Integrable systemsand moduli spaces
Boris DUBROVINSISSA, Trieste
The Hebrew University of JerusalemEinstein Institute of Mathematics
April 30, 2015
Sunday 10 May 15
Plan
1. Integrable Hamiltonian systems
2. Hamiltonian PDEs
3. KdV
4. Deligne - Mumford moduli spaces and Witten - Kontsevich solution to KdV
Mg,n
5. Construction of KdV: a recipe using Mg,n
6. More general class of integrable hierarchie (Witten’s programme) and -classes�
Sunday 10 May 15
Hamiltonian systems
x = J rH
x 2 M
a manifold (phase space), H = H(x) Hamiltonian
J operator of Poisson bracket
E.g., dimM = 2n, x = (q1, . . . , qn, p1, . . . , pn), J =
✓0 1
�1 0
◆
qi = @H@pi
pi = �@H@qi
9=
; , i = 1, . . . , n
{f, g} = hrf, J rgi
Integrability: commuting Hamiltonians{Hi, Hj} = 0, H = H1
+ completeness ) commuting flowsdx
dti= J rHi,
d
dti
✓dx
dtj
◆=
d
dtj
✓dx
dti
◆
Sunday 10 May 15
Infinite-dimensional analogue: Hamiltonian PDEs
ut
= F (u, ux
, uxx
, . . . ) = J�H
�u(x)for u = u(x, t)
a dynamical system on the space of functions u(x)
Functional (Hamiltonian)
A skew-symmetric operator J of Poisson bracket
H = H[u] =1
2⇡
Z 2⇡
0h(u, u
x
, . . . ) dx
Fréchet derivative (Euler - Lagrange operator)
�H
�u(x)=
@h
@u� d
dx
@h
@ux
+d2
dx2
@h
@uxx
� . . .
Sunday 10 May 15
Example. Korteweg - de Vries (KdV) equation
Integrability: an infinite system of commuting PDEs
(uti)tj =�utj
�ti
uti = F
i
(u, ux
, uxx
, . . . ) = J�H
i
�u(x)i = 0, 1, 2, . . .
) common solution u = u(x, t0, t1, t2, . . . )
for given Cauchy data u0(x) = u(x, 0, 0, 0, . . . )
J =@
@x
ut
= uux
+✏2
12uxxx
=@
@x
�H
�u(x), H =
Z ✓u3
6� ✏2
24u2x
◆dx
Sunday 10 May 15
Example. KdV hierarchy
ut0 = ux = @x�H0�u(x)
ut1 = uux + ✏2
12uxxx = @x�H1�u(x)
ut2 = u2
2! ux + ✏2
12 (2uxuxx + uuxxx) +✏4
240uV = @x
�H2�u(x)
. . . . . . . . . . . . . . . . . . . . . . . .
Constructions:
• Lax representation, isospectral deformations
• Infinite-dimensional Grassmannians
• Baker - Akhiezer functions
Sunday 10 May 15
Example. KdV hierarchy
ut0 = ux = @x�H0�u(x)
ut1 = uux + ✏2
12uxxx = @x�H1�u(x)
ut2 = u2
2! ux + ✏2
12 (2uxuxx + uuxxx) +✏4
240uV = @x
�H2�u(x)
. . . . . . . . . . . . . . . . . . . . . . . .
Constructions:
• Lax representation, isospectral deformations
• Infinite-dimensional Grassmannians
• Baker - Akhiezer functions
• Topology of moduli spacesSunday 10 May 15
Deligne - Mumford moduli spaces of stable algebraic curves
Mg,n = {(Cg, x1, . . . , xn)} / ⇠
Tautological line bundles
Mg,n
Li
T ⇤xiC
g
i := c1 (Li) 2 H2�Mg,n
�, i = 1, . . . , n
· ··0
1
1
M0,4 = P1
M0,3 = pt
····
0
1
1z
M1,1 = {ellipticcurves}
Sunday 10 May 15
Witten - Kontsevich solution to KdV
Then the tau-function of this solution
u(x, t0, t1, . . . ; ✏)|t=0 = x
⌧ = ⌧(t0, t1, . . . ; ✏)
such that ✏
2 @2log ⌧
@x
2= u(t; ✏) reads
where
nonzero only if p1 + · · ·+ pn = 3g � 3 + n
log ⌧(t; ✏) =X
g�0
✏2g�2Fg(t)
Fg(t) =X
n
1
n!
