Lattice Green’s Functions of the Higher-Dimensional Face-Centered Cubic Lattices Christoph Koutschan MSR-INRIA Joint Centre, Orsay, France November 7 S´ eminaires Algorithms
❩ Introduction We consider lattices in ❘ d d � � � ⊆ ❘ d n i a i : n 1 , . . . , n d ∈ ❩ i =1 for some linearly independent vectors a 1 , . . . , a d ∈ ❘ d .
Introduction We consider lattices in ❘ d d � � � ⊆ ❘ d n i a i : n 1 , . . . , n d ∈ ❩ i =1 for some linearly independent vectors a 1 , . . . , a d ∈ ❘ d . → Simplest instance is the integer lattice ❩ d − (choose a i = e i , the i -th unit vector): • d = 2 : “square lattice” • d = 3 : “cubic lattice” • d > 3 : “hypercubic lattice” The study of such lattices was inspired by crystallography in as much as the atomic structure of crystals forms such regular lattices.
Topic of this Talk Study random walks on the face-centered cubic (fcc) lattice . Consider random walks on the lattice points: • In each step move to one of the nearest neighbors. • All steps have the same probability. • A point can be visited several times. • Starting point is the origin 0 .
The fcc Lattice in 2D square lattice (= integer lattice ❩ 2 )
The fcc Lattice in 2D square lattice (= integer lattice ❩ 2 )
The fcc Lattice in 2D square lattice (= integer lattice ❩ 2 )
The fcc Lattice in 2D square lattice (= integer lattice ❩ 2 )
The fcc Lattice in 2D square lattice (= integer lattice ❩ 2 )
The fcc Lattice in 2D square lattice (= integer lattice ❩ 2 )
The fcc Lattice in 2D square lattice (= integer lattice ❩ 2 )
The fcc Lattice in 2D square lattice (= integer lattice ❩ 2 )
The fcc Lattice in 2D square lattice (= integer lattice ❩ 2 )
The fcc Lattice in 2D square lattice (= integer lattice ❩ 2 )
The fcc Lattice in 2D square lattice (= integer lattice ❩ 2 )
The fcc Lattice in 2D square lattice (= integer lattice ❩ 2 )
The fcc Lattice in 2D square lattice (= integer lattice ❩ 2 )
❩ The fcc Lattice in 2D face-centered square lattice
❩ The fcc Lattice in 2D face-centered square lattice
❩ The fcc Lattice in 2D face-centered square lattice
❩ The fcc Lattice in 2D face-centered square lattice
The fcc Lattice in 3D
The fcc Lattice in 3D
The fcc Lattice in 3D
The fcc Lattice in 3D
The fcc Lattice in 3D
The fcc Lattice in 3D
The fcc Lattice in 3D
The fcc Lattice in 3D
The fcc Lattice in 3D
The fcc Lattice in 3D
The fcc Lattice in 3D
The fcc Lattice in 3D
The fcc Lattice in 3D Densest possible packing: Kepler conjecture (Hales 2005) − → This arrangement is often encountered in nature, e.g., in aluminium, copper, silver, and gold.
The fcc Lattice in 3D It is not difficult to see that the 3D fcc lattice consists of four copies of ❩ 3 , namely � � � � � � � 1 � 1 � � � � ❩ 3 ∪ ❩ 3 + ❩ 3 + ❩ 3 + 2 , 1 2 , 0 , 1 0 , 1 2 , 1 2 , 0 ∪ ∪ . 2 2
The fcc Lattice in 3D It is not difficult to see that the 3D fcc lattice consists of four copies of ❩ 3 , namely � � � � � � � 1 � 1 � � � � ❩ 3 ∪ ❩ 3 + ❩ 3 + ❩ 3 + 2 , 1 2 , 0 , 1 0 , 1 2 , 1 2 , 0 ∪ ∪ . 2 2 From now on: Stretch the lattice by a factor 2 to avoid fractions. Then the admissible steps (nearest neighbor rule) are: { ( − 1 , − 1 , 0) , ( − 1 , 1 , 0) , (1 , − 1 , 0) , (1 , 1 , 0) ( − 1 , 0 , − 1) , ( − 1 , 0 , 1) , (1 , 0 , − 1) , (1 , 0 , 1) (0 , − 1 , − 1) , (0 , − 1 , 1) , (0 , 1 , − 1) , (0 , 1 , 1) }
The fcc Lattice in Arbitrary Dimension � d � The d -dimensional fcc lattice is composed of 1 + translated 2 copies of ❩ d . The set of permitted steps in the d -dimensional fcc lattice is � � ( s 1 , . . . , s d ) ∈ { 0 , − 1 , 1 } d : | s 1 | + · · · + | s d | = 2 , � d � i.e., there are 4 steps (called the coordination number ). 2
Lattice Green’s Functions The lattice Green’s function is the probability generating function ∞ � p n ( x ) z n P ( x ; z ) = n =0 where p n ( x ) = probability of being at position x after n steps.
Lattice Green’s Functions The lattice Green’s function is the probability generating function ∞ � p n ( x ) z n P ( x ; z ) = n =0 where p n ( x ) = probability of being at position x after n steps. → Note that c n p n ( x ) is an integer and gives the total number − of such (unrestricted) walks, where c is the coordination number of the lattice.
