Sec1:Vir
Sec2:2-dimsubalg
Sec3:The setV (r)
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Sec1:Vir
Sec2:2-dimsubalg
Sec3:The setV (r)
Contents
A quick review on the centreless Virasoro algebra
A description on two-dimensional subalgebras of d
Properties of the parameters
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Sec1:Vir
Sec2:2-dimsubalg
Sec3:The setV (r)
The centreless Virasoro algebra
The centreless Virasoro algebra (Witt algebra) d:
is the derivation Lie algebra of the Laurent polynomial algebra.
d = DerC(C[t±1]).
I d has a standard basis:{Lm := tm+1 d
dt
∣∣∣∣m ∈ Z}
satisfying
[Lm,Ln] = (n −m)Lm+n, for m,n ∈ Z.
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Sec1:Vir
Sec2:2-dimsubalg
Sec3:The setV (r)
The Virasoro algebra
The Virasoro algebra Vir
is the universal central extension of d.
Facts
I d is an infinite-dimensional simple Lie algebra.
I d has no finite-dimensional subalgebra of dimension > 4.
I d has no commutative subalgebra of dimension > 2.
S. Ng, E. J. Taft, Classification of the Lie bialgebra structures on the Witt andVirasoro algebras. J. Pure Appl. Algebra 151 (2000): 67õ88.
Y. Su, K. Zhao, Generalized Virasoro and super-Virasoro algebras and modulesof the intermediate series. J. Algebra, 252(2002):1–19.
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Sec1:Vir
Sec2:2-dimsubalg
Sec3:The setV (r)
Three dimensional subalgebras of d
Proposition
Every three dimensional subalgebra of d is of the form
spanC{L−m,L0,Lm}
for some positive integer m.
Y. Su, K. Zhao, Generalized Virasoro and super-Virasoro algebras and modulesof the intermediate series. J. Algebra, 252(2002):1–19.
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Sec1:Vir
Sec2:2-dimsubalg
Sec3:The setV (r)
Two dimensional subalgebras of d
Every two-dimensional subalgebra of d
I is non-commutative,and hence
I has a basis {X ,Y} such that
[X ,Y ] = cY .
Questions
I How do such X and Y look like?
I Can we find all two-dimensional subalgebras of d?
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Sec1:Vir
Sec2:2-dimsubalg
Sec3:The setV (r)
Known results
I Trivial examples of two-dimensional Lie subalgebras of d
z(m) := spanC{L0,Lm},
where m is a non-zero integer.I Non-trivial examples obtained by Y. Su and Z. Zhao:
Y. Su, K. Zhao, Generalized Virasoro and super-Virasoro algebras andmodules of the intermediate series. J. Algebra, 252(2002):1–19.
X := L0 + αL−m, and Y := exp(αm · adL−m
)Lnm
span a two dimensional subalgebra of d.
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Known results
I Trivial examples of two-dimensional Lie subalgebras of d
z(m) := spanC{L0,Lm},
where m is a non-zero integer.I Non-trivial examples obtained by Y. Su and Z. Zhao:
Y. Su, K. Zhao, Generalized Virasoro and super-Virasoro algebras andmodules of the intermediate series. J. Algebra, 252(2002):1–19.
X := L0 + αL−m, and Y := exp(αm · adL−m
)Lnm
span a two dimensional subalgebra of d.2015
-07-
17 A quick review on the centreless Virasoro alge-
bra
Known results
In Su and Zhao’s example,
X = L0 + αL−m = t−m+1(tm + α)ddt,
Y = exp( α
m· adL−m
)Lnm
=n+1∑k=0
(n + 1
k
)αk Lnm−km
=
(n+1∑k=0
(n + 1
k
)αk tnm−km+1
)ddt
= t−m+1(tm + α)n+1 ddt.
Sec1:Vir
Sec2:2-dimsubalg
Sec3:The setV (r)
Known results
D. Yu, C. Lu, Results for Virasoro subalgebra. Acta Math. Sinica (Chin. Ser.),49(2006):632–638.
I If X = a−mL−m + · · ·+ anLn and Y = b−mL−m + · · ·+ bnLn
satisfy [X ,Y ] = Y , then there coefficients a−m, . . . ,an,
b−m, . . . ,bn satisfy certain equation system.
I One expects to obtain two dimensional subalgebras of d
through solving these equations, but it is very difficult.I A few examples are given, for instance,
X =5
768L−2 +
548
L−1 +18
L0 − L1 − L2,
Y =1
256L−2 +
116
L−1 +38
L0 + L1 + L2,
satisfy [X ,Y ] = Y .
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Known results
D. Yu, C. Lu, Results for Virasoro subalgebra. Acta Math. Sinica (Chin. Ser.),49(2006):632–638.
I If X = a−mL−m + · · ·+ anLn and Y = b−mL−m + · · ·+ bnLn
satisfy [X ,Y ] = Y , then there coefficients a−m, . . . ,an,
b−m, . . . ,bn satisfy certain equation system.
