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The Capelli eigenvalue problem Siddhartha Sahi Rutgers University, New Brunswick NJ May 29, 2018, Kostant Conference Siddhartha Sahi (Kostant Conference, MIT) May 29, 2018 1 / 12
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Page 1: The Capelli eigenvalue problem - MathematicsThe Capelli eigenvalue problem Siddhartha Sahi Rutgers University, New Brunswick NJ May 29, 2018, Kostant Conference Siddhartha Sahi (Kostant

The Capelli eigenvalue problem

Siddhartha SahiRutgers University, New Brunswick NJ

May 29, 2018, Kostant Conference

Siddhartha Sahi (Kostant Conference, MIT) May 29, 2018 1 / 12

Page 2: The Capelli eigenvalue problem - MathematicsThe Capelli eigenvalue problem Siddhartha Sahi Rutgers University, New Brunswick NJ May 29, 2018, Kostant Conference Siddhartha Sahi (Kostant

The Capelli identity

Consider differential operators on n× n matrices; write ∂ij =∂

∂zij

The Capelli identity [Capelli 1887, Math Ann. 29] asserts

det (∑k zki∂kj + (n− i) δij ) = det (zij ) det (∂ij )

Left side ∼ matrix multiplication, right side ∼ matrix addition

Generalized in [Kostant—S. 1991, Adv Math 87]

Key idea: Matn×n → Jordan algebra N; det→ Jordan norm ϕ

Siddhartha Sahi (Kostant Conference, MIT) May 29, 2018 2 / 12

Page 3: The Capelli eigenvalue problem - MathematicsThe Capelli eigenvalue problem Siddhartha Sahi Rutgers University, New Brunswick NJ May 29, 2018, Kostant Conference Siddhartha Sahi (Kostant

The Capelli identity

Consider differential operators on n× n matrices; write ∂ij =∂

∂zij

The Capelli identity [Capelli 1887, Math Ann. 29] asserts

det (∑k zki∂kj + (n− i) δij ) = det (zij ) det (∂ij )

Left side ∼ matrix multiplication, right side ∼ matrix addition

Generalized in [Kostant—S. 1991, Adv Math 87]

Key idea: Matn×n → Jordan algebra N; det→ Jordan norm ϕ

Siddhartha Sahi (Kostant Conference, MIT) May 29, 2018 2 / 12

Page 4: The Capelli eigenvalue problem - MathematicsThe Capelli eigenvalue problem Siddhartha Sahi Rutgers University, New Brunswick NJ May 29, 2018, Kostant Conference Siddhartha Sahi (Kostant

The Capelli identity

Consider differential operators on n× n matrices; write ∂ij =∂

∂zij

The Capelli identity [Capelli 1887, Math Ann. 29] asserts

det (∑k zki∂kj + (n− i) δij ) = det (zij ) det (∂ij )

Left side ∼ matrix multiplication, right side ∼ matrix addition

Generalized in [Kostant—S. 1991, Adv Math 87]

Key idea: Matn×n → Jordan algebra N; det→ Jordan norm ϕ

Siddhartha Sahi (Kostant Conference, MIT) May 29, 2018 2 / 12

Page 5: The Capelli eigenvalue problem - MathematicsThe Capelli eigenvalue problem Siddhartha Sahi Rutgers University, New Brunswick NJ May 29, 2018, Kostant Conference Siddhartha Sahi (Kostant

The Capelli identity

Consider differential operators on n× n matrices; write ∂ij =∂

∂zij

The Capelli identity [Capelli 1887, Math Ann. 29] asserts

det (∑k zki∂kj + (n− i) δij ) = det (zij ) det (∂ij )

Left side ∼ matrix multiplication, right side ∼ matrix addition

Generalized in [Kostant—S. 1991, Adv Math 87]

Key idea: Matn×n → Jordan algebra N; det→ Jordan norm ϕ

Siddhartha Sahi (Kostant Conference, MIT) May 29, 2018 2 / 12

Page 6: The Capelli eigenvalue problem - MathematicsThe Capelli eigenvalue problem Siddhartha Sahi Rutgers University, New Brunswick NJ May 29, 2018, Kostant Conference Siddhartha Sahi (Kostant

The Capelli identity

Consider differential operators on n× n matrices; write ∂ij =∂

∂zij

The Capelli identity [Capelli 1887, Math Ann. 29] asserts

det (∑k zki∂kj + (n− i) δij ) = det (zij ) det (∂ij )

Left side ∼ matrix multiplication, right side ∼ matrix addition

Generalized in [Kostant—S. 1991, Adv Math 87]

Key idea: Matn×n → Jordan algebra N; det→ Jordan norm ϕ

Siddhartha Sahi (Kostant Conference, MIT) May 29, 2018 2 / 12

Page 7: The Capelli eigenvalue problem - MathematicsThe Capelli eigenvalue problem Siddhartha Sahi Rutgers University, New Brunswick NJ May 29, 2018, Kostant Conference Siddhartha Sahi (Kostant

Jordan algebras

(N, e, ϕ) = Jordan algebra, G = Str(N), K = StabG (e) = Aut(N)

Z = G/K ↪→ N symmetric cone; rank n, root multiplicity d

D = R,C,H, N = Hermn (D), Z = GLn (D) /Un (D), d = 1, 2, 4

Polynomials P (N) = ⊕λ∈ΛnVλ, multiplicity free G -moduleIndexed by partitions Λn = {λ ∈ Zn : λ1 ≥ · · · ≥ λn ≥ 0}Dm = ϕm (x) ϕm (∂) is a G -invariant differential operator

Theorem ([Kostant-S. 1991])

Dm acts on Vλ by the scalar cm (λ) = ∏ni=1 ∏m−1

j=0 [λi + (n− i) d − j ]

Proof involves Laplace transform on the tube domain N + iZ .

Siddhartha Sahi (Kostant Conference, MIT) May 29, 2018 3 / 12

Page 8: The Capelli eigenvalue problem - MathematicsThe Capelli eigenvalue problem Siddhartha Sahi Rutgers University, New Brunswick NJ May 29, 2018, Kostant Conference Siddhartha Sahi (Kostant

Jordan algebras

(N, e, ϕ) = Jordan algebra, G = Str(N), K = StabG (e) = Aut(N)

Z = G/K ↪→ N symmetric cone; rank n, root multiplicity d

D = R,C,H, N = Hermn (D), Z = GLn (D) /Un (D), d = 1, 2, 4

Polynomials P (N) = ⊕λ∈ΛnVλ, multiplicity free G -moduleIndexed by partitions Λn = {λ ∈ Zn : λ1 ≥ · · · ≥ λn ≥ 0}Dm = ϕm (x) ϕm (∂) is a G -invariant differential operator

Theorem ([Kostant-S. 1991])

Dm acts on Vλ by the scalar cm (λ) = ∏ni=1 ∏m−1

j=0 [λi + (n− i) d − j ]

Proof involves Laplace transform on the tube domain N + iZ .

Siddhartha Sahi (Kostant Conference, MIT) May 29, 2018 3 / 12

Page 9: The Capelli eigenvalue problem - MathematicsThe Capelli eigenvalue problem Siddhartha Sahi Rutgers University, New Brunswick NJ May 29, 2018, Kostant Conference Siddhartha Sahi (Kostant

Jordan algebras

(N, e, ϕ) = Jordan algebra, G = Str(N), K = StabG (e) = Aut(N)

Z = G/K ↪→ N symmetric cone; rank n, root multiplicity d

D = R,C,H, N = Hermn (D), Z = GLn (D) /Un (D), d = 1, 2, 4

Polynomials P (N) = ⊕λ∈ΛnVλ, multiplicity free G -moduleIndexed by partitions Λn = {λ ∈ Zn : λ1 ≥ · · · ≥ λn ≥ 0}Dm = ϕm (x) ϕm (∂) is a G -invariant differential operator

Theorem ([Kostant-S. 1991])

Dm acts on Vλ by the scalar cm (λ) = ∏ni=1 ∏m−1

j=0 [λi + (n− i) d − j ]

Proof involves Laplace transform on the tube domain N + iZ .

Siddhartha Sahi (Kostant Conference, MIT) May 29, 2018 3 / 12

Page 10: The Capelli eigenvalue problem - MathematicsThe Capelli eigenvalue problem Siddhartha Sahi Rutgers University, New Brunswick NJ May 29, 2018, Kostant Conference Siddhartha Sahi (Kostant

Jordan algebras

(N, e, ϕ) = Jordan algebra, G = Str(N), K = StabG (e) = Aut(N)

Z = G/K ↪→ N symmetric cone; rank n, root multiplicity d

D = R,C,H, N = Hermn (D), Z = GLn (D) /Un (D), d = 1, 2, 4

Polynomials P (N) = ⊕λ∈ΛnVλ, multiplicity free G -moduleIndexed by partitions Λn = {λ ∈ Zn : λ1 ≥ · · · ≥ λn ≥ 0}

Dm = ϕm (x) ϕm (∂) is a G -invariant differential operator

Theorem ([Kostant-S. 1991])

Dm acts on Vλ by the scalar cm (λ) = ∏ni=1 ∏m−1

j=0 [λi + (n− i) d − j ]

Proof involves Laplace transform on the tube domain N + iZ .

Siddhartha Sahi (Kostant Conference, MIT) May 29, 2018 3 / 12

Page 11: The Capelli eigenvalue problem - MathematicsThe Capelli eigenvalue problem Siddhartha Sahi Rutgers University, New Brunswick NJ May 29, 2018, Kostant Conference Siddhartha Sahi (Kostant

Jordan algebras

(N, e, ϕ) = Jordan algebra, G = Str(N), K = StabG (e) = Aut(N)

Z = G/K ↪→ N symmetric cone; rank n, root multiplicity d

D = R,C,H, N = Hermn (D), Z = GLn (D) /Un (D), d = 1, 2, 4

Polynomials P (N) = ⊕λ∈ΛnVλ, multiplicity free G -moduleIndexed by partitions Λn = {λ ∈ Zn : λ1 ≥ · · · ≥ λn ≥ 0}Dm = ϕm (x) ϕm (∂) is a G -invariant differential operator

Theorem ([Kostant-S. 1991])

Dm acts on Vλ by the scalar cm (λ) = ∏ni=1 ∏m−1

j=0 [λi + (n− i) d − j ]

Proof involves Laplace transform on the tube domain N + iZ .

Siddhartha Sahi (Kostant Conference, MIT) May 29, 2018 3 / 12

Page 12: The Capelli eigenvalue problem - MathematicsThe Capelli eigenvalue problem Siddhartha Sahi Rutgers University, New Brunswick NJ May 29, 2018, Kostant Conference Siddhartha Sahi (Kostant

Jordan algebras

(N, e, ϕ) = Jordan algebra, G = Str(N), K = StabG (e) = Aut(N)

Z = G/K ↪→ N symmetric cone; rank n, root multiplicity d

D = R,C,H, N = Hermn (D), Z = GLn (D) /Un (D), d = 1, 2, 4

Polynomials P (N) = ⊕λ∈ΛnVλ, multiplicity free G -moduleIndexed by partitions Λn = {λ ∈ Zn : λ1 ≥ · · · ≥ λn ≥ 0}Dm = ϕm (x) ϕm (∂) is a G -invariant differential operator

Theorem ([Kostant-S. 1991])

Dm acts on Vλ by the scalar cm (λ) = ∏ni=1 ∏m−1

j=0 [λi + (n− i) d − j ]

Proof involves Laplace transform on the tube domain N + iZ .

Siddhartha Sahi (Kostant Conference, MIT) May 29, 2018 3 / 12

Page 13: The Capelli eigenvalue problem - MathematicsThe Capelli eigenvalue problem Siddhartha Sahi Rutgers University, New Brunswick NJ May 29, 2018, Kostant Conference Siddhartha Sahi (Kostant

Jordan algebras

(N, e, ϕ) = Jordan algebra, G = Str(N), K = StabG (e) = Aut(N)

Z = G/K ↪→ N symmetric cone; rank n, root multiplicity d

D = R,C,H, N = Hermn (D), Z = GLn (D) /Un (D), d = 1, 2, 4

Polynomials P (N) = ⊕λ∈ΛnVλ, multiplicity free G -moduleIndexed by partitions Λn = {λ ∈ Zn : λ1 ≥ · · · ≥ λn ≥ 0}Dm = ϕm (x) ϕm (∂) is a G -invariant differential operator

Theorem ([Kostant-S. 1991])

Dm acts on Vλ by the scalar cm (λ) = ∏ni=1 ∏m−1

j=0 [λi + (n− i) d − j ]

Proof involves Laplace transform on the tube domain N + iZ .

