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I think you want to look at Faà di Bruno's formula, and the description of composition of formal power series, particularly the historical remarks. The linear representation of this group that allows …

I don't know where the orbit types in this case were first explicitly classified, but it is done in my paper Second order families of special Lagrangian 3-folds, Perspectives in Riemannian geometry, 6 …

One reference is in Helgason's 1984 book Groups and Geometric Analysis. The result you want appears there as Corollary 5.12. The notation he uses is $X=G/K$ is a symmetric space where $G$ is connecte …

Yes. Here is one way: Consider standard $\mathbb{R}^3$ endowed with the Lorentzian quadratic form $Q = x^2+y^2-z^2$, and let $G\simeq\mathrm{O}(2,1)\subset\mathrm{GL}(3,\mathbb{R})$ be the symmetry …

You really should have a look at F. Reese Harvey's book Spinors and Calibrations, where all of your questions are answered. For example, your 'inclusion' (1) is not correct for sufficiently large $n$. …

There are two aspects to this question, the global question and the local question. Also, the case $n=1$ is different from $n>1$. Basically, the answer is 'essentially yes, but with some caveats'. H …

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 …

The answer is 'no', though I don't know an easy way to see this without doing an explicit calculation. Here is where to look though, if you want to do the calculation yourself: Things work out a bit …

Here's another very nice (but still algebraic) interpretation that explains some of the geometry: Recall that $\operatorname{SL}(2,\mathbb{C})$ has a $2$-to-$1$ representation into $\operatorname{SL} …

As per the OP's comment, we are to assume that $\mathrm{G}_2$ and $\mathrm{F}_4$ mean the complex simple Lie groups. Let's start with $\mathrm{G}_2\subset\mathrm{SO}(7,\mathbb{C})$, in its standard re …

N.B.: I am revising my response for clarity. (The actual answer to the question asked by the OP is still the same, but I think that this re-organization, particularly at the end, makes the structure …

Here is an outline of the argument that shows that the $\mathrm{SL}_6(\mathbb{C})$-stabilizer of the generic $3$-plane $W\subset\Lambda^2(\mathbb{C}^6)$ has dimension $1$, not $0$, as (apparently) cla …

This is a trick question, right? It's not true when $m=2$ because, then $\mathrm{SO}(m{-}1)=\mathrm{SO}(1)$ is trivial, so that all polynomials on $2$-by-$2$ symmetric matrices are invariants, and th …

First, let me fix a misunderstanding: $\mathrm{Sp}(n)$ does not sit in $\mathrm{GL}(n,\mathbb{C})$, but in $\mathrm{GL}(2n,\mathbb{C})$, so I'll assume that you mean, for the second part that $A$ lie …

If all you really want to know are the irreducible representations of ${\frak{so}}(7)$ and ${\frak{so}}(8)$ up to dimension $30$, I can save you the trouble of looking up these tables: For ${\frak{so …