Thank you for your valuable suggestions, and we have corrected the typos and inappropriate notations in the modified version.
1. [This claim 2 should be removed] Indeed, we don't use retraction directly. However, we still need to bound the projection operator using retraction and vector transport based on Eq.19 and Eq.20. Thus, we must give the complete claim of retraction and vector transport.
2. [$T_{\alpha_k \eta_{i,k}}^R$ is not defined] Actually, we give its definition in Example 1, which is the vector transport constructed by retraction.
3. [The paper lacks an overview of the poof] Thanks for your suggestion. We have added an overview of the proofs in Figure 5 in the modified version.
4. [The experiments lacks the presentation of DRDGD and DPRGD] In fact, the presentation of DRDGD and DPRGD is shown in Figure 4. And their results are very close.
5. [Doubly stochastic matrix] For example, if we have 4 agents with ER $p=0.3$ graph, then the matrix satisfies
$W=\begin{bmatrix}
1/3 & 1/3 & 0 & 1/3 \\\
1/3 & 2/3 & 0 & 0 \\\
0 & 0 & 2/3 & 1/3 \\\
1/3 & 0 & 1/3 & 1/3
\end{bmatrix} $. If we have 4 agents with Ring graph, then the matrix satisfies $W=\begin{bmatrix}
1/3 & 1/3 & 0 & 1/3 \\\
1/3 & 1/3 & 1/3 & 0 \\\
0 & 1/3 & 1/3 & 1/3 \\\
1/3 & 0 & 1/3 & 1/3
\end{bmatrix}$.
6. [Is it a unit step size] It follows from the setting of DRDGD and DPRGD that a unit step size means the coefficient before $\sum_{j=1}^n W_{i j}^t x_{j,k}$ is 1.
7. [Assumption 3 (iii): what is $T_{\alpha_k \eta_{i,k}}^R$] It is a vector transport constructed by retraction, which is defined in Example 1.
8. [Lemma 1, eq (24): how do you get second and third inequalities] We miss a parenthesis in the second inequality, which has been corrected in the modified version. And $0 \leq \Vert x \Vert, \Vert y \Vert \leq 1$ is implicit in this proof.
9. [Is Theorem 2 new or is it a known result from a different paper] In fact, similar results have been given in [1], which we extended from retraction to projection operator and made it compatible with the conjugate gradient.
10. [Theorem 2 assumes that $\eta_{i,k}=0$] It's fine. Theorem 2 discusses the consensus error $\sum_{j=1}^n W_{i j}^t x_{j,k}$, and $\eta_{i,k}$ does not belong to the consensus error, so we consider setting it to 0. In particular, we still consider the role of $\eta_{i,k}$ in the part of global convergence (Appendix D).
11. [$x_{i,k_0+1} \rightarrow x_{i,k_0}$] There is no problem. It means that $x_{i,k_0+1}$ will no longer be updated and will remain at the previous point $x_{i,k_0}$ as the consensus error converges and the gradient approaches 0. A similar statement can be found in Theorem 4.2 based on [2].
12. [Second inequality in eq (36)] First, the second part comes from Eq. 35. Then, the third part is $$ \Vert \mathcal{P}\_{T\_{x\_{i,k}}\mathcal{M}}\left(\sum\_{j=1}\^n W\_{i j}\^t \eta\_{j,k-1}\right) \Vert \_{x\_{i,k}}\^2 \leq \Vert \sum\_{j=1}\^n W\_{i j}\^t \eta\_{j,k-1} \Vert \_{x\_{i,k}}\^2 \leq \Vert \eta\_{i,k-1} \Vert \_{x\_{i,k-1}}\^2$$, where $ \Vert \sum\_{j=1}\^n W\_{i j}\^t \eta\_{j,k-1} \Vert \_{x\_{i,k}}\^2 \leq \Vert \eta\_{i,k-1} \Vert \_{x\_{i,k-1}}\^2$ is a similar assumption followed with formula (4.28) in [2].
[1] Chen, Shixiang, et al. "Decentralized riemannian gradient descent on the stiefel manifold." International Conference on Machine Learning. PMLR, 2021.
[2] Sato, Hiroyuki. Riemannian optimization and its applications. Berlin: Springer, 2021.