# Generically transversally intersecting Schubert cycles

I have a question about the the proof of Pieri's formula from Harris' and Eisenbuds's lecture "3264 and All That"on page 146. Before the proof we use this terminology (see page 139):

let $$G=G(k,V)$$ be the Grassmannian of $$k$$-dimensional linear subspaces of an $$n$$-dimensional vector space $$V$$ . We start with one useful definition. As we said, Kleiman’s theorem assures us (in characteristic $$0$$) that, for a general pair of flags $$\mathcal{V}= 0=V_0 \subset V_1 \subset ... \subset V_n=V$$ and $$\mathcal{W}$$ on $$V$$ , the Schubert cycles $$\Sigma_a(\mathcal{V}),\Sigma_b(\mathcal{W})$$ intersect generically transversely.

Recall, these Schubert cycles are defined as follows:

Let $$n-k \ge a_1 \ge a_2 \ge ... \ge a_k \ge 0$$ a descending sequence considered as vector $$a=(a_1,...,a_k)$$. Then

$$\Sigma_a(\mathcal{V}):= \{\Lambda \in G \vert \dim(\Lambda \cap V_{n-k+i-a_i}) \ge i \text{ for all } i \}$$

In this case, we can actually say explicitly what “general” means:

Definition 4.4. We say that a pair of flags $$\mathcal{V}$$ and $$\mathcal{V}$$ on $$V$$ are transverse if any of the following equivalent conditions hold:

(a) $$V_i \cap W_{n-i} =0$$ for all $$i$$ .

(b) $$\dim(V_i \cap W_{n-i})= \min(0, i +j-n)$$ for all $$i, j$$ .

(c) There exists a basis $$e_1,...,e_n$$ for $$V$$ in terms of which

$$V_i =\langle e_1,...,e_i \rangle \text{ and } W_j =\langle e_{n+1-j},...,e_n \rangle$$

Lemma 4.5 Let $$\Sigma_a (\mathcal{V}), \Sigma_b (\mathcal{W}) \subset G$$ be Schubert cycles defined relative to transverse flags V and W on V . If $$\Lambda \in \Sigma_a (\mathcal{V}) \cap \Sigma_b (\mathcal{W})$$ is a general point of their intersection, then:

(a) $$\Lambda$$ does not lie in any strictly smaller Schubert cycle $$\Sigma_{a'} (\mathcal{V}) \subset \Sigma_a (\mathcal{V})$$.

(b) The flags induced by $$\mathcal{V}$$ and $$\mathcal{W}$$ on $$\Lambda$$ (that is, consisting of intersections with $$\Lambda$$ with flag elements $$V_{\alpha}$$ and $$W_{\beta}$$ ) are transverse. Note that, by the first part, the flags $$\Lambda^{\mathcal{V}}$$ and $$\Lambda^{\mathcal{W}}$$ on $$\Lambda$$ induced by $$\mathcal{V}$$ and $$\mathcal{V}$$ are, explicitly,

$$\Lambda^{\mathcal{V}}_i= \Lambda \cap V_{n-k+i-a_i}$$

and

$$\Lambda^{\mathcal{W}}_i= \Lambda \cap V_{n-k+i-b_i}$$

for $$i=1,...,k$$.

now

Proposition 4.9 (Pieri’s formula). For any Schubert class $$\sigma_a \in A(G)$$(the latter is the Chow Group) and any integer $$b$$,

$$(\sigma_b \cdot \sigma_a)= \sum_{\vert c \vert= \vert a \vert + b \text{ & } a_i \le c_i \le a_{i-1}}$$

Proof of 4.9(Pieri's formula): [...]

following the book we look at Schubert cycles $$\Sigma_a (\mathcal{V}), \Sigma_b (\mathcal{U})$$ and $$\Sigma_{c^*}(\mathcal{W})$$ defined with respect general flags $$\mathcal{V},\mathcal{U},\mathcal{W}$$. Recall that $$c^*=(c^*_1,c^*_2,..., c^*_k$$ is the dual vector to $$c$$ defined by $$c^*_i:= n-k -c_{k+1-i}$$.

By definition,

$$\Sigma_a(\mathcal{V})= \{\Lambda \vert \dim(\Lambda \cap V_{n-k+i-a_i}) \ge i \text{ for all } i \}$$

and

$$\Sigma_{c^*}(\mathcal{W})= \{\Lambda \vert \dim(\Lambda \cap W_{i+c_{k+1-i}}) \ge i \text{ for all } i \}$$

Set

$$A_i = V_{n-k+i-a_i} \cap W_{k+1-i-c_i}$$.

During the proof we define $$A:= \langle A_1,...,A_k \rangle$$ and it is claimed that by Lemma 4.5 the plane $$\Lambda$$ is spanned by it's intersections with $$A_i$$, i.e. that $$\Lambda =\langle \Lambda \cap A_i \vert i \rangle$$.

Q: why Lemma 4.5 implies that the plane $$\Lambda$$ is spanned by it's intersections with $$A_i$$? do we really need the transversality of the flags for $$\Lambda =\langle \Lambda \cap A_i \vert i \rangle$$?