β-Gaussian Ensembles and the Non-orientability of
Polygonal Glueings
Michael La Croix
Massachusetts Institute of Technology
April 6, 2013
Gaussian Ensembles
For β ∈ {1, 2, 4} an element of the β-Gaussian ensemble is constructed as
A = G + G∗
where G is n × n with i.i.d. Guassian entries selected from {R, C, H}.
Motivating Question
What is the value of E(f (A)), when f is a symmetric function of the
eigenvalues of its argument?
Example

 5n + 5n2 + 2n3 β = 1
4
n + 2n3
β=2
E(tr(A )) =
 5
5 2
3 3
β=4
4n − 2n + 4n
Michael La Croix (MIT)
β-Polygonal Glueings
April 6, 2013
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General β via Eigenvalue Density
The eigenvalues of A are all real with joint density proportional to
!
n
2
Y
X
β
λ
i
|λi − λj |β exp −
2
2
i=1
1≤i<j≤n
Theorem
For every θ, E(pθ (λ))β is a polynomial in the variables n and b =
2
β
− 1.
Example
E(p4 (λ))β = (1 + b + 3b2 )n + 5bn2 + 2n3
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A Recurrence behind the theorem
1
− 2(1+b)
p2 (x)
Set Ω := e
Integrate
2
|V (x)| 1+b , so that hf i = E(f (x)) = cb,n
R
Rn
f Ω dx.
p (x)
2
∂ j+1
∂ j+1
− 2
x1 pθ (x)Ω =
x1 pθ (x)|V (x)| 1+b e 2(1+b)
∂x1
∂x1
= (j + 1)xj1 pθ (x)Ω +
X
imi (θ)xi+j
1 pθ\i (x)Ω +
2
1+b
N j+1
X
x
p
θ (x)
x1 −xi Ω
1
−
j+2
1
1+b x1 pθ (x)Ω
i=2
i∈θ
to get
An Algebraic recurrence
Example
hpj+2 pθ i = b(j + 1) hpj pθ i + α
X
j
X
imi (θ) pi+j pθ\i +
hpl pj−l pθ i.
i∈θ
Michael La Croix (MIT)
β-Polygonal Glueings
l=0
April 6, 2013
3 / 15
What’s being counted?
Example (for β ∈ {1, 2, 4})
i
tr(A4 ) =
X
Aij Ajk Akl Ali
A ij
A li
j
A jk
i,j,k,l
l
A kl
k
n
n
E tr(A4 ) = 1!
E(A11 A11 A11 A11 ) + 2!
E(4A11 A11 A12 A21 )
1
2
n
n
+ 2!
E(A12 A21 A12 A21 ) + 3!
E(2A12 A21 A13 A31 )
2
3
n
n
+ 4!
E(A12 A23 A34 A41 ) + 3!
E(4A11 A12 A23 A31 )
4
3
n
+ 2!
E(2A11 A12 A22 A21 )
2
Michael La Croix (MIT)
β-Polygonal Glueings
April 6, 2013
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What’s being counted?
Example (for β ∈ {1, 2, 4})
i
tr(A4 ) =
X
Aij Ajk Akl Ali
A ij
A li
1
1
j
1
2
1
1
A jk
1
1
1
2
k
1
1
1
1
1
1
2
1
2
1
1
2
2
2
1
1
1
1
3
2
3
1
i,j,k,l
l
A kl
n
n
E tr(A4 ) = 1!
E(A11 A11 A11 A11 ) + 2!
E(4A11 A11 A12 A21 )
1
2
n
n
+ 2!
E(A12 A21 A12 A21 ) + 3!
E(2A12 A21 A13 A31 )
2
3
n
n
+ 4!
E(A12 A23 A34 A41 ) + 3!
E(4A11 A12 A23 A31 )
4
3
n
+ 2!
