# Subgroup $\mathrm{E}_6$ generated by $\mathrm{Spin_7}$ and $\mathrm{SL}_3$

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 $$\{(g_1,g_2,g_3) \in \mathrm{Spin}_8, \ g_1(1) = 1\}$$?

• 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). – Theo Johnson-Freyd May 31 '20 at 12:36
• @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... – Libli May 31 '20 at 15:50
• 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. – Libli May 31 '20 at 15:54
• 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 $4$. – Libli May 31 '20 at 15:56
• Is this the split $\operatorname E_6$? – LSpice Jun 4 '20 at 22:15

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 $$\mathrm{E}_6$$.