4
$\begingroup$

In Manivel's Symmetric Functions, Schubert Polynomials, and Degeneracy Loci, I am confused with some parts of his proof of Pieri's Formula. It is given as Pieri's Formula $3.2.8$ (p. $109$): If $\lambda \subset m \times n$ is a partition and we have $k$ such that $1 \leq k \leq n$, then $$\sigma_{\lambda} \cup \sigma_{k} = \sum_{\nu \subset m \times n, \nu \in \lambda \otimes k} \sigma_{\nu}.$$ According to Manivel, to prove Pieri's Formula, we must show that if $|\lambda|+|\mu|= nm-k$, then $\sigma_{\lambda} \cup \sigma_{\mu} \cup \sigma_{k}=1$ with the following inequality condition: $$n-\lambda_{m} \geq \mu_{1} \geq n-\lambda_{m-1} \geq \mu_{2} \geq \cdots \geq n-\lambda_{1} \geq \mu_{m} \text{ }(1),$$ and that $\sigma_{\lambda} \cup \sigma_{\mu} \cup \sigma_{k}=0$ otherwise. Now Manivel defines the following spaces assuming $\lambda_{i}+ \mu_{n+1-i} \leq n \text{ } \forall i$: $$A_{i}= \langle v_{1},…,v_{n+i-\lambda_{i}} \rangle =V_{n+i-\lambda_{i}},$$ $$B_{i}= \langle v_{\mu_{m+1-i}+i},…,v_{m+n} \rangle = V^{'}_{n+m+1-i-\mu_{m+1-i}},$$ $$C_{i}= \langle v_{\mu_{m+1-i}-i},…,v_{n+i-\lambda_{i}} \rangle = A_{i} \cap B_{i}.$$ But then it is claimed that the inequality condition $(1)$ is satisfied if and only if $C_{1},…, C_{m}$ are direct summands, and also if and only if the sum of their dimensions is $m+k$. I don't see how that is the case as Manivel claims it to be without proof; this is where I'm confused. What is the justification for his claim?

$\endgroup$
1
  • 2
    $\begingroup$ The condition (1) means the sets of indices of vectors defining the $C_i$'s are nonempty and disjoint; hence the $C_i \cap C_j = 0$ if $i \neq j$, so the sum is direct. The number $m+k$ is equal to the total number of such indices. I think the explanation of this is clearer in Fulton's Young Tableaux (which covers almost exactly the same material as Manivel's book). There are at least a few helpful diagrams. $\endgroup$ Mar 23, 2016 at 2:51

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.