E(2A11 A12 A22 A21 )
2
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β-Polygonal Glueings
April 6, 2013
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Expectations as Sums
Since the entries of A are independent Guassians,
X Y
E(Ai1 j1 Ai2 j2 . . . Aik jk ) =
E(Au Av )
m (u,v)∈m
summed over perfect matchings of the multiset {i1 j1 , i2 j2 , . . . , ik jk }
X
#{pairings consistent with p}
p a painting
=
X
#{paintings consistent with m}
m a matching
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Count the polygon glueings in 2 different ways
(n)1
(n)1
(n)1
(n)1
(n)1
(n)1
(n)1
(n)1
(n)1
(n)1
(n)1
(n)1
0
0
(n)2
0
0
0
0
0
(n)2
0
0
0
2 n(n-1)
0
(n)2
0
0
0
0
0
(n)2
0
0
0
0
2 n(n-1)
0
0
(n)2
0
0
(n)2
0
0
0
0
0
0
2 n(n-1)
0
(n)2
0
0
(n)2
0
0
0
0
0
0
0
2 n(n-1)
0
(n)2
(n)2
0
0
0
0
0
0
(n)2
0
0
3 n(n-1)
0
0
(n)3
0
0
0
0
0
0
0
0
0
n(n-1)(n-2)
0
(n)3
0
0
0
0
0
0
0
0
0
0
n(n-1)(n-2)
n3
n3
n
n2
n2
n
n2
n2
n2
n
n
n
Michael La Croix (MIT)
β-Polygonal Glueings
12 n
April 6, 2013
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Polygon Glueings = Maps
Identifying the edges of a polygon creates a surface.
=
Its boundary is a graph embedded in the surface.
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β-Polygonal Glueings
April 6, 2013
7 / 15
Polygon Glueings = Maps
=
=
Michael La Croix (MIT)
β-Polygonal Glueings
April 6, 2013
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Polygon Glueings = Maps
Extra polygons give extra faces (and possibles extra components)
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Graphs, Surfaces, and Maps
Example
Definition
A surface is a compact 2-manifold
without boundary. (Non-orientable
surfaces are permitted.)
Definition
A graph is a finite set of vertices
together with a finite set of edges,
such that each edge is associated
with either one or two vertices.
(It may have loops / parallel edges.)
Definition
A map is a 2-cell embedding of a
graph in a surface.
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Equivalence of Maps
Two maps are equivalent if the embeddings are homeomorphic.
Homeomorphisms are more complicated than we might think
Dehn Twists
Y-Homeomorphisms
Not Present for Photo
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April 6, 2013
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Equivalence of Maps
Two maps are equivalent if the embeddings are homeomorphic.
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Ribbon Graphs, Flags, and Rooted Maps
Example
Definition
The neighbourhood of the graph
determines a ribbon graph, and the
boundaries of ribbons determine
flags.
Definition
Automorphisms permute flags, and
a rooted map is a map together
with a distinguished orbit of flags.
Michael La Croix (MIT)
β-Polygonal Glueings
April 6, 2013
10 / 15
Ribbon Graphs, Flags, and Rooted Maps
Example
Definition
The neighbourhood of the graph
determines a ribbon graph, and the
boundaries of ribbons determine
flags.
Definition
Automorphisms permute flags, and
a rooted map is a map together
with a distinguished orbit of flags.
Michael La Croix (MIT)
β-Polygonal Glueings
April 6, 2013
10 / 15
Ribbon Graphs, Flags, and Rooted Maps
Example
Definition
The neighbourhood of the graph
determines a ribbon graph, and the
boundaries of ribbons determine
flags.
Definition
Automorphisms permute flags, and
a rooted map is a map together
with a distinguished orbit of flags.
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β-Polygonal Glueings
April 6, 2013
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Twisted Ribbons allow Non-Orientable Maps
Example (A map on the Klein Bottle)
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Twisted Ribbons allow Non-Orientable Maps
Example (A map on the Torus)
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Root Edge Deletion
A rooted map with k edges can be thought of as a sequence of k maps.
α 2
2
2 2
2
2
bn
bn
bn
n
2
bn
n
n
bn
2
Consecutive submaps differ in genus by 0, 1, or 2, and these steps are
marked by 1, b, and a = α2 to assign a weight to a glueing.
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β-Polygonal Glueings
April 6, 2013
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Algebraic and Combinatorial Recurrences agree
An Algebraic Recurrence
Example
hpj+2 pθ i = b(j + 1) hpj pθ i + α
X
j
X
imi (θ) pi+j pθ\i +
hpl pj−l pθ i.
i∈θ
l=0
A Combinatorial Recurrence
It corresponds to a combinatorial recurrence for counting polygon glueings.
i
j-l
i+j
j
j+2
Michael La Croix (MIT)
j+2
β-Polygonal Glueings
j+2
l
April 6, 2013
13 / 15
3
bn
n
3
n
n
Michael La Croix (MIT)
2
bn
2
bn
bn
2
bn
2
bn
2
bn
bn
bn
β-Polygonal Glueings
2
2
2
April 6, 2013
14 / 15
The Future
The combinatorial interpretation has a two-parameter refinement.
