PlanetPhysics/OCR2 Proofreading Test

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 $$ (22)= = \quad c_{j^{\prime\prime}k^{\prime\prime}}^{i^{\prime\prime}}=0. $$ \quad 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 $$ (23)= = \quad d\omega^{i}=\Sigma_{\rho,k}a_{\rho k}^{i}\pi^{\rho}\wedge\omega^{k}. $$ By taking the exterior derivative of (23) and using (18), we get $$ (24)= = \quad \Sigma_{\rho,k}a_{\rho k}^{i}\Pi^\rho \wedge \omega^{k}=0, $$ where we put $$ (25)= = \quad \Pi^\rho=d\pi^{\rho}+\frac{1}{2}\Sigma_{\sigma.\tau}\gamma_{\sigma\tau}^{\rho} \pi^{\sigma} \wedge \pi^{\tau}. $$ For a fixed value of $$k$$ we multiply the above equation by $$ \omega^{1}= = \wedge.=. . = \wedge= = \omega^{k-1}= = \wedge= = \omega^{k+1}\ldots= = \wedge= = \omega^{n}, $$ getting $$ \sum_{\rho}a_{\rho k}^{i}{\Pi^\rho}= = \wedge= = \omega^{1}= = \wedge.=. . = \wedge= = \omega^{n}=0, $$ or $$\Sigma_{\rho}a_{\rho k}^{i} {\Pi^\rho} \equiv 0,\ \mathrm{m}\mathrm{o}\mathrm{d}\ \omega^{j}.$$

\noindent Since the infinitesimal transformations $$X_{\rho}$$ are linearly independent, this implies that $$ \Pi^\rho\equiv 0,= = \mathrm{m}\mathrm{o}\mathrm{d}\ \omega^{j}. $$ It follows that $$\Pi^\rho$$ is of the form $$ \Pi^\rho=\Sigma_{j} \phi_{j}^{\rho} \wedge \omega^{j} $$ where $$\phi_{j}^{\rho}$$ are Pfaffian forms. Substituting these expressions into (24), we get $$ \Sigma_{\rho,j,k} (a_{\rho k}^{i}\phi_{j}^{\rho}-a_{\rho j}^{i}\phi_{k}^{\rho})\wedge\omega^{j}\wedge\omega^{k}=0. $$ It follows that $$ \Sigma_{\rho}(a_{\rho k}^{i}\phi_{j}^{\rho}-a_{\rho j}^{i}\phi_{k}^{\rho})\equiv 0,= = \mathrm{m}\mathrm{o}\mathrm{d}\ \omega^{k}. $$ Since $$G$$ has the property $$(C)$$, the above equations imply that $$ \phi_{j}^{\rho}\equiv 0,= = \mathrm{m}\mathrm{o}\mathrm{d}\ \omega^{k}. $$