Siddhartha Sahi (Kostant Conference, MIT) May 29, 2018 3 / 12

Page 14: The Capelli eigenvalue problem - MathematicsThe Capelli eigenvalue problem Siddhartha Sahi Rutgers University, New Brunswick NJ May 29, 2018, Kostant Conference Siddhartha Sahi (Kostant

Degenerate principal series

Tits-Kantor-Koecher: G [ = Conf(N) ⊃ P = G nN max. parabolic

The operators ϕm (∂) intertwine degenerate principal series IP (s)

This leads to new identities [Kostant-S.1993, Inventiones 112], e.g.

det (1+ z tz) det (∂) = a certain Pfaffi an (of differential operators)

It also leads to precise information about the principal series IP (s)[S. 1992-94] Compositio 81; Contemp Math 145; Crelle 462

Hence to explicit Hilbert spaces for unitary subquotients of IP (s)[S. 92, Invent. 101] , [Dvorsky-S. 98-99], Selecta 4, Invent. 138

Siddhartha Sahi (Kostant Conference, MIT) May 29, 2018 4 / 12

Page 15: The Capelli eigenvalue problem - MathematicsThe Capelli eigenvalue problem Siddhartha Sahi Rutgers University, New Brunswick NJ May 29, 2018, Kostant Conference Siddhartha Sahi (Kostant

Degenerate principal series

Tits-Kantor-Koecher: G [ = Conf(N) ⊃ P = G nN max. parabolicThe operators ϕm (∂) intertwine degenerate principal series IP (s)

This leads to new identities [Kostant-S.1993, Inventiones 112], e.g.

det (1+ z tz) det (∂) = a certain Pfaffi an (of differential operators)

It also leads to precise information about the principal series IP (s)[S. 1992-94] Compositio 81; Contemp Math 145; Crelle 462

Hence to explicit Hilbert spaces for unitary subquotients of IP (s)[S. 92, Invent. 101] , [Dvorsky-S. 98-99], Selecta 4, Invent. 138

Siddhartha Sahi (Kostant Conference, MIT) May 29, 2018 4 / 12

Page 16: The Capelli eigenvalue problem - MathematicsThe Capelli eigenvalue problem Siddhartha Sahi Rutgers University, New Brunswick NJ May 29, 2018, Kostant Conference Siddhartha Sahi (Kostant

Degenerate principal series

Tits-Kantor-Koecher: G [ = Conf(N) ⊃ P = G nN max. parabolicThe operators ϕm (∂) intertwine degenerate principal series IP (s)

This leads to new identities [Kostant-S.1993, Inventiones 112], e.g.

det (1+ z tz) det (∂) = a certain Pfaffi an (of differential operators)

It also leads to precise information about the principal series IP (s)[S. 1992-94] Compositio 81; Contemp Math 145; Crelle 462

Hence to explicit Hilbert spaces for unitary subquotients of IP (s)[S. 92, Invent. 101] , [Dvorsky-S. 98-99], Selecta 4, Invent. 138

Siddhartha Sahi (Kostant Conference, MIT) May 29, 2018 4 / 12

Page 17: The Capelli eigenvalue problem - MathematicsThe Capelli eigenvalue problem Siddhartha Sahi Rutgers University, New Brunswick NJ May 29, 2018, Kostant Conference Siddhartha Sahi (Kostant

Degenerate principal series

Tits-Kantor-Koecher: G [ = Conf(N) ⊃ P = G nN max. parabolicThe operators ϕm (∂) intertwine degenerate principal series IP (s)

This leads to new identities [Kostant-S.1993, Inventiones 112], e.g.

det (1+ z tz) det (∂) = a certain Pfaffi an (of differential operators)

It also leads to precise information about the principal series IP (s)[S. 1992-94] Compositio 81; Contemp Math 145; Crelle 462

Hence to explicit Hilbert spaces for unitary subquotients of IP (s)[S. 92, Invent. 101] , [Dvorsky-S. 98-99], Selecta 4, Invent. 138

Siddhartha Sahi (Kostant Conference, MIT) May 29, 2018 4 / 12

Page 18: The Capelli eigenvalue problem - MathematicsThe Capelli eigenvalue problem Siddhartha Sahi Rutgers University, New Brunswick NJ May 29, 2018, Kostant Conference Siddhartha Sahi (Kostant

Degenerate principal series

Tits-Kantor-Koecher: G [ = Conf(N) ⊃ P = G nN max. parabolicThe operators ϕm (∂) intertwine degenerate principal series IP (s)

This leads to new identities [Kostant-S.1993, Inventiones 112], e.g.

det (1+ z tz) det (∂) = a certain Pfaffi an (of differential operators)

It also leads to precise information about the principal series IP (s)[S. 1992-94] Compositio 81; Contemp Math 145; Crelle 462

Hence to explicit Hilbert spaces for unitary subquotients of IP (s)[S. 92, Invent. 101] , [Dvorsky-S. 98-99], Selecta 4, Invent. 138

Siddhartha Sahi (Kostant Conference, MIT) May 29, 2018 4 / 12

Page 19: The Capelli eigenvalue problem - MathematicsThe Capelli eigenvalue problem Siddhartha Sahi Rutgers University, New Brunswick NJ May 29, 2018, Kostant Conference Siddhartha Sahi (Kostant

Degenerate principal series

Tits-Kantor-Koecher: G [ = Conf(N) ⊃ P = G nN max. parabolicThe operators ϕm (∂) intertwine degenerate principal series IP (s)

This leads to new identities [Kostant-S.1993, Inventiones 112], e.g.

det (1+ z tz) det (∂) = a certain Pfaffi an (of differential operators)

It also leads to precise information about the principal series IP (s)[S. 1992-94] Compositio 81; Contemp Math 145; Crelle 462

Hence to explicit Hilbert spaces for unitary subquotients of IP (s)[S. 92, Invent. 101] , [Dvorsky-S. 98-99], Selecta 4, Invent. 138

Siddhartha Sahi (Kostant Conference, MIT) May 29, 2018 4 / 12

Page 20: The Capelli eigenvalue problem - MathematicsThe Capelli eigenvalue problem Siddhartha Sahi Rutgers University, New Brunswick NJ May 29, 2018, Kostant Conference Siddhartha Sahi (Kostant

Interpolation polynomials and Capelli operators

Interpolation polynomials. For µ ∈ Λn write |µ| = µ1 + · · ·+ µn.

Suppose char(F) = 0 and ρ ∈ Fn satisifies ρi+1 − ρi /∈N for all i

Theorem ([S. 1994, Prog Math 123] Kostant 65th Birthday)

For µ ∈ Λn there is a unique symm. polynomial R(ρ)µ (x) ∈ F [x1, . . . , xn ]

Sn

of degree |µ| such that R (ρ)µ (λ+ ρ) = δλµ for |λ| ≤ |µ| .

Generalized Capelli operators. Basis{Dµ

}for PD (N)G

PD (N) ≈ P (N)⊗D (N) ≈ ⊕λ,µVλ ⊗ V ∗µ ; PD (N)G ≈ ⊕µCDµ

Dµ corresponds to the identity 1µ in End(Vµ

)≈ Vµ ⊗ V ∗µ

Theorem ([ibid.])

Dµ acts on Vλ by the scalar R(ρ)µ (λ+ ρ), where ρi =

d2 (n− 2i + 1)

Siddhartha Sahi (Kostant Conference, MIT) May 29, 2018 5 / 12

Page 21: The Capelli eigenvalue problem - MathematicsThe Capelli eigenvalue problem Siddhartha Sahi Rutgers University, New Brunswick NJ May 29, 2018, Kostant Conference Siddhartha Sahi (Kostant

Interpolation polynomials and Capelli operators

Interpolation polynomials. For µ ∈ Λn write |µ| = µ1 + · · ·+ µn.

Suppose char(F) = 0 and ρ ∈ Fn satisifies ρi+1 − ρi /∈N for all i

Theorem ([S. 1994, Prog Math 123] Kostant 65th Birthday)

For µ ∈ Λn there is a unique symm. polynomial R(ρ)µ (x) ∈ F [x1, . . . , xn ]

Sn

of degree |µ| such that R (ρ)µ (λ+ ρ) = δλµ for |λ| ≤ |µ| .

Generalized Capelli operators. Basis{Dµ

}for PD (N)G

PD (N) ≈ P (N)⊗D (N) ≈ ⊕λ,µVλ ⊗ V ∗µ ; PD (N)G ≈ ⊕µCDµ

Dµ corresponds to the identity 1µ in End(Vµ

)≈ Vµ ⊗ V ∗µ

Theorem ([ibid.])

Dµ acts on Vλ by the scalar R(ρ)µ (λ+ ρ), where ρi =

d2 (n− 2i + 1)

Siddhartha Sahi (Kostant Conference, MIT) May 29, 2018 5 / 12

Page 22: The Capelli eigenvalue problem - MathematicsThe Capelli eigenvalue problem Siddhartha Sahi Rutgers University, New Brunswick NJ May 29, 2018, Kostant Conference Siddhartha Sahi (Kostant

Interpolation polynomials and Capelli operators

Interpolation polynomials. For µ ∈ Λn write |µ| = µ1 + · · ·+ µn.

Suppose char(F) = 0 and ρ ∈ Fn satisifies ρi+1 − ρi /∈N for all i

Theorem ([S. 1994, Prog Math 123] Kostant 65th Birthday)

For µ ∈ Λn there is a unique symm. polynomial R(ρ)µ (x) ∈ F [x1, . . . , xn ]

Sn

of degree |µ| such that R (ρ)µ (λ+ ρ) = δλµ for |λ| ≤ |µ| .

Generalized Capelli operators. Basis{Dµ

}for PD (N)G

PD (N) ≈ P (N)⊗D (N) ≈ ⊕λ,µVλ ⊗ V ∗µ ; PD (N)G ≈ ⊕µCDµ

Dµ corresponds to the identity 1µ in End(Vµ

)≈ Vµ ⊗ V ∗µ

Theorem ([ibid.])

Dµ acts on Vλ by the scalar R(ρ)µ (λ+ ρ), where ρi =

d2 (n− 2i + 1)

Siddhartha Sahi (Kostant Conference, MIT) May 29, 2018 5 / 12

Page 23: The Capelli eigenvalue problem - MathematicsThe Capelli eigenvalue problem Siddhartha Sahi Rutgers University, New Brunswick NJ May 29, 2018, Kostant Conference Siddhartha Sahi (Kostant

Interpolation polynomials and Capelli operators

Interpolation polynomials. For µ ∈ Λn write |µ| = µ1 + · · ·+ µn.

Suppose char(F) = 0 and ρ ∈ Fn satisifies ρi+1 − ρi /∈N for all i

Theorem ([S. 1994, Prog Math 123] Kostant 65th Birthday)

For µ ∈ Λn there is a unique symm. polynomial R(ρ)µ (x) ∈ F [x1, . . . , xn ]

Sn

of degree |µ| such that R (ρ)µ (λ+ ρ) = δλµ for |λ| ≤ |µ| .

Generalized Capelli operators. Basis{Dµ

}for PD (N)G

PD (N) ≈ P (N)⊗D (N) ≈ ⊕λ,µVλ ⊗ V ∗µ ; PD (N)G ≈ ⊕µCDµ

Dµ corresponds to the identity 1µ in End(Vµ

)≈ Vµ ⊗ V ∗µ

Theorem ([ibid.])

Dµ acts on Vλ by the scalar R(ρ)µ (λ+ ρ), where ρi =

d2 (n− 2i + 1)

Siddhartha Sahi (Kostant Conference, MIT) May 29, 2018 5 / 12

Page 24: The Capelli eigenvalue problem - MathematicsThe Capelli eigenvalue problem Siddhartha Sahi Rutgers University, New Brunswick NJ May 29, 2018, Kostant Conference Siddhartha Sahi (Kostant

Interpolation polynomials and Capelli operators

Interpolation polynomials. For µ ∈ Λn write |µ| = µ1 + · · ·+ µn.

Suppose char(F) = 0 and ρ ∈ Fn satisifies ρi+1 − ρi /∈N for all i

Theorem ([S. 1994, Prog Math 123] Kostant 65th Birthday)

For µ ∈ Λn there is a unique symm. polynomial R(ρ)µ (x) ∈ F [x1, . . . , xn ]

Sn

of degree |µ| such that R (ρ)µ (λ+ ρ) = δλµ for |λ| ≤ |µ| .

Generalized Capelli operators. Basis{Dµ

}for PD (N)G

PD (N) ≈ P (N)⊗D (N) ≈ ⊕λ,µVλ ⊗ V ∗µ ; PD (N)G ≈ ⊕µCDµ

Dµ corresponds to the identity 1µ in End(Vµ

)≈ Vµ ⊗ V ∗µ

Theorem ([ibid.])

Dµ acts on Vλ by the scalar R(ρ)µ (λ+ ρ), where ρi =

d2 (n− 2i + 1)

Siddhartha Sahi (Kostant Conference, MIT) May 29, 2018 5 / 12

Page 25: The Capelli eigenvalue problem - MathematicsThe Capelli eigenvalue problem Siddhartha Sahi Rutgers University, New Brunswick NJ May 29, 2018, Kostant Conference Siddhartha Sahi (Kostant

Interpolation polynomials and Capelli operators

Interpolation polynomials. For µ ∈ Λn write |µ| = µ1 + · · ·+ µn.

Suppose char(F) = 0 and ρ ∈ Fn satisifies ρi+1 − ρi /∈N for all i

Theorem ([S. 1994, Prog Math 123] Kostant 65th Birthday)

For µ ∈ Λn there is a unique symm. polynomial R(ρ)µ (x) ∈ F [x1, . . . , xn ]

Sn

of degree |µ| such that R (ρ)µ (λ+ ρ) = δλµ for |λ| ≤ |µ| .

Generalized Capelli operators. Basis{Dµ

}for PD (N)G

PD (N) ≈ P (N)⊗D (N) ≈ ⊕λ,µVλ ⊗ V ∗µ ; PD (N)G ≈ ⊕µCDµ

Dµ corresponds to the identity 1µ in End(Vµ

)≈ Vµ ⊗ V ∗µ

Theorem ([ibid.])

