Calabi-Yau di¤erential equations with a parameter. Gert Almkvist Introduction In this paper we give some examples of Calabi-Yau di¤erential equations whose coe¢ cients depend on a parameter a: We start with De…nition: A Calabi-Yau di¤erential equation is a 4-th order di¤erential equation with rational coe¢ cients y (4) + a3 (x)y 000 + a2 (x)y 00 + a1 (x)y 0 + a0 (x)y = 0 satisfying the following conditions. 1. It is MUM (Maximal Unipotent Monodromy), i.e. the indicial equation at x = 0 has zero as a root of order 4. It means that there is a Frobenius solution of the following form y0 = 1 + A1 x + A2 x2 + ::: y1 = y0 log(x) + B1 x + B2 x2 + :: 1 y0 log2 (x) + (B1 x + B2 x2 + :::) log(x) + C1 x + C2 x2 + ::: 2 1 1 y3 = y0 log3 (x)+ (B1 x+B2 x2 +:::) log2 (x)+(C1 x+C2 x2 +:::) log(x)+D1 x+D2 x2 +::: 6 2 It is very useful that Maple’s "formal_sol" produces the four solutions in exactly this form (though labelled 1 4 ) 2. The coe¢ cents of the equation satisfy the identity y2 =
a1 =
1 a2 a3 2
1 3 a + a02 8 3
3 a3 a03 4
1 00 a 2 3
3. Let t = y1 =y0 : Then q = exp(t) = x + c2 x2 + ::: can be solved x = x(q) = q
c2 q 2 + ::::
which is called the "mirror map". We also construct the "Yukawa coupling" de…ned by d 2 y2 K(q) = 2 ( ) dt y0 This can be expanded in a Lambert series K(q) = 1 +
1 X d=1
1
nd
d3 q d 1 qd
where the nd are called "instanton numbers". For small d the nd are conjectured to count rational curves of degree d on the corresponding CalabiYau manifold. Then the third condition is (a) y0 has integer coe¢ cients (b) q has integer coe¢ cients (c) There is a …xed integer N0 such that all N0 nd are integers
For a long time we thought that conditions 1 and 3a,b would imply conditions 2 and 3c. Recently M.Bogner and S.Reiter found the following example ( d ) = x dx 4
8x(2 + 1)2 (5
2
+ 5 + 2) + 192x2 (2 + 1)(2 + 3)(3 + 2)(3 + 4)
which satis…es conditions 1,2,3a,b but not 3c. . Here we give numerous examples of di¤erential equations depending on a parameter a which does not have to be an integer so conditions 3a,b are violated but condition 3c is still valid (Examples 1 and 2 and the Table). Examples 3-7 are strange, for only three values of a , ( a = 0; 1; 8 or 0; 1; 9 ) do we get Calabi-Yau equations. Example 1. ( # 45) For # 45=A we have An = and y0 =
P
2n n
n 2X
n k
k=0
3
An xn satis…es the equation
4
4x(2 + 1)2 (7 2 + 7 + 2) 128x2 (2 + 1)2 (2 + 3)2 P Recently I found that Y0 = Bn xn where Bn =
2n n
n X
an
j
j;k=0
n j
j k
3
2j j
is equivalent to # 45, (notation ~45) i.e. it has the same instanton numbers. Using "gfun" in Maple twice one …nds Y0 (x) = p
1 1
4ax
y0 (
x ) 1 4ax
Working out the transformation we …nd that Y0 satis…es the equation 4
2x(2 + 1)2 f(2a + 14)(
+4x2 (2 + 1)(2 + 3) (6a2 + 84a 16a(a2 + 21a
2
128)(
+ ) + a + 4g 2
+ 2 ) + 7a2 + 92a
64)x3 (2 + 1)(2 + 3)2 (2 + 5) 2
96
+16a2 (a
4)(a + 32)x4 (2 + 1)(2 + 3)(2 + 5)(2 + 7)
If a = 0 one gets back the equation for # 45. Observe that also Bn reduces to An when a = 0, since 00 = 1: Further for a = 4 and a = 32 the equation has degree 3: One checks that the Yukawa coupling and hence the instanton numbers are independent of a: In particular y0 and q need not to have integer coe¢ cents in order to the instanton numbers should satisfy condition 3c.
