I'd like to calculate directly the Christoffel symbols for $SO(3)$. Let us suppose I have the metric on $SO(3)$ given by $$g(X,Y)=1/2 [\operatorname{tr}(X^tY)].$$ I'm aware of Milnor paper and I know I can get the Christoffel symbol of a bi-invariant metric of a generic Lie Group through Koszul formula, but in this case is there a way to find the Christoffel Symbols o the connection forms directly?

Taking the standard basis $E_1,E_2,E_3$ of $\text{so}(3)$ where $[E_1,E_2]=E_3$, $[E_2,E_3]=E_1$ and $[E_3,E_1]=E_2$, if I'm not mistaking $g_{ij}$ at the identity is the identity matrix... so where do I go from there? I'm getting quite confused...

  • $\begingroup$ You compute the matrix of the $g_{i,j}$ with respect to what basis? $\endgroup$ – Mariano Suárez-Álvarez Aug 11 '16 at 17:30
  • $\begingroup$ take the standard basis $E_1,E_2,E_3$ of so(3) where $[E_1,E_2]=E_3$, $[E_2,E_3]=E_1$ and $[E_3,E_1]=E_2$, if I'm not mistaking $g_{ij}$ at the identity is the identity matrix. Anyway you're aloud to choose whatever base you like for the answer as long as you specify what you're using... $\endgroup$ – Dac0 Aug 11 '16 at 18:59
  • $\begingroup$ I know. My point is that what you wrote in the body of the question is meaningless if you do not specify what basis you used… Instead of burying that information in this comment, please add it to the question itself. $\endgroup$ – Mariano Suárez-Álvarez Aug 11 '16 at 19:42

There is no Christoffel symbols when you have only a local (or global) basis, as explained in the answer to your other question. Indeed, Christoffel symbols is not invariant under change of coordinates (i.e., it's not a tensor)

If the goal is to calculate the curvature, then one can use directly the (orthonormal) basis $\{E_1, E_2,E_3\}$. The following is true for all Lie group equipped with a bi-invariant metric (which include all compact Lie groups).

Let $\{ E_1, \cdots, E_n\}$ be an orthonormal basis on $\mathfrak g$ with respect to a metric $h$. We always identified $\mathfrak g$ with the set of left-invariant vector fields and the metric $h$ is identified with the bi-invariant metric on $G$. The structural constant are denoted by $\epsilon_{ij}^k$, that is, $$ [E_i, E_j] = \epsilon_{ij}^k E_k.$$

Since $h$ is a bi-invariant metric, the exponential map $\exp : \mathfrak g \to G$ is also the exponential mapping $T_e G \to G$ with respect to the bi-invariant metric $h$ (This is the only place we use that $h$ is bi-invariant, see here for a proof).

Lemma 1: $\nabla _XX = 0$ for all $X\in \mathfrak g$.

Proof of lemma 1: By definition of the exponential mapping $\exp(\cdot) = e^{(\cdot)}$, we have $$ e^{tX} e^{sX} = e^{(s+t)X}.$$ If we write $\gamma (t) =e^{tX}$, then $$\gamma'(t) = (e^{tX})_* X.$$ This implies that $\gamma(t)$ is the intgral curve of the vector fields $X$. Thus $\nabla_X X = 0$ since $\gamma$ is a geodesics.

Lemma 2: We have $\nabla_X Y = \frac 12 [X,Y]$.

Proof of Lemma 2: For all $X, Y\in \mathfrak g$, from Lemma 1, $$\begin{split} 0 &= \nabla_{X+Y}(X+Y) \\ &= \nabla_X X + \nabla_XY + \nabla_YX + \nabla _YY \\ &= \nabla_XY + \nabla_YX \end{split}$$ Together with the torsion free condition $$\nabla_XY - \nabla _YX = [X,Y],$$ the lemma is proved.

Note that Lemma 2 gives much more refined information about the connection than that given by the Koszul formula. Now we can somehow calculate the "Christoffel symbols" of the connection, note that since $\{E_1, \cdots E_n\}$ forms a basis of $T_gG$ for all $g\in G$, so in general $$\nabla_{E_i} E_j = T_{ij}^k(g) E_k$$ for some global functions $T_{ij}^k : G\to \mathbb R$. Lemma 2 implies that $$ T_{ij}^k =\frac 12 \epsilon_{ij}^k$$ (In particular, $T_{ij}^k$ is a constant function).

We can go on and calcaulate the curvature:

Lemma 3: The Riemann curvature tensor is given by $$ R(X, Y, Z, W) = \frac 14 h([[X,Y],Z],W)$$

Proof of Lemma 3: Using Lemma 2 and Jacobi identity, $$\begin{split} R(X, Y, Z, W) &:= h( -\nabla_X \nabla _Y Z + \nabla_Y \nabla _X Z + \nabla_{[X,Y]} Z, W) \\ &= \frac 12 h(- \nabla _X [Y,Z] + \nabla_Y [X,Z] + [[X,Y],Z],W) \\ &= \frac 12 h(-\frac 12 [X,[Y,Z]] + \frac 12 [Y,[X,Z]] + [[X,Y],Z], W) \\ &= \frac 14 h([[X,Y],Z],W). \end{split}$$

Now we can represent $R$ using the structural constant:

$$\begin{split} R_{ijkl} &= R(E_i, E_j, E_k, E_l)\\ &= \frac 14 h([[E_i, E_j], E_k], E_l) \\ &= \frac 14 h([\epsilon_{ij}^m E_m, E_k], E_l) \\ &= \frac 14 h(\epsilon_{ij}^m \epsilon_{mk}^n E_n , E_l) \\ &=\frac 14 \epsilon_{ij}^m \epsilon_{mk}^l. \end{split}$$

Going back to your case $G= SO(3)$, we have (by case by case checking) $$ R_{1212} =R_{1313} = R_{2323} = \frac 14,$$ and others are zero (we are assuming that $i < j$ and $k <l$).

  • $\begingroup$ Thank you, perfect explanation $\endgroup$ – Dac0 Aug 17 '16 at 16:22

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