# Tag Info

## Hot answers tagged differential-topology

7

I think you are asking if every closed manifold is the boundary of a compact manifold. The answer is negative. For instance, real projective plane is not the boundary of any compact 3d manifold (since it has odd Euler characteristic and the Euler characteristic of an odd-dimensional manifold is half of the Euler characteristic of its boundary: This follows ...

5

$\def\R{\mathbb{R}} \def\SL{\text{SL}} \def\SO{\text{SO}}$Often you use the group action to study $G$ and not just to study $X$. Here is an example: what does $\SL_2(\R)$ look like as a manifold? You can solve this by thinking of the group directly, but an easier way is to note that it acts transitively on the upper half plane by Mobius transformations. ...

5

I believe it's more the other way around: a group action of $G$ on a space $X$ allows to construct a new space, the quotient $G\backslash X$ (of course the quotient space is "nice" only under some technical conditions). Recognizing that a space $Y$ is actually realized as the orbit space $G\backslash X$ of some simpler space $X$ can help understand better ...

4

$$O \to Ext(H_{n-1}(M),\mathbb{Z}) \to H^n(M) \to Hom(H_n(M),\mathbb{Z}) \to 0$$ As the latter arrow is an isomorphism when M is closed, connected and orientable, it follows that $Ext(H_{n-1}(M),\mathbb{Z})=0$. You just need to understand why it implies your $n-1$ torsion group $T_{n-1}$ is $0$...

4

A trivial answer would be that given any $n$-manifold $M$, the $(n+1)$-manifold-with-boundary $M\times [0,1)$ has $M$ as its boundary. But I don't know if there's intended to be any further restrictions on the situation that make a manifold occurring as a boundary of something else more difficult.

3

Not totally sure what you're asking (particularly, are you asking about splitting in algebraic/holomorphic settings?), but the following comments may be relevant: Because of the principle that "curvature decreases in holomorphic sub-bundles and increases in quotient bundles", it's "rare" for a short exact sequence of holomorphic vector bundles to split ...

3

Here is a general class of examples. Let $G$ be a connected, simply-connected complex semisimple group with a transitive algebraic action on a smooth complex projective variety $X$. Prior to thinking about the action of $G$ on X, there is little one can say about the topology of $X$. By taking into account the action of $G$, we find that $X$ must be ...

3

Let $f:S^3 \to S^2$ be the Hopf map, and $\pi:S^3 \to \mathbf{RP}^{2}$ the composition with the projection $S^2 \to \mathbf{RP}^{2}$. The map $\pi$ sends antipodes of $S^{3}$ to the same place (since the Hopf map collepses equatorial circles), so $\pi$ factors through the projection $S^{3} \to \mathbf{RP}^{3}$. It's "clear" this expresses $\mathbf{RP}^{3}$ ...

2

This is true, provided you are using correct definition of a smooth function on a closed subset $A$; dimension of $A$ is irrelevant. Smoothness of $f$ means that there exists a smooth extension $h$ of $f$ to an open neighborhood $U$ of $A$ in $M$. Let $V, W$ be another pair of open neighborhoods of $A$, such that $$\bar{W}\subset V, \bar{V}\subset U.$$ ...

2

I'm not sure if this is what you're looking for, but there is a Galois correspondence for covering spaces and deck transformations that is analogous to the correspondence between intermediate fields and field extension automorphisms of a Galois extension.

2

If $\phi:M \to \Bbb R^n$ is in fact an isometric embedding, then the image $\text{Im} \phi$ of $M$ under $\phi$ is compact, since $M$ is compact. If the $x_i$, $1 \le i \le n$, are a set of Cartesian coordinates on $\Bbb R^n$ (where by "Cartesian" I mean compatible with the vector space structure on $\Bbb R^n$, so they are related by a nonsingular linear ...

2

Things aren't quite right here, TKM. First, the Hessian $d^2f(p)$ is a symmetric bilinear map $T_pM\times T_pM\to\Bbb R$, so your image statement is not right. Second, you're talking about $df$ as a section of $T^*M$ and you want it to be transverse to the zero section of that bundle. Here's how you should approach it: Work in local coordinates, so assume ...

1

First, $M(X,Y)$ equals $Y^X=\prod_{x\in X} Y$, the set of all functions from $X$ to $Y$ since ever function on a discrete space is continuous. Now the subbasis for the product topology on $Y^X$ consists of all $$e_x^{-1}(U)=\{(f(x))_{x\in X}\mid e_x(f)=f(x)\in U\}$$ ranging over the open subsets $U$ of $X$ and the elements $x$ of $X$. The subbasis for the ...

1

This question was essentially answered here, namely, that the cut-locus has measure zero (see the references provided in the link). I assume that your manifold $M$ is smooth, otherwise, I am not sure what notion of measure zero you would be using. I will also assume that $M$ is connected. (If not, apply this argument to each connected component.) Then, put a ...

