Understanding $G_2$ inside Spin(7)? (EDIT: problem solved)

Question

This is a rather embarrassing question, so please let me know of any duplication and I will happily remove it.

I am seeking to understand the $\mathbb Q$-split form of the algebraic group $G_2$, and the context I am working in makes it so that it would be very nice to have actually matrices to work with.

My attitude has been that since the groups and Lie algebras are split, it doesn't hurt to think of everything as being defined over $\mathbb C$. My first question is

Is this assumption okay to make?

This really only comes in with the references I have been reading to elucidate the situation.

My hope was that I could work with the embedding of Lie algebras $$ \mathfrak{g}_2 \hookrightarrow \mathfrak{so}(7)$$ (where this orthogonal lie algebra is the appropriate split form), arising from sending the short simple root of $\mathfrak{g}_2$ diagonally to the appropriate root spaces of $\mathfrak{so}(7)$.

As far as I can tell, we can't in any way exponentiate this embedding as $G_2$ does not embed into $SO(7)$, but into $Spin(7)$. But I don't know how otherwise to use this embedding to realize $G_2$ as a matrix group. My question is thus

Is there a way to utilize this embedding of Lie algebras to realize $G_2$ as a group of $7\times 7$ matrices? If not, how about the embedding into $spin(8)$?

As stated above, I am really wanting to understand the embedding as one of algebraic groups, not Lie groups. If there is any obvious confusion on my part about working with the algebraic groups versus Lie groups, please feel free to explain.

Answer 1

For what it's worth, I'll post the solution. I was incorrect that there is not an embedding $$G_2 \hookrightarrow SO_7.$$

Such an embedding exists on the level of algebraic groups, rather than just on the level of Lie algebras, arising from the Norm on the imaginary octonions, which $G_2$ leaves invariant. I had misinterpreted a comment in Dan Bump's Lie Groups book, in chapter 33, which lead me to assume that such an embedding did not exist.

My confusion came from the fact that when attempting to explicitly exponentiate the embedding on Lie algebras $$\mathfrak{g}_2 \hookrightarrow \mathfrak{so}_7,$$

one must be careful about the vectors chosen for the short root $\alpha$, which is embedded diagonally into two (orthogonal) simple root spaces of $\mathfrak{so}_7$.

Suppose the two simple roots for $\mathfrak{g}_2$ are $\alpha$ (short)and $\beta$ (long), and the 3 simple roots for $\mathfrak{so}_7$ are $\gamma_1,\gamma_2,\gamma_3$, with root vectors $e_i\in \mathfrak{so}_7$. (here I am numbering the roots in accordance with Bourbaki, so there is a double bond in the Dynkin diagram between the nodes corresponding to $\gamma_2$ and $\gamma_3$.)

We want to embedd $\mathfrak{g}_2$ in a way which corresponds to the diagram folding from type $B_3$ to type $G_2$ (see this wikipedia article), so we send $$\beta \mapsto \gamma_2$$ and $$\alpha\mapsto \gamma_1\oplus\gamma_3,$$ where I mean we map the root space of $\beta$ to the root space of $\gamma_2$ in $\mathfrak{so}_7$, and we map the root space of $\alpha$ diagonally into the root spaces of $\gamma_1$ and $\gamma_3$.

The key here is the image of the root vector $e_{\alpha}$ could be any nontrivial linear combination $ae_1+be_3$ (ie:$ab\neq 0$). This is because the two roots $\gamma_1$ and $\gamma_3$ are orthogonal in $\mathfrak{so}_7$, so anysuch linear combination will be eigenvalues of the coroot $h_{\alpha}$ (see below).

One can check (and this is the main source of the confusion which lead to my question) that if $h_{\alpha}$ is the coroot in $\mathfrak{g}_2$ for $\alpha$, and if $h_1, h_3$ are the coroots in $\mathfrak{so}_7$ corresponding to $\gamma_1$ and $\gamma_3$, then $$h_{\alpha} \mapsto h_1+2h_3.$$

This must have to do with the double bond in the Dynkin diagram, but it is unclear to me how to have expected this. Regardless, in order to embed $\mathfrak{g}_2$ into $\mathfrak{so}_7$, one needs to use this to appropriately define the roots vectors for the negative roots, the $f_{\lambda}$ for positive roots $\lambda$.

If you want your matrices to be such that $e_{\lambda}^T = f_{\lambda}$, one takes the embedding $$e_{\alpha} \mapsto e_1+\sqrt{2}e_3$$.