Lie algebras: Vector space vs linear operators

Lie algebra elements as linear operators

The fundamental representation of $A\in \mathfrak{se}(3)$ is

$$A=\left(\begin{array}{cc}0& 0\\ {\mathbf{a}}_1& {\hat{a}}_2\end{array}\right)(1)$$

which is a 4x4 matrix. We should more formally write this as

$${\rho }_{\text{fund}}(A)=\left(\begin{array}{cc}0& 0\\ {\mathbf{a}}_1& {\hat{a}}_2\end{array}\right)(2)$$

as the notion of a Lie algebra element $A$ is independent from its representation.

Recall that for a Lie algebra $\mathfrak{g}$, a representation is a mapping $\rho :\mathfrak{g}\to \mathfrak{gl}(V)$, where $V$ is some vector space and $\mathfrak{gl}(V)$ is (loosely) the space of all linear operators on $V$. In other words, a representation defines an action of $\mathfrak{g}$ on $V$.

In $(2)$ the vector space is $V={\mathbb{E}}^3$. The fundamental representation ${\rho }_{\text{fund}}(A):{\mathbb{E}}^3\to {\mathbb{E}}^3$ is the action of $A$ on ${\mathbb{E}}^3$:

$$A\cdot \mathbf{v}={\rho }_{\text{fund}}(A)(\mathbf{v})={\mathbf{a}}_1+{\hat{a}}_2\mathbf{v}(3)$$

which can also be written as

$$\left(\begin{array}{cc}0& 0\\ {\mathbf{a}}_1& {\hat{a}}_2\end{array}\right)\left(\begin{array}{c}1\\ \mathbf{v}\end{array}\right)=\left(\begin{array}{c}0\\ {\mathbf{a}}_1+{\hat{a}}_2\mathbf{v}\end{array}\right)(4)$$

There is a slight inconsistency between how ${\rho }_{\text{fund}}(A)$ appears in $(2)$ and $(3)$. In the former it is a matrix, and in the latter it appears as a function that takes $\mathbf{v}$ as its argument. Explanation: - We should see $(2)$ as the matrix form of the linear operator ${\rho }_{\text{fund}}(A)$ in a given basis of ${\mathbb{E}}^3$. - We should see $(3)$ as the linear operator form of the representation, which is basis-independent.

Henceforth, to be extra clear, we shall differentiate the two notationally. ${\rho }_{\text{fund}}(A)$ refers to the linear operator, and $[{\rho }_{\text{fund}}(A)]$ will refer to its matrix form (in a given basis of $V={\mathbb{E}}^3$). As a short-hand, we may also write $[A]\equiv [{\rho }_{\text{fund}}(A)]$.

Some remarks:

• Specifically, ${\rho }_{\text{fund}}(A)$ is the infinitesimal version of the fundamental action of $SE(3)$ on ${\mathbb{E}}^3$.
• In $(4)$, we see that the ${\mathbb{E}}^3$ element $\mathbf{v}$ has a $1$ prepended. This might seem a bit confusing, but this is necessary in order to represent translation as a linear operation. You can read more about this here and here.
• You may notice that the prepended $1$ has turned into a $0$ in the RHS of $(4)$. This is because strictly speaking the ${\rho }_{\text{fund}}(A)$ is a mapping from ${\mathbb{E}}^3$ to the tangent space $T{\mathbb{E}}^3$ (so its signature is actually ${\rho }_{\text{fund}}(A):{\mathbb{E}}^3\to T{\mathbb{E}}^3$). To see why read this again. The fact that the codomain is now the tangent space makes intuitive sense, as ${\mathbf{a}}_1$ and ${\hat{a}}_2$ correspond to an infinitesimal translation and rotation respectively (i.e. a velocity and an angular velocity). So ${\mathbf{a}}_1+{\hat{a}}_2\mathbf{v}$ should be interpreted as a tangent vector (the rate-of-change of $\mathbf{v}$ when acted upon by $A$). The only reason I got away with writing ${\rho }_{\text{fund}}(A):{\mathbb{E}}^3\to {\mathbb{E}}^3$ above is because of the isomorphism ${\mathbb{E}}^3\cong T{\mathbb{E}}^3$.
• It’s important to keep in mind that as soon as you are working with matrix-forms $[A]$ of the representation, you have implicitly assumed that you are working on some basis of the fundamental representation space $V={\mathbb{E}}^3$.

Lie algebra elements as vectors

In the above we represented Lie algebra elements as linear operators on ${\mathbb{E}}^3$. However, Lie algebras also “exist” without reference to their action on some vector space. In fact, they are vector spaces themselves.

For example, given some basis $E_i\in \mathfrak{se}(3),i=1,\dots ,6,$ we may write any Lie algebra element $A\in \mathfrak{se}(3)$ as $A=A_i{E}_i$. The components $A_i$ can also be seen as a 6-component column vector $\vec{A}=({\mathbf{a}}_1^T{\mathbf{a}}_2^T{)}^T$ in the given basis (In the paper, I often write this as $\{{\mathbf{a}}_1;{\mathbf{a}}_2\}$ instead). Why 6? That’s because $\text{dim}(SE(3))=\text{dim}(\mathfrak{se}(3))=6$.

Remark: With the risk of adding extra confusion, I should point out that the linear operators ${\rho }_{\text{fund}}(A)$ are also vectors. That is, the set $\{{\rho }_{\text{fund}}(A)|A\in \mathfrak{se}(3)\}\subset {\mathbb{R}}^{4\times 4}$ is a vector space. This is one of the (many) reasons why we often identify Lie algebras with their fundamental representations.