1

You can find all the required definitions in the context of immersions of graphs (and much more!) in J.Stallings, "Topology of finite graphs", Inventiones Math., 71 (1983) available here. An immersion in this context means that if you lift map $f$ of the circle to a map $\tilde f$ of the line to the universal cover of $R_n$ (which is a tree), the new map ...

1

Take a look at the book "Geometric Topology Localization, Periodicity, and Galois Symmetry" by Dennis Sullivan. The book (as almost everything that Sullivan wrote) is hard to read, but you can just browse it to get an idea of what it is about.

1

That's a tall order! I would wager that very few people understand Freedman's proof completely. I found this series of video lectures that he gave to be quite helpful. In particular, it will give you an idea about the sort of mathematics that is involved. Also, Freedman and Quinn's book on $4$-manifold topology is great! Even if you aren't to the point ...

1

If you have a manifold embedded in some $\mathbb R^n$, and a set of coordinates on that manifold $x^1, x^2, \ldots, x^m$, then each point on the manifold is a position vector, a function of the coordinates like so: $r = r(x^1, x^2, \ldots, x^m)$. Each partial derivative of that function, $\partial r/\partial x^1, \partial r/\partial x^2, \ldots$ is a ...

1

I can't say I'm aware of a proof that doesn't use a metric, but (in relation to Question 2), this is probably because the proof by means of a Riemannian or Hermitian metric is actually very general. applying to all paracompact locally compact Hausdorff spaces (e.g., compact Hausdorff spaces, CW complexes). If $X$ is just a locally compact Hausdorff space, ...

1

Here is a concrete perspective on tangent bundles that might help. Take an embedding of your manifold $M$ into $\mathbb{R}^N$ (by Whitney we can always do this). This gives an embedding of $TM$ into $T\mathbb{R}^N = \mathbb{R}^N\times \mathbb{R}^N$ (it's injective and you can write down the differential to check it is an immersion). This already gives you ...

1

Let $p$ denote the basepoint for the holonomy group. The Ambrose-Singer theorem says that the Lie algebra of the holonomy group of $D$ is the subspace of the Lie algebra of $G$ spanned by elements of the form $\Omega_q(X,Y)$ where $q$ is a point in the holonomy bundle at $p$, $X$ and $Y$ are horizontal tangent vectors at $q$, and $\Omega$ is the curvature ...

1

The answer is essentially "yes". More precisely, $\text{Hol}(D)$ is only well-defined up to conjugation in $G$ and the statement is that if $D$ restricts to some $H$ subbundle $Q\subset P$ then $\text{Hol}(D)$ is the conjugacy class of a subgroup of $H$. This is an immediate consequence of the definitions of the terms involved (the Theorem of ...

1

IF you mean the unit sphere in 3-space, then horizontal radial projection from an enclosing cylinder might be what you're looking for: $$(\cos t, \sin t, z) \mapsto (\sqrt{1-z^2} \cos t, \sqrt{1-z^2} \sin t, z)$$ (The map from $(t, z)$ to the cylinder is pretty clear). I'm guessing, though, that you'll be offended by the "stretch" on longitude lines, ...

1

A hyperbolic metric on n-dimensional smooth manifold M is a Riemannian metric g in M which is locally isometric to the standard Riemannian metric on the hyperbolic n-space $H^n$. Locally isometric here means that every point in M admits a neighborhood U so that the restriction of g to U is isometric to an open subset of the hyperbolic n-space. Such metric g ...

1

In the first entry, you compute the derivative of a quotient $$\dfrac{f'g - f g'}{g^2}$$ where $f = 4 x_1^2 \sum_i x_i^2$ and $g = (1 + \sum x_i^2)^2$. The $f g'$ term should be $$\left(4 x_1^2 \sum_i x_i^2\right) \left( 2 (1 + \sum_i x_i^2) 2 x_1 \right) \\ 16 x_1^3 (\sum_i x_i^2 )(1 + \sum_i x_i^2).$$ Presumably by plugging in $n = 1$ into your ...

1

Take neighborhoods $X_1$ of $x$, $Y_1$ of $f(x)$, such that $f:X_1\to Y_1$ is a diffeomorphism. Take $\phi_1:U_1\to X$, $\psi_1:V_1\to Y$ parametrizations of neighborhoods of $x,y$ (resp.). Let $X_2=\phi_1(U_1)\cap X_1$, $U=\phi_1^{-1}(X_2)$, $Y_2=\psi_1(V_1)\cap Y_1$, $V=\phi_1^{-1}(Y_2)$. Then $F:U\to V$, $u\mapsto \psi_1^{-1}( f(\phi_1(u)))$ is a ...

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