Let $\mathbb{O}$ be the octonion algebra (say over $\mathbb{R}$) and let $J_{3}(\mathbb{O})$ be the set of $3 \times 3$ hermitian matrices with octonion coefficients, that is:

$$ J_3(\mathbb{O}) = \left\{ \begin{pmatrix} \lambda_1 & a & b \\ \overline{a} & \lambda_2 & c \\ \overline{b} & \overline{c} & \lambda_3 \end{pmatrix}, \ \ \lambda_i \in \mathbb{R}, \ a,b,c \in \mathbb{O} \right\}$$

The group $\mathrm{E}_6$ is the group of linear automorphisms of $J_{3}(\mathbb{O})$ which preserve the cubic form: $$\lambda_1 \lambda_2 \lambda_3 + 2 \mathrm{Re}(a\overline{b}c) - \lambda_2 N(b)^2 - \lambda_3 N(a)^2 - \lambda_1 N(c)^2,$$ where $N$ is the norm over $\mathbb{O}$.

There are many interesting subgroups of $\mathrm{E}_6$ related to this description. $\mathrm{SL}_3(\mathbb{R})$ is one of them. The action of $\mathrm{SL}_3(\mathbb{R})$ on $J_3(\mathbb{O})$ is given by :

$$ \forall g \in \mathrm{SL}_3(\mathbb{R}), \forall A \in J_{3}(\mathbb{O}), \ g\cdot A = g A\,^{t}\! g,$$ where $^{t}\!g$ is the transpose of $g$.

The group $\mathrm{Spin_8}$ can also be seen as a subgroup of $\mathrm{E}_6$ with the action: \begin{align*} &\forall (g_1,g_2,g_3) \in \mathrm{Spin}_8,\quad \forall A = \begin{pmatrix} \lambda_1 & a & b \\ \overline{a} & \lambda_2 & c \\ \overline{b} & \overline{c} & \lambda_3 \end{pmatrix} \in J_{3}(\mathbb{O}), \\ &g\cdot A = \begin{pmatrix} \lambda_1 & g_1(a) & g_2(b) \\ \overline{g_1(a)} & \lambda_2 & g_3(c) \\ \overline{g_2(b)} & \overline{g_3(c)} & \lambda_3 \end{pmatrix}, \end{align*} where we identify $\mathrm{Spin_8}$ with $\{(g_1,g_2,g_3) \in \mathrm{SO}_8^3, \ \forall (x,y) \in \mathbb{O}, \ g_3(xy) = g_1(x)g_2(y) \}$.

It is well-known (see Harvey's Spinor and calibrations for instance) that the subgroup of $E_6$ generated by $\mathrm{SO_3}$ and $\mathrm{Spin}_8$ is $\mathrm{F}_4$. I think it is equally well-known (I don't have a reference at hand, but it seems to be an easy corollary of the previous statement) that $\mathrm{E}_6$ itself is generated by $\mathrm{SL}_3$ and $\mathrm{Spin_8}$.

Question : Is there an explicit description the subgroup of $\mathrm{E}_6$ (resp. $\mathrm{F}_4$) generated by $\mathrm{SL}_{3}$ and $\mathrm{Spin}_7$ (resp. $\mathrm{SO}_3$ and $\mathrm{Spin}_7$), where $\mathrm{Spin_7}$ is seen in $\mathrm{Spin}_8$ as $\{(g_1,g_2,g_3) \in \mathrm{Spin}_8, \ g_1(1) = 1\}$?

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    $\begingroup$ I think SL(3) contains elements that act on Spin(8) via triality. So I think those elements together with Spin(7) already generate Spin(8). $\endgroup$ May 31, 2020 at 12:36
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    $\begingroup$ @TheoJohnson-Freyd : This doesn't look possible to me. If you represent an element of $J_3(\mathbb{O})$ as a $24 \times 24$ matrix, it happens, quite miraculously, that the action of any element $g \in \mathrm{Spin}_7$ on $A \in J_3(\mathbb{O})$ can be written as $GAG^{-1}$ for a uniquely defined $G \in \mathrm{SL}_{24}$. On the other hand, it is impossible for the action of $F_4$ to be represented by matrix conjugation... $\endgroup$
    – Libli
    May 31, 2020 at 15:50
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    $\begingroup$ Indeed any element $A \in J_{3}(\mathbb{O})$ can be diagonalized (with at most $3$ different real entries on the diagonal) using a $F_4$ transformation. If the action of $F_4$ could be represented by matrix conjugation, then the dimension of the kernel (in $\mathbb{R}^{24}$) would be the same after diagonalization. $\endgroup$
    – Libli
    May 31, 2020 at 15:54
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    $\begingroup$ But the dimension of the kernel of a $24 \times 24$ diagonal matrix with at most three different real entries on the diagonal is either 0, 8, 16 or 24. While it is easily shown that there are $24 \times 24$ matrices in $J_3(\mathbb{O})$ which have kernel of dimension $4$. $\endgroup$
    – Libli
    May 31, 2020 at 15:56
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    $\begingroup$ Is this the split $\operatorname E_6$? $\endgroup$
    – LSpice
    Jun 4, 2020 at 22:15

