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No. Here's a counterexample: Let $f(r)$ satisfy the equation $f'' + 2 f^3 = 0$ with the initial conditions $f(0)=1$ and $f'(0)=0$. Then $f$ satisfies $(f')^2+f^4=1$ and is periodic with period $$ L …

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 …

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 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 …

I think this is unlikely to have a very nice answer. When $n=2$ and $n=3$, the answer is simple, but, already for $n=4$, it's not likely to be easy to give a set of generators and relations for the $ …

Note added on 26 Nov 2018: I have corrected my answer, which had a serious mistake. For simplicity of notation, let $(x,y,z) = (x_1,x_2,x_3)$. The Hessian form associated to $f_0 = {x_1}^3+{x_2}^3+{ …

Here's another way to do it that you might find useful: Recall that $\mathrm{SL}(2,\mathbb{C})$ acts on the polynomial ring $\mathbb{C}[x,y]$ by linear substitution in $x$ and $y$, making the subspa …

Here's an alternate construction of the invariant $\alpha$ that seems a little simpler (and, besides, gets used in proving the normal forms for $3$-forms in $6$-variables). The details may be found in …

In this case if your 16-dimensional $\mathrm{Spin}(9,1)$-representation is the one of highest weight $(0,0,0,0,1)$, then the $\mathrm{Spin}(9,1)$-irreducible decomposition of its symmetric square is j …

NB: Note that my $a_k$ have different signs from those defined in the question. For me, $$ (z - z_1)(z-z_2)(z-z_3) = z^3 - a_1\ z^2 + a_2\ z - a_3, $$ so that $a_k$ is the $k$-th elementary symmetri …

You may know this already, but, if not, the following may be helpful to you: The ring of invariants for binary sextics is generated in degrees 2, 4, 6, 10, and 15. (See this reference on binary sexti …

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} …

The following construction reduces your problem it to a classical and well-studied problem in invariant theory. First, I claim that there is a natural way to interpret an $n$-tuple of points in $S^2$ …