Tag Info

Hot answers tagged

7

Michael, read about characteristic classes. For example, the Poincaré dual of $c_k(E)$ (the $k$th Chern class of a complex bundle of rank $n$) is the cycle along which $n-k+1$ generic sections become linearly dependent. One can also deduce directly from the behavior of Chern classes and short exact sequences of bundles, if there is a trivial subbundle of ...


3

Not necessarily. Consider $M_1=\mathcal{O}_X(p)$, $M_2=\mathcal{O}_X(q)$ and $L=M_1^{-1}$ where $p,q\in X$. Then $H^0(M_1)=\mathbb{C}=H^0(M_2)$, but $H^0(M_1\otimes L)=H^0(\mathcal{O}_X)=\mathbb{C}$ and $H^0(M_2\otimes L)=H^0(\mathcal{O}_X(q-p))=0$. Note: Here I didn't assume you have a morphism between the line bundles, only between their global sections. ...


3

I'll write out an answer for a form of three variables. This might yield some intuition. Thing to notice is that $dx^i\wedge dx^i=0$, and that $dx^i\wedge dx^j=-dx^j\wedge dx^i$, and $f_{xy}=f_{yx}$. Thus, because $dx^i\wedge dx^i=0$ $$d^2f=f_{xy}dy\wedge dx+f_{xz}dz\wedge dx+f_{yx}dx\wedge dy+f_{yz}dz\wedge dy+f_{zx}dx\wedge dz+f_{zy}dy\wedge dz $$ ...


3

If $M$ is $\ne S^2$ we can realize it as a "domain" in the plane ${\mathbb R}^2$. Both loops then appear as smooth closed curves in the plane, intersecting transversally. Since $\alpha$ is supposed simple, by Jordan's curve theorem it separates the plane into two open regions, one of them, call it $\Omega$, compact, and $\alpha=\pm\partial\Omega$. The sign ...


3

Notice that in three dimension, a curve of constant distant to a fixed point is a curve on the sphere. So not only you shall show what you mentioned, but that torsion vanishes to ensure the curve is restricted to a plane, hence a circle in 2 dimensional subspace.


2

Using the formula $$\tilde{\kappa}=\frac{|\mathbf{\beta}'\times \mathbf{\beta}''|}{|\mathbf{\beta}'|^3}$$ and the Frenet formula: $\beta'=\kappa \mathbf{n}$, $\mathbf{n}'=-\kappa \mathbf{\beta}+\tau\mathbf{b}$. Thus $$\tilde{\kappa}=\frac{|\kappa \mathbf{n}\times(\kappa'\mathbf{n}+\kappa(-\kappa \mathbf{\beta}+\tau\mathbf{b}))|}{|\kappa ...


2

There are too many sources of different flavors to list. If you want to start learning some differential geometry, Goldberg's Curvature and Homology is a cheap Dover book with all sorts of results with the interplay of Hodge theory and Riemannian and complex geometry.


2

Since $dF_p$ is injective, $F$ is locally a immersion. With proper local coordinates, $F$ becomes $(x^1,\dots,x^m)\mapsto(x^1,\dots,x^m,0,\dots,0)$. Now any $g\in C^\infty_M(p)$ is locally a function $g:(x^1,\dots,x^m)\mapsto \mathbb R$. Consider the map $f\in C^\infty_N(F(p))\to\mathbb R$ given by: $$ f=g\circ\pi, \pi:(x^1,\dots,x^n)\mapsto(x^1,\dots,x^m) ...


2

This follows from the fact that pulling back by homotopic maps produces isomorphic bundles. Let $ p:E\to B $ and $ s:B\to E $ denote the projection and zero section, respectively. These are homotopy inverses. In particular $ sp $ is homotopic to the identity, and so $(sp)^*F=p^*s^*F =E\times_B F_B $ is isomorphic to $ F$.


2

As a follow up to the comment placed by the OP after the answer of Mark Grant, I would like to add some details. I hope the OP will find them an helpful complement to the response by Mark Grant. Let $\pi:E\longrightarrow M$ be a vector bundle, and $H$ an homotopy from $N$ to $M$ $$\begin{align}H:N\times[0,1]&\longrightarrow M\\(x,t)&\longmapsto ...


2

So $dx$ is a $1$-form on $\Omega\subset \mathbb R^3$ for example where $(x,y,z)$ define the local coordinates. It means that $dx:\Omega\rightarrow \mathcal L(\mathbb R^3,\mathbb R)$ and $dx$ is simply the differential of the map $$\begin{array}{rccl}x:&\Omega&\rightarrow &\mathbb R \\ & p=(p_1,p_2,p_3)& \mapsto& p_1\end{array}$$ ...


1

As a corollary of the hairy ball theorem, $S^2$ is not a parallelizable manifold, that is, it does not have a set of $2$ globally non-vanishing vector fields that span its tangent space at every point. The parallelizable vector fields can be used to introduce a trivialization.


1

If you consider $dt^1\wedge\cdots\wedge dt^k$ as a wedge product of $k$ 1-forms, the conclusion can't be more obvious. Recall that the wedge product of $k$ $1$-forms: $$ dt^1\wedge\cdots\wedge dt^k\triangleq\frac{k!}{1!\cdots 1!}\mathcal A(dt^1\otimes\cdots\otimes dt^k). $$ where $\mathcal A$ is the alternating operator: $$ \mathcal ...


