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I apologize if this question might seem to be a bit too elementary.

Let $\mathbb{P}^n$ be the projective space over $k$ - an algebraically closed field of characteristic 0. Let $1\leq l\leq n-1$, and denote by $G$ the grassmanian $Gr(l+1,n+1)$. Let $X\subset G$ be a set of dimension $l$ subspaces, s.t. $\forall \pi_o,\pi_1\in X,\,\pi_0\cap\pi_1\neq\emptyset$. Then my question is this: how big can $\dim(\bar{X})$ be? and if the dimension is maximal, does it imply that all of the spaces in $X$ are contained in some $\mathbb{P}^{n-1}\subset\mathbb{P}^n$? (Where the closure is taken in the Zarisky topology.)

For example, if $n=3$ and $l=1$, then having two intersecting lines $\alpha,\beta$, demanding that a third line $\gamma$ intersects them both is equivalent to saying that $\gamma\subset \mathrm{Join}(\alpha,\beta)=\mathbb{P}^2$. Thus, the maximal $\dim(X)$ in this case is 2, and $\dim(X)\geq 1$ implies that all lines in $X$ are contained in $\mathbb{P}^2$

I am particularly interested in the case where $n=2k+1,l=k$. Does $\dim(X)=k$ imply that all of the $k$-planes in $X$ are contained in some $\mathbb{P}^{2k}$?

Thanks! Edit notice: I forgot to mention that I don't want any base points for the family. And, that $X$ should be closed and irreducible in $G$.

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    $\begingroup$ At least for $n=3$, I remember something about this in the section of Griffiths-Harris: "the quadric line complex". For a general point $[\pi]$ of $X$, think of the Zariski tangent space of $\pi$ as a vector subspace of the space of global sections of the normal bundle $N_{\pi/\mathbb{P}^n}\cong \textit{Hom}(S,Q)$, where $S$ and $Q$ are the restrictions to $\pi$ of the universal subbundle, resp. quotient bundle. Your sections each vanish somewhere. You want to use this to prove that all sections have some kernel in $S$ or have image in a proper subbundle of $Q$ . . . $\endgroup$
    – Jason Starr
    Commented Dec 3, 2014 at 14:21

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