Enumerative Geometry and the Shapiro Shapiro...

Enumerative Geometryand the Shapiro–Shapiro Conjecture

Jake LevinsonUniversity of Washington

Simon Fraser UniversityNovember 20, 2019

Enumerative geometry

Some questions about 3D geometry:

1. How many lines meet four general lines?

2.

2. How many lines lie on a smooth cubic surface?

27.

3. How many rational cubic curves meet 12 lines?

80160.

Enumerative geometry

Some questions about 3D geometry:

1. How many lines meet four general lines?

2.

2. How many lines lie on a smooth cubic surface?

27.

3. How many rational cubic curves meet 12 lines?

80160.

Enumerative geometry

Some questions about 3D geometry:

1. How many lines meet four general lines? 2.

2. How many lines lie on a smooth cubic surface?

27.

3. How many rational cubic curves meet 12 lines?

80160.

Enumerative geometry

Some questions about 3D geometry:

1. How many lines meet four general lines? 2.

2. How many lines lie on a smooth cubic surface?

27.

3. How many rational cubic curves meet 12 lines?

80160.

Enumerative geometry

Some questions about 3D geometry:

1. How many lines meet four general lines? 2.

2. How many lines lie on a smooth cubic surface? 27.

3. How many rational cubic curves meet 12 lines?

80160.

(Credit: Greg Egan)

Enumerative geometry

Some questions about 3D geometry:

1. How many lines meet four general lines? 2.

2. How many lines lie on a smooth cubic surface? 27.

3. How many rational cubic curves meet 12 lines?

80160.

Enumerative geometry

Some questions about 3D geometry:

1. How many lines meet four general lines? 2.

2. How many lines lie on a smooth cubic surface? 27.

3. How many rational cubic curves meet 12 lines? 80160.

Enumerative geometry

Some questions about 3D geometry:

1. How many lines meet four general lines? 2.

2. How many lines lie on a smooth cubic surface? 27.

3. How many rational cubic curves meet 12 lines? 80160.

Prettier than 80160 twisted cubics.

Enumerative geometry

Some questions about 3D geometry:

1. How many lines meet four general lines? 2.

2. How many lines lie on a smooth cubic surface? 27.

3. How many rational cubic curves meet 12 lines? 80160.

Enumerative geometry

Some questions about 3D geometry:

1. How many lines meet four general lines? 2.

2. How many lines lie on a smooth cubic surface? 27.

3. How many rational cubic curves meet 12 lines? 80160.

To answer these questions, we study moduli spaces:

I The space of lines Gr(2, n)

I The space of curves Mg

I The space of maps of curves M0,n(P3, 3)

Enumerative geometry

Some questions about 3D geometry:

1. How many lines meet four general lines? 2.

2. How many lines lie on a smooth cubic surface? 27.

3. How many rational cubic curves meet 12 lines? 80160.

To answer these questions, we study moduli spaces:

I The space of lines Gr(2, n)

I The space of curves Mg

I The space of maps of curves M0,n(P3, 3)

Hilbert’s 15th Problem (1900): “To establish rigorously ... theenumerative calculus developed by [Schubert].”

Mumford (1983): “We take as a model for [enumeration onMg ]the enumerative geometry of the Grassmannians.”

Modern questions also go beyond enumeration

I Euler characteristic, genus, . . . for positive-dimensionalsolution spaces

I Equivariant K-theory of Grassmannians. O. Pechenik and A. Yong.Forum Math. Pi (2017).

I Explicit topology of moduli spaces (e.g. as CW-complexes)I Combinatorial equivalence of real moduli spaces. S. Devadoss.

Notices Amer. Math. Soc. (2004).

I Deforming solutionsI Genera of Brill-Noether curves and staircase paths in Young

tableaux, M. Chan; A. Lopez Martın; N. Pflueger; M. Teixidor iBigas. Trans. Amer. Math. Soc. (2018).

Goal of today

Pass from counting to describing topology.

Modern questions also go beyond enumeration

I Euler characteristic, genus, . . . for positive-dimensionalsolution spaces

I Equivariant K-theory of Grassmannians. O. Pechenik and A. Yong.Forum Math. Pi (2017).

I Explicit topology of moduli spaces (e.g. as CW-complexes)I Combinatorial equivalence of real moduli spaces. S. Devadoss.

