Construct a torsion-free connection on Lie group

Question

Let $G$ be a Lie group, and consider a connection for left-invariant vector fields on $G$ defined as

$$
\nabla_X(\sum_j\alpha_jZ_j):=\sum_j(X\alpha_j)Z_j
$$

where $\{Z_j\}$ is a global frame for left-invariant vector fields. I have computed, that the torsion and curvature of this connection is (let $X=\sum_i\alpha_iZ_i, \ Y=\sum_j\beta_jZ_j, \ Z=\sum_k\gamma_kZ_k$)

$$
T(X,Y)=\sum_{i,j}\alpha_i\beta_j[Z_i,Z_j] \ \ \ \text{and} \ \ \ R(X,Y)Z=0
$$

Now I want to construct a new connection $\tilde\nabla$ from $\nabla$, such that $\tilde\nabla$ is torsion-free. Is there any way to do it? I have no idea how to proceed.

Answer 1

Hint For any affine connection $\nabla$ on a smooth manifold $M$ and any $(2, 1)$ tensor $A$, regarded as a bundle map $\bigotimes^2 TM \to TM$, we get another affine connection $\tilde\nabla$ defined by $$\tilde\nabla_X Y := \nabla_X Y + A(X, Y)$$ (and all affine connections on $M$ arise this way). Now, compute the torsion $\tilde T$ of $\tilde\nabla$ in terms of the torsion $T$ of $\nabla$ and the modification $A$, and find conditions on $A$ such that $\tilde T = 0$.

Mechanically it might be easiest to work with the components of the tensors with respect to a local frame: $$T_{ij}^k = \Gamma_{ij}^k - \Gamma_{ji}^k - \gamma_{ij}^k .$$ As usual, $\Gamma_{ij}^k$ are the connection coefficients (characterized by $\nabla_{Z_i} Z_j = \Gamma_{ij}^k Z_k$) and $\gamma_{ij}^k$ are the commutator coefficients (characterized by $[Z_i, Z_j] = \gamma_{ij}^k Z_k$). (This process works for any connection and any local frame, but for connections like the one in the problem statement, where all of the frame components are parallel, $\Gamma_{ij}^k = 0$ and so $T_{ij}^k = -\gamma_{ij}^k$.)

Substituting gives $$\tilde T_{ij}^k = T_{ij}^k + 2 A_{[ij]}^k,$$ where as usual brackets denote skewing over the enclosed indices, $2 A_{[ij]}^k = A_{ij}^k - A_{ji}^k$. So, the condition that $\tilde T = 0$ imposes exactly that $A_{[ij]}^k = -\frac12 T_{ij}^k$. There are infinitely many solutions $A$, but the simplest choice is to choose $A$ itself to be skew in its covariant indices, i.e., $$A = -\frac12 T .$$ Indeed, precisely when $A$ is skew in its covariant indices the geodesic equations for $\nabla, \tilde\nabla$ coincide, so $$\tilde\nabla_X Y := \nabla_X Y - \frac12 T(X, Y) = \frac12 (\nabla_X Y + \nabla_Y X + [X, Y]),$$ just as well specified by the connection coefficients $$\tilde\Gamma_{ij}^k = \frac12(\Gamma_{ij}^k + \Gamma_{ji}^k + \gamma_{ij}^k) ,$$ is the unique torsion-free connection with the same geodesics as $\nabla$. In terms of a frame $(Z_i)$ whose components are all parallel with respect to $\nabla$, $\tilde\nabla$ is characterized by $\tilde \nabla_{Z_i} Z_j = \frac12 [Z_i, Z_j]$.