Dµ acts on Vλ by the scalar R(ρ)µ (λ+ ρ), where ρi =

d2 (n− 2i + 1)

Siddhartha Sahi (Kostant Conference, MIT) May 29, 2018 5 / 12

Page 26: The Capelli eigenvalue problem - MathematicsThe Capelli eigenvalue problem Siddhartha Sahi Rutgers University, New Brunswick NJ May 29, 2018, Kostant Conference Siddhartha Sahi (Kostant

Interpolation polynomials and Capelli operators

Interpolation polynomials. For µ ∈ Λn write |µ| = µ1 + · · ·+ µn.

Suppose char(F) = 0 and ρ ∈ Fn satisifies ρi+1 − ρi /∈N for all i

Theorem ([S. 1994, Prog Math 123] Kostant 65th Birthday)

For µ ∈ Λn there is a unique symm. polynomial R(ρ)µ (x) ∈ F [x1, . . . , xn ]

Sn

of degree |µ| such that R (ρ)µ (λ+ ρ) = δλµ for |λ| ≤ |µ| .

Generalized Capelli operators. Basis{Dµ

}for PD (N)G

PD (N) ≈ P (N)⊗D (N) ≈ ⊕λ,µVλ ⊗ V ∗µ ; PD (N)G ≈ ⊕µCDµ

Dµ corresponds to the identity 1µ in End(Vµ

)≈ Vµ ⊗ V ∗µ

Theorem ([ibid.])

Dµ acts on Vλ by the scalar R(ρ)µ (λ+ ρ), where ρi =

d2 (n− 2i + 1)

Siddhartha Sahi (Kostant Conference, MIT) May 29, 2018 5 / 12

Page 27: The Capelli eigenvalue problem - MathematicsThe Capelli eigenvalue problem Siddhartha Sahi Rutgers University, New Brunswick NJ May 29, 2018, Kostant Conference Siddhartha Sahi (Kostant

Connection with Jack polynomials

Let δ = (n− 1, . . . , 1, 0), let F = Q (r) and consider Rµ = R(r δ)µ

Let Ti be the ith shift operator Ti f (x) = f (x − ei )

Define D (t) = det(x δji

)−1det((xi + t) (xi + r)

δj − x δj+1i Ti

)Define D (t) = det

(x δji

)−1det(x δji

[t + rδj + xi ∂

∂xi

])

Theorem ([Knop-S. IMRN 1996])

1 The polynomial Rµ satisfies D (t)Rµ = [∏i (µi + rδi + t)]Rµ.

2 The top degree part Rµ satisfies D (t)Rµ = [∏i (µi + rδi + t)]Rµ

3 Rµ is proportional to the Jack polynomial P(1/r )µ .

For (1) we check D (t)Rµ vanishes at ν+ ρ if |ν| ≤ |µ|, ν 6= µ

(2) is a limit of (1) and implies (3) by work of Debiard-Sekiguchi.

Siddhartha Sahi (Kostant Conference, MIT) May 29, 2018 6 / 12

Page 28: The Capelli eigenvalue problem - MathematicsThe Capelli eigenvalue problem Siddhartha Sahi Rutgers University, New Brunswick NJ May 29, 2018, Kostant Conference Siddhartha Sahi (Kostant

Connection with Jack polynomials

Let δ = (n− 1, . . . , 1, 0), let F = Q (r) and consider Rµ = R(r δ)µ

Let Ti be the ith shift operator Ti f (x) = f (x − ei )

Define D (t) = det(x δji

)−1det((xi + t) (xi + r)

δj − x δj+1i Ti

)Define D (t) = det

(x δji

)−1det(x δji

[t + rδj + xi ∂

∂xi

])

Theorem ([Knop-S. IMRN 1996])

1 The polynomial Rµ satisfies D (t)Rµ = [∏i (µi + rδi + t)]Rµ.

2 The top degree part Rµ satisfies D (t)Rµ = [∏i (µi + rδi + t)]Rµ

3 Rµ is proportional to the Jack polynomial P(1/r )µ .

For (1) we check D (t)Rµ vanishes at ν+ ρ if |ν| ≤ |µ|, ν 6= µ

(2) is a limit of (1) and implies (3) by work of Debiard-Sekiguchi.

Siddhartha Sahi (Kostant Conference, MIT) May 29, 2018 6 / 12

Page 29: The Capelli eigenvalue problem - MathematicsThe Capelli eigenvalue problem Siddhartha Sahi Rutgers University, New Brunswick NJ May 29, 2018, Kostant Conference Siddhartha Sahi (Kostant

Connection with Jack polynomials

Let δ = (n− 1, . . . , 1, 0), let F = Q (r) and consider Rµ = R(r δ)µ

Let Ti be the ith shift operator Ti f (x) = f (x − ei )

Define D (t) = det(x δji

)−1det((xi + t) (xi + r)

δj − x δj+1i Ti

)

Define D (t) = det(x δji

)−1det(x δji

[t + rδj + xi ∂

∂xi

])

Theorem ([Knop-S. IMRN 1996])

1 The polynomial Rµ satisfies D (t)Rµ = [∏i (µi + rδi + t)]Rµ.

2 The top degree part Rµ satisfies D (t)Rµ = [∏i (µi + rδi + t)]Rµ

3 Rµ is proportional to the Jack polynomial P(1/r )µ .

For (1) we check D (t)Rµ vanishes at ν+ ρ if |ν| ≤ |µ|, ν 6= µ

(2) is a limit of (1) and implies (3) by work of Debiard-Sekiguchi.

Siddhartha Sahi (Kostant Conference, MIT) May 29, 2018 6 / 12

Page 30: The Capelli eigenvalue problem - MathematicsThe Capelli eigenvalue problem Siddhartha Sahi Rutgers University, New Brunswick NJ May 29, 2018, Kostant Conference Siddhartha Sahi (Kostant

Connection with Jack polynomials

Let δ = (n− 1, . . . , 1, 0), let F = Q (r) and consider Rµ = R(r δ)µ

Let Ti be the ith shift operator Ti f (x) = f (x − ei )

Define D (t) = det(x δji

)−1det((xi + t) (xi + r)

δj − x δj+1i Ti

)Define D (t) = det

(x δji

)−1det(x δji

[t + rδj + xi ∂

∂xi

])

Theorem ([Knop-S. IMRN 1996])

1 The polynomial Rµ satisfies D (t)Rµ = [∏i (µi + rδi + t)]Rµ.

2 The top degree part Rµ satisfies D (t)Rµ = [∏i (µi + rδi + t)]Rµ

3 Rµ is proportional to the Jack polynomial P(1/r )µ .

For (1) we check D (t)Rµ vanishes at ν+ ρ if |ν| ≤ |µ|, ν 6= µ

(2) is a limit of (1) and implies (3) by work of Debiard-Sekiguchi.

Siddhartha Sahi (Kostant Conference, MIT) May 29, 2018 6 / 12

Page 31: The Capelli eigenvalue problem - MathematicsThe Capelli eigenvalue problem Siddhartha Sahi Rutgers University, New Brunswick NJ May 29, 2018, Kostant Conference Siddhartha Sahi (Kostant

Connection with Jack polynomials

Let δ = (n− 1, . . . , 1, 0), let F = Q (r) and consider Rµ = R(r δ)µ

Let Ti be the ith shift operator Ti f (x) = f (x − ei )

Define D (t) = det(x δji

)−1det((xi + t) (xi + r)

δj − x δj+1i Ti

)Define D (t) = det

(x δji

)−1det(x δji

[t + rδj + xi ∂

∂xi

])Theorem ([Knop-S. IMRN 1996])

1 The polynomial Rµ satisfies D (t)Rµ = [∏i (µi + rδi + t)]Rµ.

2 The top degree part Rµ satisfies D (t)Rµ = [∏i (µi + rδi + t)]Rµ

3 Rµ is proportional to the Jack polynomial P(1/r )µ .

For (1) we check D (t)Rµ vanishes at ν+ ρ if |ν| ≤ |µ|, ν 6= µ

(2) is a limit of (1) and implies (3) by work of Debiard-Sekiguchi.

Siddhartha Sahi (Kostant Conference, MIT) May 29, 2018 6 / 12

Page 32: The Capelli eigenvalue problem - MathematicsThe Capelli eigenvalue problem Siddhartha Sahi Rutgers University, New Brunswick NJ May 29, 2018, Kostant Conference Siddhartha Sahi (Kostant

Connection with Jack polynomials

Let δ = (n− 1, . . . , 1, 0), let F = Q (r) and consider Rµ = R(r δ)µ

Let Ti be the ith shift operator Ti f (x) = f (x − ei )

Define D (t) = det(x δji

)−1det((xi + t) (xi + r)

δj − x δj+1i Ti

)Define D (t) = det

(x δji

)−1det(x δji

[t + rδj + xi ∂

∂xi

])Theorem ([Knop-S. IMRN 1996])

1 The polynomial Rµ satisfies D (t)Rµ = [∏i (µi + rδi + t)]Rµ.

2 The top degree part Rµ satisfies D (t)Rµ = [∏i (µi + rδi + t)]Rµ

3 Rµ is proportional to the Jack polynomial P(1/r )µ .

For (1) we check D (t)Rµ vanishes at ν+ ρ if |ν| ≤ |µ|, ν 6= µ

(2) is a limit of (1) and implies (3) by work of Debiard-Sekiguchi.

Siddhartha Sahi (Kostant Conference, MIT) May 29, 2018 6 / 12

Page 33: The Capelli eigenvalue problem - MathematicsThe Capelli eigenvalue problem Siddhartha Sahi Rutgers University, New Brunswick NJ May 29, 2018, Kostant Conference Siddhartha Sahi (Kostant

Connection with Jack polynomials

Let δ = (n− 1, . . . , 1, 0), let F = Q (r) and consider Rµ = R(r δ)µ

Let Ti be the ith shift operator Ti f (x) = f (x − ei )

Define D (t) = det(x δji

)−1det((xi + t) (xi + r)

δj − x δj+1i Ti

)Define D (t) = det

(x δji

)−1det(x δji

[t + rδj + xi ∂

∂xi

])Theorem ([Knop-S. IMRN 1996])

1 The polynomial Rµ satisfies D (t)Rµ = [∏i (µi + rδi + t)]Rµ.

2 The top degree part Rµ satisfies D (t)Rµ = [∏i (µi + rδi + t)]Rµ

3 Rµ is proportional to the Jack polynomial P(1/r )µ .

For (1) we check D (t)Rµ vanishes at ν+ ρ if |ν| ≤ |µ|, ν 6= µ

(2) is a limit of (1) and implies (3) by work of Debiard-Sekiguchi.

Siddhartha Sahi (Kostant Conference, MIT) May 29, 2018 6 / 12

Page 34: The Capelli eigenvalue problem - MathematicsThe Capelli eigenvalue problem Siddhartha Sahi Rutgers University, New Brunswick NJ May 29, 2018, Kostant Conference Siddhartha Sahi (Kostant

Connection with Jack polynomials

Let δ = (n− 1, . . . , 1, 0), let F = Q (r) and consider Rµ = R(r δ)µ

Let Ti be the ith shift operator Ti f (x) = f (x − ei )

Define D (t) = det(x δji

)−1det((xi + t) (xi + r)

δj − x δj+1i Ti

)Define D (t) = det

(x δji

)−1det(x δji

[t + rδj + xi ∂

∂xi

])Theorem ([Knop-S. IMRN 1996])

1 The polynomial Rµ satisfies D (t)Rµ = [∏i (µi + rδi + t)]Rµ.

2 The top degree part Rµ satisfies D (t)Rµ = [∏i (µi + rδi + t)]Rµ

3 Rµ is proportional to the Jack polynomial P(1/r )µ .

For (1) we check D (t)Rµ vanishes at ν+ ρ if |ν| ≤ |µ|, ν 6= µ

(2) is a limit of (1) and implies (3) by work of Debiard-Sekiguchi.

Siddhartha Sahi (Kostant Conference, MIT) May 29, 2018 6 / 12

Page 35: The Capelli eigenvalue problem - MathematicsThe Capelli eigenvalue problem Siddhartha Sahi Rutgers University, New Brunswick NJ May 29, 2018, Kostant Conference Siddhartha Sahi (Kostant

Connection with Jack polynomials

Let δ = (n− 1, . . . , 1, 0), let F = Q (r) and consider Rµ = R(r δ)µ

Let Ti be the ith shift operator Ti f (x) = f (x − ei )

Define D (t) = det(x δji

)−1det((xi + t) (xi + r)

δj − x δj+1i Ti

)Define D (t) = det

(x δji

)−1det(x δji

[t + rδj + xi ∂

∂xi

])Theorem ([Knop-S. IMRN 1996])

1 The polynomial Rµ satisfies D (t)Rµ = [∏i (µi + rδi + t)]Rµ.

2 The top degree part Rµ satisfies D (t)Rµ = [∏i (µi + rδi + t)]Rµ

3 Rµ is proportional to the Jack polynomial P(1/r )µ .

For (1) we check D (t)Rµ vanishes at ν+ ρ if |ν| ≤ |µ|, ν 6= µ

(2) is a limit of (1) and implies (3) by work of Debiard-Sekiguchi.

