Home » Mathematics » Algebraic geometry » Addendum to last post: the total Chern classes of tautological bundles on G(k,n)

Addendum to last post: the total Chern classes of tautological bundles on G(k,n)

I realised after finishing the last post that it’s simpler than I had thought to compute the total Chern classes of the tautological bundles appearing in the short exact sequence

0 \to \mathcal{S} \to \mathbb{C}^n \to \mathcal{Q} \to 0

on the Grassmannian G(k,n). I’m keeping it in a separate post, though, both because the last post is already very long and because I haven’t yet defined the Schubert varieties that appear in the answer.

In particular, it’s easy to compute c(\mathcal{Q}), as follows. First, \mathcal{Q} is globally generated, and all its global sections come from the trivial bundle (including global sections). So consider choosing i general global sections, that is, choosing v_1, \ldots, v_i \in \mathbb{C}^n, a “general, constant choice” of vectors from our ambient vector space. To compute the Chern class, we need to know: for which subspaces V \in G(k,n) do our sections become linearly dependent in \mathbb{C}^n / V? (Recall that \mathbb{C}^n/V is the fiber of \mathcal{Q} over V.)

Well, they are linearly dependent in C^n/V if and only if V intersects the subspace \langle v_1, \ldots, v_i \rangle nontrivially. But the set of V intersecting a given i-dimensional subspace nontrivially is precisely the Schubert variety X_{n-k+1-i}, or, more prosaically, the Schubert variety corresponding to a one-row partition of the appropriate length.

So, in fact, the total Chern class of \mathcal{Q} is the sum 1 + [X_1] + [X_2] + \ldots + [X_{n-k}], the sum of all the one-row partitions fitting in a k \times (n-k) box.

Repeating this logic for \mathcal{S}^* (thinking of it as the tautological quotient bundle on the dual Grassmannian) shows that \mathcal{S}^* is the sum 1 + [X_1] + [X_{1,1}] + \cdots + [X_{1,\ldots,1}], the sum of all the single-vertical-column partitions. To turn this into the total Chern class of \mathcal{S}, just change the plus signs to alternating signs.

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