1

Here is a slightly more systematic answer than in my comment above. Suppose we have the $(k-1)$-form $\omega=f(\vec{x})\omega_{i_1}$ where $\omega_{i_1}:=dx_{i_2}\wedge dx_{i_3}\wedge\cdots \wedge dx_{x_k}$ has been introduced for convenience. Then the exterior derivative of $\omega$ is $$d\omega=df \omega_{x_i}=\left(\frac{\partial f}{\partial ...


1

It is an application of the following fundamental property of the exterior derivative: $d(\alpha \wedge \beta) = d\alpha \wedge \beta + (-1)^{\mathrm{deg} \alpha}(\alpha \wedge d\beta)$. When $\alpha = f$ is a $0$-form (a function), this identity says that $d(f \beta) = df \wedge \beta + f d\beta$. In your situation, $f = x^{i_1}$ and $\beta = ...


1

Here is a slightly different argument (from the one in the original question) As you noticed, one direction is easy: suppose $F^*$ is surjective, and let $X$ be such that $dF(X)=0$. Then for any function $g$, there is a function $h$ such that $dh\circ dF = dg$, and hence $dg(X)=0$. Taking $g$ to be any coordinate function, as you did, you see that $X=0$, ...


1

First of all, I assume that you are asking about forms which are only locally Lipschitz, otherwise the answer will be different (the function $x^2$ on the real line is only locally Lipschitz). Then the simplest thing to do is to write your form in local coordinates and require all the component functions to be locally Lipschitz. There are alternative ...


1

$$ \begin{split} d(df)&=d(\frac{\partial f}{\partial x^j})\wedge dx^j\\ &=\frac{\partial^2 f}{\partial x^jx^k}dx^k\wedge dx^j\\ &=0 \end{split} $$ since $$ dx^k\wedge dx^j=-dx^j\wedge dx^k $$ and $$ \frac{\partial^2 f}{\partial x^jx^k}=\frac{\partial^2 f}{\partial x^kx^j} $$


1

These things are linear by definition, because the vector space structure on $\pi^{-1}(p)$ just comes from carrying the structure over via $\phi_f$. The only thing to check is that this is well-defined, which is what the condition about $GL_n$ does. In other words, if you have an arbitrary set $S$ and a vector space $V$ and a bijection $M : V \to S$ then ...


1

this is in response to your comment; since it's a bit long I write it as an answer. The idea is the following: given two points $v,w$ in the fiber $\pi^{-1}(p)$, we want to define $v+w$. Using a chart $U_f$, we borrow the vector space structure from $\mathbb{R}^n$, i.e. we take $$v +w := \phi_{f, p} ( \phi_{f, p}^{-1}(v) + \phi_{f, p}^{-1}(w))$$To check well ...


1

This is a purely local theorem. We may assume that $S$ is given in the form $$(x,y)\mapsto\bigl((x,y,f(x,y)\bigr)$$ where $f$ is $C^1$ in a neighborhood of $(0,0)$, and furthermore that $$f(0,0)=f_x(0,0)=f_y(0,0)=0\ .$$ So ${\bf 0}=(0,0,0)\in S$, and the tangent plane of $S$ at ${\bf 0}$ is the plane $z=0$. Assume now that we are given a plane ...


1

If you assume that $M$ has a riemman metric, you can build a projection $p:TM\to\nu$ this is a morphism betwenn vector bundles. Than define $\nabla_{X}Y=p([X,Y])$, you can look that $p$ is linear over each fiber, more properties of Lie Bracket you show what you want. Have you studed foliations?


1

The first equation $\beta=\hat{Φ_1}^∗(x)$ is probably just a typo. I think you mean $\beta=\hat{Φ_1}^∗(\beta)$, which is the hypothesis. The second equation is also a typo. What you want is just use the fundamental theorem of calculus: $f(1) - f(\epsilon) = \int_{\epsilon}^1 \frac{\partial f}{\partial t} dt$ (apply it to $f(t)=\hat{Φ_t}^∗(\beta)$ and then ...


1

As Mark Bennet said, it means every two regions have a nontrivial piece of boundary in common: nontrivial means containing a topological arc, i.e., not being a point. This is a standard detail in Map coloring problems, and Thurston's exercise is in the context of map coloring.


1

You can repeat the "sines and cosines" process one more time if you want. Like our friend @Bruno Joyal said: For the 1-sphere: $X(\phi)=(\cos(\phi),\sin(\phi))$, so that its squared coordinates sums up to 1. $\cos^2(\phi) + \sin^2(\phi) = 1$ For the 2-sphere: $X(\phi,\psi)=(\cos(\phi)\cdot\cos(\psi),\sin(\phi)\cdot \cos(\psi),\sin(\psi))$, so that its ...


1

a)It's almost correct: $$\lambda_{w_p}=\pi^*(w_p)=\pi^*\sum_{}c_i dx^i|_p=\sum c_i(\pi^*dx^i)_{w_p}= \sum c_i(d(\pi^*x^i))_{w_p} =\sum c_i(d\tilde{x^i})_{w_p}$$ hence in the coordinates of $T^*M$, where $w_p=(\tilde{x},c)$, you have $\lambda=\sum c_id\tilde{x^i}$. (Which many people write $\sum c_i dx_i$ using the natural identification.) b) in ...


1

I'm assuming that the Kahler form is $\omega = \displaystyle\sum_{i=1}^n\sum_{j=1}^ng_{i\bar{j}}dz^i\wedge dz^{\bar{j}}$. Note that $d\omega = \partial\omega + \bar{\partial}\omega$ and if $d\omega = 0$, then $\partial\omega = 0$ and $\bar{\partial}\omega = 0$ (because $\partial\omega$ and $\bar{\partial}\omega$ have different bidegrees). As $\omega$ is a ...



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