# A Lie Algebras and Rotations

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1. Apr 21, 2016

### observer1

The Lie Algebra is equipped with a bracket notation, and this bracket produces skew symmetric matrices.

I know that there exists Lie Groups, one of which is SO(3).

And I know that by exponentiating a skew symmetric matrix, I obtain a rotation matrix.

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First, can someone edit the following and say it correctly and with precision (I am not certain I am saying it correctly and the phrasing matters to me)

"If one exponentiatiates a MEMBER of the Lie ALgebra, one obtains a member of the Lie Group SO(3)."

Could you fix that sentence for me?

Second...

I get the idea the angular velocity matrices are skew symmetric. I get the idea that rotations are orthogonal. But does this act of exponentiating a skew symmetric matrix imply a one to one correspondence between a specific angular velocity and a rotation? What does that imply?

WHAT IS IT about the philosophy (is that a good word? theory? )of exponentiating a matrix that enables this connection between Lie Groups and Lie Algebras? What is it about the fact that we are using a function whose derivative is itself, that causes rotations and angular velocities to be connected (is connected a good word?)?

Last edited by a moderator: Apr 21, 2016
2. Apr 21, 2016

### andrewkirk

Perhaps it would be useful to identify the Lie Algebra, and to use 'element' in preference to 'member'. There's nothing wrong with 'member' but 'element' is a more common word choice. THat would give:

If one exponentiates an element of the Heisenberg Lie Algebra $H_3(\mathbb{R})$, one obtains an element of the Lie Group SO(3).

3. Apr 21, 2016

### Staff: Mentor

As far as I understood the Heisenberg Lie Algebra $H_3(\mathbb{R})$ it is the algebra of all matrices of the form \begin{bmatrix} 0 & e_{12} & e_{13} \\ 0 & 0 & e_{23} \\ 0 & 0 & 0\end{bmatrix} Therefore $e_{13}$ is a central element whereas $SO(3)$ has none. So $H_3(\mathbb{R})$ is not the Lie Algebra of $SO(3)$.

Considering the exponential map things get quite a bit more complicated. Without talking about vector fields there can be constructed a mapping from the Lie Algebra of a Lie Group into this group with certain properties concerning the tangents along a path in the group. This mapping is called exponential map and it can be shown that this is the ordinary exponential map in the case of matrix groups. It's a result and not a definition. In case of $H_3(\mathbb{R})$ one gets the Heisenberg Group of all matrices of the form \begin{bmatrix} 1 & e_{12} & e_{13} \\ 0 & 1 & e_{23} \\ 0 & 0 & 1\end{bmatrix} For short: The $"S"$ in $SO(3)$ indicates matrices with determinant 1. This translates to matrices of trace 0 as the elements of its Lie Algebra $so(3)$ where the $"s"$ indicates exactly that: trace 0.

4. Apr 21, 2016

### observer1

>>This translates to matrices of trace 0 as the elements of its Lie Algebra

How?

5. Apr 21, 2016

### Staff: Mentor

Basically it is the exponentiation rule: $1 = exp(0) = exp(trace A) = exp(\sum{a_{ii}})=\prod{exp(a_{ii})} = det(exp A)$.
If you choose nilpotent matrices for simplicity - i.e. the exponential series is finite - as in the example of the Heisenberg Algebra above you can see it at an example. Of course there's more work to do if you want to prove it in general, i.e. non upper triangular matrices.

6. Apr 21, 2016

### andrewkirk

Of what algebraic entity did you mean that $e_{13}$ is a central element? As I understand it, the Lie Algebra is a space of matrices, so $e_{13}$, being a real scalar, is not an element of that space.

EDIT: Notwithstanding that, I've realised that the Heisenberg Algebra is not the one we want. The lie algebra of $SO(3)$ is $\mathfrak{so}(3)$, which is the set of all matrices of form
$$\begin{bmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0\end{bmatrix}$$

and that is not the same as $H_3(\mathbb{R})$.

So perhaps the proposition the OP was aiming for was:

'If one exponentiates an element of the Lie Algebra $\mathfrak{so}(3)$ of 3x3 skew-symmetric matrices, one obtains an element of $SO(3)$, the Lie Group of 3D rotations.'

Last edited: Apr 21, 2016
7. Apr 21, 2016

### Staff: Mentor

I've used $e_{13}$ for both, the entry in $(1,3)$ and the matrix with only a one in $(1,3)$ and zeros elsewhere. This matrix commutes with all other matrices of $H_3(ℝ)$ and therefore spans the center of the nilpotent Lie Algebra $H_3(ℝ)$. The center of the simple Lie Algebra $so(3)$ of the Lie Group $SO(3)$ is zero.

8. Apr 22, 2016

### suremarc

No, this is not implied. Suppose $\mathbf{\hat{n}}$ is an axis of rotation, and consider the skew symmetric matrix $X$ corresponding to the angular velocity vector $2\pi\mathbf{\hat{n}}$. Clearly $\mathrm{exp}(X)=I$, because a rotation by $2\pi$ radians leaves $\mathbb{R}^3$ unchanged.

It's simpler to see this from a topological perspective: $\mathrm{SO}(3)$ is compact, whereas the tangent space $\mathbb{R}^3$ is not. Therefore there can't be a continuous bijection between the two spaces.

There's another way to recover the Lie algebra of $\mathrm{SO}(3)$ -- use the Euler decomposition, and then differentiate at the identity:
$$d_e\Big(\begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos\alpha & -\sin\alpha \\ 0 & \sin\alpha & \cos\alpha\end{bmatrix}\begin{bmatrix}\cos\beta & 0 & \sin\beta \\ 0 & 1 & 0 \\ -\sin\beta & 0 & \cos\theta\end{bmatrix}\begin{bmatrix} \cos\gamma & -\sin\gamma & 0 \\ \sin\gamma & \cos\gamma & 0 \\ 0 & 0 & 1\end{bmatrix}\Big)=\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & -d\alpha \\ 0 & d\alpha & 0\end{bmatrix}+\begin{bmatrix}0 & 0 & d\beta \\ 0 & 0 & 0 \\ -d\beta & 0 & 0\end{bmatrix}+\begin{bmatrix} 0 & -d\gamma & 0 \\ d\gamma & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}$$
which gives a basis for $\mathfrak{so}(3)$. Hopefully this gives you a better understanding of what the Lie algebra is.

There is a plethora of different interpretations of the exponential map. Among them, this StackExchange answer is my favorite. Essentially, if $X\in \mathrm{Lie}(G)$, then $I+\epsilon X$ is an "approximate" element of $G$ for small $\epsilon$.

Last edited: Apr 22, 2016
9. Apr 23, 2016

### observer1

Thank you everyone...

this has helped a lot.