4
$\begingroup$

Let say I'm in $SO(3)$. In the tangent space at the identity I define a inner product, i.e. $$g_{0}\left(X,\,Y\right)=\frac{1}{2}\mbox{Tr}\left(X^{T}Y\right),$$ Then I define a leftinvariant metric defined for every point $R$ in $SO(3)$ through the left action, i.e. $$g\left(X,\,Y\right)=g_{0}\left(L_{R^{-1}*}\left(X\right),\,L_{R^{-1}*}\left(Y\right)\right).$$ It's clearly left invariant by construction. Now let's say I take the following base $\left\{\left(E_{1}\right)_{R},...,\left(E_{n}\right)_{R}\right\} $ for the point of the tangent space in $R$ where $$\left(E_{1}\right)_{R}=L_{R*}\left(\left(E_{1}\right)_{\mathbb{1}}\right)$$ where of course $\left\{ \left(E_{1}\right)_{\mathbb{1}},...,\left(E_{n}\right)_{\mathbb{1}}\right\} $ is the canonical base at the identity. Now it should be that $$g\left(\left(E_{i}\right)_{R},\,\left(E_{j}\right)_{R}\right) = \frac{1}{2}\mbox{Tr}\left(\left(R\left(E_{i}\right)_{\mathbb{1}}\right)^{T}\left(R^{-1}\right)^{T}\left(R^{-1}\right)\left(R\left(E_{j}\right)_{\mathbb{1}}\right)\right)= g\left(\left(E_{i}\right)_{\mathbb{1}},\,\left(E_{j}\right)_{\mathbb{1}}\right),$$ So that the coefficient of the metric in the base given by $ \left\{ \left(E_{1}\right)_{R},...,\left(E_{n}\right)_{R}\right\}$ are constant. Being that the Christoffel Symbols are identically zero and I have that $\nabla_{X}Y$ is nothing else that the partial derivative in the point.

Defining the torsion $$T\left(X,Y\right)=\nabla_{X}Y-\nabla_{Y}X-\left[X,Y\right].$$

I then have $$T\left(X,Y\right)=-\left[X,Y\right],$$ and therefore in the base $\left\{ \left(E_{1}\right)_{R},...,\left(E_{n}\right)_{R}\right\}$ the coefficient of the tensor of torsion are $$T_{bc}^{a}=-\epsilon_{bc}^{a},$$ where in general since it works for every Lie Group the coefficient should be che structure constante of the lie algebra for the specific base choosen. Is this reasoning right? What did I get wrong? I found in a book the same construction to find the Curvature of the Levi-Civita connection and so -since the Levi-Civita Connection has zero torsion- I'm quite sure I've done something wrong somewhere....

$\endgroup$
3
$\begingroup$

By picking an orthonormal basis in the tangent space at the identity and extending it to a neighborhood by left translation, you've constructed a (local) orthonormal frame for your general Lie group. This is not, however, a coordinate frame unless the group is Abelian. Consequently, the (constant) inner products of your frame fields are not metric components.

$\endgroup$
5
  • $\begingroup$ Thank you for your answer. Could you expand a little bit your answer for my case showing me where's the problem? Thank you in advance $\endgroup$ – Dac0 Aug 16 '16 at 10:24
  • 1
    $\begingroup$ @Dac0 : The problem is, as stated, "you've constructed a (local) orthonormal frame for your general Lie group. This is not, however, a coordinate frame". There is no Christoffel symbol there, unless you have a coordinates system. $\endgroup$ – user99914 Aug 16 '16 at 18:05
  • $\begingroup$ But I choose the canonical base at the identity; I do have the coefficients of the metric at the identity which gave the identity matrix. So now I should have coefficient of the metric in every point following the construction above. Knowing coefficient of the metric in any point I should have my Christoffel Symbols, where's the flaw? $\endgroup$ – Dac0 Aug 16 '16 at 18:41
  • $\begingroup$ I really appreciate your explication and I upvoted the answer but is then yet I don't really get the bug... $\endgroup$ – Dac0 Aug 16 '16 at 18:42
  • $\begingroup$ @Dac0: All coordinate computations involving metrics, Christoffel symbols, and curvature tensor components rely on a coordinate frame, i.e., a set of vector fields $(\partial/\partial x_{i})$ for some local coordinate system $(x_{i})$. Your orthonormal frame is not a coordinate frame; that is, there exists no coordinate system on $SO(3)$ whose coordinate frame is left-invariant. (If there were, the fields would commute, implying $SO(3)$ is Abelian.) :) $\endgroup$ – Andrew D. Hwang Aug 16 '16 at 21:54

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.