# Tag Info

## Hot answers tagged differential-topology

4

I don't think it is true, what bout closed interval $[0,1]$, it is certainly a smooth manifold, and as you can prove it cannot have one $0-cell$. This result is true for closed smooth manifold, because you can find a Morse function on it with one minimum, i.e a critical point of index $0$. And then it corresponds to the $0-cell$ of the CW-presentation of ...

3

At most this requires the relative smoothing theorem: if $f: X \to N$ is $C^0$, and smooth on some closed subset $M$, then it is homotopic to a smooth map, with $f|_M$ fixed by this homotopy. Apply this to $X = S^n \times M$ to see that the map is a surjection on homotopy groups. (That is, we're smoothing a "sphere's worth" of continuous maps.) Now apply a ...

2

As noted in the comments, immersion needn't be injective. Even if they are injective, they needn't be embeddings: consider the injective immersion of $(0,2\pi)$ in the plane as a figure eight, so that the point $(0,0)$ is the center of the cross. Explicitly, send $t\mapsto (2 \sin t,\sin(2t))$. This is an immersion that cannot be a homeomorphism onto its ...

2

I'm confused. $H_1(\mathbb{R}P^3)=\mathbb{Z}/{2\mathbb{Z}}$, but is orientable (on edit, this answers your question 2). EDIT: Maybe it refers to the fact that $H^{m-1}(M)$ can only be torsion if the manifold $M$ is non-orientable. It is Corollary 3.28 in Hatcher.

1

The tangent space at $I_n$ is the space of antisymmetric matrices defined by $A+A^T=0$. Given $g\in SO(n)$, and let $AS(n)$ the space of antisymmetric matrices, the tangent space at $g$ is $gAS(n)$. That is the image of the tangent space $T_{I_n}SO(n)$ by the left translation defined by $g$. Suppose that $g\in SO(n)$, $A\in AS(n)$, you have ...

1

This is false. Because if you see the cellular homology, then if $X$ has only one 1-cell, $H_1(X_1,X_0)$ would be a free group of $1$ genarator, but then if yoy compute $H_1(X)$ from cellular homology, then it has only one genarator. But there are many manifold with more than one genarator of $H_1$, for example torus.

1

Yes, so proper way to formulate is to pick outward pointing vector fields on $\partial M\times I$. Let $\omega=\lambda\times\mu$ be the orientation on $M\times I$. Then the orientation on $\partial M$ is defined to be $\omega(V_{out},-)$. In one case $\lambda(V_{out})$ is positeve and so we get back $\mu$ in the other is negative and we get $-\mu$ (if you ...

1

From $$\alpha(t)=\int_0^t e^{is^2}\>ds$$ it follows that $\dot \alpha(t)=e^{it^2}$ has absolute value $1$, hence $\alpha$ is a locally isometric immersion. From ${\rm arg}\bigl(\dot\alpha(t)\bigr)=t^2$ it then follows that the curvature $$\kappa(t)={d\over dt}{\rm arg}\bigl(\dot\alpha(t)\bigr)=2t$$ is strictly monotonically increasing for ...

1

Regarding your third question, set $x(t) = \int_0^t \cos(s^2) \, ds$. The function $x$ is continuous so to show that $x$ is bounded it is enough to show that $\lim_{x \to \pm \infty} x(t) = \pm \int_0^{\infty} \cos(s^2) \, ds$ exists. We have  \int_0^{\infty} \cos(s^2) \, ds = \int_0^1 \cos(s^2) \, ds + \int_1^{\infty} \cos(s^2) \, ds = \int_0^1 \cos(s^2) ...

1

Came up with a bit simpler solution. Let $e_1(x)$ be a tangent vector field on the Klein bottle $K$. Suppose that $n_1(x)$ and $n_2(x)$ - pair of linearly indenpendent normal vector fields. Then we can choose $e_2(x)$ in the $TK_x$ so that four vectors $(e_1(x), e_2(x), n_1(x), n_2(x))$ form a positively oriented basis in $\mathbb{R}^4$. This would give us ...

1

You can understand $W_n$ as a CW-complex, so you have one $0$-cell $e^0_1$ and $n$ copies of $1$-cells $e^1_i$ ($i=1,\dots, n$) and any $m$-cell with $m>1$, so it means that if you want the fundamental group, you have $n$ diferent generators (one for each $1$-cell) and you don't have any restriction, so it means that you obtain that the fundamental group ...

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