# The Grassmannian

The first thing we need to do to simplify our life is to get out of projective space. Recall that $\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}\PP^m$ can be defined as the collection of lines through the origin in $\CC^{m+1}$. Furthermore, lines in $\PP^m$ correspond to planes through the origin in $\CC^{m+1}$, and so on.

In problem $3$ in the introduction, we are trying to find lines in $\PP^3$ with certain intersection properties. This translates to a problem about planes through the origin in $\CC^4$, which we refer simply as $2$-dimensional subspaces of $\CC^4$. We wish to know which $2$-dimensional subspaces $V$ intersect each of four given $2$-dimensional subspaces $W_1,W_2, W_3, W_4$ in at least a line. Our strategy will be to consider the algebraic varieties $Z_i$, $i=1,\ldots,4$, of all possible $V$ intersecting $W_i$ in at least a line, and find the intersection $Z_1\cap Z_2\cap Z_3\cap Z_4$. Each $Z_i$ is an example of a *Schubert variety*, a moduli space of subspaces of $\CC^m$ with specified intersection properties.

The simplest example of a Schubert variety, where we have no constraints on the subspaces, is the Grassmannian

$

\Gr^n(\CC^m)$.

*Grassmannian*$\Gr^n(\CC^m)$ is the collection of codimension-$n$ subspaces of $\CC^m$. In what follows we will set $$r=m-n,$$ so that the codimension-$n$ subspaces have dimension $r$.

We will see later that the Grassmannian has the structure of an algebraic variety, and has two natural topologies that come in handy. For this reason we will call its elements the *points* of the $\Gr^n(\CC^m)$, even though they’re “actually” subspaces of $\CC^m$ of dimension $r=m-n$. It’s the same misfortune that causes us to refer to a line through the origin as a “point in projective space.”

Now, every point of the Grassmannian is the span of $r$ independent row vectors of length $m$, which we can arrange in an $r\times m$ matrix. For instance, the following represents a point in $\Gr^3(\CC^7)$.

$$\left[\begin{array}{ccccccc}

0 & -1 & -3 & -1 & 6 & -4 & 5 \\

0 & 1 & 3 & 2 & -7 & 6 & -5 \\

0 & 0 & 0 & 2 & -2 & 4 & -2

\end{array}\right]$$

Notice that we can perform elementary row operations on the matrix without changing the point of the Grassmannian it represents. Therefore:

Each point of the Grassmannian corresponds to a unique full-rank matrix in reduced row echelon form.

Let’s use the convention that the pivots will be in order from left to right and bottom to top.

**Example.** In the matrix above we can switch the second and third rows, and then add the third row to the first to get:

$$\left[\begin{array}{ccccccc}

0 & 0 & 0 & 1 & -1 & -2 & 0 \\

0 & 0 & 0 & 2 & -2 & 4 & -2 \\

0 & 1 & 3 & 2 & -7 & 6 & -5 \\

\end{array}\right]$$

Here, the bottom left $1$ was used as the pivot to clear its column. We can now use the $2$ at the left of the middle row as our new pivot, by dividing that row by $2$ first, and adding or subtracting from the two other rows:

$$\left[\begin{array}{ccccccc}

0 & 0 & 0 & 0 & 0 & 0 & 1 \\

0 & 0 & 0 & 1 & -1 & 2 & -1 \\

0 & 1 & 3 & 0 & -5 & 2 & -3 \\

\end{array}\right]$$

Finally we can use the $1$ in the upper right corner to clear its column:

$$\left[\begin{array}{ccccccc}

0 & 0 & 0 & 0 & 0 & 0 & 1 \\

0 & 0 & 0 & 1 & -1 & 2 & 0 \\

0 & 1 & 3 & 0 & -5 & 2 & 0 \\

\end{array}\right],$$

and we are done.

In the preceding example, we were left with a reduced row echelon matrix in the form

$$\left[\begin{array}{ccccccc}

0 & 0 & 0 & 0 & 0 & 0 & 1 \\

0 & 0 & 0 & 1 & \ast & \ast & 0 \\

0 & 1 & \ast & 0 & \ast & \ast & 0 \\

\end{array}\right],$$

i.e. its leftmost $1$’s are in columns $2$, $4$, and $7$. The subset of the Grassmannian whose points have this particular form constitutes a *Schubert cell*.