Lattice Green’s Functions Of particular interest is � π � π ∞ p n ( 0 ) z n = 1 d k 1 . . . d k d � P ( 0 ; z ) = . . . 1 − zλ ( k ) . π d 0 0 n =0 that describes the return probabilities. Here λ ( k ) is the structure function , given by the discrete Fourier transform of the single-step probabilities: � p 1 ( x ) e i x · k λ ( k ) = x ∈ ❘ d (a finite sum, actually).
Example Square lattice ❩ 2 with step set { ( − 1 , 0) , (1 , 0) , (0 , − 1) , (0 , 1) } The structure function is λ ( k 1 , k 2 ) = 1 � e − ik 1 + e ik 1 + e − ik 2 + e ik 2 � = 1 2 (cos k 1 + cos k 2 ) . 4 The lattice Green’s function is � π � π P (0 , 0; z ) = 1 2 (cos k 1 + cos k 2 ) = 2 d k 1 d k 2 π K ( z 2 ) 1 − z π 2 0 0 where K ( z ) is the complete elliptic integral of the first kind.
Return Probability Question: What is the probability that a walker ever returns to the origin? The return probability R (P´ olya number) is given by 1 1 R = 1 − n =0 p n ( 0 ) = 1 − P ( 0 ; 1) . � ∞
Return Probability Question: What is the probability that a walker ever returns to the origin? The return probability R (P´ olya number) is given by 1 1 R = 1 − n =0 p n ( 0 ) = 1 − P ( 0 ; 1) . � ∞ In our 2D example: 1 R = 1 − π K (1) = 1 2 since K ( z ) diverges for z = 1 . − → It is well known that in 2D the return is certain!
Back to the fcc Lattice The trivial (but illuminating) 2D case: • step set: { ( − 1 , − 1) , ( − 1 , 1) , (1 , − 1) , (1 , 1) } • structure function: 1 � e − i ( k 1 + k 2 ) + e − i ( k 1 − k 2 ) + e i ( k 1 − k 2 ) + e i ( k 1 + k 2 ) � λ ( k 1 , k 2 ) = 4 1 � � = cos( k 1 + k 2 ) + cos( k 1 − k 2 ) = cos k 1 cos k 2 , 2 using the angle-sum identity cos( x ± y ) = cos x cos y ∓ sin x sin y . • lattice Green’s function: � π � π P (0 , 0 , z ) = 1 d k 1 d k 2 = 2 π K ( z 2 ) . π 2 1 − z cos k 1 cos k 2 0 0 − → LGF is the same as for the square lattice (as expected), but not at all obvious from the integral representation!
fcc Lattices for d > 2 � d � d d − 1 � � The structure function is λ ( k ) = cos k m cos k n . 2 m =1 n = m +1
fcc Lattices for d > 2 � d � d d − 1 � � The structure function is λ ( k ) = cos k m cos k n . 2 m =1 n = m +1 For d = 3 , the return probability is one of Watson’s integrals : √ � � − 1 � π � π � π 4 π 4 3 1 d k 1 d k 2 d k 3 = 1 − 16 R = 1 − π 3 1 − 1 9(Γ( 1 3 )) 6 3 ( c 1 c 2 + c 1 c 3 + c 2 c 3 ) 0 0 0 where c i = cos( k i ) .
fcc Lattices for d > 2 � d � d d − 1 � � The structure function is λ ( k ) = cos k m cos k n . 2 m =1 n = m +1 For d = 3 , the return probability is one of Watson’s integrals : √ � � − 1 � π � π � π 4 π 4 3 1 d k 1 d k 2 d k 3 = 1 − 16 R = 1 − π 3 1 − 1 9(Γ( 1 3 )) 6 3 ( c 1 c 2 + c 1 c 3 + c 2 c 3 ) 0 0 0 where c i = cos( k i ) . A closed form for the LGF has been found by Joyce (1998), in terms of K ( z ) and some fairly complicated algebraic functions. − → For d > 3 no such closed forms are known!
Differential Equation Approach From now on: try to compute a differential equation for the LGF, instead of a closed form!
Differential Equation Approach From now on: try to compute a differential equation for the LGF, instead of a closed form! A conjecture (“guess”) for such an equation can be made when the first terms of the Taylor expansion are known. These can be obtained by different methods, e.g. 1. rewrite and expand the d -fold integral into a multisum (Guttmann and Broadhurst) 2. count all possible walks on the lattice 3. count the excursions using multi-step guessing
Differential Equation Approach From now on: try to compute a differential equation for the LGF, instead of a closed form! A conjecture (“guess”) for such an equation can be made when the first terms of the Taylor expansion are known. These can be obtained by different methods, e.g. 1. rewrite and expand the d -fold integral into a multisum (Guttmann and Broadhurst) 2. count all possible walks on the lattice 3. count the excursions using multi-step guessing − → However, any result obtained in this way is just a conjecture !
Method 1 (Guttmann and Broadhurst) Example for d = 3 ( c i denotes cos k i ) Expand the integrand in a geometric series: 1 � z � n � ( c 1 c 2 + c 1 c 3 + c 2 c 3 ) n 3 ( c 1 c 2 + c 1 c 3 + c 2 c 3 ) = 1 − z 3 n
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