Example 2. ( # 18) For # 18 we have
n 2n X n n k
An =
4
k=0
which is the case a = 0 of Bn =
n X
an
n j
j
j;k=0
Then y0 =
P
An xn and Y0 =
P
j k
4
2j n
Bn xn are related by the transformation
Y0 (x) = y0 (x + ax2 )
and we …nd that Y0 satis…es the di¤erential equation 4
x2
+ax3 +2a2 x4 +4a3 x5
x ( 9a + 48)
4
+ (6a + 96)
3
+ (4a + 76)
2
+ (a + 38) + 4
( 33a2 + 480a + 1024) 4 + (36a2 + 768a + 4096) 3 +(a + 432a + 5824) 2 + ( 10a2 + 144a + 3456) + 56a + 720 2
(63a2 2016a 11264) 4 (78a2 + 2880a + 43008) 3 +(33a 2092a 59328) 2 + (6a2 1476a 37440) 336a 10800 2
(33a2 2304a 26624) 4 (36a2 + 3456a + 102400) 3 +(23a 3628a 152672) 2 (4a2 + 2440a + 113664) 560a 2
2
+(4a
+8a4 x6
(9a2 1548a 35328) 4 (6a2 + 2808a + 144384) 3 3532a 241872) 2 (a2 + 2240a + 200016) 560a
35280 65520
(a2 612a 28800) 4 (1440a + 130560) 3 (1948a + 245040) 2 (1272a + 218880) 336a 75600
26 a5 x7 (33a + 3696)
4
+ (102a + 19104)
27 a6 x8 (33a + 1168)
4
+ (12a + 6976) 3
3
3
+ (147a + 39816)
+ (19a + 16096)
2
2
+ (102a + 38184) + 28a + 13860
+ (14a + 16560) + 4a + 6300
210 a7 x9 (2 + 3)(26
3
+ 141
2
+ 249 + 135)
211 a8 x10 ( + 1)( + 3)(2 + 3)(2 + 5)
Also in this example are the instanton numbers independent of the parameter a: Here follows a table of other examples of di¤erential equations where the instanton numbers are independent of the parameter a:
#
degree
coe¢ cient
~3
5
An =
~3
5
An = 2
~3
8
~3
8
~3
8
~5
9
~5
9
~6
5
~18
4
~18
6
~18
10
~22
8
~45
10
~45
6
~58
4
~64
6
~69
6
~70
6
~210
7
an An = 2 P n j n An = a j P n j n An = a j P n j n An = a j P An = 2 n a n P n j n An = a j 2n P n An = n a P n j n An = a j P n j n An = a j P n j n An = a j P n j n An = a j P n j n An = a j 2n P n An = n a P n j n An = a j 2n P n An = n a P n An = 2n a n 2n P n An = n a
2n n n
P
P
P n
an an
j n j j n j j n j j 2 k j 2 k j 2 k j n j j 2 k j n j j 4 k j 4 k j 5 k j 3 k j 3 k j n j j 2 k j n j j n j j n j
j 2 2j 3 k j j 2j 4 k j j 2j 3 2j k j n 2j 2 2j j n 2j 2 n+j j n 2j 2 3j j n j 2j 3 3j k j n 2j 2 4j j 2j j 4 k 2j j n+j n
2j j 2j j j k 2k k j k j k j k
2
2j n
2 2k 2j k j (6j)! (3j)!(2j)!j! 2 2k 4j k 2j 2 2k 3j k j 2 2j 2j+2k j 2j
Now we give a general construction of transforming any Calabi-Yau equqtion to one with a parameter. Theorem. Let 1 X y0 = An x n n=0
4
be the solution of a Calabi-Yau equation. Then Y0 =
1 X
Bn xn
n=0
with n 2n X n a n j=0
Bn =
n j Aj 2j j
j
satis…es a Calabi-Yau equation with the same instanton numbers. Proof: Put x 1 y0 ( ) Y0 (x) = p 1 4ax 1 4ax which leaves the Yukawa coupling invariant. We obtain Y0 (x) = (1
