The Positive Grassmannian

14 Pages·2012·0.3 MB·English

Lecture: Alex Postnikov The Positive Grassmannian ECCO 2012, Bogot´a
Notes were taken by Zvi Rosen, and images were coded by Alejandro Morales.
Lecture notes regarding the Positive Grassmannian are available at
math.mit.edu/~apost/courses/18.318/.
1. Tuesday, June 12, 2012 / Matroids
Let v ,...,v Rk,n k. Suppose we have a linear dependence
1 n
∈ ≥
v +25v = 271v .
1 2 8
In this context, the coefficients in the equation are not important, just the variables.
Example 1.1. k = 2. See Figure 1.
v v
4 3
v
2
v v v
6 1 5
Figure 1. Rank 2 Matroid
In this example, v v v ,v v , and v are parallel classes of vectors.
1 5 6 2 3 4
|| || ||
Example 1.2. k = 3. Consider the vectors being projected towards a plane from a point, as in
Figure 2.
z
x
y
Figure 2. Vectors Intersecting the Plane
For a sample configuration, see Figure 3.
A k n matrix has column vectors v ,...,v Rk. Let us denote the maximal minor (i.e.
1 n
× ∈
determinant of submatrix) defined by these columns with ∆ where I = i ,...,i .
I 1 k
{ }
Then, v ,...,v are linearly dependent ∆ = 0.
i1 ik ⇔ I
Example 1.3.
 
1 1 3 1 1
A =  1 2 2 1 1 
1 3 1 0 0
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Lecture: Alex Postnikov The Positive Grassmannian ECCO 2012, Bogot´a
v v
2 4
v
1
v
5
v
3
v
6
Figure 3. Rank 3 Matroid
Observe that v +v = 4v .
2 3 1
(cid:12) (cid:12)
(cid:12) 1 1 3 (cid:12)
(cid:12) (cid:12)
∆123 = (cid:12) 1 2 2 (cid:12) = 0.
⇒ (cid:12) (cid:12)
(cid:12) 1 3 1 (cid:12)
On the other hand,
(cid:12) (cid:12)
(cid:12) 1 1 1 (cid:12)
(cid:12) (cid:12)
∆124 = (cid:12) 1 2 1 (cid:12) = 0
(cid:12) (cid:12) (cid:54)
(cid:12) 1 3 0 (cid:12)
implies that v ,v ,v are independent.
1 2 4
Notation 1.4. Let [n] denote the set 1,2,...,n . Let (cid:0)[n](cid:1) be the set of k-element subsets of [n].
{ } k
Definition 1.5. Let
(cid:26) (cid:18) (cid:19) (cid:27)
[n]
= I ∆ (A) = 0 .
A I
M ∈ k (cid:54)
Example 1.6 (Relations between the ∆ ’s). Consider a generic matrix of size 2 4.
I
×
Let
(cid:18) (cid:19)
a b c d
A = .
e f g h
We actually have a relation:
∆ ∆ = ∆ ∆ +∆ ∆ .
13 24 12 34 14 23
This is a consequence of Ptolemy’s Theorem regarding quadrilaterals inscribed in circles.
In general, relations of this type are called Plu¨cker relations.
Notation 1.7. Let σ S . Then,
n
∈
∆ = sgn(σ)∆ .
a1···an σ(a1)···σ(an)
If a = a for any i = j, then
i j
(cid:54)
∆ = 0.
a1···an
Lemma 1.8 (Sylvester’s Lemma).
(cid:88)
∆i1···ik ·∆j1···jk = ∆i(cid:48)1···i(cid:48)k ·∆j1(cid:48)···jk(cid:48)
where i(cid:48), ,i(cid:48) and j(cid:48), ,j(cid:48) are obtained from the original indices by switching i with some
{ 1 ··· k} { 1 ··· k} 1
j’s.
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Lecture: Alex Postnikov The Positive Grassmannian ECCO 2012, Bogot´a
More explicitly, given I,J (cid:0)[n](cid:1), and i I,
∈ k ∈
(cid:88)
∆ ∆ = ∆ ∆ .
I J (I\i)∪j (J\j)∪i
±
j∈J
Given these Plu¨cker relations, we can learn about independent sets of vectors. For example, if
12 , 23 are dependent, then ∆ = ∆ = 0. Then, since
12 23
{ } { }
∆ ∆ = ∆ ∆ +∆ ∆ ,
13 24 12 34 14 23
either ∆ or ∆ = 0, implying that at least one of these pairs is also dependent.
