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A manifold is a topological space that locally resembles Euclidean space near each point. More precisely, each point of an n-dimensional manifold has a neighbourhood that is homeomorphic to the Euclidean space of dimension n.
22
votes
Accepted
Smooth curve in $\mathbb{R}^3$ not contained in real analytic surface?
Here is an example: Let $\gamma:\mathbb{R}\to\mathbb{R}^3$ be defined by $\gamma(t) = \bigl(t,\exp(-1/t^2),0\bigr)$ for $t<0$, $\gamma(0) = (0,0,0)$ and $\gamma(t) = \bigl(t,0,\exp(-1/t^2)\bigr)$ for …
6
votes
Accepted
Is $\operatorname{Spin}(8)$ a direct product of $\operatorname{Spin}(7)$ and $S^7$?
As I suspected, the statement that the bundle $\mathrm{SO}(8)\to S^7$ is a product bundle, i.e., that
$$
\mathrm{SO}(8)\simeq S^7\times\mathrm{SO}(7)\tag1
$$
as bundles over $S^7$ is in N. Steenrod's …
12
votes
Accepted
Are there examples of Einstein manifolds with unbounded curvature?
If you don't care about completeness, here's a fairly simple way to construct such examples: Start with a compact Einstein manifold $(M^n,g)$ with Einstein constant $1$ (i.e., $\mathrm{Ric}(g) = (n{- …
14
votes
Accepted
When a Riemannian manifold is of Hessian Typ
First, the definition: A Riemannian $n$-manifold $(M^n,g)$ is of Hessian type if there exist $(n{+}1)$ functions $x^1,\ldots,x^n, u$ on $M$ such that $dx^1\wedge\cdots\wedge dx^n\not=0$ and such that …
23
votes
Accepted
Characterizing Hessians among symmetric bilinear tensors
There are local conditions, but they typically involve the curvature tensor of the underlying metric. For example, if the metric is flat, so that one can choose orthonormal coordinates $x_i$ in which …
16
votes
Accepted
ANOTHER Exterior differential system on $SO(3;\mathbb R) \times \mathbb R$
One then finds that the integral manifolds of the differential system must satisfy
$$
\begin{align}
v_5 &= -\sinh(\rho)\ (p_4{+}q_5)-\cosh(\rho)\ (p_5{+}q_4)\\\\
v_6 &= (\cosh^2(\rho){-}\cosh(\rho)\sinh … Thus, with this construction and the above formulae, we have the complete description of the (local and global) integral manifolds of the original EDS in terms of the second and third derivatives of $L …
6
votes
Accepted
special Lagrangian n-Torus has Tubular neighbourhood?
NB: As the OP pointed out, the first version of my answer was incomplete, since it didn't address the 'tubular neighborhood' part of the claim. Here's a better (but still not complete) version of an …
11
votes
Accepted
On eigenfunctions of the Laplace Beltrami operator
For $\mathrm{SU}(2)$, with the scale for the biïnvariant metric so that it becomes isometric to the unit $3$-sphere $S^3$ in Euclidean $4$-space, it is well-known what the eigenvalues of the Laplace-B …