I am studying some concepts of differential geometry in order to apply to a robotics topic.
I am a bit confused on the link between vee/hat operators and exp/log maps.
I know that:
The concepts of vee and hat operators are strictly related to the exp and log maps. They respectively allow to go from a $n$-dimensional Lie Group to a Lie algebra associated to an $n$-dimensional vector space and back. The vee operator has been associated to the symbol $\cdot^{\vee}$ while the hat operator to $\hat{\cdot}$
$\begin{equation*}
\begin{split}
\cdot^{\vee} : g \in \mathbb{R}^{n\times n} \rightarrow \mathbb{R}^n
\\
\hat{\cdot}:\mathbb{R}^n \rightarrow g \in \mathbb{R}^{n\times n}
\end{split}
\end{equation*}$
These two operators allow to move from the manifold to a tangent space to the manifold in the considered point and back.
I also know that the link between a Lie Group and its associated Lie algebra can be expressed with the exponential (exp) and logarithmic (log) operations.
QUESTION: Does it mean that the vee/hat operators are the same thing of the exp/log maps?
Thanks in advance.
Best Answer
After some time I finally understood the link between hat/vee map and exp/log map.
The answer to my question is NO. Hat/vee map is not the same thing of exp/log map. As pointed out also in some comments.
Assume we have a Lie Group $G$ and its associated Lie algebra $g$.
As I wrote in my question the vee/hat maps are defined as
$\begin{equation*} \begin{split} \cdot^{\vee} : g \in \mathbb{R}^{n\times n} \rightarrow \mathbb{R}^n \\ \hat{\cdot}:\mathbb{R}^n \rightarrow g \in \mathbb{R}^{n\times n} \end{split} \end{equation*}$
The Vee operator $(\cdot^{\vee})$ is the operator which allows to move from the Lie algebra $g$ to the real vector space of dimension equal to the one of the Lie algebra.
The Hat operator $(\hat{\cdot})$ is the operator which allows to go backward (w.r.t. the vee operator).
On the other side, the $exp$ map allows you to describe the shortest path from the identity of the group $G$ to another point. You can write this as
$ \begin{equation} exp(\mathbf{S}) \end{equation} $ where $\mathbf{S} \in g$
In the case your $G$ is the Lie Group of the rotations $SO(3)$ the exponential map allows you to go from an orientation to another one.
Another way of seeing it is that the exponential map allows you to go from a manifold to a tangent space to the manifold. The logarithmic map does the opposite.
This should be right but I am really open to discussion.