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Let $Gr(c,\infty)$ be the complex grassmannian of $c$-dimensional subspaces of the infinite dimensional complex space. Every finite dimensional grassmannian, $Gr(c,N)$, can be thought as a subspace of $Gr(c,\infty)$.

Schubert cycles $\sigma_a$, $a: a_1\geq \cdots \geq a_c\geq 0$, are a basis for the cohomology of $Gr(c,\infty)$, and by restriction, they form a basis of cohomology of $Gr(c,N)$. The restriction $\sigma_a|_{Gr(c,N)}$ is non-zero if and only if $a_1\leq N-c$.

For every two such cycles, $\sigma_a$ and $\sigma_b$, their intersection (or product, depending on the point of view) is of the form $$\sigma_a \cdot \sigma_b =\sum n(c) \sigma_{c},$$ where $c$ runs over Shubert cycles whose degree is the sum of degrees of the left side, and $n(c)$ is some non-negative integer.

Among all $\sigma_c$ on the right such that $n(c)>0$, I am looking for the one whose $c_1$ is minimum, i.e. I am looking for the minimum $N=N(a,b)$ such that $(\sigma_a\cdot \sigma_b)|_{Gr(c,N)}$ is non-zero.

Question: Is there an explicit equation or a non-trivial upper bound on $N(a,b)$, in terms of $a,b$?

If yes, then

what if I have more than two terms in the product, i.e. is there an equation or upper bound for $N(a,b,c,\cdots,d)$?

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I answer the version of your question that asks what is the smallest $n = c_1$ so that the product $\sigma_a \sigma_b$ is nonzero in $Gr(c,n+c)$.

Represent $a$ and $b$ by their Young diagrams. Rotate $b$ by $180$ degrees and pick $n$ minimal so that $a$ and the rotated $b$ do not overlap when $a$ is placed in the $c \times n$ rectangle in the top-left corner and when rotated $b$ is placed in the $c \times n$ rectangle in the bottom-right corner. (If this is not clear, look at the diagram on p.150 of Fulton's book, Young Tableaux.)

I claim that this $n$ is what you are looking for.

If we pick $m < n$, then the product is $0$ by p. 148, Lemma 3 of Fulton's book, Young Tableaux.

For the other direction, we could use the Littlewood-Richardson rule, but here is a less messy approach: if $a$ and rotated $b$ are disjoint and their union is al of the $c \times n$ rectangle, then we know that the intersection product is a single point, i.e., we get $\sigma_a \sigma_b = \sigma_d$ where $d = (n,n,\dots, n)$ ($c$ times) (equation (11) on p.149 of Fulton's book). If we make $a$ smaller but still contained in $a$, this corresponds to replacing the Schubert variety (of codimension $a$) with a larger Schubert variety. So the intersection is still nonzero.

I don't know the answer for a product of more than $2$ Schubert cycles. Part of the problem for me is that I would try to multiply the first two, then the next one, and so on, but it's unclear if there is a "greedy" approach to this, i.e., what does it mean to pick the "smallest" partition appearing in the product of the first two? (I suspect the answer depends on the shape of the next one).

edit: if you just want to pick a partition $d$ so that $\sigma_d$ appears in $\sigma_a \sigma_b$, then in the embedding into the $c \times n$ rectangle above, label all of the boxes in the $i$th row (numbered before rotating) of $b$ with $i$ and then push all boxes labeled 1 as high up as possible (staying in the same column) in the complement of $a$; push all boxes labeled 2 as high up as possible, etc. The result is a Littlewood-Richardson tableau so you get a nonzero element in the product.

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    $\begingroup$ your answer in the first case is simply $\max(a_1+b_c, \cdots, a_c+b_1)$; I am actually more interested in the product of more than two items. $\endgroup$ – Mohammad Farajzadeh-Tehrani Jan 14 '14 at 21:37
  • $\begingroup$ The induction method of Griffiths-Harris (after a generalization I do) says that if you can find $i_1+\cdots+i_m=(m-1)c+1$ such that $a^1_{i_1}+\cdots+a^m_{i_m}=N-c$, then $\sigma_{a^1}\cdots \sigma_{a^m}= \sigma_{a^1 - a^1_{i_1}} \cdots \sigma_{a^m - a^m_{i_m}}|_{Gr(c-1,N)}$ $\endgroup$ – Mohammad Farajzadeh-Tehrani Jan 14 '14 at 21:50
  • $\begingroup$ This induction argument may help us to get a bound in general. $\endgroup$ – Mohammad Farajzadeh-Tehrani Jan 14 '14 at 22:18
  • $\begingroup$ Some how very relevant to this question is the following question: we say $a$ has depth $k$ if $a_1-a_c=k$. Suppose $a$ and $b$ have depths $k_1$ and $k_2$, what can we say about the minimal depth in $\sigma_a\cdot \sigma_b$? $\endgroup$ – Mohammad Farajzadeh-Tehrani Jan 28 '14 at 15:27
  • $\begingroup$ I guess the answer should be $\max(k_1,k_2)$. $\endgroup$ – Mohammad Farajzadeh-Tehrani Jan 28 '14 at 15:49
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The point of this answer is to point out some observations that do not fully answer the case of more than two Schubert classes. Let $\lambda^1$, $\lambda^2$, ..., $\lambda^r$ be the partitions, with the parts of $\lambda^i$ being $(\lambda^i_1, \lambda^i_2, \ldots, \lambda^i_c)$. We want to know the smallest $\mu_1$ such that $\sigma(\mu)$ occurs in $\sigma(\lambda^1) \sigma(\lambda^2) \cdots \sigma(\lambda^r)$.

We have the following inequalities: $$\mu_1 \geq \sum \lambda^i_{j_i} \ \textrm{for any}\ j_1+j_2 + \cdots + j_r = r(c-1)+1. \quad (\ast)$$ $$\mu_1 \geq \left\lceil \frac{1}{r} \sum |\lambda^i| \right\rceil. \quad (\dagger)$$

When $r=2$, inequality $(\dagger)$ is the average of inequalities $(\ast)$ and is thus implied by them, but that isn't true for $r > 2$. For example, if $c=2$, $r=3$ and $\lambda^1=\lambda^2=\lambda^3=(1)$, then $(\ast)$ only forces $\mu_1 \geq 1$ but $(\dagger)$ forces $\mu_1 \geq \lceil 1.5 \rceil = 2$.

Inequalities $(\ast)$ are all of the Horn inequalities which are of the form $\mu_1 \geq \mbox{something}$. (See, for example, this survey and note particularly Section 13 for the relation to Schubert calculus.) However, the example of $c=2$, $\lambda^1=\lambda^2=\lambda^3=(1)$ shows that sometimes condition $(\dagger)$ imposes an additional condition.

My guess would be that $(\dagger)$ and $(\ast)$ are NOT enough to give the answer in general, but I don't have a counterexample yet.

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