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


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!

  • $\begingroup$ The group $SE(3)$, as a smooth manifold, is diffeomorphic to $R^3\times SO(3)$. Putting the flat metric on $R^3$ and the standard metric given by the Killing form on $SO(3)$ will yield a biinvariant Riemannian metric on $SE(3)$. Geodesics in a product $M_1\times M_2$ are "product geodesics", of the form $t\mapsto (c_1(t), c_2(t))$, where $c_1, c_2$ are geodesics. Geodesics on $SO(3)$ are easy to describe. $\endgroup$ Dec 5, 2019 at 13:27
  • $\begingroup$ Thank you for your comment Moishe, I already found a proof in a paper that the geodesics that you are describing are coinciding with the geodesics that I would receive if I chose B=0 in the approach above. Nevertheless I'm especially interessted in the case where B does not equal zero, because then the metric would somehow mix the weights of rotation and translation. $\endgroup$
    – Mathsfreak
    Dec 5, 2019 at 15:55
  • $\begingroup$ I just found out that I probably shouldn't use the map $y^{k}:=pr^{k}\circ \log\circ L_{p^{-1}}$ but $y^{k}:=pr^{k}\circ \log$ then $\partial_{i}\left(pr^{k}\circ \log\circ L_{p^{-1}}\circ L_{p}\circ \exp\right)$ changes to $\partial_{i}\left(pr^{k}\circ \log\circ L_{p^{-1}}\circ \exp\right)=\partial_{i}\left(pr^{k}\circ \log(exp(ad_{p^{-1}}(x))\cdot p)\right)$ therefore the question that arises is what is the derivative of $ad_{p^{-1}}(x)$ $\endgroup$
    – Mathsfreak
    Dec 6, 2019 at 11:04

1 Answer 1


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.


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