Tag Info

Hot answers tagged

11

Some good $($introductory$)$ sources, in general, for all things smooth manifolds: Topology from the Differentiable Viewpoint, by Milnor Differential Topology, by Guillemin-Pollack Differential Forms and Applications, by Do Carmo A Comprehensive Introduction to Differential Geometry, Vol. 1, by Spivak Introduction to Smooth Manifolds, by Lee Brian Conrad's ...


7

The key step in what I perceive to showing this is that smooth maps into a manifold $M$ that land in a smooth submanifold $N$ are also smooth maps into the submanifold. This can be checked in local coordinates, by the following argument. Given the constraint that a point $p$ lies on a $k$-dimensional submanifold $N$ of $\mathbb{R}^n$, some of its coordinates ...


6

Why do you feel this is a cop-out? It's a very elegant argument, for matrix inversion especially (it's rather obvious that matrix multiplication was going to be smooth). When you invert a matrix using LU or Gaussian elimination, for instance, you do all of these pivoting operations that sure don't look like they're going to be smooth as some choice ...


6

The fact that the zeros of $f$ are simple and positive comes from $$\det(df) = u_x^2 + u_y^2 = |f'(z)|^2 > 0.$$A straightforward computation shows that$${{df}\over{f}} = {{du + i\,dv}\over{u + iv}} = {{u\,dv + v\,du}\over{u^2 + v^2}} + i{{u\,dv - v\,du}\over{u^2 + v^2}}$$$$={1\over2}d(\log(u^2 + v^2)) + i{{u\,dv - v\,du}\over{u^2 + v^2}}.$$Hence, by ...


6

I'll explain the relationship between trivialization of the tangent bundle and having $n$ linearly independent vector fields: A vector field is a smooth map $X: M \rightarrow TM$ such that $X(p) \in TM_p$. That is, it's a smooth assignment of a tangent vector at $p$ to each point $p \in M$. Suppose we have $n = \dim(M)$ vector fields $\{X_i\}_{i=1}^n$ ...


6

It appears you are using the notation of Spivak, so I will too. $($See Theorem 3.1 and Problem 3.8 for notation and background.$)$ You have a good start on the problem. The next thing to do is to check that $\mathcal{A}$ is a $C^\infty$ atlas on ${V}$. Given two members $({V}, \varphi)$ and $({V}, \psi)$ of $\mathcal{A}$, we have that $\varphi, \psi$ are ...


5

If we have an embedding $T^k\subset\mathbb{R}^{k+1}$ then the subspace inclusion $\mathbb{R}^{k+1}\subset\mathbb{R}^{k+2}$ allows us to view $T^k$ as embedded in $\mathbb{R}^{k+2}$. The boundary of the tubular neighborhood of $T^k$ in $\mathbb{R}^{k+2}$ will then be a torus $T^{k+1}$ embedded in $\mathbb{R}^{k+2}$, and we proceed inductively.


5

Well, depends on what your definition of smooth maps between manifolds embedded in $\mathbb{R}^k$ is. If one is using the definition involving pre and post composing charts, then a priori, the multiplication, which is smooth as a map from $\mathbb{R}^{n^2} \times \mathbb{R}^{n^2} \to \mathbb{R}^{n^2}$, need not restrict to a smooth map ...


5

Here is an argument why multiplication is smooth: If you apply the definition of the derivative to a linear map $f: \mathbb R^n \to \mathbb R^n$ you see that $f' = f$. This is because the (Fréchet) derivative at $x$ is defined to be the continuous linear map $A$ such that $\lim_{h \to 0}{\|f(x+h) - f(x) - Ah\|\over \|h\|} = 0$. It is clear that the map ...


4

Every parallelizable manifold is obviously orientable, hence you get an easy to check obstruction : non-orientable manifolds are not parallelizable. This immediately shows that, for example, all even-dimensional projective spaces $\mathbb P^{2n}(\mathbb R)$ are not parallelizable. Moreover there is a perfect (albeit a bit more advanced) criterion for ...


