I want wo calculate geodesics on SE(3). I'm not sure if my definition of a Riemannian metric on this manifold is correct. Any comments on my construction will be helpful thank you in advance.
I know, that for SE(3) ist is not true, that oneparametersubgroups are geodesics https://mathoverflow.net/questions/81590/one-parameter-subgroup-and-geodesics-on-lie-group?rq=1
Here my thoughts: To be able to talk about geodesics we have to define a Riemanian metric on SE(3). A Riemanian metric is in each point on the manifold a skalarproduct. For the identity element I can define a skalarprodukt by setting
$g_{e}\left((v_{1},w_{1}),(v_{2},w_{2})\right):=\begin{pmatrix} v_{1} & w_{1} \end{pmatrix}\begin{pmatrix} A & B\\ B^\top & C \\ \end{pmatrix}\begin{pmatrix} v_{2} \\ w_{2} \end{pmatrix}$
where $(v_{1},w_{1}),(v_{2},w_{2})\in T_{e}G$ and $w$ the coordinates corresponding to the Rotation and $v$ the coordinates corresponding to the translation.
I can use Left translation $L_{p}:x\mapsto px$ in order to define a Riemanian metric from this scalarproduct on the whole SE(3) by using the pullback
$g_{p}((v_{1},w_{1}),(v_{2},w_{2})):=g_{e}(dL_{p^{-1}}((v_{1},w_{1})),dL_{p^{-1}}((v_{2},w_{2})))$
In order to calculate the geodesics I have to solve the second order differential equation called geodesic equation. Therefore I have to calculate the Christoffel symbols, therefore I have to calculate the coefficients of the metric $g_{ij}$ for any $p\in SE(3)$. For $p=e=Id$ I already know by definition
$(g_{e})_{ij}=\begin{pmatrix} A & B\\ B^\top & C \\ \end{pmatrix}$
I tried to use this article Pull-back metric
to calculate the coefficients of the pullback metric: This calculation looked as follows
I use canonical coordinates on SE(3) that means local coordinates around the identity look like $x^{k}:=pr^{k}\circ \log$ where log is the Liegrouplogarithm and $pr^k$ the Projektion on the k-th coordinate. Around any other point, the local coordinates look like $y^{k}:=pr^{k}\circ \log\circ L_{p^{-1}}$
it holds $(g_{p})_{ij}=g_{p}(e_{i},e_{j}) =g_{e}(d_{p}L_{p^{-1}}(e_{i}),d_{p}L_{p^{-1}}(e_{j})) =\frac{\partial (x^{k}\circ L_{p^{-1}})}{\partial y^{i}}\frac{\partial (x^{r}\circ L_{p^{-1}})}{\partial y^{j}}g_{e}(e_{k},e_{r}).$
Calculation of the derivatives: $\frac{\partial (x^{k}\circ L_{p^{-1}})}{\partial y^{i}}=\partial_{i}\left(x^{k}\circ L_{p^{-1}}\circ y^{-1}\right)=\partial_{i}\left(pr^{k}\circ \log\circ L_{p^{-1}}\circ L_{p}\circ \exp\right)=\delta_{ik}.$
inserting $(g_{p})_{ij}=\delta^{ik}\delta^{jr}(g_{e})_{kr}=(g_{e})_{ij}$
That would mean the coefficients are constant. Therefore the Christoffelsymbols would be zero, since for the calculation of the Christoffelsymbols I have to calculate the derivatives of g_{ij}.
Then the geodesic equation would simplify to c''=0 that means the geodesics are one parameter subgroubs and this is wrong.
Does anyone see my mistake? Thanks a lot!
I just came across this, looking for something else. An easy way to calculate the Christoffel symbols for the Levi-Civita connection on a Lie group is by using the frame of left invariant vector fields corresponding to the chosen basis $(e_1,\ldots,e_n)$ of the Lie algebra. Then $\Gamma_{ij}^k$ is expressed via the structure constants for the bracket, say $$ [e_i, e_j] = c_{ij}^k e_k $$ and the Christoffel symbols are easily found to be given as $$ \Gamma_{ij}^k = \frac12(c_{ij}^k-c_{ik}^j-c_{jk}^i) $$ For $SE(3)$ one then gets 12 out of 216 symbols to be non-zero. Note that there is no biinvariant metric on $SE(n)$ for $n\geq 2$, as you say, and the Riemannian exponential does not coincide with the Lie group exponential.