11 votes
Accepted

Homologically trivial embedding of $\mathbb{CP}^2$ into a compact closed smooth $6$-manifold?

A slight modification of the argument that $\mathbb{CP}^2$ doesn't immerse in $\mathbb{R}^6$, given by András Szűcs here, shows that there is no such $M$. Let $i : \mathbb{CP}^2 \to M$ be an immersion ...
Michael Albanese's user avatar
8 votes
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There is already a theorem that demonstrates this inequality?

Let $\Omega\subset\Bbb{R}^n$ be any non-empty open set, and $f:\Omega\to\Bbb{R}$ a $C^1$ function. Then, for any (Lebesgue-measurable) set $E\subset \Omega$, the $n$-dimensional surface area of the ...
peek-a-boo's user avatar
  • 54.6k
8 votes
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Alternative of exterior power as a tensor algebra

I can think of a couple of reasons: The definition of the alternating tensors only works over characteristic $0$ (or characteristic $> n$, if $n$ is fixed) since you are dividing by $n!$. This ...
ronno's user avatar
  • 11k
7 votes
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Standard area form on a Riemannian manifold.

This could not be the case already when $M=\mathbb R^2$ and $N=\mathbb R$ is a line through the origin. The area (i.e., length) form of $N$ could not be the pullback of any 1-form $\omega$ on $M$, as ...
Mikhail Katz's user avatar
  • 41.5k
7 votes
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Understanding $\mathbb{C}P^2$

This is best explained in the language of vector bundle, where you are observing a bunch of crucial phenomena. The Hopf fibration $S^1\rightarrow S^3\rightarrow S^2$ is the sphere bundle of the ...
Thorgott's user avatar
  • 11k
6 votes
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Are there (known to be) exotic spheres in arbitrarily high even dimension?

Yes, it is true. For $n>8$ there exist exotic $n$-spheres if $n$ is congruent modulo 192 to one of 2, 6, 8, 10, 14, 18, 20, 22, 26, 28, 32, 34, 40, 42, 46, 50, 52, 54, 58, 60, 66, 68, 70, 74, 80, ...
Connor Malin's user avatar
  • 11.6k
6 votes

Does this combinatorial identity hold?

We use the coefficient of operator $[z^i]$ to denote the coefficient of $z^i$ in a series. We obtain \begin{align*} \color{blue}{\sum_{k=1}^m}&\color{blue}{(-1)^{k-1+i}\binom{m}{k}\binom{k-1}{i}}\...
Markus Scheuer's user avatar
6 votes

Can adding a single element to a Lie group make it infinite-dimensional?

There is no bound on the dimensionality of $G$. In fact, a single element can generate a Lie group of arbitrarily large dimension. Thus the $n$-dimensional torus $\mathbb T^n$ is the closure of the ...
Robert Israel's user avatar
6 votes
Accepted

Vanishing of the second cohomology module of a pair $(M^*, \partial M^*)$

You are circling very close to a correct proof. As you say, $H^2(M,U) \cong H^2(M/U)$. The space $M/U$ is homeomorphic to $M \cup \bigsqcup D^2 / \sim$, the space where we have glued on disks to the ...
Connor Malin's user avatar
  • 11.6k
6 votes
Accepted

Why is the hyperboloid model a hyperbolic model?

The underlying set of the hyperboloid model is $$H^2 = \{(x,y,z)\in\Bbb R^3 \mid x^2+y^2-z^2=-1\mbox{ and }z>0\},$$however, the metric it inherits from $\Bbb R^3$ is not the standard Riemannian ...
Ivo Terek's user avatar
  • 77.3k
6 votes
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topology on smooth vector bundle

Since $E$ is assumed to be a smooth manifold, it is Hausdorff so any two points in $E$ have disjoint open neighborhoods. The maps $\varphi:\pi^{-1}(U)\to U\times\mathbb R^k$ are assumed to be ...
Andreas Cap's user avatar
  • 20.3k
5 votes

Explain the general idea of topology for a sophomore student

I would suggest (at least temporarily) avoiding Topology, and instead looking for a somewhat proof-oriented textbook on Real Analysis. One example is "Calculus" (volumes I and II), 2nd Ed. (...
user2661923's user avatar
  • 35.3k
5 votes
Accepted

Can a real-valued function on the sphere have exactly 2 critical points which are not antipodal?

Take the function $f(x_1, \dots, x_{d}) = x_{d}$ defined on $\mathbb S^{d-1}$. Let $\phi : \mathbb S^{d-1 }\to \mathbb S^{d-1}$ be a diffeomorphism such that $\phi (0,\cdots, 0,1) = (0,\cdots, 0,1)$ ...
Arctic Char's user avatar
  • 15.9k
5 votes
Accepted

An analog of Jordan curve theorem for various type of smooth manifolds

I do not think you have a proof for your statement. It certainly needs a lot more argument to go from the fact that the fundamental class of $X$ is the (algebraic) boundary of a chain, which is all ...
Thorgott's user avatar
  • 11k
5 votes
Accepted

Explicit Lie group embedding of $O(n)$ into $SO(n+1)$

I’ll just write up an answer with a few ways of attacking the problem, and let you fill in some of the details (which can easily be found if you search the site or any decent book). First, let me make ...
peek-a-boo's user avatar
  • 54.6k
5 votes

Order of the fundamental group from the metric of a Riemannian manifold

If a compact manifold has curvature $K\leq 0$, the Cartan-Hadamard theorem (https://en.wikipedia.org/wiki/Cartan%E2%80%93Hadamard_theorem) tells us that its universal cover is diffeomorphic to $\bf R ^...
Thomas's user avatar
  • 7,450
5 votes
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Riemannian Metric from Hamiltonian?

