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Smooth manifolds and smooth functions between them. For manifolds with additional structure, see more specific tags, such as [riemannian-geometry]. For more topological aspects, see [differential-topology].
9
votes
Accepted
Delta distribution on manifolds
We can give a complete classification of (candidates for) delta-distributions on a smooth manifold $M$ at point $p$.
Specifically, a delta-distribution is a smooth linear functional on the space of sm …
9
votes
Accepted
To what extent differentiable mappings of an affine line into a manifold determine its diffe...
This is true for infinitely differentiable curves: if a map sends smooth curves to smooth curves, then it is smooth, by a theorem of Boman from 1967:
Jan Boman. Differentiability of a Function and of …
2
votes
Accepted
Reference for the Brown-Gersten property for smooth manifolds
I typed up a proof of this result:
Numerable open covers and representability of topological stacks.
The result is proved in greater generaility for arbitrary numerable open covers of topological sp …
11
votes
1
answer
446
views
Does every smooth map of rank at most d factor through a d-manifold?
Suppose $d≥0$, $m≥0$, $n≥0$, and $\def\R{{\bf R}} f\colon \R^m→\R^n$ is a smooth map
whose rank at any point of $\R^m$ is at most $d$.
Here and below, smooth means infinitely differentiable.
Can we fi …
10
votes
Accepted
How to show that $\text{Man}(M,\mathbb{R}^n)\cong \mathbb{R}\text{-Alg}(C^\infty(\mathbb{R}^...
The proof of this fact is available in modern textbooks.
For example, see Theorem 7.16 in Jet Nestruev's Smooth Manifolds and Observables (Second Edition, 2020).
In fact, the cited book contains a lot …
2
votes
Is there any Lie groupoid structure on $Hom(\mathcal{G}, \mathcal{H})$ where $\mathcal{G}$ a...
If C is a cartesian closed category with finite limits,
then so is the category of internal groupoids in C.
Indeed, the internal hom can be constructed by replicating the usual
definitions of a functo …
8
votes
Results in “generalised smooth spaces” that did not hold in the case of smooth manifolds
There are many such results.
Consider some smooth manifolds M and N.
The internal hom Hom(M,N) is a sheaf on smooth manifolds.
We can compute its tangent bundle,
and it turns out that the tangent spac …
11
votes
When (why) did we allow manifolds to be non-Hausdorff and/or non-second countable?
The étale space construction produces non-Hausdorff and nonparacompact spaces (e.g., smooth manifolds)
in many practical examples that have nothing to do with algebraic geometry.
The étale space is …
13
votes
Accepted
Is a manifold paracompact? Should it be?
every group can be declared a Lie group if one allows the more general definition since one is permitted to have an uncountable number of connected components.
A manifold is paracompact if and o …
2
votes
Accepted
Any reference including a coordinate free proof of the inverse function theorem?
A coordinate-free proof of the inverse function
theorem in the finite-dimensional case
is provided by Theorem 19.6 in "Topological Geometry"
by Ian R. Porteous.
In general, the cited book is an expos …
28
votes
0
answers
521
views
What algebraic structure characterizes all natural operations between differential operators...
On a smooth manifold $M$ one can define various algebraic structures, natural with respect to diffeomorphisms:
the differential graded-commutative algebra $\Omega(M)$ of differential forms on $M$;
t …
5
votes
Accepted
Metrics on derived smooth manifolds
As far as I am aware, there is nothing in the literature that
treats Riemannian or pseudo-Riemannian metrics on derived smooth manifolds.
However, there is an extensive treatment of symplectic structu …
25
votes
1
answer
4k
views
Can the constant rank theorem for smooth manifolds be generalized to nonconstant rank?
The constant rank theorem says that
if $f\colon M→N$ is a smooth map whose rank equals some fixed $k≥0$ at any point of $M$, then, locally with respect to $M$ and $N$, the map $f$ assumes the easiest …
9
votes
1
answer
401
views
Reference for the Brown-Gersten property for smooth manifolds
A classical result by Brown and Gersten says that to verify the homotopy descent property for the Zariski topology
it suffices to verify it for Zariski squares and the empty cover of the empty scheme. …
10
votes
0
answers
740
views
Can any smooth triangulation of a smooth manifold be blurred?
For the purposes of this question, let's say that a blurring
of a smooth triangulation $T$ of a smooth manifold $X$
is a smooth homotopy $h\colon [0,1] \times X \to X$ such that $h_0=\operatorname{id} …