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Proving Positivity for Schubert Calculus
Clash Royale CLAN TAG#URR8PPP
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In study of the cohomology ring of the Grassmannians, which is usually known as Schubert calculus, one usually deals with a distinguished basis known as the Schubert basis $sigma_lambda$. One of the most properties of this basis is positivity, the fact that for any two basis elements $sigma_lambda$ and $sigma_nu$, the multiplication constants
$$
sigma_lambda bullet sigma_mu = sum_nu c_lambda,mu^nu sigma_nu
$$
satisfy the positivity condition
$$
c_lambda,mu^nu geq 0, ~~~~~~~~ text for all nu.
$$
Searching the literature, there seem to be a number of different proofs of this property, the relation between which is not always clear. Can people out there offer an opinion on which is the most insightful approach to proving positivity and what are the advantages/disadvantages or intuitons offered by the other approaches.
co.combinatorics rt.representation-theory enumerative-geometry schubert-calculus
asked 2 days ago
Pierre Dubois
755
add a comment |Â
up vote
10
down vote
favorite
In study of the cohomology ring of the Grassmannians, which is usually known as Schubert calculus, one usually deals with a distinguished basis known as the Schubert basis $sigma_lambda$. One of the most properties of this basis is positivity, the fact that for any two basis elements $sigma_lambda$ and $sigma_nu$, the multiplication constants
$$
sigma_lambda bullet sigma_mu = sum_nu c_lambda,mu^nu sigma_nu
$$
satisfy the positivity condition
$$
c_lambda,mu^nu geq 0, ~~~~~~~~ text for all nu.
$$
Searching the literature, there seem to be a number of different proofs of this property, the relation between which is not always clear. Can people out there offer an opinion on which is the most insightful approach to proving positivity and what are the advantages/disadvantages or intuitons offered by the other approaches.
co.combinatorics rt.representation-theory enumerative-geometry schubert-calculus
asked 2 days ago
Pierre Dubois
755
Isn't it just $c_lambda, mu^nu ge 0$, not $c_lambda, mu^nu > 0$?
– LSpice
2 days ago
Yes, this is more precise. It has been corrected. Merci!
– Pierre Dubois
2 days ago
Well positivity of the Littlewood-Richardson coefficients is obvious from the representation theory side of things. But maybe this is not what you are asking.
– Sam Hopkins
2 days ago
Same question for quantum cohomology and Gromov-Witten invariants ;)
– darij grinberg
2 days ago
add a comment |Â
up vote
10
down vote
favorite
up vote
10
down vote
favorite
In study of the cohomology ring of the Grassmannians, which is usually known as Schubert calculus, one usually deals with a distinguished basis known as the Schubert basis $sigma_lambda$. One of the most properties of this basis is positivity, the fact that for any two basis elements $sigma_lambda$ and $sigma_nu$, the multiplication constants
$$
sigma_lambda bullet sigma_mu = sum_nu c_lambda,mu^nu sigma_nu
$$
satisfy the positivity condition
$$
c_lambda,mu^nu geq 0, ~~~~~~~~ text for all nu.
$$
Searching the literature, there seem to be a number of different proofs of this property, the relation between which is not always clear. Can people out there offer an opinion on which is the most insightful approach to proving positivity and what are the advantages/disadvantages or intuitons offered by the other approaches.
co.combinatorics rt.representation-theory enumerative-geometry schubert-calculus
asked 2 days ago
Pierre Dubois
755
In study of the cohomology ring of the Grassmannians, which is usually known as Schubert calculus, one usually deals with a distinguished basis known as the Schubert basis $sigma_lambda$. One of the most properties of this basis is positivity, the fact that for any two basis elements $sigma_lambda$ and $sigma_nu$, the multiplication constants
$$
sigma_lambda bullet sigma_mu = sum_nu c_lambda,mu^nu sigma_nu
$$
satisfy the positivity condition
$$
c_lambda,mu^nu geq 0, ~~~~~~~~ text for all nu.
