Left translate back to I in SO(3)

[JTC]

Hello

I am hoping someone can explain a sentence to me. Unfortunately, I do not even recall where I read it. I wrote it down years ago and long since lost the source. (Now I think some of it is making sense, but I don't remember the source.)

Consider R(t) as an orthogonal rotation matrix.

• We know the time derivative exists: (R(t)-dot)
  ( (R(t)-dot) means R(t) with a superposed dot for time derivative: dR/dt)
• We know the inverse is the transpose: R(t)^T

And we know the product of -- R(t)^T*(R(t)-dot) -- is a skew symmetric matrix (which could be viewed as an angular velocity matrix).

Then I read this:

"We compute the time derivative of the rotation matrix -- (R(t)-dot) at R(t) -- and using R(t)^T, left translate (R(t)-dot) back to the identity I₃ in SO(3)"

What does that mean, please?

I think it has to do with the Lie algebra so(3) of SO(3), but Could someone clarify?

 

[fresh_42]

$\mathfrak{so}(3)$ is the tangent space at the identity element of $SO(3)$. The group acts via conjugation on this tangent space: $g \mapsto (X \mapsto gXg^{-1})$ which is called the adjoint representation $\operatorname{Ad}$ of $SO(3)$.

I assume this is meant by what you wrote.

 

[JTC]

$\mathfrak{so}(3)$ is the tangent space at the identity element of $SO(3)$. The group acts via conjugation on this tangent space: $g \mapsto (X \mapsto gXg^{-1})$ which is called the adjoint representation $\operatorname{Ad}$ of $SO(3)$.

I assume this is meant by what you wrote.

As a person who is still learning this, could you take this down one notch (apology to you: that is not intended to be rude, though it does seem to sound rude) and
give simpler explanation?

What do you mean by "AT THE IDENTITY ELEMENT OF SO(3)"

And what do you mean by "conjugation"

I just see a multiplication of matrices.

 

[fresh_42]

The group $SO(3)$ is a group of linear transformations and a Lie group. Their elements form an object which isn't flat space but has a curvature and we can build the tangent spaces at points of this object. Now $1 = \begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix} \in SO(3)$ and the tangent space at this point is spanned by all the tangents of all curves on $SO(3)$ through the point $1$. The only difference here to e.g. a tangent at $y=x^2$ is, that we have many possible curves through $1$ and thus many different tangents. However, they span a vector space, namely $\mathfrak{so}(3)$. Now $t \longmapsto R(t) \in SO(3), R(0)=1$ are such curves and $\left. \dot{R}(t)\right|_{t=0}$ are the slopes of those tangents. Disregarding some details here, you could imagine $SO(3)$ as a three-dimensional sphere, a ball, and $\mathfrak{so}(3)$ as a plate lying on it. The point where they touch is $1 \in SO(3)$, which serves as $0$, the origin of the plate, the vector space $\mathfrak{so}(3)$. Now, if we have a tangent $X \in \mathfrak{so}(3)$ and a rotation $r \in SO(3)$, then $rXr^{-1}$ is again a tangent in $\mathfrak{so}(3)$.

 

[JTC]

The group $SO(3)$ is a group of linear transformations and a Lie group. Their elements form an object which isn't flat space but has a curvature and we can build the tangent spaces at points of this object. Now $1 = \begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix} \in SO(3)$ and the tangent space at this point is spanned by all the tangents of all curves on $SO(3)$ through the point $1$. The only difference here to e.g. a tangent at $y=x^2$ is, that we have many possible curves through $1$ and thus many different tangents. However, they span a vector space, namely $\mathfrak{so}(3)$. Now $t \longmapsto R(t) \in SO(3), R(0)=1$ are such curves and $\left. \dot{R}(t)\right|_{t=0}$ are the slopes of those tangents. Disregarding some details here, you could imagine $SO(3)$ as a three-dimensional sphere, a ball, and $\mathfrak{so}(3)$ as a plate lying on it. The point where they touch is $1 \in SO(3)$, which serves as $0$, the origin of the plate, the vector space $\mathfrak{so}(3)$. Now, if we have a tangent $X \in \mathfrak{so}(3)$ and a rotation $r \in SO(3)$, then $rXr^{-1}$ is again a tangent in $\mathfrak{so}(3)$.

Thank you.

 

[JTC]

The group $SO(3)$ is a group of linear transformations and a Lie group. Their elements form an object which isn't flat space but has a curvature and we can build the tangent spaces at points of this object. Now $1 = \begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix} \in SO(3)$ and the tangent space at this point is spanned by all the tangents of all curves on $SO(3)$ through the point $1$. The only difference here to e.g. a tangent at $y=x^2$ is, that we have many possible curves through $1$ and thus many different tangents. However, they span a vector space, namely $\mathfrak{so}(3)$. Now $t \longmapsto R(t) \in SO(3), R(0)=1$ are such curves and $\left. \dot{R}(t)\right|_{t=0}$ are the slopes of those tangents. Disregarding some details here, you could imagine $SO(3)$ as a three-dimensional sphere, a ball, and $\mathfrak{so}(3)$ as a plate lying on it. The point where they touch is $1 \in SO(3)$, which serves as $0$, the origin of the plate, the vector space $\mathfrak{so}(3)$. Now, if we have a tangent $X \in \mathfrak{so}(3)$ and a rotation $r \in SO(3)$, then $rXr^{-1}$ is again a tangent in $\mathfrak{so}(3)$.

Actually, I do have a question. All is clear except the very last sentence. I see how we have the tangent (the R-dot). WHY do you want to pull it back to the identity?
How is multiplying R-dot by R-transpose, the same as pulling it back to the identity?

I see that R-transpose pulls back R to the identity, but what does it mean to pull back R-dot to the identity?

 

[fresh_42]

I can only guess what might have been meant. To consider the operation $r.X := r\cdot X\cdot r^{-1}= R(t) \cdot \left. \dot{R}(t)\right|_{t=0} \cdot R^{-1}(t) = R(t) \cdot \left. \dot{R}(t)\right|_{t=0} \cdot R^{\tau}(t)$ has been one possibility, and the easier of two. The other possibility is, the definition of the Lie algebra elements $X$ as left-invariant vector fields had been meant. This means: The tangent of any rotation $R$ can be calculated as the derivative of the group's left translation $L_R: S \mapsto R\cdot S$ by $R$ evaluated at a tangent at $1 \in G$, i.e. $X_R= dL_R(X_1)$ and wherever tangents of group elements are computed, they are already determined by those at the point $1$.