# Schubert cycles that intersect generically transversely.

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).

• 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. – Jake Levinson Nov 17 '19 at 17:41
• @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 $\Sigma_a(V) \cap \Sigma_b(gV) \cong g' \Sigma_a(V)\cap \Sigma_b(gV)=\Sigma_a(V') \cap \Sigma_b(W')$. – Frant Nov 17 '19 at 22:32
• It follows that $\Sigma_a(V')$ and $\Sigma_b(W')$ intersects generically transversely. Do you think it works? – Frant Nov 17 '19 at 22:34
• Yes, with one last detail which is to explain the bit about extracting a basis from a pair of transverse flags. – Jake Levinson Nov 17 '19 at 23:56
• Yes, sure, thank you! – Frant Nov 18 '19 at 9:04