Siddhartha Sahi (Kostant Conference, MIT) May 29, 2018 6 / 12

Page 36: The Capelli eigenvalue problem - MathematicsThe Capelli eigenvalue problem Siddhartha Sahi Rutgers University, New Brunswick NJ May 29, 2018, Kostant Conference Siddhartha Sahi (Kostant

Connection with Jack polynomials

Let δ = (n− 1, . . . , 1, 0), let F = Q (r) and consider Rµ = R(r δ)µ

Let Ti be the ith shift operator Ti f (x) = f (x − ei )

Define D (t) = det(x δji

)−1det((xi + t) (xi + r)

δj − x δj+1i Ti

)Define D (t) = det

(x δji

)−1det(x δji

[t + rδj + xi ∂

∂xi

])Theorem ([Knop-S. IMRN 1996])

1 The polynomial Rµ satisfies D (t)Rµ = [∏i (µi + rδi + t)]Rµ.

2 The top degree part Rµ satisfies D (t)Rµ = [∏i (µi + rδi + t)]Rµ

3 Rµ is proportional to the Jack polynomial P(1/r )µ .

For (1) we check D (t)Rµ vanishes at ν+ ρ if |ν| ≤ |µ|, ν 6= µ

(2) is a limit of (1) and implies (3) by work of Debiard-Sekiguchi.

Siddhartha Sahi (Kostant Conference, MIT) May 29, 2018 6 / 12

Page 37: The Capelli eigenvalue problem - MathematicsThe Capelli eigenvalue problem Siddhartha Sahi Rutgers University, New Brunswick NJ May 29, 2018, Kostant Conference Siddhartha Sahi (Kostant

Nonsymmetric interpolation polynomials

F = Q (q, t), Sn acts on Zn and F [x1, . . . , xn ] , generators s1, . . . sn−1

Ti = tsi − (1−t)xixi−xi+1 (1− si ), Φf (x) =

(xn − t1−n

)f(xnq , x1, . . . , xn−1

)For η ∈ Zn

+ let wη ∈ Sn be shortest such that w−1η η ∈ Λn

Write η̄i = qηi t1−wη(i ), η̄ = (η̄1, . . . , η̄n) and |η| = η1 + · · ·+ ηn

Theorem ([S. IMRN 1996], [Knop 1997 Comm Math Helv])

1 For η ∈ Zn+ there is a unique polynomial Gη = G

(r δ)η ∈ F [x1, . . . , xn ] of

degree |η| satisfying Gη (γ) = δηγ for γ ∈ Zn+ such that |γ| ≤ |η|.

2 Let Ξi = x−1i + x−1i Ti · · ·Tn−1ΦT1 · · ·Ti−1, then ΞiGη = η̄−1i Gη.

3 Top degree part Gη ∼ nonsymmetric Macdonald polynomial Eη.

4 We have ΦGη ∼ G(ηn+1,η1,...,ηn−1) and[1t−1Ti +

η̄i+1η̄i+1−η̄i

]Gη ∼ Gsi η.

Siddhartha Sahi (Kostant Conference, MIT) May 29, 2018 7 / 12

Page 38: The Capelli eigenvalue problem - MathematicsThe Capelli eigenvalue problem Siddhartha Sahi Rutgers University, New Brunswick NJ May 29, 2018, Kostant Conference Siddhartha Sahi (Kostant

Nonsymmetric interpolation polynomials

F = Q (q, t), Sn acts on Zn and F [x1, . . . , xn ] , generators s1, . . . sn−1

Ti = tsi − (1−t)xixi−xi+1 (1− si ), Φf (x) =

(xn − t1−n

)f(xnq , x1, . . . , xn−1

)

For η ∈ Zn+ let wη ∈ Sn be shortest such that w−1η η ∈ Λn

Write η̄i = qηi t1−wη(i ), η̄ = (η̄1, . . . , η̄n) and |η| = η1 + · · ·+ ηn

Theorem ([S. IMRN 1996], [Knop 1997 Comm Math Helv])

1 For η ∈ Zn+ there is a unique polynomial Gη = G

(r δ)η ∈ F [x1, . . . , xn ] of

degree |η| satisfying Gη (γ) = δηγ for γ ∈ Zn+ such that |γ| ≤ |η|.

2 Let Ξi = x−1i + x−1i Ti · · ·Tn−1ΦT1 · · ·Ti−1, then ΞiGη = η̄−1i Gη.

3 Top degree part Gη ∼ nonsymmetric Macdonald polynomial Eη.

4 We have ΦGη ∼ G(ηn+1,η1,...,ηn−1) and[1t−1Ti +

η̄i+1η̄i+1−η̄i

]Gη ∼ Gsi η.

Siddhartha Sahi (Kostant Conference, MIT) May 29, 2018 7 / 12

Page 39: The Capelli eigenvalue problem - MathematicsThe Capelli eigenvalue problem Siddhartha Sahi Rutgers University, New Brunswick NJ May 29, 2018, Kostant Conference Siddhartha Sahi (Kostant

Nonsymmetric interpolation polynomials

F = Q (q, t), Sn acts on Zn and F [x1, . . . , xn ] , generators s1, . . . sn−1

Ti = tsi − (1−t)xixi−xi+1 (1− si ), Φf (x) =

(xn − t1−n

)f(xnq , x1, . . . , xn−1

)For η ∈ Zn

+ let wη ∈ Sn be shortest such that w−1η η ∈ Λn

Write η̄i = qηi t1−wη(i ), η̄ = (η̄1, . . . , η̄n) and |η| = η1 + · · ·+ ηn

Theorem ([S. IMRN 1996], [Knop 1997 Comm Math Helv])

1 For η ∈ Zn+ there is a unique polynomial Gη = G

(r δ)η ∈ F [x1, . . . , xn ] of

degree |η| satisfying Gη (γ) = δηγ for γ ∈ Zn+ such that |γ| ≤ |η|.

2 Let Ξi = x−1i + x−1i Ti · · ·Tn−1ΦT1 · · ·Ti−1, then ΞiGη = η̄−1i Gη.

3 Top degree part Gη ∼ nonsymmetric Macdonald polynomial Eη.

4 We have ΦGη ∼ G(ηn+1,η1,...,ηn−1) and[1t−1Ti +

η̄i+1η̄i+1−η̄i

]Gη ∼ Gsi η.

Siddhartha Sahi (Kostant Conference, MIT) May 29, 2018 7 / 12

Page 40: The Capelli eigenvalue problem - MathematicsThe Capelli eigenvalue problem Siddhartha Sahi Rutgers University, New Brunswick NJ May 29, 2018, Kostant Conference Siddhartha Sahi (Kostant

Nonsymmetric interpolation polynomials

F = Q (q, t), Sn acts on Zn and F [x1, . . . , xn ] , generators s1, . . . sn−1

Ti = tsi − (1−t)xixi−xi+1 (1− si ), Φf (x) =

(xn − t1−n

)f(xnq , x1, . . . , xn−1

)For η ∈ Zn

+ let wη ∈ Sn be shortest such that w−1η η ∈ Λn

Write η̄i = qηi t1−wη(i ), η̄ = (η̄1, . . . , η̄n) and |η| = η1 + · · ·+ ηn

Theorem ([S. IMRN 1996], [Knop 1997 Comm Math Helv])

1 For η ∈ Zn+ there is a unique polynomial Gη = G

(r δ)η ∈ F [x1, . . . , xn ] of

degree |η| satisfying Gη (γ) = δηγ for γ ∈ Zn+ such that |γ| ≤ |η|.

2 Let Ξi = x−1i + x−1i Ti · · ·Tn−1ΦT1 · · ·Ti−1, then ΞiGη = η̄−1i Gη.

3 Top degree part Gη ∼ nonsymmetric Macdonald polynomial Eη.

4 We have ΦGη ∼ G(ηn+1,η1,...,ηn−1) and[1t−1Ti +

η̄i+1η̄i+1−η̄i

]Gη ∼ Gsi η.

Siddhartha Sahi (Kostant Conference, MIT) May 29, 2018 7 / 12

Page 41: The Capelli eigenvalue problem - MathematicsThe Capelli eigenvalue problem Siddhartha Sahi Rutgers University, New Brunswick NJ May 29, 2018, Kostant Conference Siddhartha Sahi (Kostant

Nonsymmetric interpolation polynomials

F = Q (q, t), Sn acts on Zn and F [x1, . . . , xn ] , generators s1, . . . sn−1

Ti = tsi − (1−t)xixi−xi+1 (1− si ), Φf (x) =

(xn − t1−n

)f(xnq , x1, . . . , xn−1

)For η ∈ Zn

+ let wη ∈ Sn be shortest such that w−1η η ∈ Λn

Write η̄i = qηi t1−wη(i ), η̄ = (η̄1, . . . , η̄n) and |η| = η1 + · · ·+ ηn

Theorem ([S. IMRN 1996], [Knop 1997 Comm Math Helv])

1 For η ∈ Zn+ there is a unique polynomial Gη = G

(r δ)η ∈ F [x1, . . . , xn ] of

degree |η| satisfying Gη (γ) = δηγ for γ ∈ Zn+ such that |γ| ≤ |η|.

2 Let Ξi = x−1i + x−1i Ti · · ·Tn−1ΦT1 · · ·Ti−1, then ΞiGη = η̄−1i Gη.

3 Top degree part Gη ∼ nonsymmetric Macdonald polynomial Eη.

4 We have ΦGη ∼ G(ηn+1,η1,...,ηn−1) and[1t−1Ti +

η̄i+1η̄i+1−η̄i

]Gη ∼ Gsi η.

Siddhartha Sahi (Kostant Conference, MIT) May 29, 2018 7 / 12

Page 42: The Capelli eigenvalue problem - MathematicsThe Capelli eigenvalue problem Siddhartha Sahi Rutgers University, New Brunswick NJ May 29, 2018, Kostant Conference Siddhartha Sahi (Kostant

Nonsymmetric interpolation polynomials

F = Q (q, t), Sn acts on Zn and F [x1, . . . , xn ] , generators s1, . . . sn−1

Ti = tsi − (1−t)xixi−xi+1 (1− si ), Φf (x) =

(xn − t1−n

)f(xnq , x1, . . . , xn−1

)For η ∈ Zn

+ let wη ∈ Sn be shortest such that w−1η η ∈ Λn

Write η̄i = qηi t1−wη(i ), η̄ = (η̄1, . . . , η̄n) and |η| = η1 + · · ·+ ηn

Theorem ([S. IMRN 1996], [Knop 1997 Comm Math Helv])

1 For η ∈ Zn+ there is a unique polynomial Gη = G

(r δ)η ∈ F [x1, . . . , xn ] of

degree |η| satisfying Gη (γ) = δηγ for γ ∈ Zn+ such that |γ| ≤ |η|.

2 Let Ξi = x−1i + x−1i Ti · · ·Tn−1ΦT1 · · ·Ti−1, then ΞiGη = η̄−1i Gη.

3 Top degree part Gη ∼ nonsymmetric Macdonald polynomial Eη.

4 We have ΦGη ∼ G(ηn+1,η1,...,ηn−1) and[1t−1Ti +

η̄i+1η̄i+1−η̄i

]Gη ∼ Gsi η.

Siddhartha Sahi (Kostant Conference, MIT) May 29, 2018 7 / 12

Page 43: The Capelli eigenvalue problem - MathematicsThe Capelli eigenvalue problem Siddhartha Sahi Rutgers University, New Brunswick NJ May 29, 2018, Kostant Conference Siddhartha Sahi (Kostant

Nonsymmetric interpolation polynomials

F = Q (q, t), Sn acts on Zn and F [x1, . . . , xn ] , generators s1, . . . sn−1

Ti = tsi − (1−t)xixi−xi+1 (1− si ), Φf (x) =

(xn − t1−n

)f(xnq , x1, . . . , xn−1

)For η ∈ Zn

+ let wη ∈ Sn be shortest such that w−1η η ∈ Λn

Write η̄i = qηi t1−wη(i ), η̄ = (η̄1, . . . , η̄n) and |η| = η1 + · · ·+ ηn

Theorem ([S. IMRN 1996], [Knop 1997 Comm Math Helv])

1 For η ∈ Zn+ there is a unique polynomial Gη = G

(r δ)η ∈ F [x1, . . . , xn ] of

degree |η| satisfying Gη (γ) = δηγ for γ ∈ Zn+ such that |γ| ≤ |η|.

2 Let Ξi = x−1i + x−1i Ti · · ·Tn−1ΦT1 · · ·Ti−1, then ΞiGη = η̄−1i Gη.

3 Top degree part Gη ∼ nonsymmetric Macdonald polynomial Eη.

4 We have ΦGη ∼ G(ηn+1,η1,...,ηn−1) and[1t−1Ti +

η̄i+1η̄i+1−η̄i

]Gη ∼ Gsi η.

Siddhartha Sahi (Kostant Conference, MIT) May 29, 2018 7 / 12

Page 44: The Capelli eigenvalue problem - MathematicsThe Capelli eigenvalue problem Siddhartha Sahi Rutgers University, New Brunswick NJ May 29, 2018, Kostant Conference Siddhartha Sahi (Kostant

Nonsymmetric interpolation polynomials

F = Q (q, t), Sn acts on Zn and F [x1, . . . , xn ] , generators s1, . . . sn−1

Ti = tsi − (1−t)xixi−xi+1 (1− si ), Φf (x) =

(xn − t1−n

)f(xnq , x1, . . . , xn−1

)For η ∈ Zn

+ let wη ∈ Sn be shortest such that w−1η η ∈ Λn

Write η̄i = qηi t1−wη(i ), η̄ = (η̄1, . . . , η̄n) and |η| = η1 + · · ·+ ηn

Theorem ([S. IMRN 1996], [Knop 1997 Comm Math Helv])

1 For η ∈ Zn+ there is a unique polynomial Gη = G

(r δ)η ∈ F [x1, . . . , xn ] of

degree |η| satisfying Gη (γ) = δηγ for γ ∈ Zn+ such that |γ| ≤ |η|.