Notices Amer. Math. Soc. (2004).

I Deforming solutionsI Genera of Brill-Noether curves and staircase paths in Young

tableaux, M. Chan; A. Lopez Martın; N. Pflueger; M. Teixidor iBigas. Trans. Amer. Math. Soc. (2018).

Goal of today

Pass from counting to describing topology.

Part 1. Counting

Counting lines and planes

How many lines meet four general lines?

Schubert calculus says: 2 solutions – enumerated by the set{1 23 4 , 1 3

2 4

}.

Caveat: in general, over C (in CP3, with multiplicity, ...).

Counting lines and planes

How many lines meet four general lines?

Schubert calculus says: 2 solutions – enumerated by the set{1 23 4 , 1 3

2 4

}.

Caveat: in general, over C (in CP3, with multiplicity, ...).

Schubert problems, k-planes, flags

The Grassmannian is the space of planes:

Gr(k , n) = {vector subspaces S ⊂ Cn : dim(S) = k}.

Simplest (codimension 1) “Schubert problem” for planes:

X (Fn−k) = {S ∈ Gr(k , n) : S ∩ Fn−k 6= 0}.

In P3: Lines meeting a given line.

Theorem. For general choices of complementary planes F (i),

Zk,n := X (F(1)n−k) ∩ · · · ∩ X (F

(k(n−k))n−k ) is finite

and counted by standard Young tableaux:

#Zk,n = #SYT ( ) =

{1 2 43 5 6 ,

1 3 52 4 6 , · · ·

}.

Schubert problems, k-planes, flags

The Grassmannian is the space of planes:

Gr(k , n) = {vector subspaces S ⊂ Cn : dim(S) = k}.

Simplest (codimension 1) “Schubert problem” for planes:

X (Fn−k) = {S ∈ Gr(k , n) : S ∩ Fn−k 6= 0}.

In P3: Lines meeting a given line.

Theorem. For general choices of complementary planes F (i),

Zk,n := X (F(1)n−k) ∩ · · · ∩ X (F

(k(n−k))n−k ) is finite

and counted by standard Young tableaux:

#Zk,n = #SYT ( ) =

{1 2 43 5 6 ,

1 3 52 4 6 , · · ·

}.

Schubert problems, k-planes, flags

The Grassmannian is the space of planes:

Gr(k , n) = {vector subspaces S ⊂ Cn : dim(S) = k}.

Simplest (codimension 1) “Schubert problem” for planes:

X (Fn−k) = {S ∈ Gr(k , n) : S ∩ Fn−k 6= 0}.

In P3: Lines meeting a given line.

Theorem. For general choices of complementary planes F (i),

Zk,n := X (F(1)n−k) ∩ · · · ∩ X (F

(k(n−k))n−k ) is finite

and counted by standard Young tableaux:

#Zk,n = #SYT ( ) =

{1 2 43 5 6 ,

1 3 52 4 6 , · · ·

}.

Schubert problems, k-planes, flags

The Grassmannian is the space of planes:

Gr(k , n) = {vector subspaces S ⊂ Cn : dim(S) = k}.

Simplest (codimension 1) “Schubert problem” for planes:

X (Fn−k) = {S ∈ Gr(k , n) : S ∩ Fn−k 6= 0}.

In P3: Lines meeting a given line.

Theorem. For general choices of complementary planes F (i),

Zk,n := X (F(1)n−k) ∩ · · · ∩ X (F

(k(n−k))n−k ) is finite

and counted by standard Young tableaux:

#Zk,n = #SYT ( ) =

{1 2 43 5 6 ,

1 3 52 4 6 , · · ·

}.

Choosing flags

Theorem. For general choices of F (i),

Zk,n = X (F(1)n−k) ∩ · · · ∩ X (F

(k(n−k))n−k )

is finite and counted by standard Young tableaux:

#Zk,n = #SYT( ) =

{1 2 43 5 6 ,

1 3 52 4 6 , · · ·

}.

General Schubert problems: consider S ∩F , for a complete flag:

F : F1 ⊂ F2 ⊂ · · · ⊂ Fn = Cn.