$$
Searching the literature, there seem to be a number of different proofs of this property, the relation between which is not always clear. Can people out there offer an opinion on which is the most insightful approach to proving positivity and what are the advantages/disadvantages or intuitons offered by the other approaches.
co.combinatorics rt.representation-theory enumerative-geometry schubert-calculus
co.combinatorics rt.representation-theory enumerative-geometry schubert-calculus
asked 2 days ago
Pierre Dubois
755
asked 2 days ago
Pierre Dubois
755
edited 2 days ago
asked 2 days ago
Pierre Dubois
755
asked 2 days ago
Pierre Dubois
755
asked 2 days ago
Pierre Dubois
755
755
Isn't it just $c_lambda, mu^nu ge 0$, not $c_lambda, mu^nu > 0$?
– LSpice
2 days ago
Yes, this is more precise. It has been corrected. Merci!
– Pierre Dubois
2 days ago
Well positivity of the Littlewood-Richardson coefficients is obvious from the representation theory side of things. But maybe this is not what you are asking.
– Sam Hopkins
2 days ago
Same question for quantum cohomology and Gromov-Witten invariants ;)
– darij grinberg
2 days ago
add a comment |Â
Isn't it just $c_lambda, mu^nu ge 0$, not $c_lambda, mu^nu > 0$?
– LSpice
2 days ago
Yes, this is more precise. It has been corrected. Merci!
– Pierre Dubois
2 days ago
Well positivity of the Littlewood-Richardson coefficients is obvious from the representation theory side of things. But maybe this is not what you are asking.
– Sam Hopkins
2 days ago
Same question for quantum cohomology and Gromov-Witten invariants ;)
– darij grinberg
2 days ago
Isn't it just $c_lambda, mu^nu ge 0$, not $c_lambda, mu^nu > 0$?
– LSpice
2 days ago
Isn't it just $c_lambda, mu^nu ge 0$, not $c_lambda, mu^nu > 0$?
– LSpice
2 days ago
Yes, this is more precise. It has been corrected. Merci!
– Pierre Dubois
2 days ago
Yes, this is more precise. It has been corrected. Merci!
– Pierre Dubois
2 days ago
Well positivity of the Littlewood-Richardson coefficients is obvious from the representation theory side of things. But maybe this is not what you are asking.
– Sam Hopkins
2 days ago
Well positivity of the Littlewood-Richardson coefficients is obvious from the representation theory side of things. But maybe this is not what you are asking.
– Sam Hopkins
2 days ago
Same question for quantum cohomology and Gromov-Witten invariants ;)
– darij grinberg
2 days ago
Same question for quantum cohomology and Gromov-Witten invariants ;)
– darij grinberg
2 days ago
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
18
down vote
accepted
I would say there are three basic reasons for / proofs of positivity.
Geometry. [Kleiman 1973] proves that the number one's trying to compute is the number of points in a transverse intersection of cycles. Ergo, a nonnegative number.
Combinatorics. Present the cohomology ring of $Gr(k,n)$ as a quotient of that of $Gr(k,infty)$, taking Schubert classes to Schubert classes or to zero. The latter ring is a polynomial ring containing the Schur polynomials in $x_1,ldots,x_k$, which give the Schubert classes. There are various combinatorial proofs of the Littlewood-Richardson rule for multiplication of Schur polynomials. Once you know this rule is correct, then you know the coefficients are nonnegative.
Representation theory. The Schur polynomials also give the characters of "polynomial" representations of $GL(k)$, multiplication corresponds to tensor product, and decomposing in the Schur polynomial basis corresponds to decomposing the representation. Here, the coefficients are dimensions of intertwining spaces, thus nonnegative.
Perhaps the principal way to judge advantage/disadvantages of these approaches is to ask how or whether they generalize beyond the original question you ask, concerning cohomology of Grassmannians. #1 generalizes in many ways, in particular to other homogeneous spaces and other cohomology theories (equivariant, $K$, quantum, etc.) #3 generalizes to representations of other groups.
