Questions about left update after LM optimization

[Joweay]

Dear author, thanks for your excellent work.I read the paper and the code recently, but some questions in my mind. firstly, it is why the position left_update left multiply dR, specifics in damping_inter function? as I think the dR is not necessary. the second,why do the transformation which transfer x_states to the pose relative to es0.

[Zale-Liu]

(1) You can see the left update in the paper. [dR & dp // 0 & 1 ] * [R & p // 0 & 1] = [dRR & dRp + dp // 0 & 1]. It is the update in SE3. (2) We just want to keep the first pose unchanged. Of course, you can use other manners to fix the first pose (e.g., set the parts corresponding to the first pose of the Hessisan matrix into identity.)

[Zale-Liu]

Of course, we recommend you use the right update. It is similar to yours.

[Joweay]

(1) You can see the left update in the paper. [dR & dp // 0 & 1 ] * [R & p // 0 & 1] = [dR_R & dR_p + dp // 0 & 1]. It is the update in SE3. (2) We just want to keep the first pose unchanged. Of course, you can use other manners to fix the first pose (e.g., set the parts corresponding to the first pose of the Hessisan matrix into identity.)

thanks for your patient reply. 1.after learning about your explanation, I understand the reason that left dR with p when do the updating. 2. why fix the first pose unchanged? the reason is the experience from the experiments or some theory consideration. 3. I also read the BALM_old code and papers. in the balm_old code, state updating adopt left update, but the position don't left multiply the dR, there maybe is a small bug, details is as following /***********************************************/ cv::solve(matA, matB, matX, cv::DECOMP_QR); for (int j = 0; j < jac_leng; j++) { dxi(j, 0) = matX.at(j, 0); }

    for (int j = 0; j < slwd_size; j++) {
        // left multiplication
        so3_poses_temp[j] = SO3::exp(dxi.block<3, 1>(6 * (j), 0)) * so3_poses[j];
        t_poses_temp[j] = t_poses[j] + dxi.block<3, 1>(6 * (j) + 3, 0);
    }

/***********************************************/

[Joweay]

(1) You can see the left update in the paper. [dR & dp // 0 & 1 ] * [R & p // 0 & 1] = [dR_R & dR_p + dp // 0 & 1]. It is the update in SE3. (2) We just want to keep the first pose unchanged. Of course, you can use other manners to fix the first pose (e.g., set the parts corresponding to the first pose of the Hessisan matrix into identity.)

thanks for your patient reply. 1.after learning about your explanation, I understand the reason that left dR with p when do the updating. 2. why fix the first pose unchanged? the reason is the experience from the experiments or some theory consideration. 3. I also read the BALM_old code and papers. in the balm_old code, state updating adopt left update, but the position don't left multiply the dR, there maybe is a small bug, details is as following /***********************************************/ cv::solve(matA, matB, matX, cv::DECOMP_QR); for (int j = 0; j < jac_leng; j++) { dxi(j, 0) = matX.at(j, 0); }

    for (int j = 0; j < slwd_size; j++) {
        // left multiplication
        so3_poses_temp[j] = SO3::exp(dxi.block<3, 1>(6 * (j), 0)) * so3_poses[j];
        t_poses_temp[j] = t_poses[j] + dxi.block<3, 1>(6 * (j) + 3, 0);
    }

/***********************************************/

sorry! the question 3 is my fault that i did not notice that the derivation is based on SO3.