Problems:

1. In bad cases, Zk,n might have multiplicity (or be infinite).

2. No canonical bijection Zk,n ↔ SYT( ) in general.

Choosing flags

Theorem. For general choices of F (i),

Zk,n = X (F(1)n−k) ∩ · · · ∩ X (F

(k(n−k))n−k )

is finite and counted by standard Young tableaux:

#Zk,n = #SYT( ) =

{1 2 43 5 6 ,

1 3 52 4 6 , · · ·

}.

General Schubert problems: consider S ∩F , for a complete flag:

F : F1 ⊂ F2 ⊂ · · · ⊂ Fn = Cn.

Problems:

1. In bad cases, Zk,n might have multiplicity (or be infinite).

2. No canonical bijection Zk,n ↔ SYT( ) in general.

Choosing flags

Theorem. For general choices of F (i),

Zk,n = X (F(1)n−k) ∩ · · · ∩ X (F

(k(n−k))n−k )

is finite and counted by standard Young tableaux:

#Zk,n = #SYT( ) =

{1 2 43 5 6 ,

1 3 52 4 6 , · · ·

}.

General Schubert problems: consider S ∩F , for a complete flag:

F : F1 ⊂ F2 ⊂ · · · ⊂ Fn = Cn.

Problems:

1. In bad cases, Zk,n might have multiplicity (or be infinite).

2. No canonical bijection Zk,n ↔ SYT( ) in general.

Tangent flags to the rational normal curve

I The rational normal curve in Pn−1:

P1 ↪→ P(Cn) = Pn−1 by

t 7→ [1 : t : t2 : · · · : tn−1]

I (Maximally) tangent flag F (t), t ∈ P1:

I Schubert problems using F (t) relate to moduli of curves:I (limit) linear series, Weierstrass points, Brill–Noether loci, . . .

I Eisenbud–Harris ’83:Solutions to these problems have the expected dimension.

Tangent flags to the rational normal curve

I The rational normal curve in Pn−1:

P1 ↪→ P(Cn) = Pn−1 by

t 7→ [1 : t : t2 : · · · : tn−1]

I (Maximally) tangent flag F (t), t ∈ P1:

I Schubert problems using F (t) relate to moduli of curves:I (limit) linear series, Weierstrass points, Brill–Noether loci, . . .

I Eisenbud–Harris ’83:Solutions to these problems have the expected dimension.

Tangent flags to the rational normal curve

I The rational normal curve in Pn−1:

P1 ↪→ P(Cn) = Pn−1 by

t 7→ [1 : t : t2 : · · · : tn−1]

I (Maximally) tangent flag F (t), t ∈ P1:

I Schubert problems using F (t) relate to moduli of curves:I (limit) linear series, Weierstrass points, Brill–Noether loci, . . .

I Eisenbud–Harris ’83:Solutions to these problems have the expected dimension.

Tangent flags to the rational normal curve

I The rational normal curve in Pn−1:

P1 ↪→ P(Cn) = Pn−1 by

t 7→ [1 : t : t2 : · · · : tn−1]

I (Maximally) tangent flag F (t), t ∈ P1:

I Schubert problems using F (t) relate to moduli of curves:I (limit) linear series, Weierstrass points, Brill–Noether loci, . . .

I Eisenbud–Harris ’83:Solutions to these problems have the expected dimension.

Schubert calculus over R?

Conjecture (Shapiro–Shapiro ’95)

For any choice of distinct real t1, . . . , tN ∈ RP1,

Zk,n = X (F (t1)n−k) ∩ · · · ∩ X (F (tN)n−k)

consists entirely of real, multiplicity-free points.

Proven in 2005 / 2009:I Schubert calculus and representations of the general linear group.

Mukhin, E.; Tarasov, V.; Varchenko, A., J. Amer. Math. Soc. (2009).

I The B. and M. Shapiro conjecture in real algebraic geometry and theBethe ansatz. Mukhin, E; Tarasov, V.; Varchenko, A., Ann. of Math.(2009).

Case k = 2 established earlier:I Rational functions with real critical points and the B. and M. Shapiro

conjecture in real enumerative geometry. Eremenko, A. and Gabrielov, A.Ann. of Math. (2002).

Schubert calculus over R?