My favorite connection between #1 and #2 is Ravi Vakil's "geometric Littlewood-Richardson rule". Probably the best connection of #2 and #3 is via the theory of crystals. In both cases one can retrodict some of the combinatorial theory. The best connection I know of #1 and #3 is Belkale's construction of a tensor invariant, given an intersection point of three Schubert cycles, such that the tensor invariants constructed form a basis.
Which approach is the most insightful... I guess I have to admit a predilection for the geometry. Certainly there are many generalizations of the original problem for which we have geometric proofs of positivity but no combinatorial proofs. Maybe the simplest one involves cohomology of $d$-step flag manifolds, where $dgeq 4$ (the $d=3$ case only solved last year, in a way that doesn't extend to $dgeq 4$).
answered 2 days ago
Allen Knutson
20.9k443126
1
It might also be worth recording the symmetric group representation theoretic meaning of the LR coefficients.
– Sam Hopkins
2 days ago
Okay, sure. That's Schur-Weyl duality applied to #3. I don't know how to apply understanding of $S_n$ (or other finite group) representation theory to retrodict the L-R combinatorics, in anything analogous to the theory of crystals on the $GL(k)$ side (but would be very interested to hear).
– Allen Knutson
yesterday
As far as I remember there is also a very beautiful proof based on something called "Mondrean Diagrams" that give an actual algorithm to expand the product in a way that only involve positive numbers without cancelations at all. It is also very geometric. Unfortunately I can not find the refference I once came across.
– S. carmeli
yesterday
1
Mondrian tableaux: homepages.math.uic.edu/~coskun/onceagain.pdf
– Sam Hopkins
yesterday
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
18
down vote
accepted
I would say there are three basic reasons for / proofs of positivity.
Geometry. [Kleiman 1973] proves that the number one's trying to compute is the number of points in a transverse intersection of cycles. Ergo, a nonnegative number.
Combinatorics. Present the cohomology ring of $Gr(k,n)$ as a quotient of that of $Gr(k,infty)$, taking Schubert classes to Schubert classes or to zero. The latter ring is a polynomial ring containing the Schur polynomials in $x_1,ldots,x_k$, which give the Schubert classes. There are various combinatorial proofs of the Littlewood-Richardson rule for multiplication of Schur polynomials. Once you know this rule is correct, then you know the coefficients are nonnegative.
Representation theory. The Schur polynomials also give the characters of "polynomial" representations of $GL(k)$, multiplication corresponds to tensor product, and decomposing in the Schur polynomial basis corresponds to decomposing the representation. Here, the coefficients are dimensions of intertwining spaces, thus nonnegative.
Perhaps the principal way to judge advantage/disadvantages of these approaches is to ask how or whether they generalize beyond the original question you ask, concerning cohomology of Grassmannians. #1 generalizes in many ways, in particular to other homogeneous spaces and other cohomology theories (equivariant, $K$, quantum, etc.) #3 generalizes to representations of other groups.
My favorite connection between #1 and #2 is Ravi Vakil's "geometric Littlewood-Richardson rule". Probably the best connection of #2 and #3 is via the theory of crystals. In both cases one can retrodict some of the combinatorial theory. The best connection I know of #1 and #3 is Belkale's construction of a tensor invariant, given an intersection point of three Schubert cycles, such that the tensor invariants constructed form a basis.
Which approach is the most insightful... I guess I have to admit a predilection for the geometry. Certainly there are many generalizations of the original problem for which we have geometric proofs of positivity but no combinatorial proofs. Maybe the simplest one involves cohomology of $d$-step flag manifolds, where $dgeq 4$ (the $d=3$ case only solved last year, in a way that doesn't extend to $dgeq 4$).
answered 2 days ago
Allen Knutson
20.9k443126
1
It might also be worth recording the symmetric group representation theoretic meaning of the LR coefficients.