Is there a corresponding matrix question?
At b = 0, we obtain glueings in 2g : 1 correspondence with orientable
glueings. Can this correspondence be made to preserve vertex degrees
as well as face degrees?
A similar recurrence describes moments of the β-Laguerre
distribution, with maps replaced by hypermaps .
For β ∈ {1, 2, 4} we can refine the combinatorial model and compute
moments of XA. Is there a model for the β-Ensembles where this
interpretation makes sense.
For β = 1 and β = 2, there is a natural duality between vertices and
faces. What operation replaces it for b-weighted glueings?
The connection with
Michael La Croix (MIT)
Jack symmetric functions
β-Polygonal Glueings
that needs to be explored.
April 6, 2013
15 / 15
The End
Thank You
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β-Polygonal Glueings
April 6, 2013
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Example
is enumerated by x32 x23 (y3 y4 y5 ) z26 .
ν = [23 , 32 ]
φ = [3, 4, 5]
= [26 ]
Return
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April 6, 2013
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Explicit Formulae
The hypermap series can be computed explicitly when H consists of
orientable hypermaps or all hypermaps.
Theorem (Jackson and Visentin - 1990)
When H is the set of orientable hypermaps,
HO
∂
p(x), p(y), p(z); 0 = t ln
∂t
X
θ∈P
!
t Hθ sθ (x)sθ (y)sθ (z) |θ|
t=1.
Theorem (Goulden and Jackson - 1996)
When H is the set of all hypermaps (orientable and non-orientable),
!
X
1
∂
t|θ|
Zθ (x)Zθ (y)Zθ (z) HA p(x), p(y), p(z); 1 = 2t ln
∂t
H2θ
θ∈P
Michael La Croix (MIT)
β-Polygonal Glueings
t=1.
April 6, 2013
17 / 15
A Generalized Series
Jack symmetric functions, Definition , are a one-parameter family, denoted
by {Jθ (α)}θ , that generalizes both Schur functions and zonal polynomials.
b -Conjecture (Goulden and Jackson - 1996)
The generalized series,
H p(x), p(y), p(z); b
X
∂
Jθ (x; 1 + b)Jθ (y; 1 + b)Jθ (z; 1 + b)
:= (1 + b)t ln
t|θ|
∂t
hJθ , Jθ i1+b
θ∈P
X X
=
cν,φ, (b)pν (x)pφ (y)p (z),
!
t=1
n≥0 ν,φ,`n
has an combinatorial
interpretation involving hypermaps. In particular
X
cν,φ, (b) =
b β(h) for some invariant β of rooted hypermaps.
h∈Hν,φ,
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For general β, integrate over eigenvalues
Definition
For a function f : Rn → R, define an expectation operator h·i by
Z
2
− 1 p (λ)
|V (λ)| 1+b f (λ)e 2(1+b) 2 dλ,
hf i1+b := c1+b
Rn
with c1+b chosen such that h1i1+b = 1.
Theorem (Okounkov - 1997)
If n is a positive integer, 1 + b is a positive real number, and θ is an
integer partition of 2n, then
E
D
(1+b)
(1+b)
(1+b)
(λ)
= Jθ
(In )[p[2n ] ]Jθ
.
Jθ
1+b
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Jack Symmetric Functions
With respect to the inner product defined by
hpλ (x), pµ (x)iα = δλ,µ
|λ|! `(λ)
α ,
|Cλ |
Jack symmetric functions are the unique family satisfying:
(P1) (Orthogonality) If λ 6= µ, then hJλ , Jµ iα = 0.
X
(P2) (Triangularity) Jλ =
vλµ (α)mµ , where vλµ (α) is a rational
µ4λ
function in α, and ‘4’ denotes the natural order on partitions.
(P3) (Normalization) If |λ| = n, then vλ,[1n ] (α) = n!.
Return
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Jack Symmetric Functions
Jack symmetric functions, are a one-parameter family, denoted by
{Jθ (α)}θ , that generalizes both Schur functions and zonal polynomials.
