Dimension of a variety parametrising all 2-planes contained in a smooth hyperquadric in $\mathbb{P}^5_{\mathbb{C}}$ Let $Q$ be a smooth hyperquadric in $\mathbb{P}^5_{\mathbb{C}}$ and let $F_2(Q)$ be the variety parametrising all 2-planes contained in $Q$. Find dim$(F_2(Q))$ knowing that each hyperquadric contains at least one plane and that dim$(G(k,n))$ is $(n-k)(k+1)$. I don't really know where to start, any help?
 A: Let me give an answer with a few more details. 
This is a nice example of an incidence correspondence argument. Let's define the variety
$$ I := \{(\Pi,Q) \mid \Pi \subset Q \} \subset \mathbf G(2,5) \times \mathbf P^{20} $$ 
consisting of pairs (2-plane,quadric) with the 2-plane contained in the quadric. The variety $I$ comes with natural projections $\pi_1: I \rightarrow \mathbf G(2,5)$ and $\pi_2: I \rightarrow \mathbf P^{20}$. 
Now if we fix a quadric $Q$, the  variety $F_2(Q)$ of 2-planes in $Q$ is precisely the fibre of $\pi_2$ over the point $[Q] \in \mathbf P^{20}$. So we want to know the dimensions of the fibres of $\pi_2$. Note that by assumption in the question, $\pi_2$ is surjective. 
The trick is now to switch to the other projection $\pi_1$. The fibre of $\pi_1$ over a point $[\Pi] \in \mathbf G(2,5)$ is the set of all quadrics containing $\Pi$. Containing a fixed plane is a linear condition on quadrics, given by the vanishing of 6 coefficients, so the fibre of $\pi_1$ over any point has dimension 14. Therefore
$$\operatorname{dim} I = \operatorname{dim} \mathbf G(2,5) + 14 = 23.$$
We also observe something else: since $I$ maps to a smooth variety with all fibres linear spaces of the same dimension, $I$ must be irreducible. 
So we have an irreducible variety $I$ of dimension 23, mapping surjectively to $\mathbf P^{20}$. By generalities on fibre dimension, the general fibre of $\pi_2$ has dimension 3.
That means that the general quadric fourfold $Q$ has a 3-dimensional family of planes. But now all smooth quadrics of a given dimension are projectively equivalent, so the same is true for any smooth quadric fourfold.
Concluding remark: once you know the dimension of a Fano variety of linear subspaces, there is a nice method to explore its geometry: namely, you describe the variety as the zero-locus of a section of a certain vector bundle on the Grassmannian. The Chern classes of this vector bundle then give information about the geometry of your Fano variety. One can argue in this way to prove, for example, that a general quintic hypersurface in $\mathbf P^4$ contains exactly 2875 lines! For details of this method, see the nice free online book 3264 And All That by Eisenbud and Harris.
A: This is a Fano variety. Have a look at this paper for instance :
http://www.math.harvard.edu/theses/senior/waldron/alex_waldron.pdf