Conjecture (Shapiro–Shapiro ’95)

For any choice of distinct real t1, . . . , tN ∈ RP1,

Zk,n = X (F (t1)n−k) ∩ · · · ∩ X (F (tN)n−k)

consists entirely of real, multiplicity-free points.

Proven in 2005 / 2009:I Schubert calculus and representations of the general linear group.

Mukhin, E.; Tarasov, V.; Varchenko, A., J. Amer. Math. Soc. (2009).

I The B. and M. Shapiro conjecture in real algebraic geometry and theBethe ansatz. Mukhin, E; Tarasov, V.; Varchenko, A., Ann. of Math.(2009).

Case k = 2 established earlier:I Rational functions with real critical points and the B. and M. Shapiro

conjecture in real enumerative geometry. Eremenko, A. and Gabrielov, A.Ann. of Math. (2002).

Schubert calculus over R?

Theorem (M–T–V ’05)

For any choice of distinct real t1, . . . , tN ∈ RP1,

Zk,n = X (F (t1)n−k) ∩ · · · ∩ X (F (tN)n−k)

consists entirely of real, multiplicity-free points.

Proven in 2005 / 2009:I Schubert calculus and representations of the general linear group.

Mukhin, E.; Tarasov, V.; Varchenko, A., J. Amer. Math. Soc. (2009).

I The B. and M. Shapiro conjecture in real algebraic geometry and theBethe ansatz. Mukhin, E; Tarasov, V.; Varchenko, A., Ann. of Math.(2009).

Case k = 2 established earlier:I Rational functions with real critical points and the B. and M. Shapiro

conjecture in real enumerative geometry. Eremenko, A. and Gabrielov, A.Ann. of Math. (2002).

Combinatorial consequences

Theorem (M–T–V ’05)

For any choice of distinct real t1, . . . , tN ∈ RP1,

Zk,n = X (F (t1)n−k) ∩ · · · ∩ X (F (tN)n−k)

consists entirely of real, multiplicity-free points.

I The cardinality of Zk,n is always exactly #SYT.

I This suggests there may be a canonical bijection

Zk,n ↔ SYT.

Combinatorial consequences

Theorem (M–T–V ’05)

For any choice of distinct real t1, . . . , tN ∈ RP1,

Zk,n = X (F (t1)n−k) ∩ · · · ∩ X (F (tN)n−k)

consists entirely of real, multiplicity-free points.

I The cardinality of Zk,n is always exactly #SYT.

I This suggests there may be a canonical bijection

Zk,n ↔ SYT.

Part 2. Labeling

Families of geometry problems

We want to study Zk,n for every possible choice of ti ’s.

Configuration space: PN(R) = {sets of distinct t1, . . . , tN ∈ R}.

Zk,n

PN(R)

Zk,n

(t1, . . . , tN)

Fiber is Zk,n = X (F (t1)) ∩ · · · ∩ X (F (tN)).

Observation: PN(R) is contractible. No monodromy!

If we can label one fiber by tableaux, we can label all of them.

Families of geometry problems

We want to study Zk,n for every possible choice of ti ’s.

Configuration space: PN(R) = {sets of distinct t1, . . . , tN ∈ R}.

Zk,n

PN(R)

Zk,n

(t1, . . . , tN)

Fiber is Zk,n = X (F (t1)) ∩ · · · ∩ X (F (tN)).

Observation: PN(R) is contractible. No monodromy!

If we can label one fiber by tableaux, we can label all of them.

Families of geometry problems

We want to study Zk,n for every possible choice of ti ’s.

Configuration space: PN(R) = {sets of distinct t1, . . . , tN ∈ R}.

Zk,n

PN(R)

Zk,n

(t1, . . . , tN)

Fiber is Zk,n = X (F (t1)) ∩ · · · ∩ X (F (tN)).

Observation: PN(R) is contractible. No monodromy!

If we can label one fiber by tableaux, we can label all of them.

Families of geometry problems

We want to study Zk,n for every possible choice of ti ’s.

Configuration space: PN(R) = {sets of distinct t1, . . . , tN ∈ R}.

Zk,n

PN(R)

Zk,n

(t1, . . . , tN)

Fiber is Zk,n = X (F (t1)) ∩ · · · ∩ X (F (tN)).

Observation: PN(R) is contractible. No monodromy!

