First note that the two-dimensional honeycomb lattice is not a Bravais lattice. It can however be represented as a triangular Bravais lattice with a basis. The Brillouin zone of the latter lattice is a parallelogram (for the relevant details, consult Sec. 2.6 of the book Principles of condensed matter physics, by Chaikin and Lubensky (Cambridge University Press, 2000)). Now, introduce a transformation that maps a parallelogram onto a square.*) Hereby you will be able to express the relevant Brillouin-zone integrals over a parallelogram (the underlying Brillouin zone) as integrals over a square. Following this, you can directly follow the discretization procedure as described by Fukui et al., which is specific to a square-shaped Brillouin zone -- becoming equivalent to a torus on identifying the opposite sides of this square on the basis of periodicity.

*) For this you can cut the triangular part of the relevant parallelogram on one side and rigidly shift it to the other, and thus form a rectangle (this is allowable because of the periodicity of the underlying integrands). Following this, a rescaling (depending on the direction, by 2√3/3, or √3/2) of the lengths along one direction is to be effected in order to turn a rectangle into a square. 

First note that the two-dimensional honeycomb lattice is not a Bravais lattice. It can however be represented as a triangular Bravais lattice with a basis. The Brillouin zone of the latter lattice is a parallelogram (for the relevant details, consult Sec. 2.6 of the book Principles of condensed matter physics, by Chaikin and Lubensky (Cambridge University Press, 2000)). Now, introduce a transformation that maps a parallelogram onto a square.*) Hereby you will be able to express the relevant Brillouin-zone integrals over a parallelogram (the underlying Brillouin zone) as integrals over a square. Following this, you can directly follow the discretization procedure as described by Fukui et al., which is specific to a square-shaped Brillouin zone -- becoming equivalent to a torus on identifying the opposite sides of this square on the basis of periodicity.

*) For this you can cut the triangular part of the relevant parallelogram on one side and rigidly shift it to the other, and thus form a rectangle (this is allowable because of the periodicity of the underlying integrands). Following this, a rescaling (depending on the direction, by 2√3/3, or √3/2) of the lengths along one direction is to be effected in order to turn a rectangle into a square. 

Hello Professor Farid 

Thank you for your timely reply. I draw a sketch to illustrate what you mentioned,  is it right? If that's true,  I think it means that I shifted the BZ to a new area and I need to choose another group of coordinates for mesh and calculation。

And I find another problem as well, since the rectangle's length/width isn't an integer,which means the number of small square is not an integer too, no matter how to choose the rescale factor, this result remains, I can't imagine the situation where the edge meshed with irrational number.

You are welcome Yuesu! Yes, that is exactly right! Wonderful.

For clarity, suppose the function f(k) is periodic with period k0. The integral of this function over [κ, κ+k0] is the same for all values of κ. This property generalises to all dimensions. In the case at hand, you can express the integral along the vertical direction as a Riemann sum, and apply the above rule to each integral along the horizontal direction. After this, you can effect the appropriate limit and write the above-mentioned Riemann sum as an integral. This proves that the construction as depicted in your diagram is correct.

Lastly. please mind the scaling factor to which I referred in my first response on this page.

Dear Professor Fraid

Before I write this reply, I wrote another long one to prove that a rectangle with irrational length-width ratio cannot be covered by finite square plaquettes, but when I read your last sentence again, I understand your stressing on the scaling factor, that's really important.... But at the same time, I have another question, if we can rescale the Brilliouin zone, why not we choose the plaquette with rectangle shape? The Berry curvature is calculated on a loop enclosed by four vertexes of one plaquette ,the local gauge is well defined on each plaquette, there is no constrain on side length of each one. Whether could I choose rectangle mesh directly? 