Proposition (Stanley - 1989)
Jack symmetric functions are related to Schur functions and zonal
polynomials by:
hJλ , Jλ i1 = Hλ2 ,
Jλ (1) = Hλ sλ ,
Jλ (2) = Zλ ,
and
hJλ , Jλ i2 = H2λ ,
where 2λ is the partition obtained from λ by multiplying each part by two.
Return
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hpj+2 pθ i = b(j + 1) hpj pθ i + (1 + b)
X
j
X
imi (θ) pi+j pθ\i +
hpl pj−l pθ i
i∈θ
l=0
Example
h1i = 1
hp0 i = n
hp2 i = b hp0 i + hp0 p0 i = bn + n2
hp1 p1 i = (1 + b) hp0 i = (1 + b)n
hp4 i = 3b hp2 i + hp0 p2 i + hp1 p1 i + hp2 p0 i = (1 + b + 3b2 )n + 5bn2 + 2n3
hp3 p1 i = 2b hp1 p1 i + (1 + b) hp2 i + hp0 p1 p1 i + hp1 p0 p1 i = (3b + 3b2 )n + (3 + 3b)n2
hp2 p2 i = b hp0 p2 i + 2(1 + b) hp2 i + hp0 p0 p2 i = 2b(1 + b)n + (2 + 2b + b2 )n2 + 2bn3 + n4
hp2 p1,1 i = b hp0 p1,1 i + 2(1 + b) hp1,1 i + hp0 p0 p1,1 i = 2(1 + b)2 n + (b + b2 )n2 + (1 + b)n3
hp1 p3 i = 3(1 + b) hp2 i = (3b + 3b2 )n + (3 + 3b)n2
hp1 p2,1 i = 2(1 + b) hp1,1 i + (1 + b) hp0 p2 i = (2 + 4b + 2b2 )n + (b + b2 )n2 + (1 + b)n3
hp1,1,1,1 i = 3(1 + b) hp0 p1,1 i = (1 + 2b + b2 )n2
Return
Michael La Croix (MIT)
β-Polygonal Glueings
April 6, 2013
22 / 15
hpj+2 pθ i = b(j + 1) hpj pθ i + (1 + b)
X
j
X
imi (θ) pi+j pθ\i +
hpl pj−l pθ i
i∈θ
l=0
Example
h1i = 1
hp0 i = n
hp2 i = b hp0 i + hp0 p0 i = bn + n2
hp1 p1 i = (1 + b) hp0 i = (1 + b)n
hp4 i = 3b hp2 i + hp0 p2 i + hp1 p1 i + hp2 p0 i = (1 + b + 3b2 )n + 5bn2 + 2n3
hp3 p1 i = 2b hp1 p1 i + (1 + b) hp2 i + hp0 p1 p1 i + hp1 p0 p1 i = (3b + 3b2 )n + (3 + 3b)n2
hp2 p2 i = b hp0 p2 i + 2(1 + b) hp2 i + hp0 p0 p2 i = 2b(1 + b)n + (2 + 2b + b2 )n2 + 2bn3 + n4
hp2 p1,1 i = b hp0 p1,1 i + 2(1 + b) hp1,1 i + hp0 p0 p1,1 i = 2(1 + b)2 n + (b + b2 )n2 + (1 + b)n3
hp1 p3 i = 3(1 + b) hp2 i = (3b + 3b2 )n + (3 + 3b)n2
hp1 p2,1 i = 2(1 + b) hp1,1 i + (1 + b) hp0 p2 i = (2 + 4b + 2b2 )n + (b + b2 )n2 + (1 + b)n3
hp1,1,1,1 i = 3(1 + b) hp0 p1,1 i = (1 + 2b + b2 )n2
Return
Michael La Croix (MIT)
β-Polygonal Glueings
April 6, 2013
23 / 15
b is ubiquitous
The many lives of b
b=0
Hypermaps
Symmetric Functions
Matrix Integrals
Moduli Spaces
Matching Systems
Michael La Croix (MIT)
Orientable
sθ
GUE
over C
Bipartite
b=1
?
Jθ (b)
?
?
?
β-Polygonal Glueings
Locally Orientable
Zθ
GOE
over R
All
April 6, 2013
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Encoding Orientable Maps
1
Orient and label the edges.
2
This induces labels on flags.
3
Clockwise circulations at each
vertex determine ν.
