5
$\begingroup$

Let $G$ be a Lie group. For $g \in G$, we can define a diffeomorphism $l_g: G \to G$ by $l_g(x)=gx$, and a bundle map ${l_g}_*:TG \to TG$. Then, I guess that we can obtain the affine connection on $G$ by reverse thinking that we could make a parallel transportation from a connection. Now, ${l_g}_*$ is the 'parallel transportation', and we would like to make the 'affine connection' from this. Is it true? If so, what is the name of this affine connection on $G$? Thanks for your help!!

$\endgroup$
  • $\begingroup$ Maybe, this problem was self-solved. By using parallel transportation ${l_g}_*$, we can make one trivialization of the tangent bundle of $G$. Then we can introduce affine connection on $G$ in the same way as we do it on $\mathbb{R}^n$. This connection corresponds to the above. $\endgroup$ – user115322 Jul 9 '14 at 13:01
3
$\begingroup$

Assuming that you are just trying to construct a connection (and not a specific connection or "the Levi-Civita" connection), then the answer is "yes." The connection you are describing is flat, but it does have torsion (unless the Lie group $G$ is Abelian).

More specifically, for a $n$-dimensional Lie group $G$, pick a left invariant frame $E_{1}, E_{2}, \ldots, E_{n}$ and define your connection by declaring $\nabla_{E_{i}}E_{j} = 0$ and extending it to all vector fields by requiring the usual linearity properties over constants and $C^{\infty}$ functions to hold. The frame is parallel with respect to the defined connection.

Note that for a curve $\gamma: \left[t_{0}, t_{1}\right] \to G$, parallel translation along $\gamma$ with respect to the given connection is defined as follows: The linear isomorphism $$ P_{t_{0}t_{1}} :T_{\gamma(t_{0})}G \to T_{\gamma(t_{1})}G $$ is defined for $v = V^{i}E_{i}\left(\gamma(t_{0})\right) \in T_{\gamma(t_{0})}$ by $P_{t_0 t_{1}}(v) = V^{i}E_{i}\left(\gamma(t_{1})\right) \in T_{\gamma(t_1)}$ (where $E_{i}(g)$ is the evaluation (or anchoring) of the vector field $E_{i}$ at the point $g$).

Regarding the curvature of the connection: The curvature endomorphism of the connection is the $(3, 1)$ tensor field $R: \chi(G) \times \chi(G) \times \chi(G) \to \chi(G)$ defined by $$ R(X, Y)Z = \nabla_{X} \nabla_{Y} Z - \nabla_{Y}\nabla_{X} Z - \nabla_{\left[X, Y\right]} Z, $$ where $\chi(G)$ denotes the $C^{\infty}$ vector fields on $G$ and $X, Y, Z \in \chi(G)$, and it should be clear from the definition of the connection that the curvature of the connection vanishes (i.e. the connection is flat). (One can also see that the connection is flat by noticing that the parrallel transport around any closed path is the identity.)

Regarding the torsion of the connection: The torsion tensor $\tau$ of the connection is the $(2, 1)$ tensor field defined by $\tau : \chi(G) \times \chi(G) \to \chi(G)$ defined by $\tau(X, Y) = \nabla_{X}Y - \nabla_{Y}X - \left[X, Y\right]$. On the frame $E_{1}, \ldots, E_{n}$, since $\nabla_{E_{i}} E_{j} = 0$, one sees that the torsion more or less corresponds to the Lie bracket.

$\endgroup$

Your Answer

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