5
$\begingroup$

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 $\left(\nabla_{e^I_\nu} e^I_\mu\right)^j=\omega^j_{\nu I} e^I_\mu$?

$\endgroup$
2
  • $\begingroup$ For those wondering, there exists a proof of this statement in Kobayahsi's foundations of differential geometry volume 1 $\endgroup$
    – Chris
    Apr 21, 2023 at 22:36
  • $\begingroup$ 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. $\endgroup$
    – mike stone
    May 25, 2023 at 16:18

1 Answer 1

1
$\begingroup$

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*}

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .