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49 votes

What is the Levi-Civita connection trying to describe?

I think that the literal answer is that the Levi-Civita connection of $g$ is trying to describe the metric $g$ and nothing else. It is the only connection-assignment that is uniquely defined by the ...
Robert Bryant's user avatar
28 votes

What is the Levi-Civita connection trying to describe?

I will try to help with the title question. I think that the real motivation for the Levi-Civita connection comes from looking at surfaces in Euclidean 3-space. Differentate one tangent vector field $...
Ben McKay's user avatar
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19 votes
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Geometric interpretation of horizontal and vertical lift of vector field

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 ...
Tobias Diez's user avatar
  • 5,592
19 votes
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When is a flow geodesic and how to construct the connection from it

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 ...
Robert Bryant's user avatar
19 votes
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Conceptual definition of the extension of a connection to 1-forms

If we denote by $\nabla$ the connection on $E\to M$, then we can define an exterior differential $d^\nabla:\Gamma(\Lambda^pM\otimes E)\to\Gamma(\Lambda^{p+1} M\otimes E) $ by $$ d^\nabla \alpha (X_0,\...
Overflowian's user avatar
  • 2,523
18 votes

What is the Levi-Civita connection trying to describe?

First, you should not dismiss the uniqueness of the connection too lightly. If you want to study a Riemannian metric per se, then you want to find invariants of it, things that are uniquely determined ...
Deane Yang's user avatar
18 votes

What is the Levi-Civita connection trying to describe?

Without loss of generality (Nash embedding theorem) we may assume the Riemannian manifold is an embedded submanifold of Euclidean space: its metric at any point is just the restriction of the ...
Jonathan Manton's user avatar
16 votes
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A non-Abelian de Rham complex?

What is being described in the main post is simply the de Rham (crossed) complex valued in a Lie group (not necessarily commutative). See, for example, Section 6.2 in Anders Kock's Synthetic Geometry ...
Dmitri Pavlov's user avatar
15 votes
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Locally Riemannian Connection

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 ...
Robert Bryant's user avatar
13 votes

Riemannian vs Non-Riemannian curvature

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 ...
Robert Bryant's user avatar
12 votes

What is the Levi-Civita connection trying to describe?

The other answers give good insight. Here's another perspective. Since the Levi-Civita connection is the unique metric and torsion-free connection, to motivate its use we need to convince ourselves ...
Gabe K's user avatar
  • 5,394
12 votes
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Embedding of a bundle with connection into a bundle with flat connection?

The paper “Existence of universal connections” by Narasimhan, M. S.; Ramanan, S. proves that the Grassmanian is universal for connections not just bundles. That is any connection in a U(n) or O(n) ...
Tom Mrowka's user avatar
  • 3,014
11 votes

When are geodesics straight lines?

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 ...
Ben McKay's user avatar
  • 25.7k
11 votes

Symplectic connections are (locally) Levi-Civita connections

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). ...
Vít Tuček's user avatar
  • 8,157
10 votes
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Question about a proof in Berthelot's crystalline book

Suppose you have a commutative ring $R$, a square zero ideal $I\subset R$, a $R$-module $M$ and an endomorphism $u$ of $M$ which is the identity modulo $IM$. Then $v:= 1_M-u$ maps $M$ to $IM$, hence $...
abx's user avatar
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9 votes
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Which manifolds are sensitive to the cocycle in the Dijkgraaf-Witten model?

The example $G = \mathbb Z/2$ and $M = \mathbb{RP}^3$ works. The inclusion $\mathbb Z/2\to\{\pm 1\}\subset\mathrm U(1)$ induces an isomorphism $H^3(B\mathbb Z/2, \mathbb Z/2)\to H^3(B\mathbb Z/2, \...
Arun Debray's user avatar
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9 votes

Alternative (easier) Proof of Ambrose Singer Holonomy theorem

I always find it easier to work with the vector bundle induced by a linear representation of the structure group. I believe this theorem is a consequence of the following loop formula (a terse proof ...
Deane Yang's user avatar
9 votes
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A consequence of Ambrose-Singer theorem on holonomy

Your first question is a bit ambiguous. Are you asking whether, for each $p\in U$, the matrices $S_k(p)$ span the Lie algebra of $\mathrm{Hol}^0_p(\nabla)$ or are you asking whether, after taking the ...
Robert Bryant's user avatar
9 votes

When do flat holomorphic connections exist?