13 24
Definition 1.9. A matroid of rank k on n elements is a non-empty subset (cid:0)[n](cid:1). Elements
M ⊂ k
I are called bases of if they satisfy the exchange axiom:
∈ M M
Exchange Axiom: I,J ,i I, there exists j J such that
∀ ∈ M ∈ ∈
(I i ) j .
\{ } ∪{ } ∈ M
Theorem 1.10. For k n matrix of rank k, is a matroid.
A
× M
Not every matroid is representable as a matrix.
Example 1.11 (Fano Matroid). See Figure 4. It satisfies the exchange axiom, but it is not
representable.
1
2 4
7
6
3 5
Figure 4. The Fano Matroid
Example 1.12 (Almost Pappus Matroid). Recall the Pappus theorem from geometry regarding
points on a pair of lines. (see Figure 5)
Including the central line posited by the theorem, the matroid is representable. Remove the
relation induced by the central line. The resulting matroid of rank 3 is not representable.
A similar thing happens when we use Desargues’ Theorem. The matroid with the predicted line
is representable, without it is not representable.
Example 1.13. Let denote a generic element. Let
∗
(cid:18) (cid:19)
A = ∗ ∗ ∗ ∗ .
0 0
∗ ∗
Then,
(cid:18) (cid:19)
[4]
= 34 .
A
M 2 \{ }
Example 1.14. Figure 5. We start with a few values of k.
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Lecture: Alex Postnikov The Positive Grassmannian ECCO 2012, Bogot´a
1 2 3
7 8 9
4 5 6
Figure 5. The Pappus Matroid
2. Wednesday, June 13, 2012
Matroid Polytopes
Definition 2.1. A convex set is a set S Rn, such that for any A,B S, the line [A,B] S.
⊂ ∈ ⊂
Definition 2.2. A convex polytope is the convex hull of, i.e. the minimal convex set containing,
some finite set of points.
Definition 2.3. Let I = i ,...,i [n]. Let v be a vector with ones in he i -th position, and
1 n I j
{ } ⊂
zeros everywhere else. Given a matroid , the matroid polytope P is the polytope whose vertices
M
M
are v for all I .
I
∈ M
Example 2.4. Let = (cid:0)[4](cid:1). Then P has six vertices: (1,1,0,0),(1,0,1,0),.... This results in
M 2 M
an octahedron (Figure 6).
1010
1001
0011
1100
0110
0101
Figure 6. Matroid Polytope for
M
The edges of the octahedron are those given by a single switch of a zero and a one. This is as a
result of the exchange axiom.
Remark 2.5. The edge [v ,v ] is an edge of the polytope iff J = (I i) j for some i,j.
I J
\ ∪
Taking only the vectors corresponding to a pyramid in the octahedron above, we obtain another
matroid polytope, as in Figure 7.
However, taking only four vectors in a tetrahedron arrangement, as in Figure 8, does not yield
a matroid polytope, since the edge connecting 1100 to 0011 does not involve a single exchange.
Positroids
Definition 2.6. A positive matroid, or positron, is a matroid represented by a k n matrix A
M ×
such that ∆ (A) 0 for all I .
I
≥ ∈ M
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Lecture: Alex Postnikov The Positive Grassmannian ECCO 2012, Bogot´a
1010
1001
0011
1100
0110
Figure 7. Pyramid as Matroid Polytope.
1010
0011
1100
0110
Figure 8. Tetrahedron, a non-Matroid Polytope.
Example 2.7 (k = 2, n = 4). Ptolemy’s relation:
∆ ∆ = ∆ ∆ +∆ ∆ .
13 24 12 34 14 23
The exchange axiom is equivalent to saying that if one term is = 0, then one other term = 0.
(cid:54) (cid:54)
On the other hand, for positroids, if one of the terms on the right-hand side is nonzero, then the
left-hand side must also be nonzero. The three diagonals of the octahedron are
D = [v ,v ] left-hand side
1 13 24
→
D = [v ,v ] first right-hand term
2 12 34
→
D = [v ,v ] second right-hand term.
3 14 23
→
Therefore,ifamatroidpolytopeP containsonediagonalthenitshouldcontainanotherdiagonal.
A positroid has the extra condition that if P contains one of the horizontal diagonals D or D ,
2 3
then it should contain the vertical diagonal D .
