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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} …

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 …

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 …

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 …

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 …

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 …

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 …

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 …

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 …

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 …

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 …

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 …