We thus have two ways of writing down a Lie algebra element $A$ explicitly (given a basis of $V={\mathbb{E}}^3$):

1. As a matrix $[A]$
2. As a column vector $\{{\mathbf{a}}_1;{\mathbf{a}}_2\}$.

Now, since $\mathfrak{se}(3)$ is a vector space, we may consider linear operators acting on $\mathfrak{se}(3)$. A set of such linear operators is given by the adjoint. For any Lie algebra element $B\in \mathfrak{se}(3)$, we have that ${\text{ad}}_B:\mathfrak{se}(3)\to \mathfrak{se}(3)$ is a linear operator acting on $\mathfrak{se}(3)$.

So if we write $\mathfrak{se}(3)$ as a vector space of 6-vectors, in a given basis, then we must have that ${\text{ad}}_B$ should be a $6\times 6$ matrix in that basis. We denote this matrix as $[{\text{ad}}_A]\in {\mathbb{R}}^{6\times 6}$.

Finally, we should also note that $\text{ad}$ is actually also a representation of $\mathfrak{se}(3)$. Its signature is $\text{ad}:\mathfrak{se}(3)\to \mathfrak{gl}(\mathfrak{se}(3))$. So in this case, we have that $V=\mathfrak{se}(3)$ (i.e. the Lie algebra is acting on itself).

Let’s tie it all together: Let $\odot$ denote the adjoint action, and let $A,B\in \mathfrak{se}(3)$, then all of the below are equivalent ways of writing the adjoint:

$$\begin{array}{rl}A\odot B& ={\text{ad}}_{A}B\\ & =[A,B]\\ & ={\rho }_{\text{fund}}^{-1}([A][B]-[B][A])\\ & =([{\text{ad}}_A]{\left(\begin{array}{c}{B}_1{B}_2{B}_3{B}_4{B}_5{B}_6\end{array}\right)}^T{)}_i{E}_i\\ & =E_i[{\text{ad}}_A{]}_{ij}{B}_j\end{array}$$

Completing the picture

What’s an easy way to simultaneously treat Lie algebra elements as matrices and vectors? One way is to write a basis for the Lie algebra using its fundamental representation. See $(1)$ again. It is quite easy to parse out what a basis for all matrices of the form $(1)$ could be:

$$\begin{array}{rl}{\tilde{E}}_1& ={\rho }_{\text{fund}}(\{(100{)}^T;(000{)}^T)=\left(\begin{array}{cccc}0& 0& 0& 0\\ 1& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right)\\ {\tilde{E}}_2& ={\rho }_{\text{fund}}(\{(010{)}^T;(000{)}^T)=\left(\begin{array}{cccc}0& 0& 0& 0\\ 0& 0& 0& 0\\ 1& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right)\\ E_3& ={\rho }_{\text{fund}}(\{(001{)}^T;(000{)}^T)=\left(\begin{array}{cccc}0& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\\ 1& 0& 0& 0\end{array}\right)\\ {\tilde{E}}_4& ={\rho }_{\text{fund}}(\{(000{)}^T;(100{)}^T)=\left(\begin{array}{cccc}0& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& -1\\ 0& 0& 1& 0\end{array}\right)\\ {\tilde{E}}_5& ={\rho }_{\text{fund}}(\{(000{)}^T;(010{)}^T)=\left(\begin{array}{cccc}0& 0& 0& 0\\ 0& 0& 0& 1\\ 0& 0& 0& 0\\ 0& -1& 0& 0\end{array}\right)\\ {\tilde{E}}_6& ={\rho }_{\text{fund}}(\{(000{)}^T;(001{)}^T)=\left(\begin{array}{cccc}0& 0& 0& 0\\ 0& 0& -1& 0\\ 0& 1& 0& 0\\ 0& 0& 0& 0\end{array}\right)\end{array}$$

We can now “forget” the difference between $A$ and ${\rho }_{\text{fund}}(A)$, identifying $A\sim {\rho }_{\text{fund}}(A)$. Now, any element $A\in \mathfrak{se}(3)$ can be written as $A=A_i{\tilde{E}}_i$, and in vector form this is $A=\{(A_1,A_2,A_3{)}^T;(A_4,A_5,A_6{)}^T\}$.

Some remarks:

• We have considered basis sets over two different spaces in the note: over the representation space $V={\mathbb{E}}^3$ and over the Lie algebra $\mathfrak{se}(3)$. In principle these can be chosen independently.
• However, the basis set ${\tilde{E}}_i\in \mathfrak{se}(3),i=1,\dots ,6$ defined above is actually induced from a given basis ${\mathbf{d}}_i\in {\mathbb{E}}^3,i=1,2,3,$ over the representation space.
  • “Weren’t the matrices ${\tilde{E}}_i$ a basis over the representation, and not the Lie algebra?”:
    • Yes but we have now identified the fundamental representation with the Lie algebra itself. So a basis over the set $\{[A]|A\in \mathfrak{se}(3)\}$ is now a basis over the Lie algebra itself.
  • “We haven’t defined ${\mathbf{d}}_i$!”
    • Yes, but as soon as you write down a $d\times d$ matrix, you implicitly assume that you are working in some basis for the $d$-dimensional vector space that the matrix acts on.
    • Recall that:
      • a vector = element of a vector space
      • a column vector = a $d\times 1$-sized matrix of components of the vector in a given basis