If we can label one fiber by tableaux, we can label all of them.

How to find combinatorics in geometry

Key idea

Degenerate the problem until it breaks into pieces.

Take (t1, . . . , tN) = (z , z2, . . . , zN) and take limz→0

.

0 1R

z → 0

z1z2z3z4

What will happen to Zk,n at z = 0?

How to find combinatorics in geometry

Key idea

Degenerate the problem until it breaks into pieces.

Take (t1, . . . , tN) = (z , z2, . . . , zN) and take limz→0

.

0 1R

z → 0

z1z2z3z4

What will happen to Zk,n at z = 0?

How to find combinatorics in geometry

Key idea

Degenerate the problem until it breaks into pieces.

Take (t1, . . . , tN) = (z , z2, . . . , zN) and take limz→0

.

0 1R

z → 0

z1z2z3z4

What will happen to Zk,n at z = 0?

Lines through 4 given lines, redux

Note:{

lines in P3}

= Gr(2, 4) ≈{[

0 1 ∗ ∗1 0 ∗ ∗

]}.

Set up Z2,4 using tangent lines from the flags F (t):

Z2,4 = X (F (z)) ∩ · · · ∩ X (F (z4)) ⊂ Gr(2, 4)

={

lines in P3 meeting 4 given (tangent) lines}.

0R

z → 0

Pr (R)

z1z2z3z4

Z2,4 ⊂ Gr(2, 4)

(two solutions)

[0 1 ≈ z ≈ z3

1 0 ≈ z4 ≈ z6

][

0 1 ≈ z ≈ z4

1 0 ≈ z3 ≈ z6

]1 23 4

1 32 4

Lines through 4 given lines, redux

Note:{

lines in P3}

= Gr(2, 4) ≈{[

0 1 ∗ ∗1 0 ∗ ∗

]}.

Set up Z2,4 using tangent lines from the flags F (t):

Z2,4 = X (F (z)) ∩ · · · ∩ X (F (z4)) ⊂ Gr(2, 4)

={

lines in P3 meeting 4 given (tangent) lines}.

0R

z → 0

Pr (R)

z1z2z3z4

Z2,4 ⊂ Gr(2, 4)

(two solutions)

[0 1 ≈ z ≈ z3

1 0 ≈ z4 ≈ z6

][

0 1 ≈ z ≈ z4

1 0 ≈ z3 ≈ z6

]1 23 4

1 32 4

Lines through 4 given lines, redux

Note:{

lines in P3}

= Gr(2, 4) ≈{[

0 1 ∗ ∗1 0 ∗ ∗

]}.

Set up Z2,4 using tangent lines from the flags F (t):

Z2,4 = X (F (z)) ∩ · · · ∩ X (F (z4)) ⊂ Gr(2, 4)

={

lines in P3 meeting 4 given (tangent) lines}.

0R

z → 0

Pr (R)

z1z2z3z4

Z2,4 ⊂ Gr(2, 4)

(two solutions)

[0 1 ≈ z ≈ z3

1 0 ≈ z4 ≈ z6

][

0 1 ≈ z ≈ z4

1 0 ≈ z3 ≈ z6

]1 23 4

1 32 4

Lines through 4 given lines, redux

Note:{

lines in P3}

= Gr(2, 4) ≈{[

0 1 ∗ ∗1 0 ∗ ∗

]}.

Set up Z2,4 using tangent lines from the flags F (t):

Z2,4 = X (F (z)) ∩ · · · ∩ X (F (z4)) ⊂ Gr(2, 4)

={

lines in P3 meeting 4 given (tangent) lines}.

0R

z → 0

Pr (R)

z1z2z3z4

Z2,4 ⊂ Gr(2, 4)

(two solutions)

[0 1 ≈ z ≈ z3

1 0 ≈ z4 ≈ z6

][

0 1 ≈ z ≈ z4

1 0 ≈ z3 ≈ z6

]1 23 4

1 32 4

Tableau labels from Plucker coordinates

Limiting matrix:[0 1 ≈ z1 ≈ z3

1 0 ≈ z4 ≈ z6

]1 23 4

Plucker coordinates (minors) on Gr(2, 4):

∅det12 = 1

det13 = O(z1)

det14 = O(z3)

det23 = O(z4)

det24 = O(z6)

det34 = O(z10)

