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?