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Application of Lie group in gtsam

2022-07-06 06:01:00 【Zhan Miao】

1. AdjointMap Definition

Why in matrix Lie group , The above two definitions can be mixed ？ Prove the following

$Ad_T \hat{x} = T \hat{x}T^{-1} \Rightarrow \\ exp(Ad_T \hat{x}) = \sum_{n = 0}^{\infty} \frac{1}{n!} \left( Ad_T\hat{x} \right)^n = \sum_{n = 0}^{\infty} \frac{1}{n!} \left( T \hat{x}T^{-1} \right)^n \\ = \sum_{n = 0}^{\infty} \frac{1}{n!} T \hat{x}^nT^{-1} = T \sum_{n = 0}^{\infty} \frac{\hat{x}^n}{n!} T^{-1} = T e^{\hat{x}} T^{-1} = Ad_T e^{\hat{x}}$

2. Li Qun's AdjointMap

How can the above formula lead to the following conclusion

AdjointMap Is the mapping from antisymmetric matrix to antisymmetric matrix

Make

$\xi' = \begin{bmatrix} \omega' \\ v' \end{bmatrix}$

Then the above formula can be written as the following formula

$\begin{bmatrix} [\omega']_{\times} & v' \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} [R\omega]_{\times} & t \times R\omega + Rv \\ 0 & 0 \end{bmatrix} \Rightarrow \begin{cases} \omega' = R \omega \\ v' = t \times R\omega + Rv \end{cases}$

Then we get the following formula

3. Local Coordinates

There is such a formula

$a e^{-\hat{x}}a^{-1} = -a e^{\hat{x}}a^{-1}$

Prove the following

$a e^{-\hat{x}}a^{-1} = exp(-a \hat x a^{-1}) \\ = \sum_{n=0}^{\infty} \frac{1}{n!} (-a \hat x a^{-1})^n \\ = \sum_{n=0}^{\infty} \frac{1}{n!} (-a \hat x^n a^{-1}) \\ = -a \sum_{n=0}^{\infty} \frac{\hat x^n}{n!} a \\ = -a e^{\hat{x}}a^{-1}$

4. ImuFactor

$\frac{\partial R_k}{\partial \theta_k} = H(\theta_k) = \sum_{k=0}^{\infty} \frac{(-1)^k}{(k+1)!}[\theta]_{\times}^k$

prove

from

have to

$\frac{\partial R_k}{\partial \theta_k} \\ = \lim_{\delta \rightarrow0} \frac{ exp([\theta + \delta]_{\times}) \ominus exp([\theta]_{\times}) }{\delta} \\ = \lim_{\delta \rightarrow0} \frac{Log \left( exp([-\theta]_{\times}) exp([\theta + \delta]_{\times}) \right) }{\delta} \\ = \lim_{\delta \rightarrow0} \frac{Log \left( exp([-\theta]_{\times}) exp([\theta]_{\times}) exp([H(\theta)\delta]_{\times}) \right) }{\delta} \\ = \lim_{\delta \rightarrow0} \frac{Log \left( exp([H(\theta)\delta]_{\times}) \right) }{\delta} \\ = H(\theta)$