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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 $A(X)$ 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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    $\begingroup$ Hint: given that V and W are transverse flags, you can recover a basis for the vector space (up to scalar). Therefore, any two transverse flags are equivalent to any other two under a change of coordinates. $\endgroup$ – Jake Levinson Nov 17 '19 at 17:41
  • $\begingroup$ @JakeLevinson Given a flag $V$, there is an element $g$ of $G$ such that $\Sigma_a(V)$ intersects $g\Sigma_b(V)=\Sigma_b(gV)$ generically transversely and $gV$, $V$ are transverse. Moreover, if $V',W'$ is couple of transverse flags, then i can recover the last by the first, by changing basis. Let $g'$ be the automorphism of $G$ which sends the basis given by $V,gV$ to the basis given by $V',W'$. By considering the automorphism of the grassmannian induced by the product by $g'$ it follows that $\Sigma_a(V) \cap \Sigma_b(gV) \cong g' \Sigma_a(V)\cap \Sigma_b(gV)=\Sigma_a(V') \cap \Sigma_b(W')$. $\endgroup$ – Frant Nov 17 '19 at 22:32
  • $\begingroup$ It follows that $\Sigma_a(V')$ and $\Sigma_b(W')$ intersects generically transversely. Do you think it works? $\endgroup$ – Frant Nov 17 '19 at 22:34
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    $\begingroup$ Yes, with one last detail which is to explain the bit about extracting a basis from a pair of transverse flags. $\endgroup$ – Jake Levinson Nov 17 '19 at 23:56
  • $\begingroup$ Yes, sure, thank you! $\endgroup$ – Frant Nov 18 '19 at 9:04

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