Examples

Example 1.
Let’s compute $\newcommand{\PP}{\mathbb{P}}
\newcommand{\CC}{\mathbb{C}}
\newcommand{\RR}{\mathbb{R}}
\newcommand{\ZZ}{\mathbb{Z}}
\DeclareMathOperator{\Gr}{Gr}
\DeclareMathOperator{\Fl}{Fl}
\DeclareMathOperator{\GL}{GL}\sigma_{(1,1)}\cdot \sigma_{(2)}$ in $H^\ast(\Gr^2(\CC^4))$. The Littlewood-Richardson rule tells us that the only possible $\nu$ must be the $2\times 2$ square $(2,2)$, but there is no way to fill $\nu/\lambda$ with two $1$’s in a semistandard way. Therefore, $$\sigma_{(1,1)}\cdot \sigma_{(2)}=0.$$

Geometrically, this makes sense: $\Omega_{(1,1)}$ is the Schubert variety consisting of all $2$-dimensional subspaces of $\CC^4$ contained in a given $3$-dimensional subspace. $\Omega_{(2)}$ is the Schubert variety of all $2$-dimensional subspaces containing a given line through $0$. For a generic choice of the given $3$-dimensional subspace and line through the origin, there is plane satisfying both conditions.

Example 2.
Let’s try the same calculation, $\sigma_{(1,1)}\cdot \sigma_{(2)}$, in $H^\ast(\Gr^3(\CC^5))$. Now the Important Box is $2\times 3$, and so the partition $\nu=(3,1)$ is a possibility. Indeed, the Littlewood-Richardson rule gives us one possible filling of $\nu/\lambda$ with two $1$’s, and so we have
$$\sigma_{(1,1)}\cdot \sigma_{(2)}=\sigma_{(3,1)}.$$

Geometrically, this also makes sense: $\Omega_{(1,1)}$ is the Schubert variety consisting of all $2$-dimensional subspaces of $\CC^5$ contained in a given $4$-dimensional subspace. $\Omega_{(2)}$ is the Schubert variety of all $2$-dimensional subspaces intersecting a given plane through $0$ in at least a line. $\Omega_{(3,1)}$, then, is the variety of all $2$-dimensional subspaces contained in a given $4$-space and also containing a given line. Clearly the first and second conditions together are equivalent to the third in $\CC^5$.

Example 3.
We can now check in the least elegant possible way that there exists a unique line in projective space passing through two given points. In other words, through two given lines through $0$ in $\Gr^2(\CC^3)$, we wish to show there is exactly one plane in the intersection of the varieties $\Omega_{(1)}$ and $\Omega_{(1)}$ (for two different flags). Our Important Box is $2\times 1$, so we have $$\sigma_{(1)}\cdot \sigma_{(1)}=\sigma_{(1,1)}.$$ Indeed, $\Omega_{(1,1)}$ consists of a single plane in $\CC^3$.

Example 4.
We can now also solve problem 3. We wish to compute the product $\sigma_{(1)}^4$ in $H^\ast(\Gr^2(\CC^4))$. We have $\sigma_{(1)}^2=\sigma_{(1,1)}+\sigma_{(2)}$ by the Littlewood-Richardson rule. Since $\sigma_{(1,1)}\cdot \sigma_{(2)}=0$ as in Example 1, we have $$\sigma_{(1)}^4=(\sigma_{(1,1)}+ \sigma_{(2)})^2=\sigma_{(1,1)}^2+\sigma_{(2)}^2=2\sigma_{(2,2)}.$$
Thus there are exactly $2$ lines intersecting four given lines in $\PP^3$.

Example 5.
Finally, let’s solve problem 4 from the first page. The statement translates to proving a relation of the form $$\sigma_{(1)}^{r(m-r)}=c\cdot \sigma_{(n,n,\cdots, n)}$$ where $c$ is the desired number of $r$-planes and the Schubert class on the right hand side refers to the class of the Important Box.

First note that some relation of this form must hold, since any partition $\nu$ on the right hand side of the product must have size $r(m-r)$ and fit in the Important Box. The Box itself is the only such partition.

To compute $c$, we notice that it is the same as the coefficient of $s_{(n,n,\ldots,n)}$ in the product of Schur functions $s_{(1)}^{rn}$ in the ring $\Lambda(x_1,x_2,\ldots)$. We now introduce some more well-known facts and definitions from symmetric function theory. (See Stanley‘s or Sagan‘s book to learn about symmetric functions in detail.)

Define the monomial symmetric function $m_\lambda$ to be the sum of all monomials in $x_1,x_2,\ldots$ having exponents $\lambda_1,\ldots,\lambda_r$. Then it is not hard to see, from the combinatorial definition of Schur functions, that $$s_\lambda=\sum_{\mu} K_{\lambda\mu} m_\mu$$ where $K_{\lambda\mu}$ is the number of semistandard Young tableaux of shape $\lambda$ and content $\mu$. The numbers $K_{\lambda\mu}$ are called the Kostka numbers, and they can be thought of as a change of basis matrix in the space of symmetric functions.

The homogeneous symmetric function $h_\lambda$ is defined to be $h_{\lambda_1}\cdots h_{\lambda_r}$ where $h_d$ is the sum of all monomials of degree $d$ for any given $d$. The homogeneous symmetric functions also form a $\CC$-basis for $\Lambda(x_1,x_2,\ldots)$, and one can then define an inner product on $\Lambda$ such that $$\langle h_\lambda,m_\mu\rangle=\delta_{\lambda\mu},$$ i.e. the $h$’s and $m$’s are orthonormal. Remarkably, the $s_\lambda$’s are orthogonal with respect to this inner product: $$\langle s_\lambda,s_\mu\rangle=\delta_{\lambda\mu},$$ and so we have
\begin{eqnarray*}
\langle h_\mu,s_\lambda\rangle &=& \langle h_\mu, \sum_{\nu} K_{\lambda\nu} m_\nu\rangle\\
&=& \sum_{\nu} K_{\lambda\nu}\langle h_\mu, m_\nu \rangle \\
&=& \sum_{\nu} K_{\lambda\nu} \delta{\mu\nu} \\
&=& K_{\lambda\mu}
\end{eqnarray*}
Thus we have the dual formula $$h_\mu=\sum_{\lambda}K_{\lambda\mu}s_\mu.$$

Returning to the problem, notice that $s_{(1)}=m_{(1)}=h_{(1)}=x_1+x_2+x_3+\cdots$. Thus $s_{(1)}^{rn}=h_{1}^{rn}=h_{(1,1,1,\ldots,1)}$ where the last vector has length $rn$. It follows from the formula above that the coefficient of $s_{(n,n,\cdots,n)}$ in $h_{(1,1,1,\ldots,1)}$ is the number of fillings of the Important Box with content $(1,1,1,\ldots,1)$, i.e. the number of Standard Young tableaux of Important Box shape.

There is a well-known and hard-to-prove theorem known as the hook length formula which will finish the computation.

Applying the hook length formula to the Important Box yields the total of $$\frac{(r(m-r))!\cdot (r-1)!\cdot (r-2)!\cdot \cdots\cdot 1!}{(m-1)!\cdot(m-2)!\cdot\cdots\cdot 1!}$$ standard fillings, as desired.