ocr 2 proofreading test | Physics Library

necessary to consider the second bundle. The curvature form of our connection is a tensorial quadratic differential form in $M$ , of type $ad(G^{\prime})$ and with values in the Lie algebra $L(G^{\prime})$ of $G^{\prime}$ . Since the Lie algebra $L(G)$ of $G$ is a subalgebra of $L(G^{\prime})$ , there is a natural projection of $L(G^{\prime})$ into the quotient space $L(G^{\prime})/L(G)$ . The image of the curvature form under this projection will be called the torsion form or the torsion tensor. If the forms $\pi^{\rho}$ in (13) define a $G$ -connection, the vanishing of the torsion form is expressed analytically by the conditions

$\displaystyle (22)$ $\displaystyle \quad c_{j^{\prime\prime}k^{\prime\prime}}^{i^{\prime\prime}}=0.$

We proceed to derive the analytical formulas for the theory of a $G$

-connection without torsion in the tangent bundle. In general we will consider such formulas in $B_{G}$ . The fact that the G-connection has no torsion simplifies (13) into the form

$\displaystyle (23)$ $\displaystyle \quad d\omega^{i}=\Sigma_{\rho,k}a_{\rho k}^{i}\pi^{\rho}\wedge\omega^{k}.$

By taking the exterior derivative of (23) and using (18), we get

$\displaystyle (24)$ $\displaystyle \quad \Sigma_{\rho,k}a_{\rho k}^{i}\Pi^\rho \wedge \omega^{k}=0,$

where we put

$\displaystyle (25)$ $\displaystyle \quad \Pi^\rho=d\pi^{\rho}+\frac{1}{2}\Sigma_{\sigma.\tau}\gamma_{\sigma\tau}^{\rho} \pi^{\sigma} \wedge \pi^{\tau}.$

For a fixed value of $k$

we multiply the above equation by

$\displaystyle \omega^{1}$ $\displaystyle \wedge.$ . . $\displaystyle \wedge$ $\displaystyle \omega^{k-1}$ $\displaystyle \wedge$ $\displaystyle \omega^{k+1}\ldots$ $\displaystyle \wedge$ $\displaystyle \omega^{n},$

getting

$\displaystyle \sum_{\rho}a_{\rho k}^{i}{\Pi^\rho}$ $\displaystyle \wedge$ $\displaystyle \omega^{1}$ $\displaystyle \wedge.$ . . $\displaystyle \wedge$ $\displaystyle \omega^{n}=0,$

$\displaystyle \Sigma_{\rho}a_{\rho k}^{i} {\Pi^\rho} \equiv 0,\ \mathrm{m}\mathrm{o}\mathrm{d}\ \omega^{j}.$

Since the infinitesimal transformations $X_{\rho}$ are linearly independent, this implies that

$\displaystyle \Pi^\rho\equiv 0,$ $\displaystyle \mathrm{m}\mathrm{o}\mathrm{d}\ \omega^{j}.$

It follows that $\Pi^\rho$ is of the form

$\displaystyle \Pi^\rho=\Sigma_{j} \phi_{j}^{\rho} \wedge \omega^{j}$

where $\phi_{j}^{\rho}$ are Pfaffian forms. Substituting these expressions into (24), we get

$\displaystyle \Sigma_{\rho,j,k} (a_{\rho k}^{i}\phi_{j}^{\rho}-a_{\rho j}^{i}\phi_{k}^{\rho})\wedge\omega^{j}\wedge\omega^{k}=0.$

It follows that

$\displaystyle \Sigma_{\rho}(a_{\rho k}^{i}\phi_{j}^{\rho}-a_{\rho j}^{i}\phi_{k}^{\rho})\equiv 0,$ $\displaystyle \mathrm{m}\mathrm{o}\mathrm{d}\ \omega^{k}.$

Since

has the property $(C)$

, the above equations imply that

$\displaystyle \phi_{j}^{\rho}\equiv 0,$ $\displaystyle \mathrm{m}\mathrm{o}\mathrm{d}\ \omega^{k}.$

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