1
Example 2.8. Let = (cid:0)[4](cid:1) 13 . The resulting polytope, displayed in Figure 9, is a matroid
M 2 \{ }
polytope but not a positroid polytope.
1001
0011
1100
0110
0101
Figure 9. Non-Positroid Polytope.
Theorem 2.9. Under cyclical shift of the indices, i.e. via the permutation (12 n), a positroid
···
remains a positroid.
Proof. Let A = [v v ...v ], and suppose ∆ (A) 0. Then A(cid:48) = [v v ...( 1)n−1v ] must also
1 2 n I 2 3 1
≥ −
have ∆ (A(cid:48)) 0 since all of the maximal minors not involving v have the same orientation, and
I 1
≥
any minor involving v will be corrected by the appropriate sign.
1
(cid:3)
Definition 2.10. A decorated permutation is a permutation π = (cid:0) 12···n (cid:1) with fixed points π = i
π1π2···πn i
colored in two colors.
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Lecture: Alex Postnikov The Positive Grassmannian ECCO 2012, Bogot´a
Example 2.11.
(cid:18) (cid:19)
123456
π =
315426
This permutation is depicted graphically in Figure 10. Let k = the number of arcs directed to the
left in the diagram.
1
6 i = 1 1 2 3 4 5 6
5 2
i = 2 2 3 4 5 6 1
4 3
π
Figure 10. Decorated Permutation.
Proposition 2.12. Positroids of rank k on n elements are in bijection with decorated permutations
of size n with k left arcs.
Exercise 2.13. Cut the circle between i 1 and i. Show that the number of left arcs is the same
−
for all i’s.
Definition 2.14. Let π be a permutation. A (weak) exceedance in π is an index i s.t. π i.
i
≥
Example 2.15.
(cid:18) (cid:19)
123456
π = .
524163
This permutation has four exceedances.
Definition 2.16. Let A = the number of permutations of size n with k exceedances π i.
kn i
≥
Theorem 2.17. The number of positroids of rank k on n elements is given by:
(cid:18) (cid:19)
n
A +nA + A +
k,n k,n−1 k,n−2
2 ···
(cid:18) (cid:19)
(cid:88) n
= A .
k,n−l
k
l≥0
These numbers A are the Eulerian numbers, i.e. the number of permutations with k 1
k,n
−
descents. A descent is an index i such that w > w .
i i+1
Example 2.18 (Permutations on 3).
Descents Exceedances
123 0 3
132 1 2
213 1 2
231 1 2
312 1 1
321 2 2
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Lecture: Alex Postnikov The Positive Grassmannian ECCO 2012, Bogot´a
A 1
11
1 1 1 1
1 1
A A
12 21
=
1 2 2 1 1 2 2 1
1 4 1
A A A
13 23 33
1 3 2 2 3 1 1 3 2 2 3 1
1 11 11 1
A A A A
14 24 34 44
Figure 11. Euler Triangle.
See Figure 11 for a Pascal-type triangle of Eulerian numbers.
Label each left branch with natural numbers starting with 1 at the leftmost, then do the same
for each right branch starting with 1 at the rightmost. The child of two values will be the sum of
each parent multiplied by the number of the branch leading from that parent to the child.
3. Thursday, June 14, 2012
Definition 3.1. Let N = number of decorated permutations of size n.
n
Let D = number of derangements, i.e. permutations without fixed points, of size n.
n
Example 3.2. For n = 3, we have (231) and (312) so D = 2.
3
Theorem 3.3.
1 1 1
D = n!(1 + + ).
n
− 1! 2! − 3! ···
1 1 1
N = n!(1+ + + + ).
n
1! 2! 3! ···
Corollary 3.4.
D 1
n
lim = .
n→∞ n! e
N
n
lim = e.
n→∞ n!
Definition 3.5. Let D (q) = the number of permutations in S with q-colored fixed points.
n n
(cid:88)
D (q) = q #fixedpointsinπ.
n
π∈Sn
Here is the proof of the Theorem:
Proof. Supposeπ hasfixedpointscoloredincolors 1,2,3,...,9 . Eraseallfixedpointswithcolors
{ }
2,3,...,9 . This gives us a usual permutation of size n r.
{ } −
n (cid:18) (cid:19)
(cid:88) n
D (q) = (q 1)r(n r)!.
n
r − −
r=0
(cid:88) n! (cid:18) (q 1) (q 1)2 (q 1)n(cid:19)
= (q 1)r(n r)! = n! 1+ − + − + + − .
r!(n r)! − − 1! 2! ··· n!