– Sam Hopkins
2 days ago
Okay, sure. That's Schur-Weyl duality applied to #3. I don't know how to apply understanding of $S_n$ (or other finite group) representation theory to retrodict the L-R combinatorics, in anything analogous to the theory of crystals on the $GL(k)$ side (but would be very interested to hear).
– Allen Knutson
yesterday
As far as I remember there is also a very beautiful proof based on something called "Mondrean Diagrams" that give an actual algorithm to expand the product in a way that only involve positive numbers without cancelations at all. It is also very geometric. Unfortunately I can not find the refference I once came across.
– S. carmeli
yesterday
1
Mondrian tableaux: homepages.math.uic.edu/~coskun/onceagain.pdf
– Sam Hopkins
yesterday
add a comment |Â
up vote
18
down vote
accepted
I would say there are three basic reasons for / proofs of positivity.
Geometry. [Kleiman 1973] proves that the number one's trying to compute is the number of points in a transverse intersection of cycles. Ergo, a nonnegative number.
Combinatorics. Present the cohomology ring of $Gr(k,n)$ as a quotient of that of $Gr(k,infty)$, taking Schubert classes to Schubert classes or to zero. The latter ring is a polynomial ring containing the Schur polynomials in $x_1,ldots,x_k$, which give the Schubert classes. There are various combinatorial proofs of the Littlewood-Richardson rule for multiplication of Schur polynomials. Once you know this rule is correct, then you know the coefficients are nonnegative.
Representation theory. The Schur polynomials also give the characters of "polynomial" representations of $GL(k)$, multiplication corresponds to tensor product, and decomposing in the Schur polynomial basis corresponds to decomposing the representation. Here, the coefficients are dimensions of intertwining spaces, thus nonnegative.
Perhaps the principal way to judge advantage/disadvantages of these approaches is to ask how or whether they generalize beyond the original question you ask, concerning cohomology of Grassmannians. #1 generalizes in many ways, in particular to other homogeneous spaces and other cohomology theories (equivariant, $K$, quantum, etc.) #3 generalizes to representations of other groups.
My favorite connection between #1 and #2 is Ravi Vakil's "geometric Littlewood-Richardson rule". Probably the best connection of #2 and #3 is via the theory of crystals. In both cases one can retrodict some of the combinatorial theory. The best connection I know of #1 and #3 is Belkale's construction of a tensor invariant, given an intersection point of three Schubert cycles, such that the tensor invariants constructed form a basis.
Which approach is the most insightful... I guess I have to admit a predilection for the geometry. Certainly there are many generalizations of the original problem for which we have geometric proofs of positivity but no combinatorial proofs. Maybe the simplest one involves cohomology of $d$-step flag manifolds, where $dgeq 4$ (the $d=3$ case only solved last year, in a way that doesn't extend to $dgeq 4$).
answered 2 days ago
Allen Knutson
20.9k443126
1
It might also be worth recording the symmetric group representation theoretic meaning of the LR coefficients.
– Sam Hopkins
2 days ago
Okay, sure. That's Schur-Weyl duality applied to #3. I don't know how to apply understanding of $S_n$ (or other finite group) representation theory to retrodict the L-R combinatorics, in anything analogous to the theory of crystals on the $GL(k)$ side (but would be very interested to hear).
– Allen Knutson
yesterday
As far as I remember there is also a very beautiful proof based on something called "Mondrean Diagrams" that give an actual algorithm to expand the product in a way that only involve positive numbers without cancelations at all. It is also very geometric. Unfortunately I can not find the refference I once came across.
– S. carmeli
yesterday
1
Mondrian tableaux: homepages.math.uic.edu/~coskun/onceagain.pdf
– Sam Hopkins
yesterday
add a comment |Â
up vote
18
down vote
accepted
up vote
18
down vote
accepted
I would say there are three basic reasons for / proofs of positivity.
Geometry. [Kleiman 1973] proves that the number one's trying to compute is the number of points in a transverse intersection of cycles. Ergo, a nonnegative number.
