Let $\mathcal{V}= 0 \subset V_1 \subset \cdots \subset V_{n-1}\subset V_n=V$, $\mathcal{W}=0 \subset W_1 \subset \cdots \subset W_{n-1} \subset W_n=W$ be two flags. We say that $\mathcal{V}$ and $\mathcal{W}$ are transverse if $V_i \cap W_{n-i}=0$ for all $i$. Now consider the following theorem:
Theorem (Kleiman's Theorem in characteristic 0)
Suppose that an irreducible algebraic group $G$ acts transitively on a variety $X$ over an algebraically closed field of characteristic $0$, and that $A \subset X$ is a subvariety.
- Is $B \subset X$ is another subvariety, then there is an open dense set of $g$ such that $gA$ is generically transverse to $B$.
- If $G$ is affine, then $[gA]=[A]$ in $A(X)$ for any $g \in G$.
Now consider two Schubert cycles $\Sigma_{a}(\mathcal{V})$, $\Sigma_{b}(\mathcal{W})$, with respect to two transverse flags $\mathcal{V},\mathcal{W}$. Can i deduce from Kleiman's theorem that the two cycles intersect generically transversely whenever $\mathcal{V}$ and $\mathcal{W}$ are transverse? If yes, why?
This is what Harris and Eisenbud claim in their book "3264 and all that" (see pag. 108, the last two lines).