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5 votes
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Specify the embedding of Lie groups (via the representation map) precisely as the embedding of two differentiable manifolds

How do we specify the embedding of a Lie group $G_1$ as a subgroup into a larger Lie group $G_2$, with $G_1 \subset G_2$ that agree with a constraint on the mapping between their representations? By ...
wonderich's user avatar
  • 10.5k
5 votes
0 answers
120 views

How does the topology of minimal surfaces depend on the radius?

Let $M^n \subset \mathbf{R}^{n+k}$ be a smooth, properly embedded minimal surface, with boundary $\partial M$. The convex hull property states that $M$ is contained inside the convex hull of its ...
Leo Moos's user avatar
  • 5,038
0 votes
1 answer
254 views

$SU(k)/SO(k)$ as a manifold, for each positive integer $k$ [closed]

The quotient space $SU(k)/SO(k)$ is also a homogeneous space constructed out of the Lie groups (special unitary $SU(k)$ and special orthogonal $SO(k)$). Because the $SO(k)$ may not be a normal ...
Марина Marina S's user avatar
19 votes
1 answer
989 views

Can the product of a 3-dimensional lens space with a circle be diffeomorphic to another such product when the lens spaces aren't diffeomorphic?

This is a question that I need to answer in order to resolve an issue for my dissertation and I am looking for a reference. Here is the precise statement of the question. Suppose we have two three-...
Tuo's user avatar
  • 293
5 votes
1 answer
371 views

Orientation reversal and restriction to submanifold of lower dimension

Let $M$ be a connected closed oriented manifold with at least one orientation-reversing homeomorphism $M\to M$. Let $S\subset M$ be a connected closed embedded submanifold of lower dimension. Let $f:M\...
malfat's user avatar
  • 59
0 votes
0 answers
179 views

Adjunction formula for non compact surfaces

Let $M$ be a non compact complex surface and S an embedded compact Riemann surface in $M$. I already know how to show the following equality of fiber bundle: $$\Omega^2_{M}|S =\Omega^1_S \otimes N^*_S$...
singularity's user avatar
0 votes
0 answers
170 views

Cohomology ring of 5-manifold generated in degrees 1, 2, 3

Is there a connected closed 5-manifold $M$ such that $\oplus_{i\geq 0} H^i(M, \mathbb{Z})$ is generated by $H^1(M, \mathbb{Z})\oplus H^2(M, \mathbb{Z})\oplus H^3(M, \mathbb{Z})$ but is not generated ...
klaus's user avatar
  • 21
-1 votes
1 answer
137 views

Manifold for which you need to specify the action on cohomology in each degree

Let $M$ be a connected closed manifold of dimension $n\geq 2$. Can it happen that for any $I\subsetneq I_n=\{1, \dots, n\}$ there are continuous maps $f, g:M\to M$ such that $f^*=g^*|_{\oplus_{i\in I}...
klaus's user avatar
  • 21
1 vote
1 answer
107 views

When given degrees generate the cohomology ring of a manifold?

Let $M$ be a connected closed manifold of dimension $n$. Suppose we have a subset $I\subset I_n=\{1, \dots, n\}$ such that for any two continuous maps $f, g:M\to M$ if $f^*=g^*|_{\oplus_{i\in I}H^i(M, ...
klaus's user avatar
  • 21
8 votes
0 answers
251 views

(Higher) flat connections and Grothendieck construction

For any (nice) topological space there is an equivalence between covering spaces and local systems $\pi_1X \to \operatorname{Set}$. We can think of it as a Grothendieck construction. If we take ...
E. KOW's user avatar
  • 834
1 vote
0 answers
151 views

Density of $G$-invariant morse functions

Let $G$ be a finite group acting on a compact manifold $M$. Let $f$ be a $G$-invariant smooth function. Can it be approximated by $G$-invariant Morse functions?
user123090's user avatar
8 votes
3 answers
914 views

Spectral sequences in algebraic topology [duplicate]

