73

Bill and Willie have (of course) given correct answers in terms of the holonomy of the given torsion-free connection $\nabla$ on the $n$-manifold $M$. However, it should be pointed out that, practically, it is almost impossible to compute the holonomy of $\nabla$ directly, since this would require integrating the ODE that define parallel transport with ...

55

Yes.
First, there's a very simple criterion for whether $\nabla$ is an orthogonal connection: look at the holonomy of $\nabla$ around closed
loops in the manifold, and ask whether they preserve a quadratic form. The set of
quadratic forms preserved by a linear transformation is a linear subspace of all quadratic
forms, so there's some linear subspace of ...

36

To start with, you need the connection to be torsion free. After that, there is a characterisation of metric connections given by Schmidt, CMP 29 (1973) 55-59, which states that the linear torsion-free connection is metric if and only if the holonomy group is a sub-group of the orthogonal group of the desired signature.

24

Many manifolds have curvature-free (i.e., flat) connections on their tangent bundles. For example, any orientable $3$-manifold $M$ is parallelizable, i.e., its tangent bundle is trivial, so it carries a flat connection (in fact, many flat connections). However, nearly all of these connections will have torsion. In fact, they all will unless $M$ is very ...

22

I presume that your variety $X$ is smooth.
Consider the additive map $\mathrm d\log \colon \mathscr O_X^*\to \Omega^1_X$ that sends $f$ to $\mathrm df/f$. It induces a map $c_1$ in cohomology from $H^1(X,\mathscr O_X^*)$
to $H^1(X,\Omega^1_X)$ — a coherent avatar of the first Chern class.
By Hodge Theory, $H^1(X,\Omega^1_X)$ is a subspace of $H^2(X,\mathbf ...

22

For question 1 : if a vector bundle admits a flat connection, it comes from a representation of the fundamental group. Thus if $TM$ admits a flat connection and $M$ is simply-connected, $TM$ is trivial. Hence any simply-connected $M$ with $TM$ nontrivial gives an example, for instance $M=\mathbb{S}^2$.

20

NB: I'm combining my previous comments into an answer, because I believe that this is better than leaving them scattered.
As another commenter has pointed out, the skew-symmetric part of the Ricci tensor is the obstruction to there being a $\nabla$-parallel volume form in the first place. To see this, consider the first Bianchi identity: $R^i_{jkl}+R^i_{...

20

I would compare at least 2 & 3, if not 4, using maps into classifying spaces.
For a space $X$ homotopic to a finite CW-complex (at least), the pullback map $Map(X,Gr_n(\mathbb C^\infty)) \to \{$isomorphism classes of $n$-plane bundles$\}$ is bijective. So we should study the induced maps on cohomology, and get definition 2.
Over $Gr_n(\mathbb C^\infty)$...

19

As the previous answer points out you have to consider local systems for a finer topology than the Zariski topology. It is natural to consider the étale topology. The category of étale local systems of finite dimensional $k$-vector spaces form a tannakian category whose group is the étale fundamental group (more precisely it is a pro-(constant finite) ...

18

I did not understand the first question
Question 1 Are there manifolds with the property that each connection on is never flat?
Because one of course can construct, on any manifold, a connection which is somewhere flat and somewhere not flat.
But of course there are manifolds that do not admit flat affine connections, the simples example is the ...

16

Note: I've decided that this answer should be rearranged a bit
so that it clearly separates the discussion of the basic properties of
the tangent bundle from the discussion of the formulae associated to a connection.
The content is the same, but I hope it's clearer.
The standard way to discuss the geometry of connections and geodesic flow 'invariantly' (...

15

Let $V$ be a complex vector bundle on a manifold $M$. Chern classes can be defined by topological means (see Milnor's book on characteristic classes), which yields elements $c_k(V) \in H^{2k}(M;\mathbb{Z})$. The normalization in the Chern-Weil theory is chosen so that the associated elements of de Rham cohomology groups $H^{2k}(M;\mathbb{R})$ agree with ...

15

"Rolling without slipping" is a powerful idea, but the phrase doesn't necessarily lead one to the intended mental model. In particular,
torsion is something that is at issue only for manifolds of dimension 3 or higher. Perhaps you can imagine taking a 3-manifold, and rolling it along a hyperplane in 4-space --- but
the metaphor becomes strained, partly ...

15

Milnor proved in [On the existence of a connection with curvature zero, Comm. Math. Helv. v 32] that bundles over a surface of genus g has flat connections iff its Euler class is less than g by an absolute value (see also Wood, Bundles with totally disconnected structure group). Sullivan in "A generalization of Milnor's inequality ...Comm. Math. Helv. v. 51" ...

15

Perhaps I can offer some information and comment on this problem. An essential part of the problem is how to interpret terms such as 'observe', 'accessible', 'identify', as the OP wants to know how to write down a computable criterion for a torsion-free connection to be the Levi-Civita connection of a Riemannian metric. Of course, this requires that one ...

