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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 …

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 …

Your quote about Cartan thinking of $B_n$ and $D_n$ as 'projective groups..." is actually Cartan describing the lowest dimensional homogeneous space of these groups (except, of course, for a few excep …

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 …

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 …

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, …

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 …

Here's a description that doesn't use octonions; instead, it uses the definition of $\mathrm{G}_2$ as the stabilizer of a $3$-form on $\mathbb{R}^7$. For simplicity, I'll do this for the split-form, …

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 …

You can work out the answers to these questions using the material in Chapter 11 of the book Spinors and Calibrations by F. Reese Harvey. You will also need to recall that, for $N\not=4$, the group o …

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 …

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{ …

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 …

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 …

When the (general) question is rephrased in less basis-dependent language, I believe that it translates to this: Let $\mathrm{U}(d)$ act on $V = \mathbb{C}^d$ in the usual way, and consider the $n$-f …