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The Grassmannian $G(k,V)$ is the set of all $k$-dimensional subspaces of a vector space of dimension $n$. It can be embedded inside $\mathbb{P}(\bigwedge^k V)\cong \mathbb{P}^{\binom{n}{k}-1}$ as a projective variety using the Plucker embedding given by $<v_1,v_2,\cdots,v_k>\ \mapsto\ v_1\wedge v_2\cdots\wedge v_n$.

When $k=2,n=4$, $G(2,4)$ can be embedded inside $\mathbb{P}^5$ and the resulting variety is called the Klein quadric. The image of all 2-dimensional subspaces containing a fixed 1-dimensional subspace is a 2-plane inside the quadric called the $\alpha$-plane and the image of all 2-dimensional subspaces contained in a fixed 3-dimensional subspace is again a 2-plane inside the quadric called the $\beta$-plane. How do we prove this?

Is there a generalization of this to $G(k,n)$? Is the image of all $k$-dimensional subspaces containing a fixed $(k-1)$-dimensional subspace a $(n-k)$-plane inside the Grassmannian variety and similarly is the image of all $k$-dimensional subspaces contained in a fixed $(k+1)$-dimensional subspace a $k$-plane inside the Grassmannian variety? Also what can we say about the images of $k$-dimensional subspaces containing a fixed $r$-dimensional subspace or contained inside a fixed $r$-dimensional subspace? I know that these must be the intersection of the Grassmannian variety with a linear subspace but when are they subspaces contained inside the Grassmannian variety?

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    $\begingroup$ I don't mean to be brief or terse, but try googling schubert cell or schubert variety. $\endgroup$ – meh Sep 23 '14 at 20:48
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    $\begingroup$ Griffiths and Harris (Principles of Algebraic Geometry) has a nice introduction to Schubert calculus, and they explicitly work out their general formulas for the case $k=2, n=4$, so this might be a particularly useful resource for the OP. $\endgroup$ – Michael Joyce Sep 23 '14 at 21:11
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Here is the answer for the Grassmannian of lines in $\mathbb{P}^3$. You can imitate this argumet in the general case of $k$-subspaces containing a fixed $(k-1)$-space or contained in a fixed $(k+1)$-space.

For any point $p\in\mathbb{P}^3$ be $\Sigma_{p}\subset\mathbb{G}(1,3)\subset\mathbb{P}^5$ be the locus parametrizing lines in $\mathbb{P}^3$ through $p$. Similarly, for any plane $H\subset\mathbb{P}^3$ be $\Sigma_{H}\subset\mathbb{G}(1,3)\subset\mathbb{P}^5$ be the locus parametrizing lines in $\mathbb{P}^3$ contained in $H$.

Let $u\in V$ be a vector representing $p\in\mathbb{P}^3$. Then the lines through $p$ are represented by $2$-vectors of the form $u\wedge v$. Let $\{u,u_1,u_2,u_3\}$ be a basis of $V$. Then we may write $v = \alpha u+\beta_1 u_1+\beta_2 u_2+\beta_3 u_3$, and $$u\wedge v = \alpha_1(u\wedge u_1)+\alpha_2(u\wedge u_2)+\alpha_3(u\wedge u_3).$$ Therefore, lines through $p$ are represented by the points of the plane spanned by $u\wedge u_1$, $u\wedge u_2$ and $u\wedge u_3$.

Now, the lines contained in the plane $H\subset\mathbb{P}^3$, by duality corresponds to the lines in $\mathbb{P}^{3^{*}}$ through the point $H^{*}$. Therefore they are parametrized by a plane in $\mathbb{G}(1,3)$ by the part above.

Now, take a plane $\Pi$ in $\mathbb{G}(1,3)$ and three points $l,r,s$ in this plane that do not lie on the same line. Let $L,R,S\subset\mathbb{P}^3$ be the corresponding lines. Since the three lines joining $l$,$r$ and $s$ are on the same plane contained in $\mathbb{G}(1,3)$ they intersect and they are contained in $\mathbb{G}(1,3)$. Therefore, the lines $L,R,S$ intersect. We have two cases.

  • $L\cap R\cap S = \{p\}$. In this case $\Pi$ parametrizes lines in $\mathbb{P}^3$ through $p$.

  • $L$, $R$ and $S$ intersect in three distinct points. Let $u,v,w$ be three representative vectors for these three points. Then $L,R,S$ are represented by $v\wedge w$, $u\wedge w$ and $u\wedge v$. Then a point on the plane $\Pi$ is of the form $$\alpha(v\wedge w)+\beta(u\wedge w)+\gamma(u\wedge v).$$ Therefore $L,R,S$ lie in the plane $\mathbb{P}(H)$, where $H = \left\langle u,v,w\right\rangle$.

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  • $\begingroup$ Thanks! I think these are special cases of Schubert varieties, is there a condition when a Schubert variety is a subspace contained in the Grassmannian? $\endgroup$ – Sivakanth Gopi Sep 26 '14 at 15:44
  • $\begingroup$ Yes, they are indeed Schubert varieties. You may find formulas for the degree of a Schubert variety here: sciencedirect.com/science/article/pii/S0001870802000981 $\endgroup$ – F_L Sep 26 '14 at 17:29

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