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The statement in GSW that you quote has to be interpreted properly. When they write, "Spinors form a representation of $\mathrm{SO}(N)$ which does not arise from a representation of $\mathrm{GL}(N,\m …

For a different viewpoint from the excellent treatments by Scott and Thurston of 3-dimensional geometries, if you are trying to get a feel for the homogeneous Riemannian $3$-manifolds (which, as noted …

The Grassmannian $$\mathrm{Gr}(2,\mathbb{C}^n) = \frac{\mathrm{SU}(n)}{\mathrm{S}\bigl(\mathrm{U}(2){\times}\mathrm{U}(n{-}2)\bigr)}$$ is the compact Hermitian symmetric space of type AIII of rank $r …

A relatively easy proof also follows from using the reflection identity: First, define the inner product associated to $F$, namely $v\ \cdot_F\ w = {\frac12}\bigl(F(v{+}w)-F(v)-F(w)\bigr)$, and then, …

Maybe the simplest argument, if you know something about compact Lie groups, is that SO(4) and SU(3) both have rank 2, i.e., they each contain a maximal torus, which is $S^1\times S^1$. Since all max …

It seems that you are asking for descriptions of the groups $\mathrm{Spin}(p,q)$ as algebraic groups. This can certainly be done explicitly for low values of $p$ and $q$, but I don't know a general p …

New answer: I now have an answer for the subgroup case that the OP originally asked about. In fact, one has the following result: Let $G$ be a connected compact Lie group and let $p = (p_1,\ldots,p …

Here is a different characterization of the subgroup $\mathrm{Ad}\bigl(\mathrm{SU}(n)\bigr)\subset\mathrm{SO}(n^2{-}1)$ that works when $n>2$. Define a skew-symmetric trilinear form $\kappa:{\frak{ …

In any case, the proof is very simple. Consider the $3$-form $$ \phi_0 = dx^1\wedge dx^2\wedge dx^3 + dx^4\wedge dx^5\wedge dx^6. $$ I claim that the subgroup $G\subset\mathrm{GL}(6,\mathbb{R})$ that …

Yes, $G_1G_2\subset\mathrm{SU}(8)$ is an algebraic set. Here is the argument: Let $G_1{\times}G_2$ act on $\mathrm{SU}(8)\subset\mathrm{End}(\mathbb{C}^8)\simeq\mathbb{C}^{64}$ by the rule $(g_1,g_2) …

I think that the answer here is just the double cover of the obvious answer for $SO(n)$, which is $U(n/2)$ when $n$ is even and $SO(n{-}1)$ when $n$ is odd. You can double-check this by consulting th …

I don't know that the partition has a name, so to speak, but it is well-understood and falls into the classification of the symmetric spaces of type A. Namely, those of type AI, which are $\mathrm{SU …

The answer is 'no', the affine symplectic group cannot appear as a Lie subgroup of any compact Lie group. The reason is that the affine symplectic group contains $\mathrm{SL}(2,\mathbb{R})$ as a Lie …

Actually, it does not look like that. Take the case $N=3$. The representation of $\mathrm{SU}(3)$ on ${\frak{so}}(8)$ breaks up into the $8$-dimensional subspace ${\frak{su}}(3)$ and an irreducible …

In addition to the above answers involving spinors and/or octonions, you might be interested in Cartan's original construction of the triality automorphisms, which is very explicit and takes just a co …