4ax)
1=2
1 X
Aj xj (1
4ax)
j
j=0
=
1 X
Aj xj (1
4ax)
j 1=2
j=0
Now (1
4ax)
j 1=2
=
1 2j j
1 X
ak
k=0
j+k j
2j + 2k k x j+k
and we …nd with n = j + k 1 n X 2n n X n Y0 (x) = x a n n=0 j=0
Example 3. ( # 70=B c) We have n 2n 3n X n An = n n k k=0 P and y0 = An xn satis…es the equation 4
3x(3 + 1)(3 + 2)(10
2
j
2
n j Aj 2j j
2k k
+ 10 + 3) + 81x2 (3 + 1)(3 + 2)(3 + 4)(3 + 5)
We observe that An is the special case a = 0 of Bn =
2n n
n X
an
j
j;k=0
5
n j
j k
2
2k k
3j n
In P this ncase, however, we were not able to …nd an equation satis…ed by Y0 = Bn x for general a: But for the special cases a = 1 and a = 9 I found the degree 6 equations # 405 and # 385 respectively. Using these three equations I managed to …nd a degree 6 equation with coe¢ cients in Z[a] specializing to the given equations for a = 0; 1; 9 4
+x2
x (16a + 270)
4
+ (32a + 540)
3
+ (20a + 411)
2
+ (4a + 141) + 18
(256a2 + 2160a + 6561) 4 + (1024a2 + 8640a + 26244) 3 +(1216a2 + 9672a + 35721) 2 + (384a2 + 2064a + 18954) 432a + 3240
22 3a2 x3 (320a + 3240)
4
3
+ (1920a + 19440)
+23 34 a2 x4 (204
4
+ 1632
4 5 3 5
3
+ 4449
2
+2 3 a x (2 + 5) (16
2
+ (4035a + 40868)
2
2
+ (3465a + 35124) + 900a + 9135
+ 4740 + 1400)
+ 80 + 35)
+24 36 a4 x6 (2 + 1)(2 + 5)(2 + 7)(2 + 11)
Let Y0 be the analytic solution of this equation. Then Y0 (x)
1 X
Bn xn
n=0
=
a(a + 1)(a + 9)
400 3 5 x + (1960a + 38739)x4 3 8
9 (11200a2 500
24229680a
230410649)x5 + :::
Here the instanton numbers are highly dependent on a and it seems like a miracle that you get only the denominator 3 for both a = 1 and a = 9 n1 = 27 + 4a n2 = 432 n2 = 18089
a (19a + 79) 2
2a (550a2 + 8719a 81
47154)
a (432544a3 + 3946009a2 10432998a + 157236129) 2304 a n4 = 68438142+ (11919160a4 +1115527462a3 +3416132103a2 72444054312a+66361748387) 112500 n3 = 997785 +
Example 4. ( # 15=B a ) We have 2n An = n
n 3n X n n k k=0
6
3
and y0 = 4
P
An xn satis…es the equation
3x(3 + 1)(3 + 2)(7
2
72x2 (3 + 1)(3 + 2)(3 + 4)(3 + 5)
+ 7 + 2)
We observe that An is the special case a = 0 of 2n n
Bn =
n X
an
j
j;k=0
n j
j k
3
3j n
In P this ncase, however, we were not able to …nd an equation satis…ed by Y0 = Bn x for general a: But for the special cases a = 1 and a = 8 I found the degree 6 equations # 336 and # 335 respectively. Using these three equations I managed to …nd a degree 6 equation with coe¢ cients in Z[a] specializing to the given equations for a = 0; 1; 8 4
x (16a + 189)
+8x2
4
+ (32a + 378)
3
+ (20a + 285)
2
+ (4a + 96) + 12