1 Answer 1


N.B.: I am revising my response for clarity. (The actual answer to the question asked by the OP is still the same, but I think that this re-organization, particularly at the end, makes the structure of the argument for the answer more clear. I was inspired to do this because some people had some difficulty following the original.) I should also say that the main idea is essentially the one that Theo Johnson-Freyd proposed in his first comment on the question.

I'll use the more usual notation

$$ J_3(\mathbb{O}) = \left\{\ \left.\begin{pmatrix} \lambda_1 & a_3 & {\overline{a_2}} \\ \overline{a_3} & \lambda_2 & a_1 \\ a_2 & \overline{a_1} & \lambda_3 \end{pmatrix}\ \right| \ \ \lambda_i \in \mathbb{R}, \ a_i \in \mathbb{O} \right\} \tag 1 $$ and the cubic form given by $$ C = \lambda_1\lambda_2\lambda_3 + 2\,\mathrm{Re}(a_1a_2a_3) - \lambda_1\,a_1\overline{a_1} - \lambda_2\,a_2\overline{a_2} - \lambda_3\,a_3\overline{a_3}\,. $$ Then $\mathrm{E}_6\subset\mathrm{GL}\bigl(J_3(\mathbb{O})\bigr)\simeq \mathrm{GL}(27,\mathbb{R})$ is the group of linear transformations of $J_3(\mathbb{O})$ that preserve the cubic form $C$ and $\mathrm{F}_4\subset\mathrm{E}_6$ is the subgroup that also fixes $I_3\in J_3(\mathbb{O})$. (Explicitly, $\mathrm{F}_4$ is a maximal compact in this noncompact real form $\mathrm{E}_6^{(-26)}$ of $\mathrm{E}_6$.)

The subgroup $\mathrm{Spin}(8)\subset{\mathrm{SO}(8)}^3$ is defined as the set of triples $g = (g_1,g_2,g_3)$ that satisfy $$ \mathrm{Re}\bigl(g_1(a_1)g_2(a_2)g_3(a_3)\bigr) = \mathrm{Re}(a_1a_2a_3) $$ for all $a_i\in\mathbb{O}$. Let $K_i\subset\mathrm{Spin}(8)$ for $1\le i\le 3$ be the subgroup that satisfies $g_i(\mathbf{1}) = \mathbf{1}$ (where $\mathbf{1}\in\mathbb{O}$ is the multiplicative identity). Each of the $K_i$ is isomorphic to $\mathrm{Spin}(7)$, any two of them generate $\mathrm{Spin}(8)$, and the intersection of any two of them is equal to the intersection of all three of them, which is a group isomorphic to $\mathrm{G}_2$, diagonally embedded in ${\mathrm{SO}(8)}^3$ as the automorphism group of the octonions.

As has already been observed, $\mathrm{SL}(3,\mathbb{R})$ acts on $J_3(\mathbb{O})$ preserving $C$ via $a\cdot A = a\,A\,^{t}a$ (usual matrix multiplication), where $a\in\mathrm{SL}(3,\mathbb{R})$ and $A\in J_3(\mathbb{O})$ are arbitrary. This is a faithful action, so, in this way, $\mathrm{SL}(3,\mathbb{R})$ will be regarded as a subgroup of $\mathrm{E}_6$.