What books/articles do you recommend for learning spectral sequences? I am interested in their applications to algebraic topology, particularly to understand the homology of fibre bundles. I have a ...
4 votes
0 answers
158 views

Postnikov square explicitly on a simplicial complex

$\DeclareMathOperator\Z{\mathbb{Z}}$ Following Wikipedia, a Postnikov square is a certain cohomology operation from a first cohomology group $H^1$ to a third cohomology group $H^3$, introduced by ...
wonderich's user avatar
  • 10.5k
4 votes
1 answer
240 views

Question about the proof of Gromov's theorem in geodesic flows

I am trying to understand the following theorem from the book Geodesic flows : Given a metric $g$ on a simply connected manifold $X$, there exists a constant $C_1>0$ such that given any pair of ...
user avatar
6 votes
2 answers
636 views

On the Euler number of an orbit space

Let $X$ be a differentiable manifold and $G$ a finite group acting differentiably on $X$. The following formula for the Euler number $\text{e}(X/G)$ of the orbit space $X/G$ appears to be well-known: \...
MathCrawler's user avatar
  • 1,020
7 votes
0 answers
350 views

Do smooth maps with nowhere-maximal rank have small image?

I’m trying to better understand the concept of “maps with small image” as used by Lipyanskiy in his construction of “geometric homology” in https://arxiv.org/abs/1409.1121. Lipyanskiy utilizes ...
Greg Friedman's user avatar
15 votes
6 answers
2k views

Any shortcuts to understanding the properties of the Riemannian manifolds which are used in the books on algebraic topology

I'm now attending a reading seminar on the algebraic topology. The seminar treats the book of Bott & Tu (Differential Forms in Algebraic Topology) and Milnor (Characteristic Classes). In those ...
gualterio's user avatar
  • 1,013
3 votes
1 answer
485 views

How to compute the index of a vectorfield defined by analytic formula?

An analytic local map (or map germ) $f: (\mathbb{R}^n, 0) \to (\mathbb{R}^n, 0) $ can be considered as a vector field with zero at the origin. Assume that the origin is an isolated zero of $f$. How ...
Gergo Pinter's user avatar
15 votes
1 answer
1k views

Atiyah's proof of the moduli space of SD irreducible YM connections

In the paper "Self-duality in Four-dimensional Riemannian Geometry" (1978), Atiyah, Hitchin and Singer present a proof that the space of self-dual irreducible Yang-Mills connections is a ...
Quaere Verum's user avatar
3 votes
0 answers
137 views

On the construction of principal $S^1$-bundles with prescribed characteristic form

I am trying to understand an argument made by Kobayashi (https://projecteuclid.org/download/pdf_1/euclid.tmj/1178245006, page 35) in his construction of a principal $S^1$-bundle with connection $1$-...
BrianT's user avatar
  • 1,227
1 vote
0 answers
293 views

How to calculate the total chern classes of CP^n [closed]

When calculating the total chern class of $\ CP^n$, we use the fact that their is a exact sequence of vector bundles over $\ CP^n$: $$\ 0\to S \to C^{n+1} \to Q \to 0$$ And identify the bundle $\ TCP^...
Sunhf's user avatar
  • 157
1 vote
1 answer
453 views

Relationship between $H^1(X, \mathbb{T})$ and complex line bundles

Let $X$ be a compact metric space and consider the sheaf cohomology group $H^1(X, \mathbb{T})$. From a class in $H^1(X, \mathbb{T})$, I can get a principal $\mathbb{T}$-bundle over $X$ and from this, ...
Francine Laporte's user avatar
13 votes
2 answers
2k views

Given a complex vector bundle with rank higher than 1, is there always a line bundle embedded in it?