(32a2 + 189a 729) 4 + (128a2 + 756a 2916) 3 +(152a2 + 849a 3969) 2 + (48a2 + 186a 2106) 36a
22 3a2 x3 (320a + 2268)
4
3
+ (1920a + 13608)
+26 32 a2 x4 (204
4
+ 1632
3
+ 4449
27 33 a3 x5 (2 + 5)2 (16
2
+ (4035a + 28604)
2
360 2
+ (3465a + 24576) + 900a + 6390
+ 4740 + 1400)
+ 80 + 35)
7 4 4 6
2 3 a x (2 + 1)(2 + 5)(2 + 7)(2 + 11) Let Y0 be the analytic solution of this equation. Then Y0 (x)
1 X
Bn xn
n=0
= a(a
1)(a + 8)
400 3 5 x + (980a + 13463)x4 3 4
3 (8400a2 125
12557640a
105341243)x5 + :::
n1 = 21 + 4a a n2 = 480 (19a + 49) 2 2a n2 = 15894 (550a2 + 6421a 70070) 81 a n3 = 894075 + (216272a3 + 1313273a2 8721217a + 812330016) 1152 a n4 = 58703151+ (5959580a4 +411594539a3 1046721283a2 34762372554a+395110345968) 56250 7
Example 5 ( # 45=A a ) The given An in Example 1 can be obtained by putting a = 0 in Bn =
2n n
n X
an
j;k=0
j
n j
j k
3
2j n
In P this ncase, however, we were not able to …nd an equation satis…ed by Y0 = Bn x for general a: But for the special cases a = 1 and a = 8 I found the degree 4 equations equivalent to # 29 and # 16 respectively. Using these three equations I managed to …nd a degree 4 equation with coe¢ cients in Z[a] specializing to the given equations for a = 0; 1; 8 4
+4x2
x (8a + 112)
4
+ (16a + 224)
3
+ (10a + 172)
2
+ (2a + 60) + 8
(20a2 + 112a 512) 4 + (80a2 + 448a 2048) 3 +(95a2 + 500a 2816) 2 + (30a2 + 104a 1536) 24a
288
27 3ax3 (2 + 1)(2 + 3)2 (2 + 5) 29 a2 x4 (2 + 1)(2 + 3)(2 + 5)(2 + 7) Let Y0 be the analytic solution of this equation. Then Y0 (x)
1 X
Bn xn = a(a
525x4
1)(a + 8)
n=0
4536 (15a + 493)x5 + ::: 25
n1 = 12 + 2a a n2 = 163 (13a + 33) 4 2a n2 = 3204 (77a2 + 774a 4505) 27 a n3 = 107582 + (279a3 2561a2 58654a + 255560) 32 2a n4 = 4203360+ (109650a4 +1150165a3 16814347a2 160500334a+763098616) 3125
Example 6. ( # 70=B c)
8
The An of # 70 is also the case a = 0 of Bn =
n X
2n n
an
n j
j
j;k=0
j k
2
2k k
n + 2j n
I have found a degree 7 equation which gives # 405 and # 385 for a = a = 9 respectively 4
4
x (52a + 270)
+x2
+ (64a + 540)
3
+ (48a + 411)
2
1 and
+ (16a + 141) + 2a + 18
(1344a2 + 10800a + 6561) 4 + (3456a2 + 32400a + 26244) 3 +(3848a + 36816a + 35721) 2 + (1744a2 + 16812a + 18954) + 268a2 + 2580a + 3240 2
22 a2 x3 (785a + 4050) 4 3 4
+2 a x
(509a + 1890)
4 4
+ 480a
3
+ (452a
26 a4 x5 (183a + 270)
4
(5a + 4176) 1080)
+ 360a
3
+28 a6 x6 (2 + 1)(20 10 7 6
3
2
+ (380a + 1962) + 140a + 1068
(575a + 366)
+ (392a + 24) 3
+ 50
2
2
2
+ (294a + 294) + 50a
+ 205a + 40a
6
6
+ 49 + 17)
2
2 a x ( + 1) (2 + 1)(2 + 3)