4

We may pass to local coordinates and assume that $M$, $N$ are open subsets of Euclidean spaces for which $p$ and $\varphi(p)$ correspond to $0$. Let $\gamma_1, \gamma_2: (-\epsilon, \epsilon) \to M$ be two curves with $\gamma_1(0) = \gamma_2(0)$ and $\gamma_1'(0) = \gamma_2'(0)$; then for any smooth function $f: M \to \mathbb{R}$ we get that $f \circ \varphi ...


4

My answer assumes only that the immersion is $C^1$. Let us denote this immersion by $x \mapsto (f(x), |f(x)|)$, where $f$ and $|f|$ are (by assumption) $C^1$ functions on the real line. It is easy to see that $f'(x) \neq 0$ for all $x \in \mathbb{R}$, for otherwise we would have also $|f|'(x) = 0$, and it would not be an immersion at that point. Let us ...


4

(a) $$f_2^{-1}if_1: \mathbb{R} \overset{f_1}{\to} \mathbb{R} \overset{i}{\to} \mathbb{R} \overset{f_2}{\to} \mathbb{R}, \text{ }x \mapsto x \mapsto x \mapsto x^{1\over3}$$is not differentiable, so $i$ is not a diffeomorphism. (b) $$f_2^{-1}\phi f_1: \mathbb{R} \overset{f_1}{\to} \mathbb{R} \overset{\phi}{\to} \mathbb{R} \overset{f_2^{-1}}{\to} \mathbb{R}, ...


4

Pick coordinate charts around $p \in M$ and $f(p) \in N$; assume that $M$ and $N$ are of dimension $m$, $n$ respectively, and $($by continuity$)$ that the points of $M$ corresponding to our chart around $p$ maps into those of our chart around $f(p)$, i.e. we may reduce to the case in which $M$, $N$ are subsets of Euclidean space, and even assume that $M$, ...


4

The map $\varphi_a$ is periodic if and only if $a$ is rational. Suppose $a = p/q$ is rational, where $\gcd(p,q) = 1$. Then $$\varphi_a (x+q) = (e^{2\pi i (x+q)} , e^{2\pi i a(x+q)}) = (e^{2\pi i x} e^{2\pi i q}, e^{2\pi i ax} e^{2\pi i p}) = (e(x), e(ax)) = \varphi_a (x)$$ so $\phi_a$ is periodic. Suppose $\phi_a$ is periodic with period $p$. Then ...


4

There are two parts to the (weak) Whitney embedding theorem: 1) Any abstract manifold can be embedded in $\mathbb{R}^N$ for some $N$. 2) Any $k$-dimensional submanifold of $\mathbb{R}^N$ can in fact be embedded in $\mathbb{R}^{2k+1}$. They prove part (2) of this. For a proof of (1), there's a nice exposition in Lee, Smooth Manifolds. Here's the idea for ...


3

Assuming that you're either talking about great circles or circles of nonzero radius, then yes, the space of circles on a sphere is a manifold. To make sure I don't screw things up, I'm going to talk about the case of great circles on a real sphere, because that has all the features you care about. In this case, each great circle corresponds nicely to a ...


3

$\mathbb{S}^{n}$ embeds in $\mathbb{R}^{n+1}$ as $$x_{0}^2+\dots+x_{n}^2=1$$ Hence $\mathbb{S}^{2}\times\mathbb{S}^{3}$ embeds in $\mathbb{R}^{7}$ as $$\begin{cases}x_0^2+x_1^2+x_2^2+x_3^2=1\\x_4^2+x_5^2+x_6^2=1\end{cases}$$ In particular $\mathbb{S}^{2}\times\mathbb{S}^{3}$ embeds in the $6-$sphere of radius $\sqrt{2}$ ...


3

No, it won't be an odd number of times. Recall that $I(\Delta,\Delta) = \chi(X)$, and $\chi(S^2) =2$. Mod-$2$ intersection numbers aren't going to cut it here. You need to update your exercise relating mod-$2$ self-intersection and cutting out by independent functions to oriented self-intersection.


