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From WiKi: Product of exponentials formula is:

$$g_{st}=e^{\hat{\xi}_1\theta_1}e^{\hat{\xi}_2\theta_2}...e^{\hat{\xi}_n\theta_n}$$

where $g_{st}\in SE(3)$,and $\xi_i$ is twist vector $\in {se}(3)$.

My question is: for programming convenient, can i use

$$g_{st}=e^{\hat{\xi}_1\theta_1+\hat{\xi}_2\theta_2+...+\hat{\xi}_n\theta_n}$$

to avoid the matrix product. If cannot, why?

I think the answer is yes, because $\hat{\xi}_1\theta_1+...+\hat{\xi}_n\theta_n$ is still in Lie algebra $se(3)$. But what is its physical explanation?

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    $\begingroup$ No, there are Lie brackets terms coming from $\xi_i$ does not commute with $\xi_j$. See, for example, Baker-Campbell-Hausdorff formula $\endgroup$ – user10354138 Nov 13 '18 at 5:44
  • $\begingroup$ @user10354138 I see, you are right, thank you. $\endgroup$ – Ben Nov 14 '18 at 8:32
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The answer is NO, follows user10354138 comment, $\xi_j$ doesn't commute.

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