The "Corollary 1" formulation in Ref. [1] is incorrectly presented. The flawless should be maintained, and the misrepresentation must be corrected.

1 Corollary 1 and its proof in Ref. [1]

In Ref. [1], Corollary 1 is incorrectly proposed. Making the letter self-contained, the corollary and its proof are rewritten as follows:

Corollary 1 The symmetric part of the deformation gradient F has no contribution to the displacement change Δu and the displacement density tensor T.

Proof Let and be the symmetric part and anti-symmetric part of the deformation gradient F, respectively. As we know, any tensor can be decomposed into a symmetric part and an antisymmetric part, i.e., F= S+ Ω. Since the curl of the symmetric tensor vanishes, curlS= S× ∇_X=0, and curlF= curl(S+ Ω)=curlΩ =Ω × ∇_X. Then, we have

(1)

curlS= 0 indicates that the symmetric part of the deformation gradient F does not have any contribution towards to the comparability conditions. In other words, the symmetric deformations are always compatible, and the incompatible deformation will make the symmetric deformation breaks down.

Since and thus we have

(2)

It must be pointed out that the statement of the above corollary and its proof are wrong and should be corrected.

2 Correction of Corollary 1 and its proof

Before presenting the corrected version of the above corollary, let us first prove a Lemma as follows:

Lemma 1 The trace of curl of a symmetric tensor is zero.

Proof Let be a symmetric tensor, i.e., A_ij=A_ji. The curl of the tensor A is defined as follows:

where e_jkm is the permutation symbol. Therefore, we can define the trace of the curl of tensor A by tr(curlA)=I:curlA =I: (A× ∇), where is the Kronecker delta. Therefore,

Since A_ij=A_ji and e_jki=-e_ikj, we have tr(curlA)=∇_kA_ije_jki=0.

Corollary 2 The symmetric part of the deformation gradient F has no contribution to the trace of displacement density tensor T.

Proof Let and be the symmetric part and the anti-symmetric part of the deformation gradient F, respectively. As we know, any tensor can be decomposed into a symmetric part and an antisymmetric part, i.e., F= S+Ω.

Since the trace of the curl of the symmetric tensor vanishes, i.e., tr(curl S)=tr(S× ∇_X)= 0, we have tr(curl F)= tr(curl(S+ Ω))=tr(curlS)+curlΩ =tr(Ω × ∇X). Then, we have the trace of the displacement density tensor T as follows:

(3)