# Lie groups, Lie algebra and left invariant vector fields

Hi I'm learning about Lie Groups to understand gauge theory (in the principal bundle context) and I'm having trouble with some concepts. Now let $a$ and $g$ be elements of a Lie group $G$, the left translation $L_{a}: G \rightarrow G$ of $g$ by $a$ are defined by : $L_{a}g=ag$ which induces a map $L_{a*}: T_{g}G \rightarrow T_{ag}G$ Let $X$ be vector field on a Lie group $G$. $X$ is said to be a left invariant vector field if $L_{a*}X|_{g}=X|_{ag}$. A vector $V \in T_{e}G$ defines a unique left-invariant vector field $X_{V}$ throughout $G$ by: $X_{V}|_{g}= L_{g*}V$, $g \in G$ Now the author gives an example of the left invariant vector field of $GL(n,\mathbb{R})$: Let $g={x^{ij}(g)}$ and $a={x^{ij}(a)}$ be elements of $GL(n,\mathbb{R})$ where $e= I_{n}=\delta^{ij}$ is the unit element. The left translation is: $L_{a}g=ag=\Sigma x^{ik}(a)x^{kj}(g)$ Now take a vector $V=\Sigma V^{ij}\frac{\partial}{\partial x^{ij}}|_{e} \in T_{e}G$ where the $V^{ij}$ are the entries of $V$. The left invariant vector field generated by $V$ is:

$X_{V|_{g}}=L_{g*}V=\Sigma V^{ij}\frac{\partial}{\partial x^{ij}}|_{e}x^{kl}(g)x^{lm}(e) \frac{\partial}{x^{km}}|_{g}= \Sigma V^{ij}x^{kl}(g) \delta^{l}_{i} \delta^{m}_{j} \frac{\partial}{\partial x^{km}}|_{g}= \Sigma x^{ki}(g)V^{ij} \frac{\partial}{\partial x^{kj}}|_{g}= \Sigma (gV)^{kj} \frac{\partial}{\partial x^{kj}}|_g$

Where $gV$ is the usual matrix multiplication. This is a bit over my head. What does it mean that one has a tangent vector at the unit element of a Lie group? Maybe solving this exercise may help with the question:

Let $c(s)=\begin{pmatrix} cos s & -sin s & 0 \\ sin s & cos s & 0 \\ 0 & 0 & 1 \end{pmatrix}$ be a curve in $SO(3)$. Find the tangent vector to this curve at $I_{3}$. And why does this induce a left invariant vector field? And btw, what is a left invariant vector field?? What does it mean geometrically? And what does it mean if a vector $V^{ij}$ has two indices?? Can one explain the example to me?

• Do you have any background in differential geometry? For example, the definition of tangent space, the definition of the differntial of a smooth map $f: M \to N$...? – user99914 Nov 4 '14 at 8:30
• Yes I know that, but I'm confused about Lie groups as manifold and what this all means visually. Why do you need a tangent vector on the unit element? What does that mean? – JonnyPython Nov 4 '14 at 11:31
• One point is that, in general, the tangent bundle of a manifold is not a trivial bundle. For example, $TS^1$ is trivial, but $TS^2$ is not (why? You probably know that story about combing a hedgehog.). Using left invariant vector fields, it's "obvious" that, for a Lie group $G$, the tangent bundle $TG$ is trivial. Incidentally, $S^1$ is a Lie group, but $S^2$ is not. – jflipp Nov 4 '14 at 15:08
• [continued] We construct a left invariant vector field on $G$ by picking any tangent vector $v$ at the unit $e \in G$ and then translate that $v \in T_eG$ to any other point $g \in G$ via the differential of the left translation $L_g$, i.e. $dL_gv \in T_gG$. Instead of with $e$, we could start with any element $h \in G$, but working with $e$ makes things simpler. That's why we consider tangent vectors at the unit. – jflipp Nov 4 '14 at 15:14
• @JonnyPython, I think you should study some simpler examples of Lie groups, starting with the abelian ones. Take the Lie groups $\mathbb{R}$ and $\mathbb{R}/\mathbb{Z}$. What are the identity elements? What are the tangent vectors there? What are the invariant vector fields? Continue with the Lie groups $\mathbb{R}^n$ and $\mathbb{R}^n/\mathbb{Z}^n$. Then treat the Lie group $\mathbb{R}^*_+$. What is the unit element there and what are the tangent vectors and the invariant vector fields? Write it as a matrix group. After that you can treat the first non-abelian example $\textrm{SO}(3)$. – guest Nov 4 '14 at 23:17