Let $\pi\colon TM\to M$ be the tangent bundle of a differentiable manifold, let $E=TM\backslash 0$ be the slit tangent bundle, and let $V_eE$ be the kernel of $\pi_*$ at $e\in E$. The set $VE=\cup_{e\in E} V_eE$ is the vertical bundle. A non-linear connection is a splitting $TE=VE\oplus HE$, where $HE$ is the horizontal bundle. Every vector $v\in TM$ can be lifted to a vector $V\in HE$, such that $\pi_*(V)=v$. Similarly one can lift closed curves of $M$ starting at $p$ to (non-)closed curves on $E$. For every closed curve $\gamma$ this gives a map $\Gamma:T_pM\to T_pM$. If the connection is linear it is well known that this holonomy is a linear map. Is the converse true?

In coordinates, if $\{x^i\}$ are coordinates on $M$ then on TM we have coordinates $\{x^i,y^i\}$ and the basis of $HE$ obtained lifting the holonomic basis is $$ \frac{\partial}{\partial x^i}-N_i^j(x,y) \frac{\partial}{\partial y^j}. $$ The coefficients $N^i_j$ are said to be the coefficients of the non-linear connection, so I am asking if they are linear in $y$ under the assumpton of linear holonomy.

  • $\begingroup$ it is definately true that if the linear transport maps are linear then the connection is linear. I this what you are asking? $\endgroup$ – Daniel Barter Aug 19 '15 at 12:16
  • $\begingroup$ No, because I was only demanding that the transport maps over closed paths be linear. $\endgroup$ – Ettore Minguzzi Aug 20 '15 at 15:44

The answer to your literal question is 'no', you can even have the holonomy be trivial (in which case, it is certainly linear) and yet the connection not be linear. This is a local question, so you can construct examples as follows: Take $M=\mathbb{R}^n$ with standard coordinates $x^i$ and let $y^i = \mathrm{d}x^i:TM\to\mathbb{R}$ be the corresponding functions on $TM$, so that $(x^i, y^i)$ define coordinates on $TM$. Now take any $n$ smooth functions $F^i:TM\to\mathbb{R}$ with the property that $$ \mathrm{d} x^1\wedge\cdots\wedge \mathrm{d} x^n\wedge \mathrm{d} F^1\wedge\cdots\wedge \mathrm{d}F^n \not=0, $$ which implies that the kernel subbundle of the map $F = (F^i):TM\to\mathbb{R}^n$ is spanned by vector fields of the form $$ X_i = \frac{\partial\ }{\partial x^i} + N_i^j(x,y)\frac{\partial\ }{\partial y^j}. $$ These then define a connection on $TM$ that is, in general, nonlinear. However, because the horizontal lifts of curves are given by lifting a curve in $M$ to a single level surface of $F$, all horizontal lifts of closed curves are closed.

Thus, the holonomy is trivial, but the connection need not be linear.


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.