# Torsion Free Spin Connection

Ok I am not exactly sure how much of this common notation/terminology, and how much is unique to the book I'm reading, so bear with me for a moment here. First we have a vector bundle $$E$$ associated to the orthonormal frame bundle of some manifold $$M$$. There is a soldering form, an isomorphism from $$TM\rightarrow E$$, given by a collection of one forms $$e^I$$ (equivalently a vector valued one form):

$$e^I=e^I_\mu dx^\mu$$

where $$I$$ is an index $$I=1,\dots,n$$. This encodes a riemannian metric on $$M$$ via:

$$g(v,u):=\langle e^I(u),e^I(v)\rangle=g_{\mu\nu}=e^I_\mu e^J_\nu \delta_{IJ}$$

A metric connection on $$E$$ then satisfies the following property: $$d^\omega\delta^{IJ}=\omega^I_K\delta^{KJ}+\omega^J_K\delta^{KI}=0$$ Which is really just the condition that $$\omega^i_j=-\omega^j_i$$, or, equivalently, that $$\omega$$ is a one form with values in $$\mathfrak{o}(n)$$. Torsion is then defined as: $$T^I:=d^\omega e^I=de^I+\omega^I_Je^J$$ I am pretty sure I am fine with all of this, but this next jump is a calculation that I haven't been able to follow: given the soldering form, there exists a unique metric and torsion free connection given by:

$$\omega^I_{\mu J}=e^{\rho I}e_J^{\sigma}\left(-C_{\mu\rho\sigma}+C_{\rho\sigma\mu}+C_{\sigma\mu\rho}\right)$$

Where:

$$C_{\mu\rho\sigma}=e_{\mu I}\partial_{[\rho}e^I_{\sigma]}$$

The object $$e^\mu_I$$ is the inverse of the soldering form defined as $$e^\mu_Ie^J_\mu=\delta^J_I$$, and $$e^\mu_Ie^I_\nu=\delta^\mu_\nu$$. I am a little confused as to what the objects $$e^{\rho I}$$ and $$e_{\mu I}$$ are.

In principal I know what this calculation is, it's essentially the equivalent of the formula for the Christoffel symbols in the levi-civita connection, however deriving this in this gauge theory esque framework as proved troublesome. I figured I should just set $$T^I$$ equal to zero and use $$\omega^I_J=-\omega^J_I$$ at some point to get the components of the connection, but this has not worked. I first wrote everything explicitly, and examined the $$i$$th component of $$T^I$$:

$$T^i=d(e^i_\mu dx^\mu)+\omega^i_{\nu j}e^j_\mu dx^\nu\wedge dx^\mu$$ Carrying out the exterior derivative of the first term we obtain:

$$T^i=\partial_\nu e^i_\mu dx^\nu\wedge dx^\mu+\omega^i_{\nu j}e^j_\mu dx^\nu\wedge dx^\mu$$ Contracting $$T^i$$ with the coordinate vector fields $$\partial_\mu$$ and $$\partial_\nu$$ we obtain: $$i_{\partial_\nu}\left(i_{\partial_\mu}T^i\right)=\partial_\nu e^i_\mu-\partial_\mu e^i_\nu+\omega^i_{\mu j}e^j_\nu-\omega^i_{\nu j}e^j_\mu$$ Setting this equal to zero I thought I could do something to solve for $$\omega^i_{\mu j}$$, but everything I think of doing doesn't pan out, which suggests to me that I'm attacking this the incorrect way.

Any advice or hints would be greatly appreciated.

Edit: I am now convinced I need to use the following coordinate invariant koszul formula:

$$2g(\nabla_X Y,Z)=Xg(Y,Z)+Yg(Z,X)-Zg(X,Y)-g(X,[Y,Z])+g(Y,[Z,X])+g(Z,[Y,X])$$

But I am not quite sure how to translate it in this frame work. When finding the components of the levi civita connection you just let $$X=\partial_i,Y=\partial_j,Z=\partial_k$$ but I am not sure what I should let $$X,Y,Z$$ be since I don't feel like I can use the coordinate vector fields. Could I just let $$e_i$$ be the standard basis vectors on $$\mathbb{R}^m$$ and then set $$X=e_\mu^ie_i, Y=e_\nu^ie_i, Z=e_\eta^ie_i$$? Or should I should let $$e^I_u$$ denote a basis vector and have $$\left(\nabla_{e^I_\nu} e^I_\mu\right)^j=\omega^j_{\nu I} e^I_\mu$$?