Dear Yuesu, the 'rescaling' to which I have referred has bearing on the fact that the geometrical construction I suggested for a honeycomb lattice gives rise to a rectangle, not a square, which you need to use in applying the procedure proposed by Fukui et al. without any modification. Of course, this rescaling rescales the measure, but so long as you perform your calculations correctly, nothing is really rescaled. I emphasise that the rescaling is only required for turning the underlying rectangle into a squire, so that the procedure by Fukui et al. is directly applicable, without modification. This is not to say that the procedure by Fukui et al. were the only conceivable procedure.

Regarding your other question, by calculating some relevant quantities, such as the Bloch wave functions, over the Brillouin zone, there is always the possibility of picking up an arbitrary phase (numerically, this phase is determined by a variety of causes over which one has no control -- for instance, in diagonalising the Hamiltonain matrix, the phase is determined by the method employed, but also by round-off error), and this is fatal for amongst others calculation of the Chern number. To calculate an integral over the Brilluoin zone, one necessarily approximates the integral by a discrete sum. With the underlying integrand in principle picking up an arbitrary phase in going from one point to the next, unless the quantity calculated is gauge invariant (such as for instance electronic charge density), one has to take special measures to remove this arbitrary phase fluctuation. This is what Fukui et al. in essence do.

Dear Professor Farid

Yes, you are right, nothing is really rescaled, rescalling  is just because of Fukui's requirement about square lattice. I tried to make a small Matlab program to carry out the calculation and have been fighting with the program for two days. I have a common used two-level Hamiltonian for trial, but what I feel confusing is that only if I set the diagonal term(or the soc term in KM model) equals to zero, it gives me the Chern number 1, if the soc gap is opened,  all of loop on plaquettes canceled-out  with each other when I summed them up.  I have considered the periodical boundary  condition, but things still goes wrong......

I think the "battle" between Fukui and me will last for a while longer, :）......

Thank you again!  Wish you have a good weekend!

Dear Yuesu, thank you for the update. 

With reference to your present calculations, two remarks. First, test whether the software you are using yields the principal branch of the logarithm function correctly. This is relevant for the field strength as defined in Eq. (8) of the paper by Fukui et al. Second, I have noticed that Fukui et al. use a one-sided finite difference for calculating the first derivative. This is not a good choice. Better is to use the centred finite difference, which is by one order more accurate. To appreciate this, let x be in the neighbourhood of 0. For f(x) a sufficiently smooth function around x=0, one has

(1) f(x) = f(0) + f ' (0) x + f '' (0) x2/2 + f ''' (0) x3/3! + ...

from which one obtains

(2) f ' (0) = [f(h) - f(0)]/h + O(h),

where h > 0 is the step size, small enough so that the expansion in Eq. (1) applies for x = h. In contrast, from Eq. (1) one obtains

(3) f ' (0)  = [f(h/2) - f(-h/2)]/h + O(h2).

Note the orders O(h) and O(h2) in Eqs (2) and (3). The fact that the third term on the right-hand side of Eq. (1) even power of x is responsible for the contribution of this term to f ' (0) not surviving the difference f(h/2) - f(-h/2). □

I have been privately asked about the transformation mediating the deformation of a parallelogram into a square. Aside from what I have mentioned earlier here above (referring to translations, etc.), deformation of a parallelogram into a square (or rectangle) is a linear transformation [1]. Explicitly, let a and b be two vectors centred at the origin of the coordinate system defining a parallelogram, and a' and b' the orthogonal vectors centred at the origin defining the desired square. One has

(1) a' = A a,

(2) b' = A b,

where A stands for a 2 × 2 matrix. With respect to the Cartesian coordinate system centred at the origin, consider the following vectors (below the superscript T denotes transpose):

(3) aT = (ax, ay), bT = (bx, 0),

(4) a'T = (0, d), b'T = (d, 0).

For the elements of the desired matrix A, as encountered in (1) and (2), one has:

(5) A11 = d/bx, A12 = -axd/aybx, A21 = 0, A22 = d/ay.

Naturally, from (1) and (2) one has

(6) a = A-1 a',

(7) b = A-1 b',

where A-1 denotes the inverse of A.

