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The group $H$ acts transitively and primitively on $\mathbb{C}=\mathbb{R}^2$. ('Primitive' means that $H$ preserves no nontrivial foliation.) It's a consequence of the classification of transitive pr …

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, …

The $\sigma_i$ as defined above satisfy $\mathrm{d}\sigma_i = -2\,\sigma_j\wedge\sigma_k$ when $(i,j,k)$ is an even permutation of $(1,2,3)$. For later use, let $E_i$ be the dual (left-invariant) fram …

Remark: I assume that you want $A$ to be a non-trivial bundle. Otherwise, of course, any parallelizable compact manifold would be an example. In particular, any compact Lie group would be an exampl …

This fails already for the split form of $\mathrm{G}_2$. Every finite-dimensional irreducible representation of its Lie algebra is a constituent of a tensor power of the $7$-dimensional representation …

I don't know a formula for $d(e^x,e^y)^2$, and I suspect that there is no simple formula, but the answer to Q2' is 'no'. The right hand side of (3) is linear in $Q^{-1}$, but it is not hard to see th …

Just so there'll be an answer: Whether every vector bundle over $G/\Gamma$ admits a flat connection depends on the group $G$ and the subgroup $\Gamma$. For example, if $G=\mathrm{SU}(2)\simeq S^3$ an …

This is something of an exercise is unwinding the definitions, but there's an interesting twist to it as well, so here's an outline of an answer: Let $G$ be a connected $3$-dimensional Lie group (not …

Here are a few brief comments, but, as you suspect, an enormous amount is known about the Laplacian on functions and forms on compact Lie groups. • Presumably, you know that the Killing form is non-de …

The answer is 'no'. In fact, a stronger statement is true: If $V$ is a finite dimensional vector space and $G\subset\mathrm{GL}(V)$ is a (connected) non-compact simple Lie group, then the only bounde …

This is more an answer to the corresponding question for the quaternions: Can one show that, for unit quaternions $p,q\in S^3\subset\mathbb{H}\simeq\mathbb{R}^4$, one has the inequality $$ \langle\lo …

Now that I have written out the completely elementary proof above for the quaternions and for $\mathrm{SO}(3)$, I feel that I should point out that the statement $|\log(ab)|\le |\log(a) + \log(b)|$, s …

I realized a problem with my 'counterexample', so I no longer claim that the desired inequality does not hold on $\mathrm{SO}(3)$. The proof that it does hold on the quaternions is still OK. I'll po …

There cannot be an 8-dimensional group $G$ acting effectively on $S^4$ by Riemannian isometries. The following argument may not be the best, but it explains why this is true. (I will assume that $G$ …

There's an easy counterexample to your guess: Let $M^6 = \mathrm{SU}(3)/\mathbb{T}^2$, where $\mathbb{T}^2\subset\mathrm{SU}(3)$ is the maximal torus (for example, the diagonal subgroup). In that ca …