## Schubert varieties and cell complex structure

To make the previous discussion rigorous, we assign to the matrices of the form

$$\left[\begin{array}{ccccccc}

0 & 0 & 0 & 0 & 0 & 0 & 1 \\

0 & 0 & 0 & 1 & \ast & \ast & 0 \\

0 & 1 & \ast & 0 & \ast & \ast & 0 \\

\end{array}\right]$$

a *partition* – a nonincreasing sequence of nonnegative integers $\lambda=(\lambda_1,\ldots,\lambda_r)$ – as follows. Cut out the “upside-down staircase” from the left of the matrix, and let $\lambda_i$ be the distance from the end of the staircase to the $1$ in each row. In the matrix above, we get the partition $\lambda=(4,2,1)$. Notice that we always have $\lambda_1\ge \lambda_2\ge \cdots \ge \lambda_r$.

By identifying the partition with its Young diagram, we can alternatively define $\lambda$ as the complement in a $r\times n$ box (recall $n=m-r$) of the diagram $\mu$ defined by the $\ast$’s, where we place the $\ast$’s in the lower right corner. For instance:

Notice that every partition $\lambda$ we obtain in this manner must fit in the $r\times n$ box. For this reason, we will call it the **Important Box**. (Warning: this terminology is not standard.)

*Schubert cell*$\Omega_{\lambda}^\circ\subset \Gr^n(\CC^m)$ is the set of points whose row echelon matrix has corresponding partition $\lambda$.

Notice that since each $\ast$ can be filled with any complex number, we have $\Omega_{\lambda}^\circ\cong \CC^{r\cdot n-|\lambda|}$. Thus we can think of the Schubert cells as forming an open cover of the Grassmannian by affine subsets.

More rigorously, the Grassmannian can be viewed as a projective variety by embedding $\Gr^n(\CC^m)$ in $\PP^{\binom{m}{r}-1}$ via the *Plücker embedding*. To do so, order the $r$-element subsets $S$ of $\{1,2,\ldots,m\}$ arbitrarily and use this ordering to label the homogeneous coordinates $x_S$ of $\PP^{\binom{m}{r}-1}$. Now, given a point in the Grassmannian represented by a matrix $M$, let $x_S$ be the determinant of the $r\times r$ submatrix determined by the columns in the subset $S$. This determines a point in projective space since row operations can only change the coordinates up to a constant factor, and the coordinates cannot all be zero since the matrix has rank $r$.

One can show that the image is an algebraic subvariety of $\PP^{\binom{m}{r}-1}$, cut out by homogeneous quadratic relations known as the *Plücker relations*. (See Miller and Sturmfels, chapter 14.) The Schubert cells form an open affine cover.

We are now in a position to define the Schubert varieties as closed subvarieties of the Grassmannian.

*standard Schubert variety*corresponding to a partition $\lambda$, denoted $\Omega_\lambda$, is the closure $\overline{{\Omega_\lambda}^\circ}$ of the corresponding Schubert cell in the Grassmannian, taken with respect to the Zariski topology. Explicitly, $$\Omega_{\lambda}=\{V\in \mathrm{Gr}^n(\CC^m)\mid \dim V\cap \langle e_1,\ldots, e_{n+i-\lambda_i}\rangle \ge i.\}$$

In general, however, we can use a different basis than the standard basis $e_1,\cdots,e_m$ for $\CC^m$. Given a *complete flag*, i.e. a chain of subspaces $$0=F_0\subset F_1\subset\cdots \subset F_m=\CC^m$$ where each $F_i$ has dimension $i$, we can define

$$\Omega_{\lambda}(F_\bullet)=\{V\in \mathrm{Gr}^n(\CC^m)\mid \dim V\cap F_{n+i-\lambda_i}\ge i.\}$$

**Remark.** The numbers $n+i-\lambda_i$ are the positions of the $1$’s in the matrix starting from the right. Combinatorially, without drawing the matrix, these numbers can be obtained by adjoining an upright staircase to the end of the $r\times n$ Important Box that $\lambda$ is contained in, and computing the distances from the right boundary of $\lambda$ to the right boundary of the enlarged figure.

