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It is not always clear what one means by 'the simplest description' of one of the exceptional Lie groups. In the examples you've given above, you quote descriptions of these groups as automorphisms o …

There's an easy argument based on the Cayley transform: If $a$ is a skew-symmetric $n$-by-$n$ real matrix, then $I_n+a$ is invertible (since $(I_n-a)(I_n+a)=I_n-a^2$ is a positive definite symmetric …

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 …

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 …

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 …

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} …

Here's a simple counterexample to Question 1: Let $d=4$ and $k=2$. Let $X\subset\mathrm{GL}_4(\mathbb{R})$ consist of a single element $J$ where $J^2=-I$. Then $J$ is not conjugate to any upper tri …

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 …

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 …

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 …

If I understand you correctly, the answer is 'no'. Because the open set $L_n\subset B_n$ is an orbit of $\mathrm{GL}^+(n,\mathbb{R})$ under the natural representation of $\mathrm{GL}^+(n,\mathbb{R} …

This was worked out explicitly by A. A. Katanova in her paper Explicit form of certain multivector invariants in Advances in Soviet Mathematics 8 (1992), pp. 87-93. According to her calculations, the …

As YCor commented, the main point is to show that the invariant polynomials separate orbits. This follows from the compactness of $\mathrm{SO}(n)$. The point is this: Because $\mathrm{SO}(n)$ is co …

The answer to your first question is 'yes, generically, the intersection is finite and transverse'. More precisely, the set $U$ of 6-dimensional subspaces that intersect $\mathrm{SO}(3)$ transversely …