4
1
2
4
3
Face circulations are the cycles
of ν.
5
6
= (1 10 )(2 20 )(3 30 )(4 40 )(5 50 )(6 60 )
ν = (1 2 3)(10 4)(20 5)(30 50 6)(40 60 )
ν = φ = (1 4 60 30 )(10 2 5 6 40 )(20 3 50 )
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Encoding Orientable Maps
1
Orient and label the edges.
2
This induces labels on flags.
3
Clockwise circulations at each
vertex determine ν.
4
Face circulations are the cycles
of ν.
1
4
2
3
5
6
= (1 10 )(2 20 )(3 30 )(4 40 )(5 50 )(6 60 )
ν = (1 2 3)(10 4)(20 5)(30 50 6)(40 60 )
ν = φ = (1 4 60 30 )(10 2 5 6 40 )(20 3 50 )
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Encoding Orientable Maps
1’
1
Orient and label the edges.
2
This induces labels on flags.
3
Clockwise circulations at each
vertex determine ν.
4
4
Face circulations are the cycles
of ν.
2
1
2’
3
4’
6’
3’
5
5’
6
= (1 10 )(2 20 )(3 30 )(4 40 )(5 50 )(6 60 )
ν = (1 2 3)(10 4)(20 5)(30 50 6)(40 60 )
ν = φ = (1 4 60 30 )(10 2 5 6 40 )(20 3 50 )
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Encoding Orientable Maps
1’
1
Orient and label the edges.
2
This induces labels on flags.
3
Clockwise circulations at each
vertex determine ν.
4
4
Face circulations are the cycles
of ν.
2
1
2’
3
4’
6’
3’
5
5’
6
= (1 10 )(2 20 )(3 30 )(4 40 )(5 50 )(6 60 )
ν = (1 2 3)(10 4)(20 5)(30 50 6)(40 60 )
ν = φ = (1 4 60 30 )(10 2 5 6 40 )(20 3 50 )
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Encoding all Maps
Equivalence classes can be encoded by perfect matchings of flags.
Start with a ribbon graph.
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Encoding all Maps
Equivalence classes can be encoded by perfect matchings of flags.
Mv
Me
Mf
Ribbon boundaries determine 3 perfect matchings of flags.
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Encoding all Maps
Equivalence classes can be encoded by perfect matchings of flags.
Mv
Me
Pairs of matchings determine, faces,
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Encoding all Maps
Equivalence classes can be encoded by perfect matchings of flags.
Mv
Mf
Pairs of matchings determine, faces, edges,
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Encoding all Maps
Equivalence classes can be encoded by perfect matchings of flags.
Me
Mf
Pairs of matchings determine, faces, edges, and vertices.
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Encoding all Maps
Mv
6’ 5 5’
6
7’
7
8’ 8
3’
3
2 2’
1
1’
Me
4’
4
Mf
Mv = {1, 3}, {10 , 30 }, {2, 5}, {20 , 50 }, {4, 80 }, {40 , 8}, {6, 7}, {60 , 70 }
Me = {1, 20 }, {10 , 4}, {2, 30 }, {3, 40 }, {5, 60 }, {50 , 8}, {6, 70 }, {7, 80 }
Mf = {1, 10 }, {2, 20 }, {3, 30 }, {4, 40 }, {5, 50 }, {6, 60 }, {7, 70 }, {8, 80 }
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Hypermaps
Generalizing the combinatorial encoding, an arbitrary triple of perfect
matchings determines a hypermap when the triple induces a connected
graph, with cycles of Me ∪ Mf , Me ∪ Mv , and Mv ∪ Mf determining
vertices, hyperfaces, and hyperedges. Example
Hypermaps both specialize and generalize maps.
Example
,→
Return
Hypermaps can be represented as face-bipartite maps.
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β-Polygonal Glueings
April 6, 2013
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Hypermaps
Generalizing the combinatorial encoding, an arbitrary triple of perfect
matchings determines a hypermap when the triple induces a connected
graph, with cycles of Me ∪ Mf , Me ∪ Mv , and Mv ∪ Mf determining
vertices, hyperfaces, and hyperedges. Example
Hypermaps both specialize and generalize maps.
Example
,→
Return
Maps can be represented as hypermaps with = [2n ].
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Example
Mf
ν = [23 ]
= [32 ]
Mv
φ = [6]
Me
Return
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