Combinatorics. Present the cohomology ring of $Gr(k,n)$ as a quotient of that of $Gr(k,infty)$, taking Schubert classes to Schubert classes or to zero. The latter ring is a polynomial ring containing the Schur polynomials in $x_1,ldots,x_k$, which give the Schubert classes. There are various combinatorial proofs of the Littlewood-Richardson rule for multiplication of Schur polynomials. Once you know this rule is correct, then you know the coefficients are nonnegative.
Representation theory. The Schur polynomials also give the characters of "polynomial" representations of $GL(k)$, multiplication corresponds to tensor product, and decomposing in the Schur polynomial basis corresponds to decomposing the representation. Here, the coefficients are dimensions of intertwining spaces, thus nonnegative.
Perhaps the principal way to judge advantage/disadvantages of these approaches is to ask how or whether they generalize beyond the original question you ask, concerning cohomology of Grassmannians. #1 generalizes in many ways, in particular to other homogeneous spaces and other cohomology theories (equivariant, $K$, quantum, etc.) #3 generalizes to representations of other groups.
My favorite connection between #1 and #2 is Ravi Vakil's "geometric Littlewood-Richardson rule". Probably the best connection of #2 and #3 is via the theory of crystals. In both cases one can retrodict some of the combinatorial theory. The best connection I know of #1 and #3 is Belkale's construction of a tensor invariant, given an intersection point of three Schubert cycles, such that the tensor invariants constructed form a basis.
Which approach is the most insightful... I guess I have to admit a predilection for the geometry. Certainly there are many generalizations of the original problem for which we have geometric proofs of positivity but no combinatorial proofs. Maybe the simplest one involves cohomology of $d$-step flag manifolds, where $dgeq 4$ (the $d=3$ case only solved last year, in a way that doesn't extend to $dgeq 4$).
answered 2 days ago
Allen Knutson
20.9k443126
I would say there are three basic reasons for / proofs of positivity.
Geometry. [Kleiman 1973] proves that the number one's trying to compute is the number of points in a transverse intersection of cycles. Ergo, a nonnegative number.
Combinatorics. Present the cohomology ring of $Gr(k,n)$ as a quotient of that of $Gr(k,infty)$, taking Schubert classes to Schubert classes or to zero. The latter ring is a polynomial ring containing the Schur polynomials in $x_1,ldots,x_k$, which give the Schubert classes. There are various combinatorial proofs of the Littlewood-Richardson rule for multiplication of Schur polynomials. Once you know this rule is correct, then you know the coefficients are nonnegative.
Representation theory. The Schur polynomials also give the characters of "polynomial" representations of $GL(k)$, multiplication corresponds to tensor product, and decomposing in the Schur polynomial basis corresponds to decomposing the representation. Here, the coefficients are dimensions of intertwining spaces, thus nonnegative.
Perhaps the principal way to judge advantage/disadvantages of these approaches is to ask how or whether they generalize beyond the original question you ask, concerning cohomology of Grassmannians. #1 generalizes in many ways, in particular to other homogeneous spaces and other cohomology theories (equivariant, $K$, quantum, etc.) #3 generalizes to representations of other groups.
My favorite connection between #1 and #2 is Ravi Vakil's "geometric Littlewood-Richardson rule". Probably the best connection of #2 and #3 is via the theory of crystals. In both cases one can retrodict some of the combinatorial theory. The best connection I know of #1 and #3 is Belkale's construction of a tensor invariant, given an intersection point of three Schubert cycles, such that the tensor invariants constructed form a basis.
Which approach is the most insightful... I guess I have to admit a predilection for the geometry. Certainly there are many generalizations of the original problem for which we have geometric proofs of positivity but no combinatorial proofs. Maybe the simplest one involves cohomology of $d$-step flag manifolds, where $dgeq 4$ (the $d=3$ case only solved last year, in a way that doesn't extend to $dgeq 4$).
answered 2 days ago
Allen Knutson
20.9k443126
edited yesterday
answered 2 days ago
Allen Knutson
20.9k443126
answered 2 days ago
Allen Knutson
20.9k443126
answered 2 days ago
Allen Knutson
20.9k443126
20.9k443126
1
It might also be worth recording the symmetric group representation theoretic meaning of the LR coefficients.
