According to the answer of Sebastan and previous edit of Ben McKay I revise my post as follows:

Assume that $E$ is a vector bundle over a manifold $M$ with a connection $\nabla$. Is there a (unique) connection $\nabla'$ on $E':=Hom (E,E)$ with the following property;

For every curve $\gamma$ which connects point $x$ to $y$, with $\nabla$ parallel transport $\phi$ and $\nabla'$ parallel transport $\psi$, we have $\psi(T)=\phi T\phi^{-1}$. Moreover if $\nabla$ is a Riemannian connection corresponding to a Riemannian metric on $E$, can we choose a Riemannian comnnection $\nabla '$ as above.In the latter we consider the natural Riemannian metric inducing by initial metric on $E$ defined by $tr(AB^{*})$ on the $E'$- bundle

  • $\begingroup$ @Ben thank you for your edit. It seems that we were editing simultanously. In my new edit I fix my errors you pointed out. $\endgroup$ – Ali Taghavi Jul 4 '16 at 11:34
  • 2
    $\begingroup$ Perhaps it should be pointed out that this is a special case of a more general fact: Each connection $\nabla$ on $E$ induces a canonical connection $\nabla^{r,s}$ on $E^{\otimes r}\otimes (E^*)^{\otimes s}$ that is compatible with all tensor products, contractions, sub-representations, etc., so all of the parallel transports are also compatible. Your case is just $(r,s)=(1,1)$. $\endgroup$ – Robert Bryant Jul 4 '16 at 17:21
  • $\begingroup$ @RobertBryant Thank you very much for your comment. And the Riemannian property would be preserved? $\endgroup$ – Ali Taghavi Jul 4 '16 at 18:38
  • $\begingroup$ Yes, certainly. $\endgroup$ – Robert Bryant Jul 4 '16 at 20:34

No, because your formula does not make sense:

$T\in Hom(E_x,X_x)$ and $\phi\in Hom(E_x,E_y)$ invertible means that $$\phi^{-1}\circ T\circ \phi$$ is not well-defined unless $x=y.$

If you define $$\psi=\phi\circ T\circ \phi^{-1},$$ then $\psi$ is actually the parallel-transport of the induced connection $\nabla^{End}$ on the endomorphism bundle which is defined by satisfying the equation $$(\nabla^{End}T)(e)=\nabla T(e)-T(\nabla e)$$ for all $T\in\Gamma(Hom(E,E))$ and sections $e\in\Gamma(E).$

For a proof of this property consider $v,w\in E_x$ with $T(v)=w,$ and denote the corresponding parallel sections (along the given curve $\gamma$ from $x$ to $y$) by $v(t)$ and $w(t).$ Then, $\nabla_{\gamma'} v(t)=0$ and $\nabla_{\gamma'}w(t)=0.$ Hence, by uniqueness of solutions of ODE's, the parallel endomorphism field $T(t)$ along $\gamma$ satisfies $$T(t)(v(t))=w(t).$$ As $v\in E_x$ is arbitrary, this is equivalent to the equation $\psi=\phi\circ T\circ \phi^{-1}.$

The construction is compatible with metrics, as the same standard arguments for tensor products, dual bundles and corresponding connections carry over to the Riemannian/hermitian situation, i.e., the induced metric on the endomorphism bundle is parallel with respect to $\nabla^{End}.$ Note also that $\nabla^{End}$ is the unique connection whose parallel transport satisfies the equation $\psi=\phi\circ T\circ \phi^{-1}.$

  • $\begingroup$ thank you for your answer.I ,ll fix the typos in my post as you pointed out. $\endgroup$ – Ali Taghavi Jul 4 '16 at 11:37
  • $\begingroup$ May ellaborate why this connection work or please give a reference.moreover what about the Riemannian part of my question?Does your connection automatically work? $\endgroup$ – Ali Taghavi Jul 4 '16 at 11:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.