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13

Diffeomorphisms are the isomorphisms in the category of smooth manifolds, while isometries are the isomorphisms in the category of Riemannian manifolds. That is to say, diffeomorphisms are under no obligation to preserve the extra structure (metric, and all the geometry that comes with it) that the manifolds might have. They do preserve the differentiable ...


10

You're getting at the fundamental difference between differential geometry and differential topology. When we are considering manifolds (surfaces are a $2$d case) we don't only care about the set, we care about all the structure. So, what structure are we preserving? A homeomorphism is an isomorphism of spaces equipped with topologies A diffeomorphism is ...


7

When we have a "new" category, like when we pass from (Abelian) groups to rings, we require that the maps between the objects have properties that preserve the structure. For instance, a(n) (Abelian) group homomorphism preserves the binary operation: $\phi:A\to B$ has $\phi(x+y)=\phi(x)+\phi(y)$. A ring homomorphism has the additional requirement that $\phi(...


5

Not every complex $n$-form on $M$ defines an almost complex structure. If $\Omega$ is such a form, it has to satisfy two additional conditions: $\Omega$ is decomposable, meaning that in a neighborhood of each point it can be written in the form $\Omega = \zeta^1 \wedge\dots \wedge \zeta^n$ for complex-valued $1$-forms $\zeta^1,\dots,\zeta^n$; and $\Omega\...


5

No, it's not true at all. For example, there is an infinite number of diffeomorphism classes of manifolds with the same homology groups as $S^3$. Besides, your examples are wrong. Characteristic classes are elements of cohomology groups. Moreover, you cannot deduce what the characteristic classes are just by knowing the homology groups. They are extra ...


4

While all three answers are clear, I think a counter-example could be pedagogical. It is easy to think of homeomorphisms between smooth manifolds that are not diffeomorphisms: any bijective, continuous mapping whose coordinate representation for local maps are non-smooth functions should do. A pair of smooth manifolds that are homeomorphic, but not ...


4

In the case of discrete groups acting on the spaces $\mathbb R^n$, $\mathbb H^n$, or $\mathbb S^n$ (Euclidean spaces, hyperbolic spaces, and spheres), the Poincaré polyhedron theorem is a very nice tool for verifying properness. There are many features of the Poincare Polyhedron Theorem that I cannot go into here, and for the general statement it's ...


4

$V_2(\mathbb{C}^4)$ is an $S^5$-bundle over $S^7$. But every $S^5$-bundle over $S^7$ is trivial: $S^5$-bundles over $S^7$ are constructed by gluing two trivial $S^5$-bundle over $D^7$ along the boundary $\partial D^7=S^6$. So they are classified by $$ [S^6,\operatorname{Aut}S^5]=[S_6,SO(6)]=\pi_6(SO(6))=0. $$ We can also see this by identify $\mathbb{C}^...


3

The two topologies are the same and coincide with the original topology on $M$. There are many ways to prove this. Here's one quick way using a bit of machinery: by the Whitney embedding theorem we can embed $M$ as a closed submanifold of $\mathbb{R}^n$ for some $n$. Then, every closed subset of $\mathbb{R}^n$ is the vanishing set of a smooth function, ...


3

Note that by definition, $ d(x, y) \ge \| x-y\|_E$ for all $x, y\in M$, so $$ \frac{d(x, y)}{\|x-y\|_E} \ge 1, \ \ \ \forall x, y\in M, x\neq y.$$ So it suffices to show that $$\tag{1}\limsup_{y\to x} \frac{d(x,y)}{\|x-y\|_E} \le 1$$ We assume that $M$ has dimension $k <n$. Using a rotation and translation, we assume that $x = 0\in M$ and $T_xM = \...


3

Hausdorff tells you that you can always find non-intersecting opens under certain conditions. It doesn't say that all opens are non-intersecting.


3

The formula $k(s)=\lvert\alpha''(s)\rvert$ works for a regular curve $\alpha(s)$ parametrised by arclength $s$. It doesn't work if you parametrise your curve differently, as you discovered. On the other hand, $k(t)=\dfrac{\lvert\alpha'(t)\times\alpha''(t)\rvert}{\lvert\alpha'(t)\rvert^3}$ works for all regular parametrisation of a space curve.


2

OK, this is a long computation. First, the Koszul formula gives \begin{align*}\newcommand{\grad}{\operatorname{grad}_g}\newcommand{\Hess}{\operatorname{Hess}_g} 2g(\nabla_XY,Z)&=Xg(Y,Z)+Yg(X,Z)-Zg(X,Y)\\ &\quad+g([X,Y],Z)-g([X,Z],Y)-g([Y,Z],X)\\ 2\tilde{g}(\tilde{\nabla}_XY,Z)&=X\tilde{g}(Y,Z)+Y\tilde{g}(X,Z)-Z\tilde{g}(X,Y)\\ &\quad+\tilde{...


2

Parametrize it by $x$. We have the parametrization $\gamma(x) = (x,-7/3x)$, for $x\in [0,3]$. Then $$\int_{\pi(A)}dx = \int_0^3 d(x\circ \gamma) = \int_0^3 dx = 3.$$ Since you haven't specified the orientation, the answer could also be $-3$. This should make sense intuitively, since the line wraps around the torus in the $x$ direction 3 times. (If you ...


2

We say that a surface is doubly ruled when, through each of its points, there are two distinct lines that lie on the surface.


2

What your highlighted passage should say is that $\dfrac{\psi(1/w)}{w^4}$ must have a removable singularity at $w=0$. But here's a different approach for you. Since $\Psi$ is a global quadratic differential, $\sqrt{\Psi} = \sqrt\psi\,dz$ is a (not necessarily well-defined) holomorphic differential. Be that as it may, the quantity $\sqrt\Psi\wedge\overline{\...


