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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.

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up vote 8 down vote accepted

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 $n\geq 2$, $m\geq 2$.

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A quibble: you need to get rid of the center, so use special linear groups. – Victor Protsak Sep 4 '12 at 18:26
Interestingly, the first version of this answer used special linear groups but then I edited it because I thought it would be more elegant the other way. I guess that wasn't such a great idea. – Will Sawin Sep 4 '12 at 18:30

Counter-example for $(1)$: $G = SO(2,2)$ with $V$ the standard four dimensional rep. To be concrete, $G$ preserves the quadratic form $q(w,x,y,z) = w^2+x^2-y^2-z^2$. $U$ is the two dimensional subspace $\mathrm{Span}((1,0,1,0), (0,1,0,1))$ and $W = \mathrm{Span}((1,0,1,0), (0,1,0,-1))$.

Geometrically, inside the $3$-sphere $S:=(\mathbb{R}^4 \setminus \{ 0 \})/\mathbb{R}_{+}$, the quadratic $w^2+x^2=y^2+z^2$ cuts out a torus $T$, and $G$ takes this torus to itself. (One can see this by intersecting $w^2+x^2=y^2+z^2$ with $w^2+x^2+y^2+z^2=1$ to get $w^2+x^2=y^2+z^2=1/2$, clearly a torus.) The images of $U$ and $W$ in $S$ are two circles which have intersection number $2$ on the torus, so you can't move them not to intersect.

An algebraic geometer would work in $\mathbb{RP}^3$ rather than $S$, and would say that $T$ is a quadric surface and $U$ and $W$ belong to opposite rulings of the surface.

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Thanks! I have accepted Will Sawin's answer, but, as I now remember, your example was the one I was thinking of many years ago. – Alex Eskin Sep 5 '12 at 10:43
You're welcome. Mine is a special case of his: $SO(2,2)$ is $SL_2(\mathbb{R}) \times SL_2(\mathbb{R}) / \pm (\mathrm{Id}, \mathrm{Id})$. You should also be able to get examples with $SO(m,m)$, but they are aren't as pretty as Will's, and they don't give the tight bounds that his does. – David Speyer Sep 5 '12 at 11:29

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