Does a codimension 1 subspace of a representation of lie group intersect all orbits? Let $G$ be a nice lie groups, and $V$ a complex irred representation.
I am interested in understanding for which codim 1 subspaces $U \subset V$, we have $G \cdot U = V$ (does there exist such $U$?). Let me emphasize that in $G U = V$ on the left I mean pointwise, not the span (which correlates with the title of the question by moving $G$ to the right)
Specifically in my case, I care about $SO(3)$ and its irreducible representations.

Let me do a very particular case.
In the case of $G = SO(3)$, and $V$ the three-dimensional complex representation, I claim the subspace perpendicular to $u = (1,0,0)$ satisfies every orbit meets it.
First to visualize $V$, we take $SO(3)$ acting on ${\RR}^3$ via rotations, and tensor with $\CC$.
Consider a given $v = (x,y,z)$. We think of it as $v = v_0 + iv_1$ both in $\RR^3$, so we want to rotate both $v_0, v_1$ to a given plane which is possible.
In fact here is a general argument for all subspaces! Given $u$ that defines a subspace, we use the above construction to make it (under an action) to be of the form (also multiplying by a const) $$(0, 1, \gamma)$$.
Now given $v$, we first rotate it to be of the form $(x,y,0)$. Next, we consider rotating using the stabilizer of $(0,0,1)$, writing $$x = x_r + i*x_i $$$$ y = y_r + i*y_i$$, the action considers $$(x_r,y_r)$$, $$(x_i, y_i)$$, rotates those, and then takes the two $y$ coordinates of the two vectors to form a complex numbers which will be the inner product with our $$(0, 1, \gamma)$$. It's easy to see that (Assuming it's never zero, which will happen generically); this traces a path in $\CC - 0$ which has winding number $1$.
Thus our map $F_u: SO(3) \to \CC - 0$ given by $g \to <gv,>$  restricts at $S^1$ (the $S^1$ here is the stabilizer mentioned above) to a map with winding number $1$. This means that $F_u$ must vanish; otherwise it'd give a nontrivial map of fundamental groups but $SO(3)$ has fundamental group $Z/2$.
In fact I have a proof that shows that for $SO(3)$, for $V_{4n+3}$ every codim $1$ subspace covers everything via the orbit!