The comment of HYL should be an answer. Since the OP has asked for explicit counterexamples, I will give an example that 1) does not imply 2): Consider a compact Riemann surface $\Sigma$ of genus $g\...
Sebastian's user avatar
  • 6,745
8 votes
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Multiplication in Deligne cohomology: explicit formula for $p=q=1$

There is indeed a "universal" holomorphic bundle with connection on $\mathbb{C}^\times \times \mathbb{C}^\times$ which induces the bundles $r(f,g)$ defined in Esnault-Viehweg. This universal ...
François Brunault's user avatar
8 votes

history of geometric mechanics

Darboux was perhaps the first to argue fully explicitly that “the general problem of mechanics is nothing but the generalisation to an arbitrary number of variables of the problem of the study of ...
Viktor Blasjo's user avatar
8 votes

A non-Abelian de Rham complex?

Unfortunately, things are subtle and terrible (at least, compared to the abelian case). I wrote a bit about this in an unpublished article. One of the surprising, terrible features is that you can ...
Matt Noonan's user avatar
  • 3,984
8 votes

Obstructions to the existence of a flat connection on a vector bundle

A $d$-dimensional flat real vector bundle $E→M$ is classified by a map $\def\B{{\sf B}}\def\GL{{\rm GL}}M→\B\GL(d)_δ$, where $\GL(d)_δ$ is the orthogonal group equipped with the discrete topology. ...
7 votes

Generalized Dirac operators

A very good place to read about this is the 3. chapter of the book "Heat kernels and Dirac operators" by Nicole, Getzler & Vergne. Up to the wrong sign in your second definition, 1. and 2. are ...
Sebastian's user avatar
  • 6,745
7 votes

When is a flow geodesic and how to construct the connection from it

Here is a geometric way that turns out to be equivalent to Robert's answer (i.e., to the Klein-Grifone-Foulon approach to connections associated to a second order ordinary differential equation of a ...
alvarezpaiva's user avatar
  • 13.2k
7 votes
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Is there such a connection on the punctured plane?

Yes. Take the Levi-Civita connection of any conformal metric $g = e^{2u}(dx^2+dy^2)$ of positive curvature, say. Then, by (local) Gauss-Bonnet, the holonomy around any smooth closed loop $\gamma$ is ...
Robert Bryant's user avatar
7 votes
Accepted

The automorphism group of a symplectic symmetric space

The affine group of $(M,\nabla)$ is a Lie group $G$ by Kobayashi's theorem that shows that the automorphism group of any affine connection is a Lie group (see Kobayashi and Nomizu's Foundations of ...
Robert Bryant's user avatar
7 votes
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Does $\nabla g=\omega(\cdot) g$ imply $\nabla$ is metric w.r.t a conformal rescaling of $g$?

The answer is 'no'. For example, just take $M$ to be $\mathbb{R}^n$ (for $n>1$), and $E = M\times \mathbb{R}^r$ for some $r>1$. Let $\omega$ be any $1$-form on $M$, and define a connection $\...
Robert Bryant's user avatar
7 votes

Yang-Mills over surfaces

In general it is not in the center of $\frak g$. The curvature of a Yang-Mills connection in this dimension is parallel which means it commutes with the image of holonomy representation which need ...
Tom Mrowka's user avatar
  • 3,014
7 votes
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The purpose of connections in differential geometry

If you are interested in local-to-global results, i.e., collecting local info about the manifold and then patch it together to get a global info then you need tools for the patching part of the ...

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