3

The image does not actually need to be a $2$-manifold, depending on the specific nature of the functions. Also, the approach you suggest relies on the fact $2$-manifold cannot also be a $3$-manifold. It is of course true that a manifold has a unique dimension, but showing this, in the most general setting, is not a triviality. It is not too hard to show the ...


3

The proof is basically that, near a preimage of $(0,0)$ the map $f$ has to be close to its linear approximation, which is a non-trivial line because the map is an immersion. But no line is close to the set $\{(t,|t|): t \in \mathbb{R}\}$ near $(0,0)$. Here are the gory details for one way of making this precise: Suppose it is. Then we have $f: \mathbb{R} ...


3

First, note that a homeomorphism $f$ of compact smooth manifolds is homotopic to a diffeomorphism if and only if one can approximate $f$ arbitrarily well by diffeomorphisms. For $n \leq 3$, this paper of Munkres claims as a corollary that a homeomorphism $f: M \to N$ of smooth manifolds may be approximated arbitrarily well by a diffeomorphism. This settles ...


2

In dimensions 2 and 3 every homeomorphism is isotopic to a diffeomorphism (this should be in Moise's book "Geometric topology in dimensions 2 and 3", it also follows from Kirby and Siebenmann's work). In dimension 4 there are self-homeomorphisms of simply-connected smooth compact manifolds which are not homotopic to diffeomorphisms. This follows e.g. from ...


2

As referenced in the comments, this MathOverflow question gives the answer as no: every compact $3$-manifold embeds into $\Bbb R^5$, as proved by Wall. The Immersion theorem gives the optimal result for immersions: every compact $n$-manifold embeds into $\Bbb R^{2n-\alpha(n)}$, where $\alpha(n)$ is the number of $1$s in the $2$-adic expansion of $n$. ...


2

One way to do this is to use differential topology, and to prove $S^2$ and $T^2$ are not diffeomorphic, which is technically a weaker statement (though it does happen to be equivalent in dimension two). I'll make comments on homeomorphism after. Intersection numbers mod 2 can be defined in the smooth category using transversality and are well-defined up to ...


2

Original Answer Here is a hint. For the two obvious cases of division algebras over $\mathbb{R}$, namely the complex numbers and the quaternions, we know that the set of unit vectors (more generally, the set of non-zero vectors) is a Lie group. Moreover, for any Lie group, we know that they are parallelizable---that is, the tangent bundle to the Lie group ...


2

I figured it out myself. Actually we notice composite of g and g inverse is identity. Using chain rule on composite of g and g inverse we can get the conclusion.


2

You know that $S^1$ embeds in $\mathbb{R}^2$, but also $S^1\times\mathbb{R}$ embeds in $\mathbb{R}^2$ (consider the homeomorphism of the cylinder to $\mathbb{R}^2\setminus(0,0)$). Hence $S^1\times\mathbb{R^2}$ embeds in $\mathbb{R}^3$, but since $S^1$ embeds in $\mathbb{R}^2$ you have that $S^1\times S^1$ embeds in $\mathbb{R}^3$. You can then proceed by ...


2

I'm going to echo Incnis Mrsi's comment: The first concept you describe is called homotopy equivalence of spaces. Two spaces are then homotopy equivalent if there are compositions of continuous maps $X \overset{f}{\to} Y \overset{g}{\to} X$ homotopic to $\operatorname{id}_X$ and $Y \overset{g}{\to} X \overset{f}{\to} Y$ homotopic to $\operatorname{id}_Y$. As ...


2

Use the $g$ you already have from the previous exercise. What do you know about $f((-1,0))$ compared to $f((1,0))$? What does this tell you about $g$? EDIT Further hint: First, \begin{align*} (\cos(g(t+\pi)),\sin(g(t+\pi)))&=f(\cos(t+\pi),\sin(t+\pi)) = f(-\cos t,-\sin t) = -f(\cos t,\sin t)\\ &= -(\cos(g(t)),\sin(g(t))). \end{align*} Next, if ...



Only top voted, non community-wiki answers of a minimum length are eligible