Skip to main content
Search type Search syntax
Tags [tag]
Exact "words here"
Author user:1234
user:me (yours)
Score score:3 (3+)
score:0 (none)
Answers answers:3 (3+)
answers:0 (none)
isaccepted:yes
hasaccepted:no
inquestion:1234
Views views:250
Code code:"if (foo != bar)"
Sections title:apples
body:"apples oranges"
URL url:"*.example.com"
Saves in:saves
Status closed:yes
duplicate:no
migrated:no
wiki:no
Types is:question
is:answer
Exclude -[tag]
-apples
For more details on advanced search visit our help page
Results tagged with
Search options not deleted user 13972

Complex, contact, Riemannian, pseudo-Riemannian and Finsler geometry, relativity, gauge theory, global analysis.

82 votes
Accepted

When can a connection Induce a Riemannian metric for which it is the Levi-Civita connection?

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, practica …
Robert Bryant's user avatar
81 votes

Beautiful descriptions of exceptional groups

It is not always clear what one means by 'the simplest description' of one of the exceptional Lie groups. In the examples you've given above, you quote descriptions of these groups as automorphisms o …
Robert Bryant's user avatar
63 votes
Accepted

"The complex version of Nash's theorem is not true"

The failure is actually more profound than you might guess at first glance: There are conformal metrics on the Poincare disk that cannot (even locally) be isometrically induced by embedding in $\ma …
Robert Bryant's user avatar
61 votes

Theorem of Bryant in higher dimensions

First, the hypotheses of the theorem I proved require that $Y$ be compact and oriented, in addition to requiring that $g$ be real-analytic. Second, the method I used (the Cartan-Kähler Theorem) exten …
Robert Bryant's user avatar
54 votes

Does the curvature determine the metric?

It should not be surprising that, for a $2$-dimensional manifold, the Gauss curvature $K:M\to\mathbb{R}$ does not determine a unique metric $g$. After all, the former is locally one function of $2$ v …
Robert Bryant's user avatar
50 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 m …
Robert Bryant's user avatar
48 votes
Accepted

Finding a 1-form adapted to a smooth flow

If I understand correctly, there is already a counterexample on the torus: On the $xy$-plane $\mathbb{R}^2$, let $X$ be the vector field $$ X = \sin x\,\frac{\partial\ }{\partial x} + \cos x\,\frac{\ …
Robert Bryant's user avatar
48 votes
Accepted

Hsiung on the Complex Structure of $S^6$

While it's good to have a source, such as Datta's paper that points out the error, I find that his explanation of why the key equation is wrong is not as clear as it could be. In fact, with a little …
Robert Bryant's user avatar
45 votes
Accepted

Are there some other notions of "curvature" which measure how space curves?

in addition to these excellent examples of non-local curvature quantities and their extensions to the non-smooth setting (which I am not sure the OP was anticipating), I might add the 'original' non-l …
45 votes
Accepted

Vector bundles vs principal $G$-bundles

The difference is that, for a vector bundle, there is usually no natural Lie group action on the total space that acts transitively on the fibers. The fact that all of the fibers are, individually Li …
43 votes
Accepted

Does the Pfaffian have a geometric meaning?

The thing you are missing is one further geometric property of the $(2n{-}1)$-form $\Pi$ that Chern constructs on the unit sphere bundle $\mathsf{S}(M)$ of the oriented $2n$-manifold $M$: The fact tha …
Robert Bryant's user avatar
40 votes

deRham cohomology of $S^n$ without Mayer-Vietoris

Have you done any integration theory? (I assume you have, otherwise you wouldn't necessarily know what the deRham cohomology does for you.) The fastest proof I know is: Take a closed $k$-form $\o …
Robert Bryant's user avatar
40 votes
Accepted

Isometry group of a homogeneous space

Here is an algorithm to compute the Lie algebra of the group of isometries of a homogeneous space $G/H$ endowed with a $G$-invariant (pseudo-)Riemannian metric $g$. It is phrased in terms of essentia …
Robert Bryant's user avatar
37 votes
Accepted

Surfaces filled densely by a geodesic

Any surface of revolution in $3$-space with poles will have this property. The reason is that, in this case, any geodesic either goes through a pole (i.e., a point where the axis of revolution meets …
Robert Bryant's user avatar
34 votes

Riemannian surfaces with an explicit distance function?

NB (3/1/13): I revised this answer to make it more complete (and, to be frank, more accurate). My original answer did not take into account the difference between the cut locus and the conjugate loc …
Robert Bryant's user avatar

1
2 3 4 5
33
15 30 50 per page