2 Let Ξi = x−1i + x−1i Ti · · ·Tn−1ΦT1 · · ·Ti−1, then ΞiGη = η̄−1i Gη.

3 Top degree part Gη ∼ nonsymmetric Macdonald polynomial Eη.

4 We have ΦGη ∼ G(ηn+1,η1,...,ηn−1) and[1t−1Ti +

η̄i+1η̄i+1−η̄i

]Gη ∼ Gsi η.

Siddhartha Sahi (Kostant Conference, MIT) May 29, 2018 7 / 12

Page 45: The Capelli eigenvalue problem - MathematicsThe Capelli eigenvalue problem Siddhartha Sahi Rutgers University, New Brunswick NJ May 29, 2018, Kostant Conference Siddhartha Sahi (Kostant

Nonsymmetric interpolation polynomials

F = Q (q, t), Sn acts on Zn and F [x1, . . . , xn ] , generators s1, . . . sn−1

Ti = tsi − (1−t)xixi−xi+1 (1− si ), Φf (x) =

(xn − t1−n

)f(xnq , x1, . . . , xn−1

)For η ∈ Zn

+ let wη ∈ Sn be shortest such that w−1η η ∈ Λn

Write η̄i = qηi t1−wη(i ), η̄ = (η̄1, . . . , η̄n) and |η| = η1 + · · ·+ ηn

Theorem ([S. IMRN 1996], [Knop 1997 Comm Math Helv])

1 For η ∈ Zn+ there is a unique polynomial Gη = G

(r δ)η ∈ F [x1, . . . , xn ] of

degree |η| satisfying Gη (γ) = δηγ for γ ∈ Zn+ such that |γ| ≤ |η|.

2 Let Ξi = x−1i + x−1i Ti · · ·Tn−1ΦT1 · · ·Ti−1, then ΞiGη = η̄−1i Gη.

3 Top degree part Gη ∼ nonsymmetric Macdonald polynomial Eη.

4 We have ΦGη ∼ G(ηn+1,η1,...,ηn−1) and[1t−1Ti +

η̄i+1η̄i+1−η̄i

]Gη ∼ Gsi η.

Siddhartha Sahi (Kostant Conference, MIT) May 29, 2018 7 / 12

Page 46: The Capelli eigenvalue problem - MathematicsThe Capelli eigenvalue problem Siddhartha Sahi Rutgers University, New Brunswick NJ May 29, 2018, Kostant Conference Siddhartha Sahi (Kostant

Intertwiners and Macdonald conjectures

Intertwiners: Φ̄f = xnf(xnq , x1, . . . , xn−1

)and Sη

i =1t−1Ti +

η̄i+1η̄i+1−η̄i

Then Sηi Eη ∼ Esi η and Φ̄Eη ∼ E(ηn+1,η1,...,ηn−1)

Theorem ([Knop-S. 97 Invent 128], [S. 96 IMRN], [K. 97 CMH])

1 Jack polynomials J(1/r )λ have coeffi cients in Z+ [1/r ]

2 Macdonald polynomials J(q,t)λ have coeffs in Z [q, t]

These are the “type A”conjectures of Macdonald. “Proof”: Checkfor Eη by induction using intertwiners, then for Jλ by symmetrization.

The same approach proves Macdonald’s conjectures for arbitrary rootsystems [Cherednik 1997, Selecta 3], and also for “type BC"Koorwinder polynomials [S. 1999, Annals 150]

Siddhartha Sahi (Kostant Conference, MIT) May 29, 2018 8 / 12

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Intertwiners and Macdonald conjectures

Intertwiners: Φ̄f = xnf(xnq , x1, . . . , xn−1

)and Sη

i =1t−1Ti +

η̄i+1η̄i+1−η̄i

Then Sηi Eη ∼ Esi η and Φ̄Eη ∼ E(ηn+1,η1,...,ηn−1)

Theorem ([Knop-S. 97 Invent 128], [S. 96 IMRN], [K. 97 CMH])

1 Jack polynomials J(1/r )λ have coeffi cients in Z+ [1/r ]

2 Macdonald polynomials J(q,t)λ have coeffs in Z [q, t]

These are the “type A”conjectures of Macdonald. “Proof”: Checkfor Eη by induction using intertwiners, then for Jλ by symmetrization.

The same approach proves Macdonald’s conjectures for arbitrary rootsystems [Cherednik 1997, Selecta 3], and also for “type BC"Koorwinder polynomials [S. 1999, Annals 150]

Siddhartha Sahi (Kostant Conference, MIT) May 29, 2018 8 / 12

Page 48: The Capelli eigenvalue problem - MathematicsThe Capelli eigenvalue problem Siddhartha Sahi Rutgers University, New Brunswick NJ May 29, 2018, Kostant Conference Siddhartha Sahi (Kostant

Intertwiners and Macdonald conjectures

Intertwiners: Φ̄f = xnf(xnq , x1, . . . , xn−1

)and Sη

i =1t−1Ti +

η̄i+1η̄i+1−η̄i

Then Sηi Eη ∼ Esi η and Φ̄Eη ∼ E(ηn+1,η1,...,ηn−1)

Theorem ([Knop-S. 97 Invent 128], [S. 96 IMRN], [K. 97 CMH])

1 Jack polynomials J(1/r )λ have coeffi cients in Z+ [1/r ]

2 Macdonald polynomials J(q,t)λ have coeffs in Z [q, t]

These are the “type A”conjectures of Macdonald. “Proof”: Checkfor Eη by induction using intertwiners, then for Jλ by symmetrization.

The same approach proves Macdonald’s conjectures for arbitrary rootsystems [Cherednik 1997, Selecta 3], and also for “type BC"Koorwinder polynomials [S. 1999, Annals 150]

Siddhartha Sahi (Kostant Conference, MIT) May 29, 2018 8 / 12

Page 49: The Capelli eigenvalue problem - MathematicsThe Capelli eigenvalue problem Siddhartha Sahi Rutgers University, New Brunswick NJ May 29, 2018, Kostant Conference Siddhartha Sahi (Kostant

Intertwiners and Macdonald conjectures

Intertwiners: Φ̄f = xnf(xnq , x1, . . . , xn−1

)and Sη

i =1t−1Ti +

η̄i+1η̄i+1−η̄i

Then Sηi Eη ∼ Esi η and Φ̄Eη ∼ E(ηn+1,η1,...,ηn−1)

Theorem ([Knop-S. 97 Invent 128], [S. 96 IMRN], [K. 97 CMH])

1 Jack polynomials J(1/r )λ have coeffi cients in Z+ [1/r ]

2 Macdonald polynomials J(q,t)λ have coeffs in Z [q, t]

These are the “type A”conjectures of Macdonald. “Proof”: Checkfor Eη by induction using intertwiners, then for Jλ by symmetrization.

The same approach proves Macdonald’s conjectures for arbitrary rootsystems [Cherednik 1997, Selecta 3], and also for “type BC"Koorwinder polynomials [S. 1999, Annals 150]

Siddhartha Sahi (Kostant Conference, MIT) May 29, 2018 8 / 12

Page 50: The Capelli eigenvalue problem - MathematicsThe Capelli eigenvalue problem Siddhartha Sahi Rutgers University, New Brunswick NJ May 29, 2018, Kostant Conference Siddhartha Sahi (Kostant

Intertwiners and Macdonald conjectures

Intertwiners: Φ̄f = xnf(xnq , x1, . . . , xn−1

)and Sη

i =1t−1Ti +

η̄i+1η̄i+1−η̄i

Then Sηi Eη ∼ Esi η and Φ̄Eη ∼ E(ηn+1,η1,...,ηn−1)

Theorem ([Knop-S. 97 Invent 128], [S. 96 IMRN], [K. 97 CMH])

1 Jack polynomials J(1/r )λ have coeffi cients in Z+ [1/r ]

2 Macdonald polynomials J(q,t)λ have coeffs in Z [q, t]

These are the “type A”conjectures of Macdonald. “Proof”: Checkfor Eη by induction using intertwiners, then for Jλ by symmetrization.

The same approach proves Macdonald’s conjectures for arbitrary rootsystems [Cherednik 1997, Selecta 3], and also for “type BC"Koorwinder polynomials [S. 1999, Annals 150]

Siddhartha Sahi (Kostant Conference, MIT) May 29, 2018 8 / 12

Page 51: The Capelli eigenvalue problem - MathematicsThe Capelli eigenvalue problem Siddhartha Sahi Rutgers University, New Brunswick NJ May 29, 2018, Kostant Conference Siddhartha Sahi (Kostant

Intertwiners and Macdonald conjectures

Intertwiners: Φ̄f = xnf(xnq , x1, . . . , xn−1

)and Sη

i =1t−1Ti +

η̄i+1η̄i+1−η̄i

Then Sηi Eη ∼ Esi η and Φ̄Eη ∼ E(ηn+1,η1,...,ηn−1)

Theorem ([Knop-S. 97 Invent 128], [S. 96 IMRN], [K. 97 CMH])

1 Jack polynomials J(1/r )λ have coeffi cients in Z+ [1/r ]

2 Macdonald polynomials J(q,t)λ have coeffs in Z [q, t]

These are the “type A”conjectures of Macdonald. “Proof”: Checkfor Eη by induction using intertwiners, then for Jλ by symmetrization.

The same approach proves Macdonald’s conjectures for arbitrary rootsystems [Cherednik 1997, Selecta 3], and also for “type BC"Koorwinder polynomials [S. 1999, Annals 150]

Siddhartha Sahi (Kostant Conference, MIT) May 29, 2018 8 / 12

Page 52: The Capelli eigenvalue problem - MathematicsThe Capelli eigenvalue problem Siddhartha Sahi Rutgers University, New Brunswick NJ May 29, 2018, Kostant Conference Siddhartha Sahi (Kostant

Intertwiners and Macdonald conjectures

Intertwiners: Φ̄f = xnf(xnq , x1, . . . , xn−1

)and Sη

i =1t−1Ti +

η̄i+1η̄i+1−η̄i

Then Sηi Eη ∼ Esi η and Φ̄Eη ∼ E(ηn+1,η1,...,ηn−1)

Theorem ([Knop-S. 97 Invent 128], [S. 96 IMRN], [K. 97 CMH])

1 Jack polynomials J(1/r )λ have coeffi cients in Z+ [1/r ]

2 Macdonald polynomials J(q,t)λ have coeffs in Z [q, t]

These are the “type A”conjectures of Macdonald. “Proof”: Checkfor Eη by induction using intertwiners, then for Jλ by symmetrization.

The same approach proves Macdonald’s conjectures for arbitrary rootsystems [Cherednik 1997, Selecta 3], and also for “type BC"Koorwinder polynomials [S. 1999, Annals 150]

Siddhartha Sahi (Kostant Conference, MIT) May 29, 2018 8 / 12

Page 53: The Capelli eigenvalue problem - MathematicsThe Capelli eigenvalue problem Siddhartha Sahi Rutgers University, New Brunswick NJ May 29, 2018, Kostant Conference Siddhartha Sahi (Kostant

Binomial coeffi cients

As before, let Rµ = R(r δ)µ , Jµ = J

(1/r )µ , define (λ

µ) = Rµ (λ+ rδ)

[Okounkov-Olshanski 1997, MRL 4] Jλ(1+x )Jλ(1)

= ∑µ (λµ)Jµ(x )Jµ(1)

Macdonald analog [Ok. MRL 4 ]: Nonsymmetric [S., Duke 94.]

[Kaneko SIAM J. 24] Explicit formula aλµ for (λµ) if λ\µ single box.

A = Λn ×Λn matrix with entries{aλµ λ\µ single box0 otherwise

.

Theorem ([S. 2011, IMRN])

The binomial coeffi cients (λµ) are entries of B = exp (A) .

Macdonald analog [S. Contemp Math 2012]

Siddhartha Sahi (Kostant Conference, MIT) May 29, 2018 9 / 12

Page 54: The Capelli eigenvalue problem - MathematicsThe Capelli eigenvalue problem Siddhartha Sahi Rutgers University, New Brunswick NJ May 29, 2018, Kostant Conference Siddhartha Sahi (Kostant

Binomial coeffi cients

As before, let Rµ = R(r δ)µ , Jµ = J

(1/r )µ , define (λ

µ) = Rµ (λ+ rδ)

[Okounkov-Olshanski 1997, MRL 4] Jλ(1+x )Jλ(1)

= ∑µ (λµ)Jµ(x )Jµ(1)

Macdonald analog [Ok. MRL 4 ]: Nonsymmetric [S., Duke 94.]

[Kaneko SIAM J. 24] Explicit formula aλµ for (λµ) if λ\µ single box.

A = Λn ×Λn matrix with entries{aλµ λ\µ single box0 otherwise

.

Theorem ([S. 2011, IMRN])

The binomial coeffi cients (λµ) are entries of B = exp (A) .

Macdonald analog [S. Contemp Math 2012]

Siddhartha Sahi (Kostant Conference, MIT) May 29, 2018 9 / 12

Page 55: The Capelli eigenvalue problem - MathematicsThe Capelli eigenvalue problem Siddhartha Sahi Rutgers University, New Brunswick NJ May 29, 2018, Kostant Conference Siddhartha Sahi (Kostant

Binomial coeffi cients

As before, let Rµ = R(r δ)µ , Jµ = J

(1/r )µ , define (λ

µ) = Rµ (λ+ rδ)

[Okounkov-Olshanski 1997, MRL 4] Jλ(1+x )Jλ(1)

= ∑µ (λµ)Jµ(x )Jµ(1)

Macdonald analog [Ok. MRL 4 ]: Nonsymmetric [S., Duke 94.]