**Example.** The Schubert variety $\Omega_{\square}(F_\bullet)\subset \Gr^{2}(\CC^4)$ is the collection of $2$-dimensional subspaces $V\subset \CC^4$ for which $\dim V\cap F_2\ge 1$, i.e. $V$ intersects another $2$-dimensional subspace (namely $F_2$)in at least a line.

By choosing four different flags $F^{(1)}_{\bullet},F^{(2)}_{\bullet},F^{(3)}_{\bullet},F^{(4)}_{\bullet}$, problem 3 becomes equivalent to finding the intersection of the Schubert varieties $$\Omega_{\square}(F^{(1)}_\bullet)\cap \Omega_{\square}(F^{(2)}_\bullet)\cap \Omega_{\square}(F^{(3)}_\bullet)\cap \Omega_{\square}(F^{(4)}_\bullet).$$

## The CW complex structure

The Schubert varieties also give a CW complex structure on the Grassmannian for each complete flag as follows. Given a fixed flag, define the $0$-skeleton $X_0$ to be the $0$-dimensional Schubert variety $\Omega_{(n^r)}$. Define $X_2$ to be $X_0$ along with the $2$-cell (since we are working over $\CC$ and not $\RR$) formed by removing a corner square from the rectangular partition $(n^r)$, and the attaching map given by the closure in the Zariski topology on $\Gr^n(\CC^m)$. Continue in this manner to define the entire cell structure, $X_0\subset X_2\subset\cdots \subset X_{2nr}$.

This gives the second topology on the Grassmannian, and the one which is easier to work with in computing its cohomology.

Hey Maria, thanks for posting this! I’ve been looking to learn a little bit about Schubert calculus, and this looks like a fantastic introduction – I’ll try to get around to reading it in detail this weekend.

Anyway – do you suggest any references for a systematic treatment of the subject? Preferably something that isn’t too technical (everything you use in your post would be fine, I think; I think I can fake an understanding of algebraic geometry up to a point).

Hey Carl, glad you found this useful! I have a few references scattered throughout the post, but I’d say the main one I learned from is Fulton’s book, “Young tableaux”. I’d say that one is pretty accessible for those who don’t have a strong background in algebraic geometry/topology, because he neatly pushes all the technical details about cohomology into the appendix. But then, the appendix is also detailed enough if you do want to understand those details.

I’ve also heard good things about the new book by Eisenbud and Harris, “3264 and all that.” It can be found online here: http://isites.harvard.edu/fs/docs/icb.topic720403.files/book.pdf

and it certainly goes into much more detail on the geometric side of things, because it’s building up a more general theory than just intersections of Schubert varieties. I plan to read (parts of) it at some point soon.

One combinatorial way to look into the Schubert decomposition is:

1.We associate a matrix to each k-subset as you said.

2.Then in each matrix consider the set of subsets of columns (numbered from left to right) which are bases.

3.Put a total order on such sets by considering the induce lexicographic order.

Each schubert cell then correspond to all subspaces with an specified subset of the columns being the maximal basis. So in the example of page 2 we want all matrices with the subset (2,4,7) being the biggest basis lexicographically. This is a complicated way of saying the same but now:

– If we consider ALL possible orders of the columns (schubert is very dependent on ordering from left to right) and intersect them all, in the common refinement we would be grouping subspaces by specifying the complete set of bases. This is the Matroid Strata decomposition. Bad thing is, Mnevs Universality theorem says that such cells can be “as bad as possible”, and while Im never sure of the precise meaning, it certainly means that working with them is much worse than working with the schubert cells which are just affine spaces.

-If we consider all cyclic orders and intersect them, the cells in the refinement are called the Positroid decomposition, and there are lots of combinatorics associated to them. In particular they have something called a Le diagram associated

-Lauren Williams and Kelli Talaska consider something similar, the Go diagrams, to give a refinement of the positroid decomposition. The open cells are not bad, but their closures and how the intersect was (or maybe still is) a bit mysterious.

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Hi Maria, thanks for the post and the link of the video lecture about Schubert calculus in your other post, they have been so helpful!

I have one question which might be stupid. When we want to find which Schubert cell a point of the Grassmannian belongs to, we cut off the upside-down staircase from the matrix, all entries in the staircase should be 0, right? Why must the first column of the matrix be a zero column?