2

The higher right derived functors generalize cohomology. If you take the map $X$ to a point, then the right derived functors are exactly the cohomology of the the global sections functor. In particular, cohomology of the constant sheaf $\mathbb Z$ gives singular/cw/de-rham cohomology. You should think of the higher derived functors as a way to patch ...


2

This is an example of a Monge patch. It's easy to see that $f$ is injective, $f$ is onto $\Phi:=f(U)$ ($\Phi$ was never defined, so I assume that it is $f(U)$; note that it will be a subset of a hyperbolic paraboloid), and $f$ is smooth. As you noted, $f_u\times f_v\neq 0$ on $U$, and so $f$ is regular. So, we've shown everything that we need to. As for ...


2

Note $$ \int_M L(f)f=\lVert\nabla f\rVert^2_{L^2(M)}\geq 0 $$ with equality if and only if $\nabla f=0$ almost everywhere. So any smooth minimizer $f$ has zero derivative everywhere, hence locally constant. But such functions are eigenfunctions of $L$ with eigenvalue $0$.


2

Yes, it is enough. For a vector field to be Hamiltonian all you need is to find a Hamiltonian (in your case $f$) such that your field is obtained as $$ X=Idf $$ where $I$ is the identification mapping between 1-forms and vector fields via the symplectic structure. In your case, that matrix has rows $(0,1)$ and $(-1,0)$ and the coordinates of $df$ are $(\...


2

It means two things: The function $F\colon\, V\longrightarrow W$ is a differentiable function which has an inverse $F^{-1}\colon\, W\longrightarrow V$. The function $F^{-1}\colon W\longrightarrow V$ is differentiable, too.


2

$$\begin{cases} f(a)\cos(a) = f(b)\cos(b)\\ f(a)\sin(a) = f(b)\sin(b) \end{cases}$$ because $\gamma$ is closed. Multiplying side by side $$ f(a)^2\cos(a)\sin(a)=f(b)^2\cos(b)\sin(b)$$ Multiplying now by $2$ one has $$ f(a)^2\sin(2a)=f(b)^2\sin(2b)$$


2

Just so we're on the same page, regarding notation etc I'll state the IFT: Inverse Function Theorem: Let $E,F$ be real Banach spaces, $f:E \to F$ be a $C^k$ map such that at a point $x_0 \in E$, the differential $df_{x_0}$ is an invertible element of $L(E,F)$. Then, there is an open set $U \subset E$ containing $x_0$, an open set $V \subset F$ ...


2

Yes, it's just polar coordinates. Define $\phi:B\to \mathbb R$ as follows: If $x>0, y\ge0,$ then $\phi(x,y)=\tan^{-1}(y/x).$ If $x\le0,y>0$ then $\phi(x,y)=-\tan^{-1}(x/y)+\frac{\pi}{2}.$ If $x<0,y\le0$ then $\phi(x,y)=\tan^{-1}(y/x)+\pi.$ If $x>0,y\le 0$ then $\phi(x,y)=-\tan^{-1}(x/y)+\frac{3\pi}{2}.$ Then, $\phi$ is smooth on $B$. Now, ...


1

No, it is not necessarily countable. Let $E$ be any closed subset of $[0,1]$. There is a smooth function $f$ on $\mathbb R$ such that $f^{-1}(0) = E$ (take a suitable infinite series of "bump functions", each nonzero on one of the intervals in $E^c$). Take $\psi$ so that $\psi(\gamma(t)) = f(t)$; it doesn't matter what $\psi$ does off this line.


1

The Wikipedia page says that the action is transitive and the space $X=G/H$ is a smooth manifold of dimension $\dim X=\dim G-\dim H$. Why? You can take a look at the chapter $21$ on Quotient Manifolds of John.M.Lee's book Introduction to Smooth Manifolds. You might be interested in the Quotient Manifolds Theorem (th. 21.10) which is used to prove this ...


1

Let's try to make the question a bit more precise. Define $$ S_1 = \{ f_1(\theta, v) \, | \, (\theta, v) \in (0,2\pi) \times \mathbb{R} \} \subseteq \mathbb{R}^3_{x,y,z}, \\ S_2 = \{ f_2(\phi, u) \, | \, (\phi, u) \in (0,2\pi) \times \mathbb{R} \} \subseteq \mathbb{R}^3_{a,b,c}. $$ Then $S_1,S_2$ are both parametric surfaces in $\mathbb{R}^3$. In order to ...


1

I'm not familiar with the specific construction you're studying, I don't have a direct answer, but this will be too long for a comment. If $E \to M$ is a vector bundle with connection $\nabla$, you can define a exterior derivative ${\rm d}^\nabla\colon \Omega^k(M;E) \to \Omega^{k+1}(M;E)$ acting on $E$-valued differential forms, with the aid of $\nabla$. ...


1

If I got your question, you can move $\omega_2 \wedge d \omega_3$ on the RHS, so the RHS becomes $d \omega_2 \wedge \omega_3-\omega_2 \wedge d \omega_3$. This is equal to $d(\omega_2 \wedge \omega_3)$, since $d(\omega_2 \wedge \omega_3)=d\omega_2 \wedge \omega_3 + (-1)^1 \omega_2 \wedge d\omega_3=d \omega_2 \wedge \omega_3-\omega_2 \wedge d \omega_3$ ($\...


1

That's a rather baroque proof! If we write $x=(x_0,\ldots,x_n)$ and $y=(y_0,\ldots,y_n)$ then the finitely many equations are the $x_iy_j-x_jy_i=0$ or $0\le i< j\le n$. By continuity, the subset defined by one equation $x_iy_j-x_jy_i=0$ is a closed subset. The subset defined by all of them is the intersection of these closed subset, and is also closed. (...


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