Xtp1 . . . tpn
Z
Mg,n
p11 . . . pn
n
Sunday 10 May 15
In physics literature (Witten et al.):
the tau-function = partition function of 2D quantum gravity
log ⌧(t) =DeP
i�0 ti⌧iE
⌧0 = 1, ⌧1, ⌧2, . . . observables
time variables of KdV hierarchy = coupling constants
Correlators
h⌧p1 . . . ⌧pni =Z
Mg,n
p11 . . . pn
n , p1 + · · ·+ pn = 3g � 3 + n
Sunday 10 May 15
Moreover, Hamiltonian densities Hp
[u] =
Zhp
(u, ux
, . . . ; ✏) dx
are two-point correlation functions
hp = hh⌧0⌧p+1ii = ✏
2 @2log ⌧(t)
@x @tp+1h�1 = u
h0 = u
2
2 + ✏
2
12uxx
h1 = u
3
3! +✏
2
24
�u2x
+ 2uuxx
�+ ✏
4
240uxxxx
h2 = u
4
4! +✏
2
24
�uu2
x
+ u2uxx
�+ ✏
4
480
�3u2
xx
+ 4ux
uxxx
+ 2uuxxxx
�+ ✏
6
6720u(6)
etc. Hence@hp�1
@tq=
@hq�1
@tptau-symmetry
NB: change of densities hp
7! hp
+ @x
(. . . )
does not change the PDEsSunday 10 May 15
Construction of KdV: a recipe using Mg,n
Start from KdVε=0 vt0 = v
x
vt1 = v v
x
vt2 = v
2
2! vx. . . . . . . . .
vtk = v
k
k! vx
(change notations: u(x, t) 7! v(x, t) )
Verify commutativity:
Exercise. Prove
=@
@x
v
k+1
(k + 1)!
(vtk)tl =@
@x
@
@tl
v
k+1
(k + 1)!=
@
2
@x
2
v
k+l+1
k! l! (k + l + 1)= (k $ l)
Derive the WK solution at g=0: v(t) =X
n�1
1
n
X
k1+···+kn=n�1
tk1
k1!. . .
tkn
kn!
@
nv
@tk1 . . . @tkn
=@
n
@x
n
v
k1+···+kn+1
k1! . . . kn!(k1 + · · ·+ kn + 1)
Sunday 10 May 15
Next, to recover full KdV do a substitution v ! u
u = v + ✏2@2x
2
4 1
24
log vx
+ ✏2✓
vxxxx
1152v2x
� 7 vxx
vxxx
1920v3x
+
v3xx
360v4x
◆+O �
✏4�
| {z }
3
5
�F
Then(“quasitriviality”, B.D., Y.Zhang).
• KdVε=0(v)=0 KdV(u)=0
• Operator of Poisson bracket unchangedJ =@
@x
• New Hamiltonian densities
hp
(u, ux
, . . . ) = h0p
(v) + ✏2@x
@tp+1�F
are differential polynomials in u satisfying tau-symmetry
Sunday 10 May 15
E.g., start with vt
= v vx
, plug
u = v +✏2
24
(log vx
)
xx
+O �✏4�
ut
= vt
+✏2
24
✓vxt
vx
◆
xx
+ · · · = v vx
+✏2
24
✓v v
xx
+ v2x
vx
◆
xx
+ . . .
then
= v vx
+
✏2
24
[(v (log vx
)
x
)
xx
+ vxxx
] = v vx
+
✏2
24
[(v (log vx
)
xx
)
x
+ 2vxxx
]
= uux
+✏2
12uxxx
+O �✏4�
(cf. N.Ibragimov, V.Baikov, R.Gazizov, ’89)
Sunday 10 May 15
Geometrical meaning of the substitution (Dijkgraaf, Witten 1990): expressing higher genera via genus zero:
�F =X
g�1
✏2g�2Fg
⇣v, v
x
, . . . , v(3g�2)⌘
plug the genus zero solution
vx
= vx
(t)v = v(t) =X
n�1
1
n
X
k1+···+kn=n�1
tk1
k1!. . .
tkn
kn!, etc. Then
Fg
⇣v(t), v
x
(t), . . . , v(3g�2)(t)⌘=
X 1
n!
Xtk1 . . . tkn
Z
Mg,n
k11 . . . kn
n
= Fg
(t), g � 1
Sunday 10 May 15
How to find the substitution? Solve loop equation
L(�)e�F = 0 8�
L(�) =X
k�0
�Ak(�)� ✏2Bk(�)
� @
@v(k)� ✏2
2
X
k,l�0
Ckl(�)@2
@v(k)@v(l)+
1
16(v � �)2=
X
m��1
Lm
�m+2
Ak
(�) = @k
x
✓1
v � �
◆+
kX
j=1
✓kj
◆@j�1x
✓1pv � �
◆@k�j+1x
✓1pv � �
◆
Bk
(�) = � 1
16@k+2x
✓1
(v � �)2
◆
Ckl
(�) = @k+1x
✓1pv � �
◆@l+1x
✓1pv � �
◆
Commutation relations [Lm,Ln] = (m� n)Lm+n, m, n � �1
Sunday 10 May 15
Digression: on computation of intersection numbers
h⌧k1 . . . ⌧kni =Z
Mg,n
k11 . . . kn
n
Let M(z) =1
2
0
B@�P1
g=1(6g�5)!!