[Kaneko SIAM J. 24] Explicit formula aλµ for (λµ) if λ\µ single box.

A = Λn ×Λn matrix with entries{aλµ λ\µ single box0 otherwise

.

Theorem ([S. 2011, IMRN])

The binomial coeffi cients (λµ) are entries of B = exp (A) .

Macdonald analog [S. Contemp Math 2012]

Siddhartha Sahi (Kostant Conference, MIT) May 29, 2018 9 / 12

Page 56: The Capelli eigenvalue problem - MathematicsThe Capelli eigenvalue problem Siddhartha Sahi Rutgers University, New Brunswick NJ May 29, 2018, Kostant Conference Siddhartha Sahi (Kostant

Binomial coeffi cients

As before, let Rµ = R(r δ)µ , Jµ = J

(1/r )µ , define (λ

µ) = Rµ (λ+ rδ)

[Okounkov-Olshanski 1997, MRL 4] Jλ(1+x )Jλ(1)

= ∑µ (λµ)Jµ(x )Jµ(1)

Macdonald analog [Ok. MRL 4 ]: Nonsymmetric [S., Duke 94.]

[Kaneko SIAM J. 24] Explicit formula aλµ for (λµ) if λ\µ single box.

A = Λn ×Λn matrix with entries{aλµ λ\µ single box0 otherwise

.

Theorem ([S. 2011, IMRN])

The binomial coeffi cients (λµ) are entries of B = exp (A) .

Macdonald analog [S. Contemp Math 2012]

Siddhartha Sahi (Kostant Conference, MIT) May 29, 2018 9 / 12

Page 57: The Capelli eigenvalue problem - MathematicsThe Capelli eigenvalue problem Siddhartha Sahi Rutgers University, New Brunswick NJ May 29, 2018, Kostant Conference Siddhartha Sahi (Kostant

Binomial coeffi cients

As before, let Rµ = R(r δ)µ , Jµ = J

(1/r )µ , define (λ

µ) = Rµ (λ+ rδ)

[Okounkov-Olshanski 1997, MRL 4] Jλ(1+x )Jλ(1)

= ∑µ (λµ)Jµ(x )Jµ(1)

Macdonald analog [Ok. MRL 4 ]: Nonsymmetric [S., Duke 94.]

[Kaneko SIAM J. 24] Explicit formula aλµ for (λµ) if λ\µ single box.

A = Λn ×Λn matrix with entries{aλµ λ\µ single box0 otherwise

.

Theorem ([S. 2011, IMRN])

The binomial coeffi cients (λµ) are entries of B = exp (A) .

Macdonald analog [S. Contemp Math 2012]

Siddhartha Sahi (Kostant Conference, MIT) May 29, 2018 9 / 12

Page 58: The Capelli eigenvalue problem - MathematicsThe Capelli eigenvalue problem Siddhartha Sahi Rutgers University, New Brunswick NJ May 29, 2018, Kostant Conference Siddhartha Sahi (Kostant

Binomial coeffi cients

As before, let Rµ = R(r δ)µ , Jµ = J

(1/r )µ , define (λ

µ) = Rµ (λ+ rδ)

[Okounkov-Olshanski 1997, MRL 4] Jλ(1+x )Jλ(1)

= ∑µ (λµ)Jµ(x )Jµ(1)

Macdonald analog [Ok. MRL 4 ]: Nonsymmetric [S., Duke 94.]

[Kaneko SIAM J. 24] Explicit formula aλµ for (λµ) if λ\µ single box.

A = Λn ×Λn matrix with entries{aλµ λ\µ single box0 otherwise

.

Theorem ([S. 2011, IMRN])

The binomial coeffi cients (λµ) are entries of B = exp (A) .

Macdonald analog [S. Contemp Math 2012]

Siddhartha Sahi (Kostant Conference, MIT) May 29, 2018 9 / 12

Page 59: The Capelli eigenvalue problem - MathematicsThe Capelli eigenvalue problem Siddhartha Sahi Rutgers University, New Brunswick NJ May 29, 2018, Kostant Conference Siddhartha Sahi (Kostant

Binomial coeffi cients

As before, let Rµ = R(r δ)µ , Jµ = J

(1/r )µ , define (λ

µ) = Rµ (λ+ rδ)

[Okounkov-Olshanski 1997, MRL 4] Jλ(1+x )Jλ(1)

= ∑µ (λµ)Jµ(x )Jµ(1)

Macdonald analog [Ok. MRL 4 ]: Nonsymmetric [S., Duke 94.]

[Kaneko SIAM J. 24] Explicit formula aλµ for (λµ) if λ\µ single box.

A = Λn ×Λn matrix with entries{aλµ λ\µ single box0 otherwise

.

Theorem ([S. 2011, IMRN])

The binomial coeffi cients (λµ) are entries of B = exp (A) .

Macdonald analog [S. Contemp Math 2012]

Siddhartha Sahi (Kostant Conference, MIT) May 29, 2018 9 / 12

Page 60: The Capelli eigenvalue problem - MathematicsThe Capelli eigenvalue problem Siddhartha Sahi Rutgers University, New Brunswick NJ May 29, 2018, Kostant Conference Siddhartha Sahi (Kostant

Okounkov polynomials and Quadratic Capelli operators

[Okounkov 1998, Transf Groups 3] Let δ = (n− 1, . . . , 1, 0) andρτ,α = τδ+ α1; there is a unique even symmetric polynomial Pτ,α

λ (x)of degree 2 |λ| such that Pτ,α

λ

(µ+ ρτ,α

)= δλµ for all |µ| ≤ |λ| .

D = R,C,H, N = Hermn (D) ,Z = Gn/Kn = GLn (D) /Un (D)Generalized Capelli operator Dµ ∈ PD (N)G = PD (Z )G

Lifts to Kn-invariant operator D̃µ on Grassmannian Y = Grn (Dm)via the double fibration Y ←− X −→ Z , where X = Mat∗n×m (D)Assume n ≤ m/2 then R (Y ) ≈ ⊕λ∈ΛnWλ as Km-module.

Theorem ([S.-Salmasian 2018, Ann ENS, to appear])

D̃µ acts on Wλ by the scalar Pτ,αλ

(µ̃+ ρτ,α

), where µ̃ = (−µn, . . . ,−µ1) ,

τ = d/2 and α = (1− nτ) /2, with d = dimR (D) = 1, 2, 4.

Generalizes [S. 2013, Rep Theory 17] and [S.-Zhang 2017, IMRN]

Siddhartha Sahi (Kostant Conference, MIT) May 29, 2018 10 / 12

Page 61: The Capelli eigenvalue problem - MathematicsThe Capelli eigenvalue problem Siddhartha Sahi Rutgers University, New Brunswick NJ May 29, 2018, Kostant Conference Siddhartha Sahi (Kostant

Okounkov polynomials and Quadratic Capelli operators

[Okounkov 1998, Transf Groups 3] Let δ = (n− 1, . . . , 1, 0) andρτ,α = τδ+ α1; there is a unique even symmetric polynomial Pτ,α

λ (x)of degree 2 |λ| such that Pτ,α

λ

(µ+ ρτ,α

)= δλµ for all |µ| ≤ |λ| .

D = R,C,H, N = Hermn (D) ,Z = Gn/Kn = GLn (D) /Un (D)

Generalized Capelli operator Dµ ∈ PD (N)G = PD (Z )G

Lifts to Kn-invariant operator D̃µ on Grassmannian Y = Grn (Dm)via the double fibration Y ←− X −→ Z , where X = Mat∗n×m (D)Assume n ≤ m/2 then R (Y ) ≈ ⊕λ∈ΛnWλ as Km-module.

Theorem ([S.-Salmasian 2018, Ann ENS, to appear])

D̃µ acts on Wλ by the scalar Pτ,αλ

(µ̃+ ρτ,α

), where µ̃ = (−µn, . . . ,−µ1) ,

τ = d/2 and α = (1− nτ) /2, with d = dimR (D) = 1, 2, 4.

Generalizes [S. 2013, Rep Theory 17] and [S.-Zhang 2017, IMRN]

Siddhartha Sahi (Kostant Conference, MIT) May 29, 2018 10 / 12

Page 62: The Capelli eigenvalue problem - MathematicsThe Capelli eigenvalue problem Siddhartha Sahi Rutgers University, New Brunswick NJ May 29, 2018, Kostant Conference Siddhartha Sahi (Kostant

Okounkov polynomials and Quadratic Capelli operators

[Okounkov 1998, Transf Groups 3] Let δ = (n− 1, . . . , 1, 0) andρτ,α = τδ+ α1; there is a unique even symmetric polynomial Pτ,α

λ (x)of degree 2 |λ| such that Pτ,α

λ

(µ+ ρτ,α

)= δλµ for all |µ| ≤ |λ| .

D = R,C,H, N = Hermn (D) ,Z = Gn/Kn = GLn (D) /Un (D)Generalized Capelli operator Dµ ∈ PD (N)G = PD (Z )G

Lifts to Kn-invariant operator D̃µ on Grassmannian Y = Grn (Dm)via the double fibration Y ←− X −→ Z , where X = Mat∗n×m (D)Assume n ≤ m/2 then R (Y ) ≈ ⊕λ∈ΛnWλ as Km-module.

Theorem ([S.-Salmasian 2018, Ann ENS, to appear])

D̃µ acts on Wλ by the scalar Pτ,αλ

(µ̃+ ρτ,α

), where µ̃ = (−µn, . . . ,−µ1) ,

τ = d/2 and α = (1− nτ) /2, with d = dimR (D) = 1, 2, 4.

Generalizes [S. 2013, Rep Theory 17] and [S.-Zhang 2017, IMRN]

Siddhartha Sahi (Kostant Conference, MIT) May 29, 2018 10 / 12

Page 63: The Capelli eigenvalue problem - MathematicsThe Capelli eigenvalue problem Siddhartha Sahi Rutgers University, New Brunswick NJ May 29, 2018, Kostant Conference Siddhartha Sahi (Kostant

Okounkov polynomials and Quadratic Capelli operators

[Okounkov 1998, Transf Groups 3] Let δ = (n− 1, . . . , 1, 0) andρτ,α = τδ+ α1; there is a unique even symmetric polynomial Pτ,α

λ (x)of degree 2 |λ| such that Pτ,α

λ

(µ+ ρτ,α

)= δλµ for all |µ| ≤ |λ| .

D = R,C,H, N = Hermn (D) ,Z = Gn/Kn = GLn (D) /Un (D)Generalized Capelli operator Dµ ∈ PD (N)G = PD (Z )G

Lifts to Kn-invariant operator D̃µ on Grassmannian Y = Grn (Dm)via the double fibration Y ←− X −→ Z , where X = Mat∗n×m (D)

Assume n ≤ m/2 then R (Y ) ≈ ⊕λ∈ΛnWλ as Km-module.

Theorem ([S.-Salmasian 2018, Ann ENS, to appear])

D̃µ acts on Wλ by the scalar Pτ,αλ

(µ̃+ ρτ,α

), where µ̃ = (−µn, . . . ,−µ1) ,

τ = d/2 and α = (1− nτ) /2, with d = dimR (D) = 1, 2, 4.

Generalizes [S. 2013, Rep Theory 17] and [S.-Zhang 2017, IMRN]

Siddhartha Sahi (Kostant Conference, MIT) May 29, 2018 10 / 12

Page 64: The Capelli eigenvalue problem - MathematicsThe Capelli eigenvalue problem Siddhartha Sahi Rutgers University, New Brunswick NJ May 29, 2018, Kostant Conference Siddhartha Sahi (Kostant

Okounkov polynomials and Quadratic Capelli operators

[Okounkov 1998, Transf Groups 3] Let δ = (n− 1, . . . , 1, 0) andρτ,α = τδ+ α1; there is a unique even symmetric polynomial Pτ,α

λ (x)of degree 2 |λ| such that Pτ,α

λ

(µ+ ρτ,α

)= δλµ for all |µ| ≤ |λ| .

D = R,C,H, N = Hermn (D) ,Z = Gn/Kn = GLn (D) /Un (D)Generalized Capelli operator Dµ ∈ PD (N)G = PD (Z )G

Lifts to Kn-invariant operator D̃µ on Grassmannian Y = Grn (Dm)via the double fibration Y ←− X −→ Z , where X = Mat∗n×m (D)Assume n ≤ m/2 then R (Y ) ≈ ⊕λ∈ΛnWλ as Km-module.

Theorem ([S.-Salmasian 2018, Ann ENS, to appear])

D̃µ acts on Wλ by the scalar Pτ,αλ

(µ̃+ ρτ,α

), where µ̃ = (−µn, . . . ,−µ1) ,

τ = d/2 and α = (1− nτ) /2, with d = dimR (D) = 1, 2, 4.

Generalizes [S. 2013, Rep Theory 17] and [S.-Zhang 2017, IMRN]

Siddhartha Sahi (Kostant Conference, MIT) May 29, 2018 10 / 12

Page 65: The Capelli eigenvalue problem - MathematicsThe Capelli eigenvalue problem Siddhartha Sahi Rutgers University, New Brunswick NJ May 29, 2018, Kostant Conference Siddhartha Sahi (Kostant

Okounkov polynomials and Quadratic Capelli operators

[Okounkov 1998, Transf Groups 3] Let δ = (n− 1, . . . , 1, 0) andρτ,α = τδ+ α1; there is a unique even symmetric polynomial Pτ,α

λ (x)of degree 2 |λ| such that Pτ,α

λ

(µ+ ρτ,α

)= δλµ for all |µ| ≤ |λ| .

