Lie Groups, Lie Algebra and Vectorfields

[JonnyMaddox]

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 this 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?

 

[homeomorphic]

I'm too busy to give a full reply, but I think you might be having some issues with thinking of a vector as a more flexible concept. Specifically, for a matrix Lie group, at tangent vector will be a matrix. The matrix is a being thought of as a vector--that's why it has two indices. Matrices can be added and multiplied by scalars, so abstractly, they can be thought of as vectors. For example, for SO(3). the tangent "vectors" to the identity will be skew-symmetric matrices. You take a path through the identity matrix and differentiate it. When you differentiate a path of matrices, the result is a matrix. Because these paths go through the identity element, which doesn't move anything, the resulting differentiated matrices can be thought of as infinitesimal symmetries. Because they are infinitesimally close to the identity, they only move things by an infinitesimal amount. What happens when you move every point on a manifold by an infinitesimal amount? Well, looking at each point, you get a vector, so altogether, a vector field. The vector fields arising this way are the left invariant ones, and what I've said illustrates how they are the same thing as the tangent vectors at the identity.

 

[Fredrik]

Hi I'm learning about Lie Groups to understand gauge theory (in the principal bundle context) and I'm having trouble with some concepts.

Do you understand the basics of differential geometry? (Manifolds, coordinate systems, tangent and cotangent spaces, the bases for $T_pM$ and $T^*_pM$ associated with the coordinate system x, the pushforward concept (i.e. the exact definition of $(L_g)_*$)).

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?

Lie groups are both manifolds and groups. Since they are manifolds, there's a tangent space associated with each point. (If the manifold is n-dimensional, its tangent spaces are n-dimensional too ). The identity is no exception.

Maybe solving this exercise may help with this 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 $I_{3}$.

In general, if C is a curve in the manifold, the tangent vector of the curve at the point $C(t)$ is the tangent vector $\dot C(t)$ defined by $(\dot C(t))(f)=(f\circ C)'(t)$ for all smooth maps $f:M\to\mathbb R$ (where M is the manifold). If $x$ is a coordinate system defined on a set that includes the point $C(t)$, then we can write
\begin{align}
&(\dot C(t))(f) =(f\circ C)'(t)= (f\circ x^{-1}\circ x\circ C)'(t) =(f\circ x^{-1})_{,i}(x(C(t))) (x\circ C)^i{}'(t)\\
&=(x\circ C)^i{}'(t) \frac{\partial}{\partial x^i}\bigg|_{C(t)} f
\end{align} Here ${}_{,i}$ denotes partial differentiation with respect to the $i$th variable. This result tells us that the components of $\dot C(t)$ in the coordinate system x, are the same as the components of $(x\circ C)'(t)$ with respect to the standard basis on $\mathbb R^n$.

And why does this induce a left invariant vector field?

If G is a Lie group, then every smooth map $\phi:G\to G$ does. If $g\in G$, then we can define $\phi_*:T_gG\to T_{\phi(g)}G$ by $(\phi_* v)(f)=v(f\circ\phi)$ for all smooth $f:G\to G$. These "pushforward" maps (one for each choice of g) are used to define the pushforward of a vector field X in the following way: $(\phi_* X)_{\phi(g)}=\phi_* X_g$. The $\phi_*$ defined here is a map that takes vector fields to vector fields.

A left-invariant vector field is one that for all $g\in G$ is unchanged by the application of $(L_g)_*$, i.e. it's a vector field X such that $(L_g)_* X=X$ for all $g\in G$. Note that this means that for all $h\in G$, we have
$$X_{gh}=((L_g)_*X)_{gh}=(L_g)_*X_h.$$ In particular, when h=e, we have $X_g=(L_g)_* X_e$. This suggests that given a vector $V\in T_eG$, the vector field $X_V$ defined by $(X_V)_g=(L_g)_*V$ is going to be left-invariant. It's not hard to verify that it is. This gives us a map from $T_eG$ to the set of left-invariant vector fields on $G$.

Edit: This map is invertible, since for every left-invariant vector field $X$, we have $X_e\in T_eG$. This enables us to use the Lie bracket on the set of left-invariant vector fields (i.e. the commutator operation) to define a Lie bracket on $T_eG$. For all $U,V\in T_eG$, we define $[U,V]$ by $[U,V]=[X_U,X_V]_e$. The bracket on the right is the commutator of two vector fields. Do you understand the definition of the commutator?

And what does it mean if a vector $V^{ij}$ has two indices??

A vector is just an element of a vector space. The elements of a vector space can be a matrices, functions, etc. So there's nothing odd about two indices on a vector. It's very natural when the vectors are matrices. What may appear a bit odd here is the use of two indices on the basis vectors $\frac{\partial}{\partial x^{ij}}\big|_p$. This is however very natural when we're dealing with a coordinate system on a set of matrices. If the Lie group is a group of n×n matrices, the manifold may be n²-dimensional, and in that case, a coordinate system should take each matrix to an n²-tuple of real numbers. But that n²-tuple can of course be viewed as an n-tuple of n-tuples, i.e. as an n×n matrix. So we choose to use two indices that run from 1 to n instead of a single index that runs from 1 to n².

Edit: If the group isn't n²-dimensional, then it can still be viewed as a submanifold of GL(n,ℝ), which is n²-dimensional. If we use two indices when we're talking about such a group, then I guess what we have done is to use a coordinate system on GL(n,ℝ) instead of a coordinate system on the original group.

 

[Fredrik]

Do you understand these things now?