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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.

  1. Is $B \subset X$ is another subvariety, then there is an open dense set of $g$ such that $gA$ is generically transverse to $B$.
  2. If $G$ is affine, then $[gA]=[A]$ in $A(X)$ (the Chow Ring of $X$) 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).

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By generic transversality, there is an open dense set $U$ in $\mathrm{GL}_n$ such that $\Sigma_a(V)$ is transversal to $\Sigma_b(gV)=g\Sigma_b(V)$. There is another open dense set $V$ in $\mathrm{GL}_n$ such that $gV$ is transversal to $V$. It follows that there is a $g$ such that $\Sigma_a(V)$ is transverse to $\Sigma_b(W)$ with $W=gV$ also transverse to $V$.

Any pair of transverse flags $V$ and $W$ determines a basis (upto scalar multiple) as $V_i\cap W_{n-i+1}=k\cdot e_i$. Conversely, this basis determines $V$ (respectively $W$) as the increasing (respectively decreasing) span of the $e_i$'s. Thus, if $(V,W)$ is one pair of transverse flags, it can be taken to another pair $(V',W')$ by a change of basis.

Obviously, such a change of basis does not affect the transversality of the intersection of $\Sigma_a(V)$ and $\Sigma_b(W)$ as it (the change of basis) applies to both factors.

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  • $\begingroup$ So i can find an element $g$ such that $\Sigma_b(g\mathcal{V})$ intersects $\Sigma_a(\mathcal{V})$ generically transversely for each couple $(a,b)$ and $\mathcal{V}$ and $g\mathcal{V}$ are transverse. If $\mathcal{V}$, $\mathcal{W}$ are transverse flags how can i conclude that $\Sigma_a(\mathcal{V})$ and $\Sigma_b(\mathcal{W})$ intersects generically transversely?. $\endgroup$
    – klerk
    Nov 15, 2019 at 23:11
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    $\begingroup$ By para 2 and 3, any two pairs of transverse flags can be taken to each other by a change of basis. $\endgroup$
    – Kapil
    Nov 16, 2019 at 4:16

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