Action of $SL_{n+1}$ on couples of linear spaces in $\mathbb{P}^n$. - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T18:29:46Z http://mathoverflow.net/feeds/question/77363 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/77363/action-of-sl-n1-on-couples-of-linear-spaces-in-mathbbpn Action of $SL_{n+1}$ on couples of linear spaces in $\mathbb{P}^n$. IMeasy 2011-10-06T14:45:16Z 2011-10-06T15:53:34Z <p>Let $SL_{n+1}$ act on $\mathbb{P}^n$ in the natural way. Suppose I take two linear subspaces $\mathbb{P}^m$ and $\mathbb{P}^{n-m}$, with $m &lt; n$, that intersect in one point. Is the action of $SL_{n+1}$ transitive on the set of such couples of linear subspaces?</p> http://mathoverflow.net/questions/77363/action-of-sl-n1-on-couples-of-linear-spaces-in-mathbbpn/77368#77368 Answer by Anton Fonarev for Action of $SL_{n+1}$ on couples of linear spaces in $\mathbb{P}^n$. Anton Fonarev 2011-10-06T15:53:34Z 2011-10-06T15:53:34Z <p>Sure it does. Denote your variety by $X$. Consider this action on the corresponding vector space $V$. Then your projective subspaces correspond to linear subspaces $V_1$ and $V_2$ of dimension $m+1$ and $n-m+1$ that intersect transversely. There is an obvious projection from the variety $Y$ of all frames in $V$ to $X$ (take the linear hull of the first $m+1$ and the last $n-m+1$ vectors). Now, up to a multiplication of all the vectors in the frame by a common scalar, $SL_{n+1}$ acts transitively on $Y$. Thus, it acts transitively on $X$.</p>