1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ Do you have any background in differential geometry? For example, the definition of tangent space, the definition of the differntial of a smooth map $f: M \to N$...? $\endgroup$ – user99914 Nov 4 '14 at 8:30
  • $\begingroup$ 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? $\endgroup$ – JonnyPython Nov 4 '14 at 11:31
  • $\begingroup$ 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. $\endgroup$ – jflipp Nov 4 '14 at 15:08
  • $\begingroup$ [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. $\endgroup$ – jflipp Nov 4 '14 at 15:14
  • $\begingroup$ @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 $\textrm{SO}(3)$. $\endgroup$ – guest Nov 4 '14 at 23:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.