24g�1·(g�1)!z�6g+4 �2
P1g=0
(6g�1)!!24g·g! z�6g
2P1
g=06g+16g�1
(6g�1)!!24g·g! z�6g+2
P1g=1
(6g�5)!!24g�1·(g�1)!z
�6g+4
1
CA
(cf. Faber - Zagier series). Then (M.Bertola, B.D., Di Yang, 2015)
Xh⌧k1 . . . ⌧kni
(2k1 + 1)!!
z2k1+21
. . .(2kn + 1)!!
z2kn+2n
= � 1
n
X
r2Sn
TrM(zr1) · · ·M(zrn)Qnj=1(z
2rj � z2rj+1
)� �n,2
z21 + z22(z21 � z22)
2, n � 2
Sunday 10 May 15
Proof uses Lax representation for KdV hierarchyn -th equation , Ltn = [An, L]
L = @2x
+ 2u Lax operator, An
=
1
(2n+ 1)!!
⇣L
2n+12
⌘
+
(set ✏ = 1) Eigenfunctions L = z2 satisfy
tn = An , n � 0
At t = 0 u = u0(x) = x
000 + 2x 0 = z
2 0 ⇠ Airy equation
Use
Ai(z) ⇠ e�⇣
2p⇡z1/4
1X
k=0
(6k � 1)!!
(2k � 1)!!
(�216 ⇣)�k
k!, ⇣ =
2
3z3/2, z ! 1, | arg z| < ⇡
Sunday 10 May 15
More general integrable Hamiltonian tau-symmetric hierarchies conjecturally depend on an infinite number of parameters s1, s2, . . . (a deformation of KdV)
(B.D., S.-Q.Liu, Y.Zhang, D.Yang, 2014)
Construction uses Hodge potential
ut
= uux
+ ✏2⇣u
xxx
12� s1ux
uxx
⌘
+ ✏4� s160
u(5) + s21
✓uxx
uxxx
+1
5ux
u4
◆� 4s31
5
�2u
x
u2xx
+ u2x
uxxx
�
�s26
�2u
x
u2xx
+ u2x
uxxx
�i+O(✏6)
Hg(t, s) =X
n�0
1
n!
Xtp1 . . . tpn
Z
Mg,n
eP
k�1 skchk(E) p11 . . . pn
n
Hodge bundle
Mg,n
E
H0 (T ⇤Cg)
(even components of Chern character vanish ch2i (E) = 0)
Sunday 10 May 15
Clearly Hg(t, 0) = Fg(t) Redenote s2k�1 7! � B2k
(2k)!sk
Thm.1) H0 = F0, for g � 1 Hg = (Fg +�Hg)v=v(t),v
x
=vx
(t),...
where�H
g
2 Qhs1, . . . , sg; v, v
±1x
, vxx
, . . . , v(3g�3)i
�H1 = �1
2s1v
E.g.,
�H2 = s1
✓11v2
xx
480v2x
� vxxx
40vx
◆+
7
40s21vxx �
✓s3110
+s248
◆v2x
RecallF1 =
1
24
log vx
, F2 =
v(4)
1152v2x
� 7vxx
vxxx
1920v3x
+
v3xx
360v4x
Sunday 10 May 15
2) The substitution
v 7! u = v + ✏2@2x
X
g�1
✏2g�2Hg
h0p
=vp+2
(p+ 2)!7! h
p
= h0p
+ ✏2@x
@tp+1
X
g�1
✏2g�2Hg
transforms KdVε=0 to a new integrable Hamiltonian tau-symmetric hierarchy
=X
g�0
✏2gh[g]p
, h[g]p
2 Qhs1, . . . , sg;u, ux
, . . . , u(2g+2)i
utp =@
@x
�H[u; s]
�u(x), H[u; s] =
Zhp dx, p = 0, 1, 2, . . .
@hp�1
@tq=
@hq�1
@tpSunday 10 May 15
E.g., for t = t1 one obtains a deformation of KdV
depending on the parameters s1, s2, . . .
ut
= uux
+ ✏2⇣u
xxx
12� s1ux
uxx
⌘
+ ✏4� s160
u(5) + s21
✓uxx
uxxx
+1
5ux
u(4)
◆� 4s31
5
�2u
x
u2xx
+ u2x
uxxx
�
�s26
�2u
x
u2xx
+ u2x
uxxx
�i+O(✏6)
Sunday 10 May 15
Conjecture. This is a universal deformation of KdVε=0 in the class of Hamiltonian tau-symmetric integrable hierarchies. For
Example 1. For sk = � B2k
2k(2k � 1)
s2k�1, for k � 1
one obtains intermediate long wave eq. (A.Buryak)
ut
= uux
+X
g�1
✏2gsg�1 |B2g|(2g)!
u2g+1
Example 2. For sk = (4k � 1)B2k
2k(2k � 1)s2k�1, k � 1
one obtains (?) Volterra equation (=discrete KdV)
ut
=1
✏
⇣eu(x+✏) � eu(x�✏)
⌘(checked up to ✏12)
s = 0 obtain KdV
Sunday 10 May 15
Thank you!
Sunday 10 May 15
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