−
(cid:3)
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Lecture: Alex Postnikov The Positive Grassmannian ECCO 2012, Bogot´a
We also have recurrence relations: D = nD +( 1)n,N = nN +1.
n n−1 n n−1
−
Returning to matroids, consider a k n matrix A with ∆ (A) maximal minors. Row operations
I
×
do not change ; therefore, we can reduce it to a simple form – the row-echelon form.
A
M
Example 3.6.
 
1 0 0 0 0
∗ ∗ ∗ ∗
 0 0 1 0 0 0 
 ∗ ∗ ∗ 
 0 0 0 1 0 0 
 ∗ ∗ ∗ 
 0 0 0 0 0 0 1 0 
∗
0 0 0 0 0 0 0 0 1
In this example, we call columns 1,3,4,7,9 pivot columns.
∆ = 1 = 0.
1,3,4,7,9
⇒ (cid:54)
I = 1,3,4,7,9 is the lexicographically minimal base of the matroid .
A
{ } M
Leaving out the pivot columns, we get a matrix that looks like a Young diagram:
 
∗ ∗ ∗ ∗
 0 
 ∗ ∗ ∗ 
 0 
 ∗ ∗ ∗ 
 0 0 0 
∗
0 0 0 0
The vertical steps in the Young diagram represent the pivot columns of the matrix that were
excluded.
Consider the positroid given by the matrix
M
A = [v v ] ∆ (A) 0.
1 n I
··· ≥
Example 3.7 (k = 2). For k = 2, we have ∆ 0 i < j requires v to be counterclockwise
ij j
≥ ⇒
from v (with the positive x-axis as the boundary).
i
Let the vectors v ,...,v be situated in the upper half plane according to Figure 12.
1 8
v
5
v
v 4
7 v
3
v v v v v
10 9 8 2 1
Figure 12. Vector Positroid.
After row reducing and deleting the pivot columns, we have a matrix that looks like:
(cid:18) (cid:19)
r q p t 0 0 z x
0 0 0 s 0 u y
Better, we write it as:
(cid:18) (cid:19)
0 0
• • • • • •
0 0 0 0
• • •
A blocked zero is one with a dot over it. Everything to the left of a blocked zero is zero. This
translates the geometric rule from above into a combinatorial rule: the pattern of having a 0 with
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Lecture: Alex Postnikov The Positive Grassmannian ECCO 2012, Bogot´a
a dot above and a dot to the left is forbidden. This is indeed the positroid condition for all values
of k.
Example 3.8 (k = 3). Consider the affine representation of a matroid in R3. (See Figure 13 for
an example.)
v
i
v v
j k
Figure 13. Affine Representation of a Matroid.
∆ 0,i < j < k any triangle has to have increasing index in counter-clockwise direction.
ijk
≥ ⇒
This implies that the set of all vertices must be on the boundary of a convex polygon.
See Figure 14 for an example, along with the corresponding matrix shape.
v vv
4 33
v v v
5 6 2
v v
v 8 9 v
7 1
Figure 14. Representation of Positroid.
The geometric condition again translates to the blocked zero condition.
Definition 3.9. A hook diagram is a filling of a Young diagram by dots and zeros without the
forbidden pattern described in the examples.
Another characterization: At every dot in the diagram draw a line downward and a line to the
right; there should be a dot at every point of intersection.
Theorem 3.10. Positroids of rank k on n elements are in bijection with hook diagrams that fit
inside a k n k rectangle.
× −
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Lecture: Alex Postnikov The Positive Grassmannian ECCO 2012, Bogot´a
Example 3.11 (k = 2, n = 4).
Young Diagram #Hook Diagrams
1
∅
2
4
4
8
16 2 = 14
−
Thetotalnumberofhookdiagramsis33,whichwecanalsoobtainfromexaminationofthematroid
polytope.
See Figure 15 for a description of the bijection between hook diagrams and decorated permuta-
tions. Alternatively, a trick with mirrors is demonstrated in Figure 16.
1 9 1 9
2 2
8 8
3 3
7 7
4 4
6 6
5 5 5 5
6 6
4 4
7 7
3 3
8 8
2 2
9 1 9 1
1 2 3 4 5 6 7 8 9
3 1 6 2 4 8 5 7 9
π = 5 3 1 4 7 6 9 2 8
Figure 15. Pipe Dreams.
1
2
5 4 3
6
7
8
9
Figure 16. Mirrors.
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