Preface The purpose of my question - on high level - is to understand exceptional symmetric spaces. My latest idea is to embed them into Lie group. There is quite nice embedding of 32-dimensional $E_{III}$ and 40-dimensional $E_{II}$ in $E_6$. It would be nice to see similar embedding for $E_I$ and $E_{IV}$.

Another thing which strikes me is that symmetric spaces $E_I$, $E_{II}$, $E_{III}$, $E_{IV}$ correspond to 2-dimensional projective space over $\mathbb C \otimes \mathbb O$. Can we analyze 1-dimensional space over this algebra ? In such case we obtain symmetric spaces $G_{5,5}^+\times S^1$, $G_{4,6}^+$, $G_{2,8}^+$, $S^9\times S^1$. These conlusion I present without proof. The dimensions of last symmetric spaces are lower by 16 than the first ones. Can we see the same analogy when considering symmetric spaces corresponding to $\mathbb C \subset \mathbb H$ ? I am asking, because on this wikipedia page symmetric space $E_{II}$ is presented as "space of symmetric subspaces isometric to $\mathbb C\otimes \mathbb HP^2$" (which is $G_{2,4}\mathbb C$ according to me).

As a last statement in my preface I would like to add that I am not interested in abstract definitions. I look at Lie groups as matrix groups preferably. I also accept Lie group as group acting on some geometrical object called "Rosenfeld projective plane". (It seems this term is not accepted by all mathematical community). Therefore I would preferably see embedding of symmetric space in Lie group $E_6$ in this language.

Formulating the questions

Question 1 Is it possible to embed symmetric spaces 42-dimensional $E_I$ and 26-dimensional $E_{IV}$ into compact Lie group $E_6$ ? Please provide construction of such embedding.

As it was answered in this Conjugacy classes of involutions in compact simple Lie group question, $E_{II}$ and $E_{III}$ can be embedded in $E_6$ as components if $\{x^2=1\}$ set.

Next I would like to continue in $Spin_{10}\times Spin_2$ group considered as automorphism of $\mathbb C\otimes \mathbb OP^1$. The plan is to find nice embeddding of $E_{IV}$ in $E_6$ first. Define nine letters in Clifford algebra $C_9$. Let $e_k$=$$\begin{pmatrix} L_k & \\ & R_k \end{pmatrix} $$ where $L_k$ and $R_k$ are left and right multiplications by imaginary base octonions, k=1..7. Let $e_8$=$$\begin{pmatrix} & S \\ -S & \end{pmatrix} $$ and $ e_9$=$$\begin{pmatrix} & iS \\ iS & \end{pmatrix} $$. $S$ is octonion conjugation. Let $V_{10}$=$<1,e_1,...,e_9>$ and $S^9$ be unit sphere in $V_{10}$.

Let $E_{III}^1$ = $\{iv\bar w: v\perp w \in S^9 \}$. It can be treated as $P^1$ (shortcut for $\mathbb C \otimes \mathbb OP^1$). Note that elements of this set does not depend on selection of $v,w$ from given plane of $V_{10}$. So $P^1$=$G_{2,8}^+$ topologically. In this definition $i$=$e_1...e_9$ is center of Clifford algebra $C_9$. Products of perpendicular elements from $P^1$ is set $E_{II}^1$=$E_{II}\cap P^1$=$\{v\bar wx\bar y: v\perp w\perp x \perp y \in S^9 \}$=$G_{4,6}^+$. The dimension of $E_*^1$ is 16 less then dimension of $E_*$ where *=I,II,III,IV.

The candidate for $E_I^1$ is $G_{5,5}^+\times S^1$, $E_{IV}^1$=$S^9\times S^1$. To start the prove let's ask question.

Question 2: Can $S^9$ be treated as embedding of $Spin_{10}/Spin_9$ into $Spin_{10}$ ?

To analyze dimension problem let's ask following question.

Question 3: Is dimension of $\{\mathbb CP^1\subset\mathbb HP^1\}$ four dimensions lower than $\{\mathbb CP^2\subset\mathbb HP^2\}$ ?

The first space seems to be 6-dimensional $G_{2,3}$. I don't know dimension of second space (please help), but it should be 10-dimensional $:)$

EDIT: Thank you for re-opening my question. Using first comment from Robert Bryant I will answer my first question. The Cartan's embedding uses outer involution. However I would like to find out more about the geometry before I post the answer. It will add more flavor to the answer. I might post similar questions for $E_7$ and $E_8$. Again I need to understand more the geometry.

EDIT October 2016 I paste following pictures illustrating at least how dimensions of exceptional symmetric spaces and corresponding maximal subgroups can be calculated. I hope there is more in the pictures. I leave the reader to figure out what they contain. First picture show compact $E_6$ and $E_{II}$

Compact $E_6$ and $E_{II}$

and remaining $E_{IV}$, $E_I$, $E_{III}$.

enter image description here

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    $\begingroup$ I'm not sure what you are asking. Cartan showed that every (irreducible) symmetric space can be isometrically embedded (up to a constant factor) into its group of symmetries. Moreover, this embedding is canonical: Let $\sigma:G\to G$ be the involution whose fixed subgroup is $K\subset G$, then the symmetric space $G/K$ has an embedding $\iota:G/K\to G$ given by $\iota(gK) = g\sigma(g)^{-1}$. This $\iota$ is equivariant with respect to the left $G$-action and isometric when $G$ is given its bi-invariant metric (with the appropriate constant scale factor). $\endgroup$ – Robert Bryant Sep 5 '16 at 17:13
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    $\begingroup$ I have asked, because on my previous question Mikhail Borovoi commented that EI and EIV cannot be embedded into $E_6$ as G-subvarieties. So posted this question to see the answer. I have not finished my question, because I planned to add some proposal, how EI and EIV could be embedded into $E_6$. I would like to see it in whole "magic square" picture. It is nice to hear that Cartan already embedded symmetric spaces into its group of symmetric. In such case I would like to see some nice embeddings as there are such for EII and EIII. $\endgroup$ – Marek Mitros Sep 5 '16 at 18:50
  • $\begingroup$ @RobertBryant: I don't think that your embedding $\iota$ is equivariant with respect to the left $G$-action. $\endgroup$ – Mikhail Borovoi Sep 5 '16 at 19:57
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    $\begingroup$ @MikhailBorovoi: What do you mean, 'not equivariant'? The embedding $\iota$ satisfies $$\iota\bigl(h(gK)\bigr) = h\,\iota(gK)\,\sigma(h)^{-1}$$ for all $h,g\in G$, so it intertwines the left action of $G$ on $G/K$ with the $\sigma$-twisted conjugate action of $G$ on itself in the range. Meanwhile, the $\sigma$-twisted conjugate action is clearly isometric with respect to the bi-invariant metric on $G$, so the embedding $\iota$ is certainly isometric (up to a constant scale factor, which can be taken to be $1$). $\endgroup$ – Robert Bryant Sep 6 '16 at 1:41
  • $\begingroup$ @RobertBryant: Going back to Cartan. Can we find $\sigma$ which fix subgroup $Sp_4$ or $F_4$ of $E_6$ ? $\endgroup$ – Marek Mitros Sep 6 '16 at 8:31

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