13

NB: In what follows, to save typing, I will be working on a manifold $M$, but I will write $T$, $T^*$, etc. to denote the bundles $TM$, $T^*M$, etc. and let $M$ be understood.
It seems that the OP wants to be able to test whether a $(1,3)$-tensor $R$ with the symmetries of a Riemann curvature tensor is actually the curvature of a Riemannian metric. By the ...

12

Yes, certainly. The model for BG as you described it has a canonical, universal connection
for its $SO(n)$ bundle: just the induced Riemannian connection from Euclidean space.
As you
move an $n$-dimensional plane in $\mathbb E^{n+m}$, the induced connection
is the limit of compositions
of orthogonal projections between nearby planes. In the limit, these ...

12

The $(+)$/0/$(-)$-connections on a Lie group $G$ are the left-invariant connections (I will identify connections with covariant derivatives) defined on left-invariant vector fields by $\nabla_XY=c[X,Y]$ with respectively $c=1,\frac{1}{2},0$. Their name is related to the fact that their torsion is $T(X,Y)=d[X,Y]$ with respectively $d=1,0,-1$.
The name is ...

12

Ad 1: Yes, there is. The formula is
$$\nabla^*(X^\flat \otimes u) = - \nabla_X u -\mathrm{div}(X) \cdot u,$$
as can easily seen by local computation. Here, $X$ is a vector field and $X^\flat$ is the dual one form w.r.t. the metric.
Note that unless you have a scalar product on the bundle $E$, too, the dual operator will be an operator $\Gamma^\infty(T^*M \...

11

I'm not sure what you're looking for, but maybe this will help. I am completely avoiding any mention of coordinates or bases.
A connection on a vector bundle determines and is determined by the corresponding covariant derivative on sections of $V$.
Connections on vector bundles $V$ and $W$ determine a connection on $V\otimes W$. In terms of covariant ...

11

I like the question. Below is a somewhat sketchy version of how I see this.
I think the importance of tensors and contraction of tensors originates from trying to do basic differential geometry or vector calculus from a co-ordinate-free point of view. The most basic objects are curves and velocity vectors of curves. The key observation is that the set of ...

11

Special examples of Robert Bryants answer are Lie groups.
On any Lie group, the left trivialization of the tangent bundle corresponds to a flat connection whose torsion is essentially the Lie bracket, whereas the right trivialization corresponds to another connection without curvature whose torsion is essentially the negative of the Lie bracket.
On the ...

11

Of course, yes. Lie derivative is defined for any geometric object (= when it is defined what happends when we change a coordinate system): take the flow $\phi_t$ of the vector field, consider the pullback $\phi_t^*\Gamma$ of your geometric object $\Gamma$ and define Lie derivative as the $\tfrac{d}{dt}$-derviative at $t=0$ of $\phi_t^*\Gamma$.
For ...

11

I find the following viewpoint helpful to translate between the different incarnation of a connection.
To every vector bundle $\pi: E \to M$ (in your case $E = TM$) we have an associated exact sequence of vector bundles (sometimes called the Atiyah sequence, at least in the principal bundle case):
$$ 0 \to V E \to TE \to \pi^* TM \to 0 $$
Here $VE$ denotes ...

11

The geodesics are straight lines, in geodesic normal coordinates, just when the associated projective connection is flat. See Kobayashi and Nagano, On projective connections, Journal of Mathematics and Mechanics, vol. 13, no. 2, 1964. If an affine connection is projectively flat, then the Weyl and Cotton tensors vanish, as these are projective connection ...

10

A mathematical reason is as follows.
On the one hand, the Laurent series for the modified Bessel functions of the first kind $I_k$ can be deduced from the Laurent series for the Bessel functions of the first kind $J_k$ given here. It reads
$$
\sum_{k\in\mathbb{Z}}I_k(x)t^k=\mathrm{e}^{(x/2)(t+1/t)}.
$$
On the other hand, the characteristic function of a ...

10

This question has been definitively answered in the following paper:
http://nzjm.math.auckland.ac.nz/images/f/f4/When_is_a_Connection_Metric_Connection%3F.pdf

10

The covariant constant 1-forms are parallel. The value of a covariant constant 1-form is determined throughout each connected component by its value at any one point: just take that value and parallel transport along any path to any other point. This recipe will succeed in producing a 1-form just when the 1-form you start with is invariant under the holonomy ...

10

For any Riemannian metric $g$ on a symplectic manifold $(M, \omega)$ there exists (canonical) almost complex structure $J$ making these threetensors compatible (any one can be defined by other 2). compatible with $\omega$, which means that $\omega(\_, J\_)$ defines a Riemannian metric $\widetilde{g}$. For such a compatible triple and the Levi-Civita ...

9

The standard connection is the Levi-Civita connection of the flat metric. So if you have an embedding such that the given connection is the (projection of the) flat connection then you can also induce a metric on the embedded manifold. Hence the given metric was already a Levi-Civita connection of a Riemannian metric on your manifold. Thus the question is ...

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