[1] https://mathinsight.org/linear_transformations_map_parallelograms_parallelepipeds

My God!! What a kind of wrong things said here in this thread! The Chern number only measures the non triviality of given topology through the curvature (gaussian, etc), and fortunately we have nowadays the Berry phase to associate to the bands the possibility to calculate their curvature. In most of the materials the gauge transformation of this phase is chosen to be Abelian U(1) and in such a case you have only -1 or +1 for the Chern number associated to one of the singularities of the non trivial band functional space, but there are more possibilities as SU(3) or even SU(3).

This is not related which saying that we have two Bravais lattices (pseudo spin concept) or other crazzy things. I think that people who is ignorant of one question it should avoid to answer threads because the level of RG deserves a respect.

Dear Yuesu,

I answer you here in your thread for giving also information to people who could try to follow this subject and not in my private message. When you ask for a concrete calculation it is only a question to apply the formalism, but if you ask for a general subject as the graphene it is impossible to help you (at least this is my case). Let me to suggest you one reference where you have the main ingredients that you are looking for

https://arxiv.org/pdf/1901.09748.pdf

Another solution is that you put in contact with the authors of the paper that you presented for entering in more details

Those who do not know a word about Fukui's paper under discussion (intellectually so lazy as not even to have checked the paper, which has been explicitly cited above, before calling others names) should keep their quiet, or their utter ignorance will be made more manifest!

The Chern number and its calculation is independent of Fukui's paper and less if the explanations given are deeply wrong or misunderstood, based in such a paper. Less if the explanations given are deeply wrong or misunderstood based in such a paper

Idle prattle is no scientific argumentation! Those who wilfully shed negative light on the responses by others here on RG, must tell what the correct response must have been -- hitting and running is cowardly, to say the least, not befitting someone who purports to be a scientist!

Lastly, for those ignorant enough not to know what Chern number is, and less how it is calculated in practice, the Chern number as calculated by Fukui is neither +1 nor -1, but -2, coinciding with the known correct value for the system considered. This should indisputably demonstrate the depth of ignorance of those who go though the pages of RG in pursuit of making a reputation by defaming others. Being ignorant is one thing, being an ignorant ruffian quite another!

Question

How to do Chern Number calculation?

First answer (4 Recommendations)

First note that the two-dimensional honeycomb lattice is not a Bravais lattice. It can however be represented as a triangular Bravais lattice with a basis. The Brillouin zone of the latter lattice is a parallelogram (for the relevant details, consult Sec. 2.6 of the book Principles of condensed matter physics, by Chaikin and Lubensky (Cambridge University Press, 2000)). Now, introduce a transformation that maps a parallelogram onto a square.*) Hereby you will be able to express the relevant Brillouin-zone integrals over a parallelogram (the underlying Brillouin zone) as integrals over a square. Following this, you can directly follow the discretization procedure as described by Fukui et al., which is specific to a square-shaped Brillouin zone -- becoming equivalent to a torus on identifying the opposite sides of this square on the basis of periodicity.

*) For this you can cut the triangular part of the relevant parallelogram on one side and rigidly shift it to the other, and thus form a rectangle (this is allowable because of the periodicity of the underlying integrands). Following this, a rescaling (depending on the direction, by 2√3/3, or √3/2) of the lengths along one direction is to be effected in order to turn a rectangle into a square. 

Do you think that this is the answer when the Chern number of a torus is always zero?

Daniel Baldomir: You clearly don't know what Brillouin zone (BZ) is! This is not surprising, since you demonstrably don't know what Chern number is! The torus to which I had referred concerns the k-space, which is not the Berry connection -- here the Berry connection is defined over the BZ. The sheer stupidity of the issues that you raise is on a par with objecting to the statements that the derivative of sin(x) is equal to 1 at x=0 and equal to 0 at x=π/2 on account of the fact that the real x-axis has no curvature! Forget about torus and the Chern number, and just take 1/z integrated in the anti-clockwise direction along a circle centred at the origin in the complex z-plane. Your objection is on a par with objecting to the statement that the latter integral is equal to 2πi, and this objection on account of a circle being a closed curve -- with no end-points, or boundary (in the same way that a torus is a closed space and has no boundary), or the complex z-plane being flat.