Tableau labels from Plucker coordinates

Limiting matrix:[0 1 ≈ z1 ≈ z3

1 0 ≈ z4 ≈ z6

]1 23 4

Plucker coordinates (minors) on Gr(2, 4):

∅ 1

1 2

13

1 23

1 23 4

det12 = z0

det13 = O(z1)

det14 = O(z1+2)

det23 = O(z1+3)

det24 = O(z1+2+3)

det34 = O(z1+2+3+4)

Combinatorics and geometry

Theorem. This procedure gives a bijection Zk,n ↔ SYT( ).(Purbhoo ’09, Speyer ’14)

And: moving {ti} on RP1 changes the labels by known algorithms!

RP10 ∞

“tableau promotion”

RP10 ∞

“tableau evacuation”

And more:

I Topology and genus when dim(Z ) = 1 (Levinson, Gillespie–L)

I Orthogonal Grassmannians (Purbhoo, Gillespie–L–Purbhoo)

I Vector bundles on M0,n (Kamnitzer, Rybnikov)

Combinatorics and geometry

Theorem. This procedure gives a bijection Zk,n ↔ SYT( ).(Purbhoo ’09, Speyer ’14)

And: moving {ti} on RP1 changes the labels by known algorithms!

RP10 ∞

“tableau promotion”

RP10 ∞

“tableau evacuation”

And more:

I Topology and genus when dim(Z ) = 1 (Levinson, Gillespie–L)

I Orthogonal Grassmannians (Purbhoo, Gillespie–L–Purbhoo)

I Vector bundles on M0,n (Kamnitzer, Rybnikov)

Combinatorics and geometry

Theorem. This procedure gives a bijection Zk,n ↔ SYT( ).(Purbhoo ’09, Speyer ’14)

And: moving {ti} on RP1 changes the labels by known algorithms!

RP10 ∞

“tableau promotion”

RP10 ∞

“tableau evacuation”

And more:

I Topology and genus when dim(Z ) = 1 (Levinson, Gillespie–L)

I Orthogonal Grassmannians (Purbhoo, Gillespie–L–Purbhoo)

I Vector bundles on M0,n (Kamnitzer, Rybnikov)

Part 3. Topology!

A challenge and a new approach

Theorem (M–T–V ’05, ’09)

For t1, . . . , tN ∈ RP1, Zk,n consists of real, multiplicity-free points.

Challenge for geometers:

I M–T–V proof uses integrable systems, the Bethe ansatz

I Subsequent geometry work used M–T–V as black box.

I Many open generalizations of interest!

It turns out there is a topological / geometric approach.

(+)1 2 5

3 4 6, (−)

1 3 5

2 4 6, · · ·

Oriented Young tableaux.

A challenge and a new approach

Theorem (M–T–V ’05, ’09)

For t1, . . . , tN ∈ RP1, Zk,n consists of real, multiplicity-free points.

Challenge for geometers:

I M–T–V proof uses integrable systems, the Bethe ansatz

I Subsequent geometry work used M–T–V as black box.

I Many open generalizations of interest!

It turns out there is a topological / geometric approach.

(+)1 2 5

3 4 6, (−)

1 3 5

2 4 6, · · ·

Oriented Young tableaux.

A challenge and a new approach

Theorem (M–T–V ’05, ’09)

For t1, . . . , tN ∈ RP1, Zk,n consists of real, multiplicity-free points.

Challenge for geometers:

I M–T–V proof uses integrable systems, the Bethe ansatz

I Subsequent geometry work used M–T–V as black box.

I Many open generalizations of interest!

It turns out there is a topological / geometric approach.

(+)1 2 5

3 4 6, (−)

1 3 5

2 4 6, · · ·

Oriented Young tableaux.

Generalization: complex conjugate points on P1

Before: Defined Zk,n using real points ti ∈ RP1.

Now: Zk,n =

n1⋂i=1

X (ti )︸ ︷︷ ︸real

∩n2⋂j=1

X (tj) ∩ X (tj)︸ ︷︷ ︸complex conjugate pairs

.

Mixed configuration space:For a partition µ = (2n2 , 1n1), let

P(µ) =

{sets of

n1 distinct points on R,n2 complex conjugate pairs on C \ R

}⊆ Pr (C) (r = 2n2 + n1).