– Sam Hopkins
2 days ago
Okay, sure. That's Schur-Weyl duality applied to #3. I don't know how to apply understanding of $S_n$ (or other finite group) representation theory to retrodict the L-R combinatorics, in anything analogous to the theory of crystals on the $GL(k)$ side (but would be very interested to hear).
– Allen Knutson
yesterday
As far as I remember there is also a very beautiful proof based on something called "Mondrean Diagrams" that give an actual algorithm to expand the product in a way that only involve positive numbers without cancelations at all. It is also very geometric. Unfortunately I can not find the refference I once came across.
– S. carmeli
yesterday
1
Mondrian tableaux: homepages.math.uic.edu/~coskun/onceagain.pdf
– Sam Hopkins
yesterday
add a comment |Â
1
It might also be worth recording the symmetric group representation theoretic meaning of the LR coefficients.
– Sam Hopkins
2 days ago
Okay, sure. That's Schur-Weyl duality applied to #3. I don't know how to apply understanding of $S_n$ (or other finite group) representation theory to retrodict the L-R combinatorics, in anything analogous to the theory of crystals on the $GL(k)$ side (but would be very interested to hear).
– Allen Knutson
yesterday
As far as I remember there is also a very beautiful proof based on something called "Mondrean Diagrams" that give an actual algorithm to expand the product in a way that only involve positive numbers without cancelations at all. It is also very geometric. Unfortunately I can not find the refference I once came across.
– S. carmeli
yesterday
1
Mondrian tableaux: homepages.math.uic.edu/~coskun/onceagain.pdf
– Sam Hopkins
yesterday
1
1
It might also be worth recording the symmetric group representation theoretic meaning of the LR coefficients.
– Sam Hopkins
2 days ago
It might also be worth recording the symmetric group representation theoretic meaning of the LR coefficients.
– Sam Hopkins
2 days ago
Okay, sure. That's Schur-Weyl duality applied to #3. I don't know how to apply understanding of $S_n$ (or other finite group) representation theory to retrodict the L-R combinatorics, in anything analogous to the theory of crystals on the $GL(k)$ side (but would be very interested to hear).
– Allen Knutson
yesterday
Okay, sure. That's Schur-Weyl duality applied to #3. I don't know how to apply understanding of $S_n$ (or other finite group) representation theory to retrodict the L-R combinatorics, in anything analogous to the theory of crystals on the $GL(k)$ side (but would be very interested to hear).
– Allen Knutson
yesterday
As far as I remember there is also a very beautiful proof based on something called "Mondrean Diagrams" that give an actual algorithm to expand the product in a way that only involve positive numbers without cancelations at all. It is also very geometric. Unfortunately I can not find the refference I once came across.
– S. carmeli
yesterday
As far as I remember there is also a very beautiful proof based on something called "Mondrean Diagrams" that give an actual algorithm to expand the product in a way that only involve positive numbers without cancelations at all. It is also very geometric. Unfortunately I can not find the refference I once came across.
– S. carmeli
yesterday
1
1
Mondrian tableaux: homepages.math.uic.edu/~coskun/onceagain.pdf
– Sam Hopkins
yesterday
Mondrian tableaux: homepages.math.uic.edu/~coskun/onceagain.pdf
– Sam Hopkins
yesterday
add a comment |Â
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Isn't it just $c_lambda, mu^nu ge 0$, not $c_lambda, mu^nu > 0$?
– LSpice
2 days ago
Yes, this is more precise. It has been corrected. Merci!
– Pierre Dubois
2 days ago
Well positivity of the Littlewood-Richardson coefficients is obvious from the representation theory side of things. But maybe this is not what you are asking.
– Sam Hopkins
2 days ago
Same question for quantum cohomology and Gromov-Witten invariants ;)
– darij grinberg
2 days ago