As for your last question: the general rule applies here as elsewhere that those who answer questions with questions are not genuine. You had rudely accused me of making wrong statements on this page, and when I asked you to present the correct statements, you came up with questions, and in doing so copying my text without appropriately putting it inside the quotation marks (this amounts to a misappropriation of my text, which is my intellectual property, and not yours thereby to copy and paste as you please)!

As I have told you earlier, and more than once, I have absolutely nothing to discuss with you; you demonstrably don't know the rudiments of the subjects you propound on the pages of RG! You have shown not even to know how the Bohr radius is defined in solid state physics, ..., and now what the BZ and the Chern number are. I am very sorry to bring up these scandals of first order, but you seem compulsive in creating conditions where one is forced to remind you of your glaring professional shortcomings! Stop throwing mud at me at every occasion, since it will ultimately only cost you your personal dignity. It doesn't give me any pleasure to address you, or anyone else for that matter, in this way.

That is enough and I think that any clever reader can understand what we have said. For seeing if our knowledge is credible in this issue, everybody can visit our publications in RG.

I wish a good day for all !!!

Daniel Baldomir: For the record, as for what "we" have said, demonstrably you said nothing of any substance! You came here just to throw some mud, and failed to say a word indicative of your knowledge of the subject matter. The little you said proved that what you deem to know has no bearing on reality.

You can follow

https://mathinsight.org/linear_transformations_map_parallelograms_parallelepipeds

Wise an polite opinion. I wish you the best and that you take care in the next answer in RG. Have a good day!

Moaid K Hussain : It is not clear to me why you have posted the above link, specifically because it is a link which I have cited in my comment on this page dated 22 May 2019. It is also not clear who the "You" (in "You can follow") refers to.

Principles of condensed matter physics, by Chaikin and Lubensky (Cambridge University Press, 2000)

https://www.amazon.es/Principles-Condensed-Matter-Physics-Paperback/dp/0521794501/ref=sr_1_1?hvadid=80195661345477&hvbmt=bp&hvdev=c&hvqmt=p&keywords=principles+of+condensed+matter&qid=1564234285&s=gateway&sr=8-1

Daniel Baldomir: Why are you cluttering the page with irrelevancies?! This seems to be your camouflage strategy, cluttering the RG pages where you have lost an argument so as to make the page unreadable. For clarity, the book by Chaikin & Lubensky contains not a single reference to Chern numbers, nor to the Berry connection and assorted concepts.

It is your reference on this issue and I think the idea of Moaid is good perhaps people don't know what important is for your answer.

Question

How to do Chern Number calculation?

First answer (4 Recommendations)

First note that the two-dimensional honeycomb lattice is not a Bravais lattice. It can however be represented as a triangular Bravais lattice with a basis. The Brillouin zone of the latter lattice is a parallelogram (for the relevant details, consult Sec. 2.6 of the book Principles of condensed matter physics, by Chaikin and Lubensky (Cambridge University Press, 2000)).T Now, introduce a transformation that maps a parallelogram onto a square.*) Hereby you will be able to express the relevant Brillouin-zone integrals over a parallelogram (the underlying Brillouin zone) as integrals over a square. Following this, you can directly follow the discretization procedure as described by Fukui et al., which is specific to a square-shaped Brillouin zone -- becoming equivalent to a torus on identifying the opposite sides of this square on the basis of periodicity.

*) For this you can cut the triangular part of the relevant parallelogram on one side and rigidly shift it to the other, and thus form a rectangle (this is allowable because of the periodicity of the underlying integrands). Following this, a rescaling (depending on the direction, by 2√3/3, or √3/2) of the lengths along one direction is to be effected in order to turn a rectangle into a square. 

Do you think that this is the answer when the Chern number of a torus is always zero?