Freudenthal magic square:

$$\begin{array}{r|c|c|c|c|} &\mathbb{R}&\mathbb{C}&\mathbb{H}&\mathbb{O}\\\hline \mathbb{R}&\mathfrak{so}(1,2)&\mathfrak{su}(1,2)&\mathfrak{sp}(1,2)&\mathfrak{f}_4(\mathfrak{so}_9)\\\hline \mathbb{C}&\mathfrak{su}(1,2)&\mathfrak{su}(1,2) \oplus \mathfrak{su}(1,2)&\mathfrak{su}(2,4)&\mathfrak{e}_6(\mathbb{R}\oplus\mathfrak{so}_{10})\\\hline \mathbb{H}&\mathfrak{sp}(1,2)&\mathfrak{su}(2,4)&\mathfrak{so}(4,8)&\mathfrak{e}_7(\mathfrak{su}_{2}\oplus\mathfrak{so}_{12})\\\hline \mathbb{O}&\mathfrak{f}_4(\mathfrak{so}_9)&\mathfrak{e}_6(\mathbb{R}\oplus\mathfrak{so}_{10})&\mathfrak{e}_7(\mathfrak{su}_{2}\oplus\mathfrak{so}_{12})&\mathfrak{e}_8(\mathfrak{so}_{16})\\\hline \end{array}$$

(Notation is hopefully understandable: in the case of exceptional Lie algebras, the real form is indicated by putting their maximal compact subalgebra in parentheses. For example, $\mathbb{P}^2(\mathbb{H}\otimes\mathbb{H}) = \mathit{SO}_{12}/\mathit{S}(\mathit{O}_4\times \mathit{O}_8)$ and $\mathbb{P}^2(\mathbb{C}\otimes\mathbb{O}) = \mathit{E}_6/(\mathit{U}_1\cdot\mathit{Spin}_{10})$. I am unsure about finite coverings, which is why I only put Lie algebras in this table and why I put quotation marks around "the" above; but this is irrelevant for my question.)

This table is found, e.g., in Hans Freudenthal's 1964 report "Lie Groups in the Foundations of Geometry" (Advances in Math. 1, 145–190), ¶4.16 on p. 172. It is also found in Robert Goss's 2015 Ph.D. thesis The topology of the higher projective planes in §2.3 but with an incorrect claim as to its construction.

When the set of $3\times 3$ Hermitian matrices with coordinates in $\mathbb{K}\otimes\mathbb{L}$ is a Jordan algebra, then $\mathbb{P}^2(\mathbb{K}\otimes\mathbb{L})$ can be defined as the set of trace $1$ idempotents of that algebra. But as far as I understand, there is no such construction of $\mathbb{P}^2(\mathbb{K}\otimes\mathbb{L})$ from $\mathbb{K}\otimes\mathbb{L}$ in the case of $\mathbb{H}\otimes\mathbb{O}$ and $\mathbb{O}\otimes\mathbb{O}$. (Rosenfeld's book Geometry of Lie Groups seems to suggest that there is one, but I find it very difficult to understand what this book actually says.) Instead, the way we know which particular real form of the magic square to use is that it gives a symmetric space of the desired real dimension $2pq$ (with $p := \dim_{\mathbb{R}}(\mathbb{K})$ and $q := \dim_{\mathbb{R}}(\mathbb{L})$) and desired real rank $\min(p,q)$.

Anyway, my question is:

How can we construct the real form $\tilde L_3(\mathbb{K},\mathbb{L})$ given in the above table from $\mathbb{K}$ and $\mathbb{L}$ using an algebraic construction?

I emphasize that the difficult part is getting the correct real form: the complexification is the usual magic square of Lie algebras. There are a number of constructions given in Barton & Sudbery's article "Magic squares and matrix models of Lie algebras", but unless I missed something, they do not produce the above tables (for $\mathbb{K},\mathbb{L} \in \{\mathbb{R},\mathbb{C},\mathbb{H},\mathbb{O}\}$, their $L_3(\mathbb{K},\mathbb{L})$ defined in theorem 4.4 gives a compact real form; and if we replace $\mathbb{K}$ and $\mathbb{L}$ by the split form of the algebra, the result, shown near the end of section 7 of their paper, produces a different table from the one above).

So, is there a way to produce the above table without resorting to filtering real forms by their real rank?

$\endgroup$
1
$\begingroup$

Alberto Elduque obtained all real forms of exceptional Lie algebras in his variant of Freudenthal square in A new look at Freudenthal's magic square, Non-associative algebra and its applications, 149–165, Lect. Notes Pure Appl. Math., 246.

There is also a PhD thesis (pdf link) of Yongdong Huang that shows how to get all Riemannian symmetric spaces out of a magic square construction.