When I read the GTM 082, "The Splitting Principle of the complex vector bundle", I see that in the proof we split off one subbundle at a time by pulling back to the projectivization of a ...
Sunhf's user avatar
  • 157
3 votes
0 answers
194 views

The first Stiefel Whitney class v.s. fermion eta invariant v.s. spin structure v.s. $H^1(M,\mathbb{Z}_2)$

$H^1(M,\mathbb{Z}_2)$ specifies the 1st cohomology class of manifold $M$ (can be regarded as spacetime) with $\mathbb{Z}_2$ coefficient, it is often to see that we say the 1st Stiefel Whitney class $$...
annie marie cœur's user avatar
9 votes
2 answers
405 views

Differential refinement of homology

Differential cohomology is a refinement of ordinary cohomology by differential data. It's construction comes down to the observation that $H^2(M, \mathbb{Z})$ is isomorphic to the space of isomorphism ...
Tobias Diez's user avatar
  • 5,824
9 votes
1 answer
372 views

A strong form of Mostow rigidity without geometrization?

Mostow rigidity theorem says that two closed hyperbolic manifolds with isomorphic fundamental groups are isometric. Here is my question: suppose that $M$ and $N$ are two closed 3-manifolds such that $...
cellular's user avatar
  • 855
7 votes
1 answer
363 views

Inverse Hodge and inverse Betti problems for Kähler manifolds

A Betti sequence is a map $\mathbb{Z}_{\geq 0}\to \mathbb{Z}_{\geq 0}$. A Betti sequence $b$ is realizable if there is a connected closed Kähler manifold $M$ such that $b(k)=b_k(M)$. A Hodge diamond ...
user avatar
3 votes
1 answer
257 views

Deformation equivalent Hodge structures

An HH type is the oriented homotopy type of a closed simply-connected Kähler manifold together with the Hodge structure on cohomology. Two HH types are deformation equivalent if they are represented ...
user avatar
5 votes
1 answer
394 views

Arbitrary torsion in cohomology of Kähler manifolds

Fix an integer $i\geq 3$ and a finite abelian group $G$. Is there a connected closed Kähler manifold $M$ such that $H^i(M, \mathbb{Z})\approx \mathbb{Z}^n\oplus G$ for some integer $n\geq 0$?
user avatar
3 votes
0 answers
55 views

Infinitely many deformation equivalent Hodge diamonds II

Let $S$ be a connected smooth complex-analytic space. Let $\phi:X\to S$ be a proper holomorphic submersion with connected fibers. Can the fibers of $\phi$ have infinitely many distinct Hodge diamonds? ...
user avatar
5 votes
0 answers
130 views

Minimal sum of Betti numbers of Kähler manifold with trivial canonical bundle

Let $M$ be a closed Kähler manifold of real dimension $2n$. Suppose the canonical bundle of $M$ is holomorphically trivial. Is it true that $\sum_{i=0}^{2n} b_i(M)=n+3\implies n=1$?
user avatar
3 votes
0 answers
86 views

Exotic smooth structures on HK manifolds

An HK manifold is a closed simply-connected Kähler manifold $M$ such that $H^0(M, \Omega_M^2)=\mathbb{C}\omega$, where $\omega$ is a holomorphic 2-form on $M$ which is nowhere degenerate as a skew-...
user avatar
12 votes
2 answers
767 views

Unique almost complex structure up to diffeomorphism

For which closed smooth manifolds does the action of the diffeomorphism group on the set of almost complex structures have exactly one orbit? For example it is true for $S^2$.
user avatar
3 votes
0 answers
71 views

Holomorphic homeomorphisms

Let $M$ be a connected closed smooth manifold. Consider the group $\mathrm{Homeo}(M)$ of homeomorphisms $M\to M$ endowed with the $C^0$-topology. If $M$ has a symplectic structure some people study ...
user avatar
3 votes
1 answer
124 views

Homotoping diffeomorphism to a $J$-holomorphic one

Let $M$ be a closed simply-connected smooth manifold. Assume $M$ admits at least one almost complex structure. Is any diffeomorphism $M\to M$ homotopic as a continuous map to a $J$-holomorphic ...
user avatar
6 votes
1 answer
463 views

Antiholomorphic involution with a fixed point

Let $M$ be a connected closed complex manifold. Assume it has an antiholomorphic involution. Must it have an antiholomorphic involution with a fixed point?
user avatar
18 votes
0 answers
496 views

