Why VASP calculates the elastic constant for another trigonal space group?

Question

Please, help me to understand the problem. I calculated the elastic constants for the beta PHB (Space group P3221 (154)), it has to show the reduced elastic constant matrix (six), however, VASP calculated me 7 constants for the trigonal symmetry point group 3 (I used IBRION=6, ISIF =3, NFREE=2 and POMIT =0.015). I really don't have any idea why maybe anyone has the same problem? I attached my POSCAR file of beta PHB below. Thank you in advance!

My iNCAR:

PREC    = Accurate
ENCUT   = 520
EDIFF   = 1e-6
IBRION  = 6
NFREE   = 2
ISIF    = 3
ISYM    = 2
POTIM   = 0.03
NSW     = 1
ISMEAR  = 0
SIGMA   = 0.1
ALGO    = Normal
IALGO   = 48
ISTART  = 0
LPLANE  = .TRUE.
LCHARG  = .FALSE.
LWAVE   = .FALSE.

My POSCAR :

betaPHB                                 
   1.00000000000000     
     9.8451449602803063   -0.0000000000135996    0.0000000000000000
    -4.9225724801283759    8.5261456395240547    0.0000000000000000
     0.0000000000000000    0.0000000000000000    4.7787902416829580
   C    O    H 
    12     6    18
Direct
  0.4473348233320940  0.0098405107170283  0.6760647908616504
 -0.0098405007170275  0.4374943136150696  0.3427314468616541
  0.5625056563849357  0.5526651466679172  0.0093981238616490
  0.3366105966273176  0.9865190814127033  0.9194381388236403
  0.0134809465872924  0.3500915452146104  0.5861048248236396
  0.6499084547853901  0.6633893733726792  0.2527714818236401
  0.3529941479661700  0.9450538256463469  0.4095367091398003
  0.0549461813536503  0.4079403223198091  0.0762033731398051
  0.5920596476801820  0.6470058220338273  0.7428700221398052
  0.2595370428953800  0.0867656901155276  0.8956333322755682
  0.9132342808844670  0.1727713537798490  0.5623000192755711
  0.8272286462201509  0.7404629871046158  0.2289666612755733
  0.4289822938404916  0.0307095082417344  0.1799468951856006
  0.9692905017582659  0.3982727945987476  0.8466135811855999
  0.6017272054012526  0.5710177061595080  0.5132802081855978
  0.2217465615454436  0.8287633736506727  0.4032896651433156
  0.1712366113493296  0.3929831728947801  0.0699563291433201
  0.6070168271052196  0.7782534534545532  0.7366229781433196
  0.5360727369674801  0.1339552292303300  0.6484256472182774
  0.8660447627696694  0.4021175007371532  0.3150923032182820
  0.5978824992628468  0.4639272630325195  0.9817589602182746
  0.5084713038609513  0.9437122479440904  0.7170050087321858
  0.0562877230559037  0.5647590559168535  0.3836716647321826
  0.4352409730831384  0.4915286961390490  0.0503383407321812
  0.2476974100787215  0.8620678575193049  0.9347039188362016
  0.1379321714807079  0.3856295665594249  0.6013706048362009
  0.6143704334405750  0.7523025749212737  0.2680372618362011
  0.1875979130778683  0.0556088716245804  0.7051825810070412
  0.9443911253754161  0.1319890374532876  0.3718492370070449
  0.8680109625467131  0.8124020869221317  0.0385159130070434
  0.1827452577710241  0.0681168022337463  0.0754234213029564
  0.9318831687662399  0.1146284565372744  0.7420901073029557
  0.8853715284627206  0.8172547562289704  0.4087567643029560
  0.3478634616025427  0.2118811433998376  0.8853498234737521
  0.7881188266001669  0.1359823182027122  0.5520165094737514
  0.8640177107972945  0.6521365383974570  0.2186831514737536

Answer 1

Elastic constants can be calculated using two approach, one with Energy-strain relationship and another with stress-strain relationship. The key ingredient in above two methods is energy or stress at particular mode of distortion. For example in cubic crystal which have 3 fold symmetry about body diagonal need three independent elastic parameters to define second-order elastic constants. VASP algorithm IBRION=6 also do the same. Some time problem of not getting elastic constants correctly depends on many parameters. Specially with large systems which consists of many number of atoms along with different atomic numbers, full volume relaxation results in difference between input and output structure. For example you can try relaxation of 4 atom aluminum unit cell and re-relaxed its supercell again with same K-space density and kinetic energy cutoff. You may get minor difference in output configuration with ISIF=3(full volume and position relaxation). You can also simply check your input file symmetry using python spglib library or just use vaspkit. The question arises if there is minor change in symmetry, can we get same number of elastic constants as it should be. The answer is check your elastic constants values, is two elastic constants are apart by small amount, you can always say those are equal. Other way to calculate elastic constants without VASP andIBRION=6 is just distort crystal manually based on symmetry and do only position relaxation(ISIF=2) and then fit energy and distorsion by second order polynomial.