I'm inspired by Yuhao's question. The functor that takes a scheme S to the set of k-dimensional vector subbundles of C^n x S (understanding "subbundle" to mean that the quotient by it is another vector bundle) is represented by the Grassmannian G(k,n). What functor is represented by the Schubert subvarieties of G(k,n)?

The Schubert variety associated to a sequence of numbers d = (d_0,d_1,...,d_n) is the collection of k-planes that meet the standard i-plane in a subspace of dimension $\geq$ d_i. (The d for which the associated Schubert variety is nonempty are naturally parameterized by k-element subsets of {1,2,...,n}, or more usefully by partitions that fit in a k x (n-k) box.)

It's tempting to say that the functor represented by a Schubert variety should take a scheme S to the set of subbundles of C^n x S that, fiber-by-fiber, meet the standard i-plane in a subspace of dimension $\geq$ d_i. But I don't know what this should mean on a non-reduced scheme.

One way to look for a moduli interpretation for the functor represented by a variety X is to find one or several "universal" or "tautological" families of things over X. Then one hopes that families of things of the same type over another scheme S will be pulled back along maps from S to X. The natural-looking tautological objects over a Schubert variety are the families of intersections $V \cap \mathbf{C}^i$, where V runs through k-planes belonging to the Schubert variety. These families are not equidimensional, and in particular not flat; this might be an obstacle.

...

It's really clear what's going on from Steven's answer, but I decided to follow it to the end and write down a functor. Beyond saying that these things are zero loci of interesting sections of vector bundles, this point of view probably doesn't much illuminate Schubert varieties. But here it is.

Steven's answer is cleanest if we regard the Grassmannian and its subvarieties as parameterizing quotients rather than subspaces of C^n. So a point in the Grassmannian G(k,n) is an isomorphism class of surjective maps C^n --> V, or equivalently of tuples (V,v_1,...,v_n) where V is a k-dimensional vector space and v_1,...,v_n span V.

The equivalence class of (V,v_1,...,v_n) belongs to the Schubert variety corresponding to r = (r_1,...,r_n) if for each i the first i terms on the list of vs span a subspace of dimension $\leq$ r_i. As in the answer we can use exterior powers to make sense of these rank conditions over a scheme S. So an S-valued point of the Schubert variety corresponding to r is an equivalence class of tuples (E,s_1,...,s_n) where

- E is a vector bundle
- s_1,...,s_n is a list of sections of E
- these sections generate E in each fiber, and given any i and any (r_i + 1)-tuple of sections chosen from (s_1,...,s_i), we have

$$s_{j_1} \wedge s_{j_2} \wedge \cdots \wedge s_{j_{r_i + 1}} = 0$$

For example, if S = Spec(C[x]/x^2), and E is the rank 2 free sheaf on S spanned by e_1 and e_2, then (x.(ae_1 + be_2), x.(ce_1 + de_2), e_1, e_2) is an S-valued point of the Schubert variety corresponding to r = (1,1,2,2). The complex numbers a,b,c, and d parameterize the 4-dimensional Zariski tangent space to the singular point on this 3-dimensional variety.

There's a loose end. I think it's clear that the functor $$S \mapsto \{(E,s_1,\ldots,s_n)\}/\sim$$ represents a subscheme of the Grassmannian with the same set-theoretic support as a Schubert variety. Why does it represent the actual Schubert variety? It would be enough to know that the representing object is reduced. But I don't even know how to express "X is reduced" in terms of the functor of points Hom(-,X).