Suppose $G$ is a semisimple $\mathbb{R}$-algebraic group with finite center, and suppose $G$ acts irreducibly on a vector space $V$. Suppose $U \subset V$ and $W \subset V$ are subspaces. 


$\mathbf{Question 1}.$ 
Is it always possible to find $g \in G$ such that $g U$ and $W$ are in general position, i.e. 
$$\dim(g U \cap W) = \max(0, \dim U + \dim W - \dim V)?$$


I believe the answer to this question is no. I remember thinking about this when I was a graduate student, and there was some simple example, but I no longer remember what the example was. So the real question is: construct a simple counterexample to this assertion. 


$\mathbf{Question 2}.$ It is possible to show using the Weyl unitary trick and orthogonality of characters that there is always $g \in G$ such that
$$\dim(gU \cap W) \leq \frac{(\dim U)(\dim W)}{\dim V}.$$ 
(Actually John Stalker showed this to me when we were graduate students). The question is: can one get a better bound? 


Basically, I am mostly curious if this circle of questions was ever studied, and what keywords do I need to look it up. 



Separating subspaces in an irreducible representation
Let $G=SL\_n(\mathbb R) \times SL\_m(\mathbb R)$. Let $V$ be the irreducible representation given by taking the defining representation $V\_n$ of the first tensor the defining representation $V\_m$ of the second. Let $U$ be a copy of of $V\_n$ inside $V\_n \otimes V\_m$ and let $W$ be a copy of $V\_m$. The intersection of $gU$ and $W$ is always $1$-dimensional, showing the bound in question $2$ is tight and constructing a simple counterexample to question $1$ when $n\geq 2$, $m\geq 2$. 