(Base case: µ = (1N), all real ti .)

Generalization: complex conjugate points on P1

Before: Defined Zk,n using real points ti ∈ RP1.

Now: Zk,n =

n1⋂i=1

X (ti )︸ ︷︷ ︸real

∩n2⋂j=1

X (tj) ∩ X (tj)︸ ︷︷ ︸complex conjugate pairs

.

Mixed configuration space:For a partition µ = (2n2 , 1n1), let

P(µ) =

{sets of

n1 distinct points on R,n2 complex conjugate pairs on C \ R

}⊆ Pr (C) (r = 2n2 + n1).

(Base case: µ = (1N), all real ti .)

Generalization: complex conjugate points on P1

Before: Defined Zk,n using real points ti ∈ RP1.

Now: Zk,n =

n1⋂i=1

X (ti )︸ ︷︷ ︸real

∩n2⋂j=1

X (tj) ∩ X (tj)︸ ︷︷ ︸complex conjugate pairs

.

Mixed configuration space:For a partition µ = (2n2 , 1n1), let

P(µ) =

{sets of

n1 distinct points on R,n2 complex conjugate pairs on C \ R

}⊆ Pr (C) (r = 2n2 + n1).

(Base case: µ = (1N), all real ti .)

Topological and algebraic degrees

How many real points in Zk,n for (t1, . . . , tN) ∈ P(µ)?

I Upper bound from (algebraic) degree = #SYT( ).

I Lower bound from topological degree... (=?):

Zk,n

P(µ)

+

−

+

+

+

Algebraic degree: 3

Topological degree: 1

I We use a careful twist of standard orientation on Gr(k, n).

Topological and algebraic degrees

How many real points in Zk,n for (t1, . . . , tN) ∈ P(µ)?

I Upper bound from (algebraic) degree = #SYT( ).

I Lower bound from topological degree... (=?):

Zk,n

P(µ)

+

−

+

+

+

Algebraic degree: 3

Topological degree: 1

I We use a careful twist of standard orientation on Gr(k, n).

Topological and algebraic degrees

How many real points in Zk,n for (t1, . . . , tN) ∈ P(µ)?

I Upper bound from (algebraic) degree = #SYT( ).

I Lower bound from topological degree... (=?):

Zk,n

P(µ)

+

−

+

+

+

Algebraic degree: 3

Topological degree: 1

I We use a careful twist of standard orientation on Gr(k, n).

Topological and algebraic degrees

How many real points in Zk,n for (t1, . . . , tN) ∈ P(µ)?

I Upper bound from (algebraic) degree = #SYT( ).

I Lower bound from topological degree... (=?):

Zk,n

P(µ)

+

−

+

+

+

Algebraic degree: 3Topological degree: 1

I We use a careful twist of standard orientation on Gr(k, n).

Topological and algebraic degrees

How many real points in Zk,n for (t1, . . . , tN) ∈ P(µ)?

I Upper bound from (algebraic) degree = #SYT( ).

I Lower bound from topological degree... (=?):

Zk,n

P(µ)

+

−

+

+

+

Algebraic degree: 3Topological degree: 1

I We use a careful twist of standard orientation on Gr(k, n).

The topological degree of Zk ,n

Character table of S4. (χλ(µ))

λ, µ (4) (3, 1) (22) (2, 12) (14)

1 1 1 1 1

1 0 -1 -1 3

0 -1 2 0 2

-1 0 -1 1 3

-1 1 1 -1 1

For Sk(n−k), let be the k × (n − k) rectangle, µ = (2n2 , 1n1).

Theorem (L, Purbhoo ‘19)

There is an orientation, the character orientation, such that thefamily Zk,n has topological degree χ (µ) over P(µ).

The topological degree of Zk ,n

Character table of S4. (χλ(µ))

λ, µ (4) (3, 1) (22) (2, 12) (14)

1 1 1 1 1

1 0 -1 -1 3

0 -1 2 0 2

-1 0 -1 1 3

-1 1 1 -1 1

For Sk(n−k), let be the k × (n − k) rectangle, µ = (2n2 , 1n1).