D = R,C,H, N = Hermn (D) ,Z = Gn/Kn = GLn (D) /Un (D)Generalized Capelli operator Dµ ∈ PD (N)G = PD (Z )G

Lifts to Kn-invariant operator D̃µ on Grassmannian Y = Grn (Dm)via the double fibration Y ←− X −→ Z , where X = Mat∗n×m (D)Assume n ≤ m/2 then R (Y ) ≈ ⊕λ∈ΛnWλ as Km-module.

Theorem ([S.-Salmasian 2018, Ann ENS, to appear])

D̃µ acts on Wλ by the scalar Pτ,αλ

(µ̃+ ρτ,α

), where µ̃ = (−µn, . . . ,−µ1) ,

τ = d/2 and α = (1− nτ) /2, with d = dimR (D) = 1, 2, 4.

Generalizes [S. 2013, Rep Theory 17] and [S.-Zhang 2017, IMRN]

Siddhartha Sahi (Kostant Conference, MIT) May 29, 2018 10 / 12

Page 66: The Capelli eigenvalue problem - MathematicsThe Capelli eigenvalue problem Siddhartha Sahi Rutgers University, New Brunswick NJ May 29, 2018, Kostant Conference Siddhartha Sahi (Kostant

Okounkov polynomials and Quadratic Capelli operators

[Okounkov 1998, Transf Groups 3] Let δ = (n− 1, . . . , 1, 0) andρτ,α = τδ+ α1; there is a unique even symmetric polynomial Pτ,α

λ (x)of degree 2 |λ| such that Pτ,α

λ

(µ+ ρτ,α

)= δλµ for all |µ| ≤ |λ| .

D = R,C,H, N = Hermn (D) ,Z = Gn/Kn = GLn (D) /Un (D)Generalized Capelli operator Dµ ∈ PD (N)G = PD (Z )G

Lifts to Kn-invariant operator D̃µ on Grassmannian Y = Grn (Dm)via the double fibration Y ←− X −→ Z , where X = Mat∗n×m (D)Assume n ≤ m/2 then R (Y ) ≈ ⊕λ∈ΛnWλ as Km-module.

Theorem ([S.-Salmasian 2018, Ann ENS, to appear])

D̃µ acts on Wλ by the scalar Pτ,αλ

(µ̃+ ρτ,α

), where µ̃ = (−µn, . . . ,−µ1) ,

τ = d/2 and α = (1− nτ) /2, with d = dimR (D) = 1, 2, 4.

Generalizes [S. 2013, Rep Theory 17] and [S.-Zhang 2017, IMRN]

Siddhartha Sahi (Kostant Conference, MIT) May 29, 2018 10 / 12

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Shimura operators and Okounkov polynomials

G/K = Hermitian symmetric space, a = Cartan subspace

Σ = Σ (g, a) type BCn; ρ (Σ) = ρτ,α = τδ+ α1, some τ and α.

η : D (G/K )→ S (a)W0 ≈ even symmetric polynomials.

g = k+ p+ + p−, p+ ≈ (p−)∗ , U (p+) ≈ S (p+) ≈ ⊕λ∈ΛnVλ

1λ ∈ End (Vλ) ≈ Vλ ⊗ V ∗λ ⊂ U (p+)⊗U (p−)mult7→ U (g)K

Defines Lλ ∈ D (G/K ) [Shimura 1990, Ann Math 132]

Theorem ([S.- Zhang 2018, Math Res Lett, to appear])

We have η (Lλ) = Pτ,αλ with τ and α as above.

Siddhartha Sahi (Kostant Conference, MIT) May 29, 2018 11 / 12

Page 68: The Capelli eigenvalue problem - MathematicsThe Capelli eigenvalue problem Siddhartha Sahi Rutgers University, New Brunswick NJ May 29, 2018, Kostant Conference Siddhartha Sahi (Kostant

Shimura operators and Okounkov polynomials

G/K = Hermitian symmetric space, a = Cartan subspace

Σ = Σ (g, a) type BCn; ρ (Σ) = ρτ,α = τδ+ α1, some τ and α.

η : D (G/K )→ S (a)W0 ≈ even symmetric polynomials.

g = k+ p+ + p−, p+ ≈ (p−)∗ , U (p+) ≈ S (p+) ≈ ⊕λ∈ΛnVλ

1λ ∈ End (Vλ) ≈ Vλ ⊗ V ∗λ ⊂ U (p+)⊗U (p−)mult7→ U (g)K

Defines Lλ ∈ D (G/K ) [Shimura 1990, Ann Math 132]

Theorem ([S.- Zhang 2018, Math Res Lett, to appear])

We have η (Lλ) = Pτ,αλ with τ and α as above.

Siddhartha Sahi (Kostant Conference, MIT) May 29, 2018 11 / 12

Page 69: The Capelli eigenvalue problem - MathematicsThe Capelli eigenvalue problem Siddhartha Sahi Rutgers University, New Brunswick NJ May 29, 2018, Kostant Conference Siddhartha Sahi (Kostant

Shimura operators and Okounkov polynomials

G/K = Hermitian symmetric space, a = Cartan subspace

Σ = Σ (g, a) type BCn; ρ (Σ) = ρτ,α = τδ+ α1, some τ and α.

η : D (G/K )→ S (a)W0 ≈ even symmetric polynomials.

g = k+ p+ + p−, p+ ≈ (p−)∗ , U (p+) ≈ S (p+) ≈ ⊕λ∈ΛnVλ

1λ ∈ End (Vλ) ≈ Vλ ⊗ V ∗λ ⊂ U (p+)⊗U (p−)mult7→ U (g)K

Defines Lλ ∈ D (G/K ) [Shimura 1990, Ann Math 132]

Theorem ([S.- Zhang 2018, Math Res Lett, to appear])

We have η (Lλ) = Pτ,αλ with τ and α as above.

Siddhartha Sahi (Kostant Conference, MIT) May 29, 2018 11 / 12

Page 70: The Capelli eigenvalue problem - MathematicsThe Capelli eigenvalue problem Siddhartha Sahi Rutgers University, New Brunswick NJ May 29, 2018, Kostant Conference Siddhartha Sahi (Kostant

Shimura operators and Okounkov polynomials

G/K = Hermitian symmetric space, a = Cartan subspace

Σ = Σ (g, a) type BCn; ρ (Σ) = ρτ,α = τδ+ α1, some τ and α.

η : D (G/K )→ S (a)W0 ≈ even symmetric polynomials.

g = k+ p+ + p−, p+ ≈ (p−)∗ , U (p+) ≈ S (p+) ≈ ⊕λ∈ΛnVλ

1λ ∈ End (Vλ) ≈ Vλ ⊗ V ∗λ ⊂ U (p+)⊗U (p−)mult7→ U (g)K

Defines Lλ ∈ D (G/K ) [Shimura 1990, Ann Math 132]

Theorem ([S.- Zhang 2018, Math Res Lett, to appear])

We have η (Lλ) = Pτ,αλ with τ and α as above.

Siddhartha Sahi (Kostant Conference, MIT) May 29, 2018 11 / 12

Page 71: The Capelli eigenvalue problem - MathematicsThe Capelli eigenvalue problem Siddhartha Sahi Rutgers University, New Brunswick NJ May 29, 2018, Kostant Conference Siddhartha Sahi (Kostant

Shimura operators and Okounkov polynomials

G/K = Hermitian symmetric space, a = Cartan subspace

Σ = Σ (g, a) type BCn; ρ (Σ) = ρτ,α = τδ+ α1, some τ and α.

η : D (G/K )→ S (a)W0 ≈ even symmetric polynomials.

g = k+ p+ + p−, p+ ≈ (p−)∗ , U (p+) ≈ S (p+) ≈ ⊕λ∈ΛnVλ

1λ ∈ End (Vλ) ≈ Vλ ⊗ V ∗λ ⊂ U (p+)⊗U (p−)mult7→ U (g)K

Defines Lλ ∈ D (G/K ) [Shimura 1990, Ann Math 132]

Theorem ([S.- Zhang 2018, Math Res Lett, to appear])

We have η (Lλ) = Pτ,αλ with τ and α as above.

Siddhartha Sahi (Kostant Conference, MIT) May 29, 2018 11 / 12

Page 72: The Capelli eigenvalue problem - MathematicsThe Capelli eigenvalue problem Siddhartha Sahi Rutgers University, New Brunswick NJ May 29, 2018, Kostant Conference Siddhartha Sahi (Kostant

Shimura operators and Okounkov polynomials

G/K = Hermitian symmetric space, a = Cartan subspace

Σ = Σ (g, a) type BCn; ρ (Σ) = ρτ,α = τδ+ α1, some τ and α.

η : D (G/K )→ S (a)W0 ≈ even symmetric polynomials.

g = k+ p+ + p−, p+ ≈ (p−)∗ , U (p+) ≈ S (p+) ≈ ⊕λ∈ΛnVλ

1λ ∈ End (Vλ) ≈ Vλ ⊗ V ∗λ ⊂ U (p+)⊗U (p−)mult7→ U (g)K

Defines Lλ ∈ D (G/K ) [Shimura 1990, Ann Math 132]

Theorem ([S.- Zhang 2018, Math Res Lett, to appear])

We have η (Lλ) = Pτ,αλ with τ and α as above.

Siddhartha Sahi (Kostant Conference, MIT) May 29, 2018 11 / 12

Page 73: The Capelli eigenvalue problem - MathematicsThe Capelli eigenvalue problem Siddhartha Sahi Rutgers University, New Brunswick NJ May 29, 2018, Kostant Conference Siddhartha Sahi (Kostant

Shimura operators and Okounkov polynomials

G/K = Hermitian symmetric space, a = Cartan subspace

Σ = Σ (g, a) type BCn; ρ (Σ) = ρτ,α = τδ+ α1, some τ and α.

η : D (G/K )→ S (a)W0 ≈ even symmetric polynomials.

g = k+ p+ + p−, p+ ≈ (p−)∗ , U (p+) ≈ S (p+) ≈ ⊕λ∈ΛnVλ

1λ ∈ End (Vλ) ≈ Vλ ⊗ V ∗λ ⊂ U (p+)⊗U (p−)mult7→ U (g)K

Defines Lλ ∈ D (G/K ) [Shimura 1990, Ann Math 132]

Theorem ([S.- Zhang 2018, Math Res Lett, to appear])

We have η (Lλ) = Pτ,αλ with τ and α as above.

Siddhartha Sahi (Kostant Conference, MIT) May 29, 2018 11 / 12

Page 74: The Capelli eigenvalue problem - MathematicsThe Capelli eigenvalue problem Siddhartha Sahi Rutgers University, New Brunswick NJ May 29, 2018, Kostant Conference Siddhartha Sahi (Kostant

Lie superalgebras

Consider symmetric pairs (g, k) ∼ Jordan superalgebra N

Type Q: (q (n)× q (n) , q (n)) and (gl (n|n) , p (n)), Type A: restIn Type A can associate a deformed root system of type gl (m|n) withparamter θ à la [Sergeev-Veselov 2004, Comm Math Phys 245]S (N) is comp. reducible mult-free g-module unless θ is singular.Summands: Type A ∼ (m, n)-hook partitions, Type Q ∼ strictOne can define Capelli operators and ask for eigenvalues.

Theorem ([S.-Salmasian-Serganova 2018])

1 Type Q: Eigenvalues ∼ Okounkov-Ivanov shifted Q-functions2 Type A: ∼ Sergeev-Veselov shifted super Jack polynomials

The two kinds of polynomials are defined in [Ivanov 2001, J. MathSci 107] and [Sergeev-Veselov 2005 Adv Math 192]Earlier: (gl (m|n)× gl (m|n) , gl (m|n)) and (gosp (m|2n) , gl (m|n))[S.-Salmasian 2016 Adv Math 303], (q (n)× q (n) , q (n))[Alldridge-S.-Salmasian 2018, Contemp Math]

Siddhartha Sahi (Kostant Conference, MIT) May 29, 2018 12 / 12

Page 75: The Capelli eigenvalue problem - MathematicsThe Capelli eigenvalue problem Siddhartha Sahi Rutgers University, New Brunswick NJ May 29, 2018, Kostant Conference Siddhartha Sahi (Kostant

Lie superalgebras

Consider symmetric pairs (g, k) ∼ Jordan superalgebra NType Q: (q (n)× q (n) , q (n)) and (gl (n|n) , p (n)), Type A: rest

In Type A can associate a deformed root system of type gl (m|n) withparamter θ à la [Sergeev-Veselov 2004, Comm Math Phys 245]S (N) is comp. reducible mult-free g-module unless θ is singular.Summands: Type A ∼ (m, n)-hook partitions, Type Q ∼ strictOne can define Capelli operators and ask for eigenvalues.

