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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 $m < n$, that intersect in one point. Is the action of $SL_{n+1}$ transitive on the set of such couples of linear subspaces?

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Fix your favorite two such subspaces $H$ and $H'$. (Mine are the span of $\{ e_0, e_1, \dots, e_m \}$ and the span of $\{ e_m, e_{m+1}, \dots, e_n \}$ for some choice of ordered basis of your underlying vector space $V$). You want to show that the $PGL_{n+1}$-orbit of this point is the variety of pairs of subspaces that intersect in a point. You know that the orbit is an irreducible subvariety of this irreducible variety, so all you need is for them to have the same dimension. You can compute the dimension of an orbit once you identify the stabilizer of a point. –  Michael Joyce Oct 6 '11 at 15:47
    
On second thought, proving that your variety is irreducible is most easily done by showing its an orbit variety, so I retract my previous approach. Instead, fix an $n+1$-diml v.s. $V$ and let $W_1, W_2$ be subspaces of dimension $m+1$ and $n-m+1$ that intersect in a $1$-diml subspace. You can find a basis, call it $\mathscr{V} = \{v_0, \dots v_n\}$ as in the above comment so that $W_1$ and $W_2$ have the form of the $H$ and $H'$ above. Then change of basis from $\mathscr{E}$ to $\mathscr{V}$ gives you an element of $PGL_{n+1}$ that takes one pair of such subspaces to the other. –  Michael Joyce Oct 6 '11 at 15:58
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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 $X$.

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