Orientation-reversing homotopy equivalence

If there is an orientation-reversing homotopy equivalence on a closed simply-connected orientable manifold is there an orientation-reversing homeomorphism? It is not true, for instance, that if there ...
user avatar
3 votes
0 answers
98 views

Non-diffeomorphic surface bundles over homeomorphic 4-manifolds

For a smooth manifold $M$ an $M$-surface is the total space of a smooth surface bundle over $M$. Let $M_1$ and $M_2$ be two homeomorphic closed simply-connected smooth 4-manifolds. Can there be an $...
user avatar
3 votes
0 answers
162 views

Exotic smooth structures on Calabi-Yau manifolds

A Calabi-Yau manifold is a simply-connected closed Kähler manifold with holomorphically trivial canonical bundle and $h^{2, 0}=0$. If two Calabi-Yau manifolds are homeomorphic are they diffeomorphic?
user avatar
13 votes
0 answers
319 views

Exotic smooth structures on Fano manifolds

If two Fano projective manifolds are homeomorphic are they diffeomorphic? There are examples with one manifold being Fano and the other of general type (Barlow surfaces). Moreover, the number of ...
user avatar
4 votes
1 answer
378 views

Chern/Hodge numbers of the conjugate complex manifold

Let $(M, J)$ be a complex projective manifold. Can $(M, -J)$ have different Chern/Hodge numbers?
user avatar
5 votes
1 answer
241 views

Fano manifold becoming general type upon conjugation

Let $(M, J)$ be a Fano projective manifold. Can $(M, -J)$ be general type? For complex curves and surfaces Kodaira dimension is diffeomorphism invariant so this cannot happen.
user avatar
2 votes
1 answer
652 views

Complex manifold defined over $\mathbb{R}$

Let $M$ be a connected closed complex manifold with an antiholomorphic involution. Must there be an atlas and a choice of a reference point in each chart such that the transition functions are ratios ...
user avatar
4 votes
1 answer
447 views

Complex projective manifold with an antiholomorphic involution

Let $M$ be a complex projective manifold with an antiholomorphic involution. Can $M$ be defined by equations with real coefficients then?
user avatar
9 votes
2 answers
753 views

Deformation equivalent vs diffeomorphic to projective manifold

Let $M$ be a closed complex manifold that is not deformation equivalent to a complex projective manifold. Can $M$ be orientedly diffeomorphic to a complex projective manifold? What if $M$ is moreover ...
user avatar
3 votes
0 answers
109 views

Kähler manifolds deformation equivalent to projective manifolds

Let $M$ be a closed non-projective Kähler manifold. There are three possibilities there is a proper holomorphic submersion $f:X\to \Delta$ with $f^{-1}(0)\cong M$ such that the projective fibers ...
user avatar
2 votes
1 answer
128 views

Infinitely many deformation equivalent Hodge diamonds

Let $S$ be a connected open complex manifold. Let $\phi:X\to S$ be a proper holomorphic submersion with connected fibers. Can the fibers of $\phi$ have infinitely many distinct Hodge diamonds? An ...
user avatar
10 votes
2 answers
499 views

Symplectic structure on the square of a 3-manifold

Let $M$ be a connected closed orientable 3-manifold. Assume $M$ is not the direct product of a surface and the circle. Can there be a symplectic or Kähler manifold homeomorphic to $M\times M$? I think ...
user avatar
7 votes
1 answer
634 views

Hodge diamonds of complex threefolds

There is no closed complex curve or surface with $h^{1, 0}-h^{0, 1}=1$. Now consider threefolds. Can this condition be satisfied? Is Serre duality in fact the only restriction on the Hodge diamond?
user avatar
11 votes
1 answer
379 views

Smooth structure on direct product

Let $M$ be the $E_8$ manifold. Is there a closed manifold $N$ such that $M\times N$ is smoothable? What is the smallest possible dimension of $N$?
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