Meanwhile, by its very definition, $g = (g_1,g_2,g_3)\in\mathrm{Spin}(8)$ acts on $A\in J_3(\mathbb{O})$ via

$$ g\cdot \begin{pmatrix} \lambda_1 & a_3 & \overline{a_2} \\ \overline{a_3} & \lambda_2 & a_1 \\ a_2 & \overline{a_1} & \lambda_3 \end{pmatrix} = \begin{pmatrix} \lambda_1 & g_3(a_3) & {\overline{g_2(a_2)}} \\ \overline{g_3(a_3)} & \lambda_2 & g_1(a_1) \\ g_2(a_2) & \overline{g_1(a_1)} & \lambda_3 \end{pmatrix}\tag 2 $$ and this faithful action preserves $C$ as well, so $\mathrm{Spin}(8)$ will also be regarded as a subgroup of $\mathrm{E}_6$.

Now, as mentioned, $\mathrm{SO}(3)$ and $\mathrm{Spin}(8)$ together generate $\mathrm{F}_4\subset \mathrm{E}_6$. Consequently (since there is no connected Lie group that lies properly between $\mathrm{F}_4$ and $\mathrm{E}_6$), it follows easily that $\mathrm{SL}(3,\mathbb{R})$ and $\mathrm{Spin(8)}$ together generate $\mathrm{E}_6$.

We want to show that $\mathrm{SO}(3)$ and $K_1\simeq\mathrm{Spin}(7)$ also suffice to generate $\mathrm{F}_4$ while $\mathrm{SL}(3,\mathbb{R})$ and $K_1\simeq\mathrm{Spin}(7)$ suffice to generate $\mathrm{E}_6$.

To do this, let $h\in\mathrm{SO}(3)\subset\mathrm{SL}(3,\mathbb{R})$ be $$ h = \begin{pmatrix}0&-1&0\\-1&0&0\\0&0&-1\end{pmatrix} = h^{-1} = {}^th. $$ Then we have $$ h\cdot\begin{pmatrix} \lambda_1 & a_3 & \overline{a_2} \\ \overline{a_3} & \lambda_2 & a_1 \\ a_2 & \overline{a_1} & \lambda_3 \end{pmatrix} = \begin{pmatrix} \lambda_2 & \overline{a_3} & a_1 \\ a_3 & \lambda_1 & \overline{a_2} \\ \overline{a_1} & a_2 & \lambda_3 \end{pmatrix}. $$ Consequently, for $g = (g_1,g_2,g_3)\in \mathrm{Spin}(8)$, computation yields $$ h(g_1,g_2,g_3)h = \bigl(\ cg_2c,\ cg_1c,\ cg_3c\ \bigr)\in\mathrm{Spin}(8), \tag 3 $$ where $c:\mathbb{O}\to\mathbb{O}$ is conjugation, i.e., $c(a) = \overline{a}$. (Thus, conjugation by $h$ gives an involution of $\mathrm{Spin}(8)$ that, together with the order $3$ homomorphism $k(g_1,g_2,g_3) = (g_2,g_3,g_1)$, generates a group of automorphisms of $\mathrm{Spin}(8)$ isomorphic to $S_3$ that maps isomorphically onto $\mathrm{Out}\bigl(\mathrm{Spin}(8)\bigr)$. I imagine that this is what Theo Johnson-Freyd had in mind with his initial comment on this question.)

Note that $g_i(\mathbf{1}) =\mathbf{1}$ implies that $cg_ic = g_i$. Consequently, from the above formula $(3)$, it follows that if $g\in K_1$, then $hgh\in K_2$.

Thus the subgroup of $\mathrm{E}_6$ generated by $\mathrm{SO}(3)$ and $K_1 \simeq\mathrm{Spin}(7)$ contains $K_2$ and, hence, $\mathrm{Spin}(8)$ (since $K_1$ and $K_2$ generate $\mathrm{Spin}(8)$). Thus, this group is $\mathrm{F}_4$. Similarly, the subgroup of $\mathrm{E}_6$ generated by $\mathrm{SL}(3,\mathbb{R})$ and $K_1 \simeq\mathrm{Spin}(7)$ contains $K_2$ and, hence, $\mathrm{Spin}(8)$. Thus, this group is $\mathrm{E}_6$, as desired.

Similar arguments (using similar choices of $h$) suffice to show that, for any of $i= 1$, $2$, or $3$, the subgroup of $\mathrm{E}_6$ generated by $\mathrm{SO}(3)$ and $K_i$ is $\mathrm{F}_4$, while $\mathrm{SL}(3,\mathbb{R})$ and $K_i$ generate $\mathrm{E}_6$.


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