$\endgroup$
  • $\begingroup$ Obtaining every real form is not what I'm after: the point is to obtain precisely the "right" ones in a uniform way from $\mathbb{R}$, $\mathbb{C}$, $\mathbb{H}$, $\mathbb{O}$. But the paper by Elduque you mention references a remark in another paper by the same author ("The Magic Square and Symmetric Compositions", spec. rk. 3.2) which appears to describe the same construction as in the answer by Marek Mitros and give the right real forms, at least for the exceptional algebras, (contd.) $\endgroup$ – Gro-Tsen Mar 16 '17 at 14:36
  • $\begingroup$ (contd.) namely the forms of $F_4$, $E_6$, $E_7$ and $E_8$ with respective Cartan index $-20$, $-14$, $-5$ and $8$, merely by changing two signs in the Barton-Sudbery construction. So this would indeed be the construction I was hoping for (assuming it also works in the non-exceptional cases; I'll try to check that). $\endgroup$ – Gro-Tsen Mar 16 '17 at 14:38
  • $\begingroup$ PS: On the other hand, Huang's thesis doesn't seem too relevant here. It's not really an algebraic construction, and I seem to understand that the octonionic grassmanians are defined in an ad hoc fashion. (I find the details fairly sketchy; there is a more readable paper by Huang and Leung in Math. Ann. in 2011, but even there I'm not completely sure what it is they actually construct.) $\endgroup$ – Gro-Tsen Mar 16 '17 at 14:45
  • $\begingroup$ Thanks for the update. What I meant was that if Elduque claims to obtain all real forms then what remains is to check whether he obtained the crucial ones in "one table". $\endgroup$ – Vít Tuček Mar 16 '17 at 15:20
1
$\begingroup$

I suggest using Barton-Sudbery method for $\mathbb K \otimes \mathbb L$ with one adjustment. In space $Tri (\mathbb K) +Tri (\mathbb L)+\mathbb K\otimes \mathbb L +\mathbb K\otimes \mathbb L+\mathbb K\otimes \mathbb L$ change sign in result of bracket of two spaces $\mathbb K \otimes \mathbb L$. I tried this method in 2008 when I created GAP script for generation of all algebras in magic square. However I did not finish my work, so I do not give 100% guarantee.

Here is example with description whan sign should be changed. I show the difference on $so(8,1)$ and $so_9$ Lie algebras example. Let $e_{i,j}$ be matrix having $-1$ on $i,j$ coordinate and $1$ on $j,i$ and zeros otherwise. Let $f_{i,j}$ be matrix having $1$ on $i,j$ and $j,i$ coordinates and zeros otherwise. Let $V_8$ be space generated by $e_{i,9}$ or $f_{i,9}$ for $i=1..8$. One can easily verify following results.

  1. $[e_{i,j}, e_{k,j}]=e_{i,k}$
  2. $[f_{i,j}, f_{k,j}]=-e_{i,k}$
  3. $e_{i,j}$ for $1 \leqslant i < j \leqslant 9$ generate $so_9$
  4. $e_{i,j}$ for $1 \leqslant i < j \leqslant 8$ and $f_{i,9}$, $i=1..8$ generate $so_{8,1}$

As one can see the $so_{8,1}$ can be obtained from $so_9$ by changing sign on results of $[V_8,V_8]$ brackets.

The same method can be applied to Barton-Sudbery Lie algebras by changing sign on $[A,A]$, $[B,B]$ brackets where $A, B, C$ are three components of $\mathbb K \otimes \mathbb L$.

In my opinion there is no good definition of exceptional Lie groups. One would expect Lie group as being symmetry group of something. For example $F_4$ is defined as isometry group of $\mathbb O P^2$ projective plane over octonions. How to define $\mathbb O P^2$ ? We can use Jordan algebra do define it in algebraic way. How to define it in geometric way ? Next question is how to define "bioctonionic projective plane" $\mathbb C \otimes \mathbb OP^2$ ? Baez in his "The Octonions" defined it as quotient of $E_6$ by the subgroup. Lie group $E_6$ is defined from it's Lie algebra. Lie algebra $e_6$ is defined in abstract way as bracket on some 78-dimensional vector space.

Similar for $E_7$ and $E_8$ there is missing nice definition we expect. Rosenfeld seemed to had some ideas but as you say - it is difficult to read and understand. Freudenthal worked on "octonionic projective geometry" but for me is also difficult to read and understand.

All things are difficult before they are easy (Thomas Fuller). Exceptional Lie groups looks difficult, so it means they are not well understood yet.

$\endgroup$
  • $\begingroup$ I had considered this modification and I thought it was too simple to work; but it seems you're right: see my comments on the answer by Vít Tuček: apparently Alberto Elduque considered the variation on the Barton-Sudbery construction that you mention, and it seems that it might indeed give the "right" real forms in every case. $\endgroup$ – Gro-Tsen Mar 16 '17 at 14:47

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.