Theorem (L, Purbhoo ‘19)

There is an orientation, the character orientation, such that thefamily Zk,n has topological degree χ (µ) over P(µ).

The topological degree of Zk ,n

Character table of S4. (χλ(µ))

λ, µ (4) (3, 1) (22) (2, 12) (14)

1 1 1 1 1

1 0 -1 -1 3

0 -1 2 0 2

-1 0 -1 1 3

-1 1 1 -1 1

For Sk(n−k), let be the k × (n − k) rectangle, µ = (2n2 , 1n1).

Theorem (L, Purbhoo ‘19)

There is an orientation, the character orientation, such that thefamily Zk,n has topological degree χ (µ) over P(µ).

Signed Young tableaux

Theorem (L, Purbhoo ‘19)

There is an orientation, the character orientation, such that thefamily Zk,n has topological degree χ (µ) over P(µ).

Murnaghan–Nakayama rule for χλ(µ), µ = (2n2 , 1n1):

χλ(µ) =∑T

(−1)# (T ) :µ-domino tableaux (+)

1 2 4

3 5 6, (−)

1 3 4

2 5 6, · · ·

shape(T ) = λ.

I Special case: µ = (1N), no dominos χ (1N) = #SYT.

I Corollary: Shapiro–Shapiro Conjecture.

Signed Young tableaux

Theorem (L, Purbhoo ‘19)

There is an orientation, the character orientation, such that thefamily Zk,n has topological degree χ (µ) over P(µ).

Murnaghan–Nakayama rule for χλ(µ), µ = (2n2 , 1n1):

χλ(µ) =∑T

(−1)# (T ) :µ-domino tableaux (+)

1 2 4

3 5 6, (−)

1 3 4

2 5 6, · · ·

shape(T ) = λ.

I Special case: µ = (1N), no dominos χ (1N) = #SYT.

I Corollary: Shapiro–Shapiro Conjecture.

Signed Young tableaux

Theorem (L, Purbhoo ‘19)

There is an orientation, the character orientation, such that thefamily Zk,n has topological degree χ (µ) over P(µ).

Murnaghan–Nakayama rule for χλ(µ), µ = (2n2 , 1n1):

χλ(µ) =∑T

(−1)# (T ) :µ-domino tableaux (+)

1 2 4

3 5 6, (−)

1 3 4

2 5 6, · · ·

shape(T ) = λ.

I Special case: µ = (1N), no dominos χ (1N) = #SYT.

I Corollary: Shapiro–Shapiro Conjecture.

Labeling Zk ,n by signed Young tableaux

Proof sketch:

I Label limit fibers by tableaux.

I Track +/− signs along a network of paths.

Case 1: 1

2←→ 1

2/ 1 2 ←→ 1 2

++ + + + +

++

+−

− −

P(µ = (16)) P(µ′ = (2, 14))

P(µ) P(µ′)

Z

Z1 2 43 5 6

1 3 42 5 6

1 2 4

3 5 6

1 3 4

2 5 6

(+1)

(−1)

Labeling Zk ,n by signed Young tableaux

Proof sketch:

I Label limit fibers by tableaux.

I Track +/− signs along a network of paths.

Case 1: 1

2←→ 1

2/ 1 2 ←→ 1 2

++ + + + +

++

+−

− −

P(µ = (16)) P(µ′ = (2, 14))

P(µ) P(µ′)

Z

Z1 2 43 5 6

1 3 42 5 6

1 2 4

3 5 6

1 3 4

2 5 6

(+1)

(−1)

Labeling Zk ,n by signed Young tableaux

Proof sketch:

I Label limit fibers by tableaux.

I Track +/− signs along a network of paths.

Case 2: 32 ←→ 2

3

+ + + + +

++

+−

−

P(µ = (16)) P(µ = (2, 14))

P(µ) P(µ′)

Z

Z

1 3 42 5 6

1 2 43 5 6

(∅) (+1)

Open questions

I (Representation theory).Do all SN characters χλ(µ) give topological degrees of realSchubert problems?

I (Complex geometry).Explicit geometry over P(µ) for µ 6= (1N)?

I (Stable curves).How does the geometry look over the moduli space M0,N?

I M0,N(R) is non-orientable!

Many interesting relationships to find between geometry andcombinatorics.

Thank you!