Let Y0 be the analytic solution of this equation. Then Y0 (x)
1 X
Bn xn
n=0
= a(a + 1)(a + 9)
5 (16417a + 159804)x4 + ::: 16
160x3
n1 = 27 + 4a a n2 = 432 (19a + 79) 2 2a (2454a2 + 25613a + 4718) n3 = 18089 27 a n4 = 997785 (2225923a3 + 36831814a2 172780203a + 24052584) 1536 a n4 = 68438142+ (435050488a4 2613465314a3 20667728a2 1492027527876a 303900011517) 150000 Let Y7 be the solution for the equation of degree 7 above and Y6 the corresponding solution for the degree 6 equation in Example 2. Then Y7 (x) = p
1 1
4ax 9
Y6 (
1
x ) 4ax
is valid for a = 0; 1: 9 but not for general a: Example 7. ( # 58=A c) 2 P n 2 2k Here An = 2n which is the case a = 0 of n k k n X
2n n
Bn =
an
n j
j
j;k=0
j k
2
2k k
n+j n
I have found a degree 2 equation which gives # 29 and # 41 for a = and a = 9 respectively 4
+4x2
4
2x (12a + 80) +(375
2
+ (24a + 160)
3
+ (19a + 124)
2
1
+ (7a + 44) + a + 6
(68a2 + 640a + 576) 4 + (272a2 + 2560a + 2304) 3 + 3520a + 3168) 2 + (206a2 + 1920a + 1728) + 39a2 + 360a + 324
Let Y0 be the analytic solution of this equation. Then Y0 (x)
1 X
Bn xn
n=0
400 3 x 9
= a(a + 1)(a + 9)
245 (133a + 1020)x4 + ::: 36
n1 = 16 + 2a a n2 = 142 (13a + 51) 4 11056 2a (1733a2 + 20174a 7155) n3 = 3 243 a (29113a3 + 850014a2 + 6679721a 3133692) n4 = 121470 1152 a n4 = 4971792+ (11411401a4 +98002158a3 1662950443a2 17315801988a+6524464212) 56250 Remark: We have for # 41= AZ-numbers in [4]) n X
2n n
( ) where ( ) has as coe¢ cients ( so called
n 3k
n + k (3k)! n k!3
( 27)n+k
n + 3k (4k)! 4k k!4
( 1)n+k 3n
k=0
= ( 1)n
n X
3k
k=0
10
=
n X
( 9)n+j
j;k=0
n j
j k
2
2k k
n+j n
See also [5]. References. 1.G.Almkvist, W.Zudilin, Di¤erential equations, mirror maps and zeta values, in: Mirror Symmetry V, N.Yui, S.-T. Yau, and J.D. Lewis (eds), AMS/IP Stud. Adv. Math. 38 (International Press & Amer. Math. Soc., Providence,RI, 2007) 481-515; arXiv: math/0402386 2.G.Almkvist, C.van Enckevort, D.van Straten, W.Zudilin, Tables of CalabiYau equations, arXiv: math/0507430. 3.G.Almkvist, D.van Straten, W.Zudilin, Generalizations of Clausen’s formula and algebraic transformations of Calabi-Yau di¤erential equations, preprint 2009-38 MPI Bonn. 4.H.H.Chan, H.Verrill, The Apéry numbers,the Almkvist-Zudilin numbers 1 and a new series for , Math. Res. Lett.16 (2009),no.3, 405-420. 5.H.H.Chan, Y.Tanigawa, Yifan Yang, and W.Zudilin, New analogues of Clausen’s identities arising from the theory of modular forms, Preprint 2009.
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