shiftk and bse kernel

[lawson]

I have questions about how YAMBO handles shifts of the k-space grid.
When I use the default shiftk in abinit for an 8x8x8 grid:
ngkpt 8 8 8
nshiftk 1
shiftk 0.5 0.5 0.5

I get an 16x16x16 grid in YAMBO:
Grid dimensions : 16 16 16

However, if I use an unshifted grid:
ngkpt 8 8 8
nshiftk 1
shiftk 0.0 0.0 0.0

I get an 8x8x8 grid in YAMBO. This difference of course means a factor of 8x in the
dimension of the BSE kernel.

Could you explain the reason for this difference?

A related question is: I am trying to reproduce Eric Shirley's BSE results for diamond. He
uses an 8x8x8 grid shifted by 1/64,2/64,3/64 (PRB 57, pR9385, (1998)). When I try this
symmetry breaking grid in YAMBO, I get an error:
<02s> [03] Transferred momenta grid
[ERROR] STOP signal received while in :[03] Transferred momenta grid
[ERROR][RL indx] 2 equivalent points in the rlu grid found

Could you explain this error and why Eric's shift won't work in YAMBO?

Below is the abinit input file I am using for diamond.

Thanks for any help!

John


###########################################################################
# Diamond
# Generation of the KSS file
###########################################################################

# Number of datasets: 2-step calculation
ndtset 2

# Definition of the unit cell: fcc
acell 3*3.54 Angstrom
rprim 0.0 0.5 0.5
0.5 0.0 0.5
0.5 0.5 0.0

# Definition of the system
ntypat 1
znucl 6
natom 2
typat 1 1
xred 0.0 0.0 0.0
0.25 0.25 0.25

ecut 20
nband 8
kptopt 1

#CASE 1
ngkpt 8 8 8
nshiftk 1
shiftk 0.0 0.0 0.0

#CASE 2
#ngkpt 8 8 8
#nshiftk 1
#shiftk 0.5 0.5 0.5

#CASE 3
#ngkpt 8 8 8
#nshiftk 1
#shiftk 1/64 2/64 3/64

# Step 1: determine the ground state
iscf1 5
toldfe1 1.0d-8
nstep1 100
prtden1 1

# Step 2: obtain the Kohn-Sham band structure
symmorphi2 0 # Disallow non-symmorphic operations (required)
iscf2 -2
tolwfr2 1.0d-18
nstep2 100
getden2 1
kssform2 3 # Format of the KSS file
nbandkss2 8 # Number of bands to output to the KSS file
istwfk2 29*1