For $a:a\_1\geq \cdots\geq a\_c$, let $\sigma\_a$ be the corresponding Schubert cycle over $Gr(c,\infty)$. We say $a$ is of depth $k$ if $a\_1-a\_c=k$ ($c>1$). Let $a$ and $b$ be of depth $k\_1$ and $k\_2$, respectively. Is the following true?


Among all Schubert cycles inside the expansion of $\sigma\_a\cdot \sigma\_b$, there is one whose depth is at most $\max(k\_1,k\_2)$.


One need to use Littlewood-Richardson rule cleverly!



Depth of Schubert cycles
Yes. Let $(d\_1, d\_2, \ldots, d\_c)$ be the partition obtained by sorting $(a\_1+b\_c, a\_2+ b\_{c-1}, \cdots, a\_c+b\_1)$ into decreasing order. The Parthasarathy-Ranga Rao-Varadarajan conjecture, now proved by [Kumar](http://www.ams.org/mathscinet-getitem?mr=943925) (see also [Knutson and Tao](http://arxiv.org/abs/math/9807160)), has as a special case that $\sigma\_d$ has nonzero coefficient in $\sigma\_a \cdot \sigma\_b$.


Now, the depth of $d$ is
$$\max\_{1 \leq i,j \leq c} (a\_i+b\_{c+1-i}) - (a\_j + b\_{c+1-j}).$$
Whenever $i<j$, we have
$$(a\_i+b\_{c+1-i}) - (a\_j + b\_{c+1-j}) = (a\_i - a\_j) - (b\_{c+1-j} - b\_{c+1-i}) \leq a\_i - a\_j \leq k\_a.$$
Similarly, whenever $i > j$, we have 
$$(a\_i+b\_{c+1-i}) - (a\_j + b\_{c+1-j}) \leq k\_b.$$


By the way, the PRV conjecture can be used to give an alternate proof of the result you asked about [here](https://mathoverflow.net/questions/154537/upper-bound-for-the-product-of-schubert-cycles), that the smallest possible value for $d\_1$ is $\max(a\_1+b\_c, a\_2+b\_{c-1}, \ldots, a\_c+b\_1)$.


