Separating subspaces in an irreducible representation - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T14:46:18Z http://mathoverflow.net/feeds/question/106366 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/106366/separating-subspaces-in-an-irreducible-representation Separating subspaces in an irreducible representation Alex Eskin 2012-09-04T17:06:18Z 2012-09-04T18:27:48Z <p>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. </p> <p>$\mathbf{Question 1}.$ Is it always possible to find $g \in G$ such that $g U$ and $W$ are in general position, i.e. $$\dim(g U \cap W) = \max(0, \dim U + \dim W - \dim V)?$$</p> <p>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. </p> <p>$\mathbf{Question 2}.$ 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? </p> <p>Basically, I am mostly curious if this circle of questions was ever studied, and what keywords do I need to look it up. </p> http://mathoverflow.net/questions/106366/separating-subspaces-in-an-irreducible-representation/106368#106368 Answer by David Speyer for Separating subspaces in an irreducible representation David Speyer 2012-09-04T17:32:40Z 2012-09-04T17:32:40Z <p>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 $W = \mathrm{Span}((1,0,1,0), (0,1,0,-1))$.</p> <p>Geometrically, inside the $3$-sphere <code>$S:=(\mathbb{R}^4 \setminus \{ 0 \})/\mathbb{R}_{+}$</code>, 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.</p> <p>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.</p> http://mathoverflow.net/questions/106366/separating-subspaces-in-an-irreducible-representation/106369#106369 Answer by Will Sawin for Separating subspaces in an irreducible representation Will Sawin 2012-09-04T17:34:51Z 2012-09-04T18:27:48Z <p>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 $n\geq 2$, $m\geq 2$. </p>