I One expects to obtain two dimensional subalgebras of d
through solving these equations, but it is very difficult.I A few examples are given, for instance,
X =5
768L−2 +
548
L−1 +18
L0 − L1 − L2,
Y =1
256L−2 +
116
L−1 +38
L0 + L1 + L2,
satisfy [X ,Y ] = Y .
2015
-07-
17 A quick review on the centreless Virasoro alge-
bra
Known results
In Yu and Lu’s example,
X =
(5
768t−1 +
548
+18
t − t2 − t3)
ddt,
Y =
(1
256t−1 +
116
+38
t + t2 + t3)
ddt.
Then we can deduce that
X − 53
Y = −16
t(16t2 + 16t + 3
) ddt
= − 196
t(
t +14
)(t +
34
)ddt,
Y = t−1(
t +14
)4 ddt.
Sec1:Vir
Sec2:2-dimsubalg
Sec3:The setV (r)
Contents
A quick review on the centreless Virasoro algebra
A description on two-dimensional subalgebras of d
Properties of the parameters
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Sec1:Vir
Sec2:2-dimsubalg
Sec3:The setV (r)
Main Idea
I Find all solutions to the equation
[X ,Y ] = Y
in the Lie algebra d.
I Write X = tF (t) ddt and Y = tG(t) d
dt .
I Equivalently, find all solutions to the ODE:
tF (t)G′(t)− tG(t)F ′(t) = G(t)
in the Laurent polynomial algebra C[t±1].
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An attempt using ODE
I The ODE
tF (t)G′(t)− tG(t)F ′(t) = G(t)
is separable:G′(t)G(t)
=1 + tF ′(t)
tF (t).
I Given arbitrary F (t) ∈ C[t±1], one can easily obtain a G(t)
G(t) = exp(∫ t
a
1 + uF ′(u)
uF (u)du)
as a complex function.
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Problems of the ODE method
For a F (t) ∈ C[t±1],
G(t) = exp(∫ t
a
1 + uF ′(u)
uF (u)du)
is not necessarily a Laurent polynomial.
For instance,
I If F (t) = 1, then G(t) = ta is a Laurent polynomial.
I If F (t) = t , then G(t) = ta + exp
(1a −
1t
)is not a Laurent
polynomial.
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Sec1:Vir
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Two dimensional subalgebras of d: new examples
LetX := t(t − a1) · (t − an)
ddt
and Y := t−r1−...−rk+n−k+1(t − a1)r1+1 · · · (t − an)
rk+1 ddt,
where n > k , a1, . . . ,an ∈ C are distinct non-zero numbers,
and r1, . . . , rk are positive integers.
Then, X and Y span a two-dimensional subalgebra of dif (a1, . . . ,an) satisfies
r1 · a1 + · · ·+ rk · ak = ak+1 + · · ·+ an,
r1 · a21 + · · ·+ rk · a2
k = a2k+1 + · · ·+ a2
n,
· · ·
r1 · an−11 + · · ·+ rk · an−1
k = an−1k+1 + · · ·+ an−1
n .
Moreover, such an equation system has a componentwise
non-zero solution only ifr1 + · · ·+ rk > n.
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Sec1:Vir
Sec2:2-dimsubalg
Sec3:The setV (r)
Notation
I For n > k ∈ N, let
Γ(n, k) := {(r1, . . . , rn) ∈ Nk × {−1}n−k |r1 + · · ·+ rn > k}.
I For r = (r1, . . . , rn) ∈ Γ(n, k), we denote
V (r)× :=
(a1, . . . ,an) ∈ (C×)n
∣∣∣∣∣∣n∑
j=1
rjaij = 0, for i = 1, . . . ,n − 1
.
I Σ := {(n, k , r,a)|n > k , r ∈ Γ(n, k),a ∈ V (r)}.
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Sec1:Vir
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Sec3:The setV (r)
Main Result I
Proposition
I Every quadruple µ := (n, k , r,a) ∈ Σ determinesa two-dimensional subalgebra of d:
s(µ) = spanC
{t(t − a1) · · · (t − an)
ddt, t−r1−···−rn+1(t − a1)
r1+1 · · · (t − an)rn+1 d
dt
}.
I s(µ) is not equal to any z(m) = spanC{L0,Lm}.
I Given µ = (n, k , r,a) and µ′ = (n′, k ′, r′,a′) ∈ Σ,
s(µ) = s(µ′) iff n = n′, k = k ′ and there is a permutation
σ ∈ Sn such that rσ(i) = r ′i and aσ(i) = a′i for i = 1, . . . ,n.
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Concrete Examples I
I n = k = 1, r = r , then V (r)× = C×.
