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I have the following question. Let $G \subset SO(n)$ be a Lie Group and $M$ be a smooth manifold of dimension $n$. Furthermore let $P$ be a $G$-structure on $M$ i.e. $P$ is a principal subbundle of the principal $GL(n,\mathbb{R})$-bundle $F$ (the frame bundle) on $M$. Set now $S=F/G$ which carries a smooth manifold structure with the quotient mapping $\tau : F \rightarrow S$ and $\overline{\pi}:S\rightarrow M$ and $\pi:F\rightarrow M$ the obvious bundle projections. It is known that $G$-Structures on $M$ are in one to one correspondence with sections of $\overline{\pi}:S\rightarrow M$. Since $G \subset SO(n)$ it is also known that every $G$-Structure has an underlying Riemannian metric, denote it by $g_{\sigma}$, for a section $\sigma : M \rightarrow S$ and an orientation that is defined by the condition that $u:T_{x}M \rightarrow \mathbb{R}^{n}$ be an oriented isometry for all $u \in P_{x}$ and all $x \in M$. What I just wrote is only some preliminary work. Now to my real question on the understanding of the definition of a torsion-free $G$-Structure. The definition is as follows: A $G$-Structure $P$ on $M$ and a the corresponding section $\sigma : M \rightarrow S$ are said to be torsion-free if $P$ is parallel with respect to the Levi-Civita connection of the underlying Riemannian metric.

I dont understand this definition. How can a principal bundle be parallel with respect to the Levi-Civita connection of the underlying Riemanninan metric, i.e. how can one understand $\nabla P = 0$, where $\nabla$ is the Levi-Civita connection of $g_{\sigma}$ on $M$ ? How can one differentiate a principal bundle with a connection that "lives" on $M$ ? I hope someone can explain this to me.

Greetings monica

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This condition is easy to say if you think of the Levi-Civita connection as an Ehresmann connection, i.e. as a distribution on $F$. Then, $P$ being parallel just means that the distribution is tangent to it. –  Paul Reynolds Apr 9 '13 at 19:11
    
Here's how I think about it. Parallel transport in $TM$ along a path from $p$ to $q$ in $M$ with respect to the LC connection preserves the Riemannian metric, so it sends an orthonormal frame at $p$ to an orthonormal frame at $q$. The $G$-structure just picks out a preferred collection of orthonormal frames at each point (on which $G$ acts freely and transitively), and one can ask: will parallel transport send a preferred orthonormal frame at $p$ to a preferred orthonormal frame at $q$? If the answer is yes for all $p$ and $q$ then the $G$-structure is parallel. –  Paul Siegel Apr 10 '13 at 2:19

1 Answer 1

up vote 5 down vote accepted

The bundle $P$ is made out of frames, being a subbundle of the frame bundle $F$. So each point in $P$ is a basis of a tangent space of $M$. We can take any metric on $M$, and use it to parallel transport the vectors of that basis. In general, for an arbitrary metric, such a parallel transport will take these vectors into a basis that does not belong to $P$. We don't want to write out something like $\nabla P$. We just think geometrically about carrying tangent vectors to $M$ along curves in $M$ by parallel transport. In order to see whether a given $G$-structure $P$ is torsion-free, we can pick a local section of $P$, which we think of as some 1-forms $\omega_i$. We differentiate them, to produce the connection 1-forms $\omega_{ij}=-\omega_{ji}$, so that $d \omega_i = - \omega_{ij} \wedge \omega_j$. These 1-forms $\omega_{ij}$ will split into a part valued in the Lie algebra of $P$ and a part valued in the orthogonal complement to that Lie algebra inside $SO(n)$. That orthogonal-valued bit is the torsion of the $G$-structure: if it vanishes, then parallel transport of these $\omega_i$ will act as an element of $G$ on the $\omega_i$. Therefore $P$ is taken to itself under parallel transport.

For more explanation, try Dominic Joyce's book Riemannian holonomy groups and calibrated geometry or Andreas Cap and Jan Slovak Parabolic geometries I: background and general theory.

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If you really want to make a symbol $\nabla P$, it probably would mean the torsion, as described above, as a section of an associated vector bundle of $P$, $P \times_G W$ where $W$ is the obvious representation of $G$. I think the theory of $G$-structures is clearer if we don't assume that $G$ is a subgroup of $SO(n)$, and use the definition of torsion via Spencer cohomology. –  Ben McKay Apr 9 '13 at 10:24
    
you consider the connection 1-forms as a matrix (locally) $(\omega_{ij})$. Why does its value split into a part of the Lie Algebra of $G$ and one into the Orthogonal complement of the Lie Algebra of $G$ inside $SO(n)$? Why does it make sense to define the torsion like this (as the orthogonal complement)? –  monica Apr 9 '13 at 13:37
    
the above splitting you mentioned is orthogonal in $SO(n)$ with respect to what metric? Bi-invariant metric in $SO(n)$ ? –  monica Apr 9 '13 at 14:08
    
what do you mean by "parallel transport of these $\omega_{i}$ will act as a $P$-transformation on the $\omega_{i}$"? Why does this hold? –  monica Apr 9 '13 at 14:24
    
Think of $\Omega=(\omega_{ij})$ as an antisymmetric matrix, whose entries are 1-forms. Antisymmetric matrices are precisely the elements of the Lie algebra of $SO(n)$. Split this Lie algebra into the Lie algebra of $G$ and its orthogonal complement (using the Killing form on the Lie algebra of $SO(n)$). I should have said that parallel transport of these $\omega_i$ acts as a $G$ transformation. –  Ben McKay Apr 9 '13 at 15:59

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