Left-Invariant Vector Fields

Section 3.2 Left-Invariant Vector Fields

The definition of a Lie group as a group that is also a manifold has some important consequences. We show here how to connect the structure of the Lie group to that of its Lie algebra, which can be identified with the tangent plane of the Lie group at the identity element.

So let \(G\) be a matrix Lie group, with identity element \(1\text{.}\) The elements of \(G\) are thus \(n\times n\) matrices, that is, elements of \(\RR^{n\times n}\text{.}\) So we take the extrinsic point of view, discussed in Appendix A.1, which amounts to thinking of \(G\) as a parametric surface in \(\RR^{n\times n}\text{,}\) whose tangent vectors are just coordinate derivatives of the parameterization.

Let \(X\) be a tangent vector to \(G\) at the origin. Our first task is to use the group structure to extend \(X\) to a vector field on \(G\text{.}\) Any element \(P\in G\) acts on elements \(Q\in G\) by left multiplication \(L_P\text{,}\) that is, we have the map

\begin{align} L_P: G\amp\longrightarrow G\notag\\ Q\amp\longmapsto PQ .\tag{3.2.1} \end{align}

We can therefore extend \(X=X_1\) from a vector at a single point (\(1\)) to a vector field \(X\) on all of \(G\) by defining

\begin{equation} X_P = L_{P*} X_1 .\tag{3.2.2} \end{equation}

As discussed in Example A.1.2 and Example A.1.4, in the extrinsic description it is straightforward to show that, first, \(X_1\) is itself an \(n\times n\) matrix (since tangent vectors are derivatives of suitably parameterized matrices) and, second, the action of \(L_{P*}\) is, like \(L_P\) itself, given by matrix multiplication, that is, that

\begin{equation} L_{P*} X_1 = P X_1 .\tag{3.2.3} \end{equation}

Now that we have a vector field, we can consider its integral curves. An integral curve of a vector field is a curve whose tangent vectors agree everywhere with the given vector field. So if \(\alpha(u)\) is a curve in \(G\text{,}\) that is, a parametric matrix with parameter \(u\text{,}\) the tangent vector at the point \(\alpha(u)\) is just \(\alpha'(u)\text{.}\) Thus, for \(\alpha(u)\) to be an integral curve of \(X\text{,}\) we must have

\begin{equation} \alpha'(u) = X_{\alpha(u)} .\tag{3.2.4} \end{equation}

Finally, suppose that the curve starts at the identity, so \(\alpha(0)=1\text{.}\) Putting the pieces together, we have

\begin{equation} \alpha'(u) = X_{\alpha(u)} = \alpha(u) X_1 .\tag{3.2.5} \end{equation}

It is straightforward to solve this differential equation, yielding

\begin{equation} \alpha(u) = e^{uX}\tag{3.2.6} \end{equation}

where we have reverted to writing \(X\) for \(X_1\text{.}\)

We have shown that, at least near the identity element, a Lie group consists of one-parameter families of transformations, satisfying

\begin{align} \alpha(0) \amp= 1 ,\tag{3.2.7}\\ \alpha(a+b) \amp= \alpha(a)\alpha(b) .\tag{3.2.8} \end{align}

In more abstract (non-matrix) settings, this construction can also be used to define \(\exp(X)\text{,}\) namely as \(\alpha(1)\text{.}\)