We obtain two-dimensional subalgebras:
spanC
{t(t − a)
ddt, t−r+1(t − a)r+1 d
dt
}.
I n = k = 2, r = (r1, r2) ∈ N2, then V (r)× = {(ar2,−ar1)|a ∈ C×}.We obtain two-dimensional subalgebras:
spanC
{t(t − ar2)(t + ar1)
ddt, t−r1−r2+1(t − ar2)r1+1(t + ar1)r2+1 d
dt
}.
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Sec3:The setV (r)
Concrete Examples II
I n = k ∈ N and r = (r , . . . , r) ∈ Γ(n, k).
Let ζn be a primitive n-th root of unity,
then (ζn, ζ2n , . . . , ζ
nn ) ∈ V (r)×.
We obtain two-dimensional subalgebra
spanC
{t(tn − 1)
ddt, t−rn+1(tn − 1)r+1 d
dt
}.
I n = 3, k = 2, r = (r ,1,−1) with r > 2.
Then V (r)× = {a(2,1− r ,1 + r)|a ∈ C×}.We obtain two-dimensional subalgebras
spanC
{t(t − 2)(t − 1 + r)(t − 1− r)
ddt, t−r+1(t − 2)r+1(t − 1 + r)2 d
dt
}.
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Sec1:Vir
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Sec3:The setV (r)
Main Result II
TheoremEvery two-dimensional subalgebra of d is equal to
either z(m) = span{L0,Lm} for some non-zero integer m
or s(µ) for some µ ∈ Σ.
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Sec1:Vir
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Sketch of Proof I
I For X = (tα1 + ·+ tαs )t ddt with α1 < · · · < αs, we write
deg1 X = αs, and deg2 X = α1.
I If [X ,Y ] = cY then [X − aY ,Y ] = cY for any a ∈ C.
Hence, we may assume deg2 X 6= deg2 Y .
I [X ,Y ] = cY and deg2 X 6= deg2 Y implies that
deg2 X = 0 and deg1 X = deg1 Y .
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Sketch of Proof II
I We may assume both X and Y are monic, and then write
X = tF (t)ddt
and Y = ts+1G(t)ddt,
where F (t),G(t) ∈ C[t ], F (0) 6= 0 and G(0) 6= 0.
I [X ,Y ] = cY implies thatI every root of G(t) is a root of F (t);I F (t) has no multiple root;I G(t) has no simple root.
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Sec3:The setV (r)
Sketch of Proof III
I We may further write
X = t(t−a1) · · · (t−an)ddt, and Y = ts+1(t−a1)r1+1 · · · (t−ak )rk+1,
and denote rk+1 = · · · = rn = 0.
I deg1 X = deg1 Y implies that s = −(r1 + · · ·+ rn).
I Consider the automorphism of d:
ω : t l+1 ddt7→ −t−l+1 d
dt.
ω(X ) and ω(Y ) also span a two-dimensional subalgebra of
d, from which we obtain
r1 + · · ·+ rn > k .
I Using [X ,Y ] = cY again, we deduce (a1, . . . ,an) ∈ V (r)×.
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Sec1:Vir
Sec2:2-dimsubalg
Sec3:The setV (r)
Contents
A quick review on the centreless Virasoro algebra
A description on two-dimensional subalgebras of d
Properties of the parameters
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Sec1:Vir
Sec2:2-dimsubalg
Sec3:The setV (r)
The algebraic set V (r)
Our description of two-dimensional subalgebras of d depends
on componentwise nonzero points in V (r), where V (r) is an
affine algebraic set defined by
r1x1 + · · ·+ rnxn = 0,
r1x21 + · · ·+ rnx2
n = 0,
· · ·
r1xn−11 + · · ·+ rnxn−1
n = 0.
(∗)
QuestionCan we find all componentwise nonzero solutions to (*)?
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The projective variety V (r)
Observing that (*) is homogeneous and hence determines a
projective variety V (r).
Results on V (r):
I Let a be a componentwise nonzero solution to (*) and a its
canonical image in V (r). Then a has multiplicity 1 in V (r).
I The subset V (r)× of V (r) consisting of all componentwise
nonzero elements contains at most (n − 1)! elements.
I Let r = (r1, . . . , rk ,−1, . . . ,−1). If
ri > n − k + 1, i = 1, . . . , k ,
then V (r)× contains exactly (n − 1)! elements.24 / 26
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A conjecture
Conjecture
Let r = (r1, . . . , rk ,−1, . . . ,−1) ∈ Nk × {−1}n−k . If
r1 + · · ·+ rk > n,
then V (r)× is non-empty, i.e., the equation system
r1x1 + · · ·+ rnxn = 0,
r1x21 + · · ·+ rnx2
n = 0,
· · ·
r1xn−11 + · · ·+ rnxn−1
n = 0.
(∗)
has a componentwise nonzero solution.
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Sec1:Vir
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Thank you!
Questions?
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