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Riemannian Geometry is a subfield of Differential Geometry, which specifically studies "Riemannian Manifolds", manifolds with "Riemannian Metrics", which means that they are equipped with continuous inner products.

4 votes
Accepted

Applications of Hessian operator in the Riemann manifold. Simple samples $S_{2}(f)$

On the torus $T^2$ with the coordinates $x,y$ and the flat metric $g= dx^2 + dy^2$ take any function $f(x)$. Its hessian is given, after raising the index, by the (1,1)-tensor $f''(x) dx\otimes \fra …
Vladimir S  Matveev's user avatar
1 vote

Hilbert's Theorem relevance to positive curvature

May be I misunderstood your question; I reformulate it as follows: whether there exists a regular embedding of a complete surface of constant positive curvature in $R^3$ and whether this surface can …
Vladimir S  Matveev's user avatar
12 votes

Does positively curved sphere admit an isometric embedding as hypersurface in Euclidean space?

The answer of j.c. given prior to mine is of course correct but let me give a trivial reason why (in big dimensions) every Riemannian metric after an arbitrary small perturbation is not isometr …
Vladimir S  Matveev's user avatar
10 votes
Accepted

Smoothing of the distance function on a Riemannian manifold

You function $d_p$ is a Lipschitz function (w.r.t. to the Riemannian distance) with the Lipschitz constant $1$ and it is possibly, for every $\varepsilon>0$, to $\varepsilon$-approximate it by a smoot …
Vladimir S  Matveev's user avatar
8 votes

Questions on smoothness of Riemann metrics

I confirm the Anton's answer (No, and the phenomenon is essentially local), but I suggest another explanation which works for C^1 2-dimensional metrics. We will look for a counterexample in the clas …
Vladimir S  Matveev's user avatar
8 votes
Accepted

What does it mean that the Hessian is proportional to the metric?

It is known (say, Y. Tashiro, Complete Riemannian manifolds and some vector fields, Trans.Amer.Math.Soc. 117(1965) 251– 275; I am not sure that Tashiro is the first who proved it and there were many …
Vladimir S  Matveev's user avatar
5 votes

Reference for when a metric on a four-manifold is Kahler?

I reformulate you question as follows: suppose we a given a Riemannian metric. How can one decide whether there exists a complex structure $J$ such that the metric is Kähler w.r.t. this complex struc …
Vladimir S  Matveev's user avatar
5 votes
Accepted

Length spectrum for Riemannian metrics in the projective plane

The answer is positive; in fact any smooth manifold has two nonisometric metrics with conjugate geodesic flows. A construction is in C. Croke, B. Kleiner, Conjugacy and rigidity for manifolds with a p …
Vladimir S  Matveev's user avatar
1 vote

Riemannian metric and Volume form for $SE(n)$ and/or $E(n)$

Of course any left invariant metric with respect the group of orientation preserving euclidean isometries (I believe that it what you call SE(n) ) is a multiple of the standard Euclidean metric and …
Vladimir S  Matveev's user avatar
1 vote

Tensor contraction and Covariant Derivative

It is a natural requirement and is more-ore less equivalent to the natural analog of the leibniz rule. Let us consider the following example: $$ d g(v,u) = \nabla g(u, v)+ g(\nabla u, v)+ g(u, \nabla …
Vladimir S  Matveev's user avatar
5 votes

The necessary and sufficient condition for $\textbf{global}$ conformal flatness of a n-dim (...

I add a small $\varepsilon$ to Robert's answer, which is a simple explanation and a simple example to what he said concerning the 2 dim case. Conformal structure of signature (1,1) on the surface …
Vladimir S  Matveev's user avatar
3 votes

Curvature of singular Riemannian metric

Under stronger regularity assumptions, an analog of the curvature exists in the weak sense, i.e., in the sense of generalized functions. The stronger regularity assumption is that the metric (in your …
Vladimir S  Matveev's user avatar
18 votes

Is there a global obstruction for a diffeomorphism to be an isometry?

The answer is ``no'', the pointwise condition is not enough. The example exists in dimension 1 already and can be generalized and made arbitrary weird for all dimensions. Consider a smooth functi …
Vladimir S  Matveev's user avatar
5 votes

"Famous" 2d Riemannian manifolds with non-constant curvature

My favorite 2D metrics of nonconstant curvature admitting a Killing vector field are the so-called Darboux-superintegrable metrics. Their definition is: the space of Killing tensors of degree two (i. …
Vladimir S  Matveev's user avatar
23 votes

Manifolds admitting flat connections

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 connectio …
Vladimir S  Matveev's user avatar

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