I wouldn’t phrase it as “the Riemannian metric is given by reading off the coefficients of $\ddot{q}_i$” because that makes it seem like the Riemannian metric is some secondary piece of information, ...
peek-a-boo's user avatar
  • 54.6k
5 votes
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Why is the signature of a manifold homotopy invariant?

The definition you've given for the signature is phrased in terms of de Rham cohomology and is indeed not obviously homotopy invariant. However, the pairing you've given has a "lift" to ...
Connor Malin's user avatar
  • 11.6k
5 votes
Accepted

Why can we replace the euclidean space with the upper half space as model space for a manifold

This is because $H^n$ contains the open half-space $\mathbb R^n_+ = \{(x_i) \in \mathbb R^n \mid x_n > 0 \}$ which is open in $\mathbb R^n$. It is homeomorphic (even diffeomorphic) to $\mathbb R^n$...
Paul Frost's user avatar
  • 75.1k
5 votes
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Connection between (1,1) tensors and maps of vector fields

This is a matter of linear algebra, more so than one of differential geometry, so let's just consider a vector space $V$. There is an isomorphism $V\otimes V^*\cong\text{End}(V)$ (assuming $V$ is of ...
Quaere Verum's user avatar
  • 2,761
5 votes

Understanding $\mathbb{C}P^2$

Your reasoning goes wrong in your point number $1$. Just because you have fibered $S^3$ by disjoint $S^1$'s, you cannot assert without justification that you can extend this to a fibering of $\mathbb ...
Lee Mosher's user avatar
  • 119k
4 votes
Accepted

A vector bundle which has an orientation-reversing isomorphism has a subbundle of rank $1$?

Let us say that a vector bundle is reversible if it has an orientation-reversing automorphism. I claim the following, which provides a negative answer to the question: Proposition: Up to isomorphism, ...
Derived Cats's user avatar
  • 2,632
4 votes

Reference: the number of smooth structures on a topological $n$-manifold, $n\geq 5$ is finite

There are not really as many structures on $T^n$ as that: typically one mods out $H^3(T^n;\mathbb{Z}/2)$ by the action of $GL(n,\mathbb{Z}/2)$. This is discussed by Wall in chapter 15A of Surgery on ...
Dave Davidson's user avatar
4 votes
Accepted

dimension of total space of a smooth vector bundle

I'll break this down for you so that you are able to justify your claim with a rigorous argument. Step 1: For any open subset $U$ of the base manifold $M$, $\pi^{-1}(U)$ is an open subset of the total ...
Kenny Wong's user avatar
  • 32.1k
4 votes
Accepted

Is the non-trivial $S^3$ bundle over the torus $T^2$ an $H$-space?

Here's one way to construct this manifold. Note that $\pi_2(BSO(4)) \cong \pi_1(SO(4)) \cong \mathbb{Z}_2$, so up to isomorphism, there is only one non-trivial real rank four vector bundle $E \to S^2$ ...
Michael Albanese's user avatar
4 votes
Accepted

What does $C^1([0,1])$ mean?

You are close, but not quite there (or perhaps you just took too many shortcuts in writing $1$ and $2$). Replace $1$ by this: $1'$: $f$ is differentiable on $(0,1)$, the right derivative exists at $0$...
Lee Mosher's user avatar
  • 119k
4 votes
Accepted

Can a cubic polynomial in two real variables have exactly three isolated critical points?

Yes. The cubic polynomial $$ c(x,y) := 3x^3 + x^2y + y^3 + x^2 + 2xy + y^2 $$ has 3 isolated critical points at $(0,0)$, $(-\frac 6 {13}, - \frac{12}{13})$ and $(\frac 2 {19}, - \frac{8}{57})$. The ...
cs89's user avatar
  • 3,331
4 votes
Accepted

Definition of single k-slice

The meaning here is completely literal: "single $k$-slice" means "one $k$-slice". If you like, the word "single" can be omitted, so the statement is just that $S\cap U$ ...
Eric Wofsey's user avatar
4 votes
Accepted

Is a vector bundle which is trivial on a smooth submanifold trivial in a neighbourhood?

Yes, if $M$ is paracompact there exists such an open subset $N\subset U\subset M$ on which $E$ is trivial. Here is why: Let $\theta$ be the trivial bundle of the same rank as $E$. There exists a ...
Georges Elencwajg's user avatar
4 votes

Vector field determining a family of codimension 1 submanifolds perpendicular to the field

Turn the vector field $V$ into a $1$-form $\omega$ in the obvious way. Then you will get integrability iff $d\omega\wedge\omega=0$. (In the case $n=3$ you could state this directly as $\text{curl}\,V\...
Ted Shifrin's user avatar

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