• For those wondering, there exists a proof of this statement in Kobayahsi's foundations of differential geometry volume 1 Apr 21, 2023 at 22:36
• It's a bit late to comment, but I think that there is a sign error in your last term. If the $X$, $Y$, $Z$ are orthonormal the expression should be antisymmetric in $Y$, and $Z$ and yours is not. I get a minus sign before $g(Z,[Y,X])$ when I work through the derivation. May 25, 2023 at 16:18

I think that you're looking for this

$$T^a \equiv de^a + \omega^a{}_b \wedge e^b = 0 \Rightarrow$$

$$\frac{1}{2} e^a_{[\mu,\nu]} dx^{\mu} \wedge dx^{\nu} + \omega^a_{\mu b} dx^{\mu} \wedge e^b{}_{\nu}dx^{\nu} = 0$$

$$\frac{1}{2} e^a_{[\mu,\nu]} dx^{\mu} \wedge dx^{\nu} + \frac{1}{2} \omega^a_{[\mu b} e^b{}_{\nu]} dx^{\mu} \wedge dx^{\nu} = 0 \Rightarrow$$

$$e^a_{[\mu,\nu]} + \omega^a_{[\mu b} e^b{}_{\nu]} = 0 \Rightarrow$$

$$e^{\mu}{}_c e^{\nu}{}_d ( e^a_{[\mu,\nu]} + \omega^a_{[\mu b} e^b{}_{\nu]} ) = 0$$

Now we are going to use a magic trick, we take the quantity

$$e^{\mu}{}_c e^{\nu}{}_d e^a_{[\mu,\nu]} + e^{\mu}{}_a e^{\nu}{}_c e^d_{[\mu,\nu]} - e^{\mu}{}_d e^{\nu}{}_a e^c_{[\mu,\nu]} =$$

$$-e^{\mu}{}_c e^{\nu}{}_d ( \omega^a_{\mu b} e^b{}_{\nu} - \omega^a_{\nu b} e^b{}_{\mu}) - e^{\mu}{}_a e^{\nu}{}_c ( \omega^d_{\mu b} e^b{}_{\nu} - \omega^d_{\nu b} e^b{}_{\mu}) + e^{\mu}{}_d e^{\nu}{}_a ( \omega^c_{\mu b} e^b{}_{\nu} - \omega^c_{\nu b} e^b{}_{\mu}) =$$

$$- e^{\mu}{}_c \omega^a_{\mu d} + e^{\nu}{}_d \omega^a_{\nu c} - e^{\mu}{}_a \omega^d_{\mu c} + e^{\nu}{}_c \omega^d_{\nu a} + e^{\mu}{}_d \omega^c_{\mu a} - e^{\nu}{}_a \omega^c_{\nu d} =$$

$$\require{cancel} - e^{\mu}{}_c \omega^a_{\mu d} + \cancel{e^{\nu}{}_d \omega^a_{\nu c}} - \cancel{e^{\mu}{}_a \omega^d_{\mu c}} - e^{\nu}{}_c \omega^a_{\nu d} - \cancel{e^{\mu}{}_d \omega^a_{\mu c}} + \cancel{e^{\nu}{}_a \omega^d_{\nu c}} =$$

$$-2 e^{\mu}{}_c \omega^a_{\mu d}$$

And so we proved that

$$\begin{equation*} \boxed{\omega^{ab}_{\mu}= \frac{1}{2} e^{\nu a}e^b_{[\nu,\mu]} - \frac{1}{2} e^{\nu b}e^a_{[\nu,\mu]} - \frac{1}{2} e^{\kappa a} e^{\lambda b}(e_{\lambda c , \kappa} - e_{\kappa c, \lambda } ) e^c{}_{\mu} } \end{equation*}$$