Theorem ([S.-Salmasian-Serganova 2018])

1 Type Q: Eigenvalues ∼ Okounkov-Ivanov shifted Q-functions2 Type A: ∼ Sergeev-Veselov shifted super Jack polynomials

The two kinds of polynomials are defined in [Ivanov 2001, J. MathSci 107] and [Sergeev-Veselov 2005 Adv Math 192]Earlier: (gl (m|n)× gl (m|n) , gl (m|n)) and (gosp (m|2n) , gl (m|n))[S.-Salmasian 2016 Adv Math 303], (q (n)× q (n) , q (n))[Alldridge-S.-Salmasian 2018, Contemp Math]

Siddhartha Sahi (Kostant Conference, MIT) May 29, 2018 12 / 12

Page 76: The Capelli eigenvalue problem - MathematicsThe Capelli eigenvalue problem Siddhartha Sahi Rutgers University, New Brunswick NJ May 29, 2018, Kostant Conference Siddhartha Sahi (Kostant

Lie superalgebras

Consider symmetric pairs (g, k) ∼ Jordan superalgebra NType Q: (q (n)× q (n) , q (n)) and (gl (n|n) , p (n)), Type A: restIn Type A can associate a deformed root system of type gl (m|n) withparamter θ à la [Sergeev-Veselov 2004, Comm Math Phys 245]

S (N) is comp. reducible mult-free g-module unless θ is singular.Summands: Type A ∼ (m, n)-hook partitions, Type Q ∼ strictOne can define Capelli operators and ask for eigenvalues.

Theorem ([S.-Salmasian-Serganova 2018])

1 Type Q: Eigenvalues ∼ Okounkov-Ivanov shifted Q-functions2 Type A: ∼ Sergeev-Veselov shifted super Jack polynomials

The two kinds of polynomials are defined in [Ivanov 2001, J. MathSci 107] and [Sergeev-Veselov 2005 Adv Math 192]Earlier: (gl (m|n)× gl (m|n) , gl (m|n)) and (gosp (m|2n) , gl (m|n))[S.-Salmasian 2016 Adv Math 303], (q (n)× q (n) , q (n))[Alldridge-S.-Salmasian 2018, Contemp Math]

Siddhartha Sahi (Kostant Conference, MIT) May 29, 2018 12 / 12

Page 77: The Capelli eigenvalue problem - MathematicsThe Capelli eigenvalue problem Siddhartha Sahi Rutgers University, New Brunswick NJ May 29, 2018, Kostant Conference Siddhartha Sahi (Kostant

Lie superalgebras

Consider symmetric pairs (g, k) ∼ Jordan superalgebra NType Q: (q (n)× q (n) , q (n)) and (gl (n|n) , p (n)), Type A: restIn Type A can associate a deformed root system of type gl (m|n) withparamter θ à la [Sergeev-Veselov 2004, Comm Math Phys 245]S (N) is comp. reducible mult-free g-module unless θ is singular.

Summands: Type A ∼ (m, n)-hook partitions, Type Q ∼ strictOne can define Capelli operators and ask for eigenvalues.

Theorem ([S.-Salmasian-Serganova 2018])

1 Type Q: Eigenvalues ∼ Okounkov-Ivanov shifted Q-functions2 Type A: ∼ Sergeev-Veselov shifted super Jack polynomials

The two kinds of polynomials are defined in [Ivanov 2001, J. MathSci 107] and [Sergeev-Veselov 2005 Adv Math 192]Earlier: (gl (m|n)× gl (m|n) , gl (m|n)) and (gosp (m|2n) , gl (m|n))[S.-Salmasian 2016 Adv Math 303], (q (n)× q (n) , q (n))[Alldridge-S.-Salmasian 2018, Contemp Math]

Siddhartha Sahi (Kostant Conference, MIT) May 29, 2018 12 / 12

Page 78: The Capelli eigenvalue problem - MathematicsThe Capelli eigenvalue problem Siddhartha Sahi Rutgers University, New Brunswick NJ May 29, 2018, Kostant Conference Siddhartha Sahi (Kostant

Lie superalgebras

Consider symmetric pairs (g, k) ∼ Jordan superalgebra NType Q: (q (n)× q (n) , q (n)) and (gl (n|n) , p (n)), Type A: restIn Type A can associate a deformed root system of type gl (m|n) withparamter θ à la [Sergeev-Veselov 2004, Comm Math Phys 245]S (N) is comp. reducible mult-free g-module unless θ is singular.Summands: Type A ∼ (m, n)-hook partitions, Type Q ∼ strict

One can define Capelli operators and ask for eigenvalues.

Theorem ([S.-Salmasian-Serganova 2018])

1 Type Q: Eigenvalues ∼ Okounkov-Ivanov shifted Q-functions2 Type A: ∼ Sergeev-Veselov shifted super Jack polynomials

The two kinds of polynomials are defined in [Ivanov 2001, J. MathSci 107] and [Sergeev-Veselov 2005 Adv Math 192]Earlier: (gl (m|n)× gl (m|n) , gl (m|n)) and (gosp (m|2n) , gl (m|n))[S.-Salmasian 2016 Adv Math 303], (q (n)× q (n) , q (n))[Alldridge-S.-Salmasian 2018, Contemp Math]

Siddhartha Sahi (Kostant Conference, MIT) May 29, 2018 12 / 12

Page 79: The Capelli eigenvalue problem - MathematicsThe Capelli eigenvalue problem Siddhartha Sahi Rutgers University, New Brunswick NJ May 29, 2018, Kostant Conference Siddhartha Sahi (Kostant

Lie superalgebras

Consider symmetric pairs (g, k) ∼ Jordan superalgebra NType Q: (q (n)× q (n) , q (n)) and (gl (n|n) , p (n)), Type A: restIn Type A can associate a deformed root system of type gl (m|n) withparamter θ à la [Sergeev-Veselov 2004, Comm Math Phys 245]S (N) is comp. reducible mult-free g-module unless θ is singular.Summands: Type A ∼ (m, n)-hook partitions, Type Q ∼ strictOne can define Capelli operators and ask for eigenvalues.

Theorem ([S.-Salmasian-Serganova 2018])

1 Type Q: Eigenvalues ∼ Okounkov-Ivanov shifted Q-functions2 Type A: ∼ Sergeev-Veselov shifted super Jack polynomials

The two kinds of polynomials are defined in [Ivanov 2001, J. MathSci 107] and [Sergeev-Veselov 2005 Adv Math 192]Earlier: (gl (m|n)× gl (m|n) , gl (m|n)) and (gosp (m|2n) , gl (m|n))[S.-Salmasian 2016 Adv Math 303], (q (n)× q (n) , q (n))[Alldridge-S.-Salmasian 2018, Contemp Math]

Siddhartha Sahi (Kostant Conference, MIT) May 29, 2018 12 / 12

Page 80: The Capelli eigenvalue problem - MathematicsThe Capelli eigenvalue problem Siddhartha Sahi Rutgers University, New Brunswick NJ May 29, 2018, Kostant Conference Siddhartha Sahi (Kostant

Lie superalgebras

Consider symmetric pairs (g, k) ∼ Jordan superalgebra NType Q: (q (n)× q (n) , q (n)) and (gl (n|n) , p (n)), Type A: restIn Type A can associate a deformed root system of type gl (m|n) withparamter θ à la [Sergeev-Veselov 2004, Comm Math Phys 245]S (N) is comp. reducible mult-free g-module unless θ is singular.Summands: Type A ∼ (m, n)-hook partitions, Type Q ∼ strictOne can define Capelli operators and ask for eigenvalues.

Theorem ([S.-Salmasian-Serganova 2018])

1 Type Q: Eigenvalues ∼ Okounkov-Ivanov shifted Q-functions2 Type A: ∼ Sergeev-Veselov shifted super Jack polynomials

The two kinds of polynomials are defined in [Ivanov 2001, J. MathSci 107] and [Sergeev-Veselov 2005 Adv Math 192]Earlier: (gl (m|n)× gl (m|n) , gl (m|n)) and (gosp (m|2n) , gl (m|n))[S.-Salmasian 2016 Adv Math 303], (q (n)× q (n) , q (n))[Alldridge-S.-Salmasian 2018, Contemp Math]

Siddhartha Sahi (Kostant Conference, MIT) May 29, 2018 12 / 12

Page 81: The Capelli eigenvalue problem - MathematicsThe Capelli eigenvalue problem Siddhartha Sahi Rutgers University, New Brunswick NJ May 29, 2018, Kostant Conference Siddhartha Sahi (Kostant

Lie superalgebras

Consider symmetric pairs (g, k) ∼ Jordan superalgebra NType Q: (q (n)× q (n) , q (n)) and (gl (n|n) , p (n)), Type A: restIn Type A can associate a deformed root system of type gl (m|n) withparamter θ à la [Sergeev-Veselov 2004, Comm Math Phys 245]S (N) is comp. reducible mult-free g-module unless θ is singular.Summands: Type A ∼ (m, n)-hook partitions, Type Q ∼ strictOne can define Capelli operators and ask for eigenvalues.

Theorem ([S.-Salmasian-Serganova 2018])

1 Type Q: Eigenvalues ∼ Okounkov-Ivanov shifted Q-functions

2 Type A: ∼ Sergeev-Veselov shifted super Jack polynomials

The two kinds of polynomials are defined in [Ivanov 2001, J. MathSci 107] and [Sergeev-Veselov 2005 Adv Math 192]Earlier: (gl (m|n)× gl (m|n) , gl (m|n)) and (gosp (m|2n) , gl (m|n))[S.-Salmasian 2016 Adv Math 303], (q (n)× q (n) , q (n))[Alldridge-S.-Salmasian 2018, Contemp Math]

Siddhartha Sahi (Kostant Conference, MIT) May 29, 2018 12 / 12

Page 82: The Capelli eigenvalue problem - MathematicsThe Capelli eigenvalue problem Siddhartha Sahi Rutgers University, New Brunswick NJ May 29, 2018, Kostant Conference Siddhartha Sahi (Kostant

Lie superalgebras

Consider symmetric pairs (g, k) ∼ Jordan superalgebra NType Q: (q (n)× q (n) , q (n)) and (gl (n|n) , p (n)), Type A: restIn Type A can associate a deformed root system of type gl (m|n) withparamter θ à la [Sergeev-Veselov 2004, Comm Math Phys 245]S (N) is comp. reducible mult-free g-module unless θ is singular.Summands: Type A ∼ (m, n)-hook partitions, Type Q ∼ strictOne can define Capelli operators and ask for eigenvalues.

Theorem ([S.-Salmasian-Serganova 2018])

1 Type Q: Eigenvalues ∼ Okounkov-Ivanov shifted Q-functions2 Type A: ∼ Sergeev-Veselov shifted super Jack polynomials

The two kinds of polynomials are defined in [Ivanov 2001, J. MathSci 107] and [Sergeev-Veselov 2005 Adv Math 192]Earlier: (gl (m|n)× gl (m|n) , gl (m|n)) and (gosp (m|2n) , gl (m|n))[S.-Salmasian 2016 Adv Math 303], (q (n)× q (n) , q (n))[Alldridge-S.-Salmasian 2018, Contemp Math]

Siddhartha Sahi (Kostant Conference, MIT) May 29, 2018 12 / 12

Page 83: The Capelli eigenvalue problem - MathematicsThe Capelli eigenvalue problem Siddhartha Sahi Rutgers University, New Brunswick NJ May 29, 2018, Kostant Conference Siddhartha Sahi (Kostant

Lie superalgebras

Consider symmetric pairs (g, k) ∼ Jordan superalgebra NType Q: (q (n)× q (n) , q (n)) and (gl (n|n) , p (n)), Type A: restIn Type A can associate a deformed root system of type gl (m|n) withparamter θ à la [Sergeev-Veselov 2004, Comm Math Phys 245]S (N) is comp. reducible mult-free g-module unless θ is singular.Summands: Type A ∼ (m, n)-hook partitions, Type Q ∼ strictOne can define Capelli operators and ask for eigenvalues.

Theorem ([S.-Salmasian-Serganova 2018])

1 Type Q: Eigenvalues ∼ Okounkov-Ivanov shifted Q-functions2 Type A: ∼ Sergeev-Veselov shifted super Jack polynomials

The two kinds of polynomials are defined in [Ivanov 2001, J. MathSci 107] and [Sergeev-Veselov 2005 Adv Math 192]

Earlier: (gl (m|n)× gl (m|n) , gl (m|n)) and (gosp (m|2n) , gl (m|n))[S.-Salmasian 2016 Adv Math 303], (q (n)× q (n) , q (n))[Alldridge-S.-Salmasian 2018, Contemp Math]

Siddhartha Sahi (Kostant Conference, MIT) May 29, 2018 12 / 12

Page 84: The Capelli eigenvalue problem - MathematicsThe Capelli eigenvalue problem Siddhartha Sahi Rutgers University, New Brunswick NJ May 29, 2018, Kostant Conference Siddhartha Sahi (Kostant

Lie superalgebras

Consider symmetric pairs (g, k) ∼ Jordan superalgebra NType Q: (q (n)× q (n) , q (n)) and (gl (n|n) , p (n)), Type A: restIn Type A can associate a deformed root system of type gl (m|n) withparamter θ à la [Sergeev-Veselov 2004, Comm Math Phys 245]S (N) is comp. reducible mult-free g-module unless θ is singular.Summands: Type A ∼ (m, n)-hook partitions, Type Q ∼ strictOne can define Capelli operators and ask for eigenvalues.

Theorem ([S.-Salmasian-Serganova 2018])

1 Type Q: Eigenvalues ∼ Okounkov-Ivanov shifted Q-functions2 Type A: ∼ Sergeev-Veselov shifted super Jack polynomials

The two kinds of polynomials are defined in [Ivanov 2001, J. MathSci 107] and [Sergeev-Veselov 2005 Adv Math 192]Earlier: (gl (m|n)× gl (m|n) , gl (m|n)) and (gosp (m|2n) , gl (m|n))[S.-Salmasian 2016 Adv Math 303], (q (n)× q (n) , q (n))[Alldridge-S.-Salmasian 2018, Contemp Math]

Siddhartha Sahi (Kostant Conference, MIT) May 29, 2018 12 / 12