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3

A differential $k$-form on a smooth manifold $M$ is a skew-symmetric $C^{\infty}(M)$-linear map $\omega :\Gamma(TM)^k \to C^{\infty}(M)$. The exterior derivative of $\omega$ is a skew-symmetric $C^{\infty}(M)$-linear map $d\omega : \Gamma(TM)^{k+1} \to C^{\infty}(M)$ where $d\omega(V_0, \dots, V_k)$ is equal to $$\sum_{i=0}^k(-1)^iV_i(\omega(V_0, \dots, ...


4

The text here uses the Einstein summation convention, defined below: In any local coframe on a manifold, here we'll use $(dx^i)$, defined, say, on $U \subseteq M$, we can write (the restriction to $U$ of) a $1$-form $\omega$ as a linear combination of coframe elements, namely, as $$\omega = \omega_1 \,dx^1 + \cdots + \omega_n \,dx^n = \sum_{i = 1}^n \omega_i ...


5

In a coordinate chart $(U, (x^1, \dots, x^n))$ on an $n$-dimensional manifold, every one-form $\omega \in \Omega^1(U)$ can be written as $\omega = \omega_1 dx^1 + \dots + \omega_n dx^n$ where $\omega_1, \dots, \omega_n : U \to \mathbb{R}$ are smooth functions. So we can write $\omega$ as $$\omega = \sum_{i=1}^n\omega_i dx^i.$$ Using Einstein notation, we ...


4

As mollyerin says, you need to be more careful. I would use the following argument. Let $u,v:D\to\mathbb{R}$ be the real and imaginary parts of $f$, and let $x,y,$ be real coordinates on $D$. We have $$f(z)dz=(u(z)+iv(z))(dx+idy)=u(z)dx-v(z)dy+i(u(z)dy+v(z)dx).$$Since $f$ is holomorphic, the Cauchy-Riemann equations imply that both real and imaginary part ...


1

It doesn't make sense to integrate $d(\gamma^*\omega)$ on $B_2$ since $\gamma^*\omega$ is not defined on $B_2$ and consequently neither is $d(\gamma^*\omega)$. Put differently an interior point of $B_2$ is not related in any way to any point of $X$. Thus your use of Stokes is indeed unwarranted.


1

That all looks pretty decent to me. As an alternative, you could just say that given a 1-form $\alpha$ on the first factor and a function $f$ on the second factor, you can define a 1-form on the product, $$ \beta_p(v) = f(p) \alpha(D\pi_1(p)[v]) $$ where $\pi_1$ is the projection on the first factor. A natural name for $\beta$ is $\alpha \times f$. So the ...


0

To achieve a contradiction, show by some other means that the integral is nonzero. Since $d\theta \wedge d\phi$ is in fact the standard volume form on $T$, this shouldn't be too difficult.


4

You are using that $\partial B^2 = S^1$, but by doing so you assumed already that your $S^1$ is homotopically trivial (it bounds a disk). So you have used the condition (simply connectedness) already. Your first answer is correct (module the fact that you can find a smooth disk $B^2$) For your second question, as $g$ is a constant map, then $g^*\omega = ...


5

To be (as requested) nitpicky and (I hope) meticulous: Step $2$ suggests that $d\theta$ is the differential of an angle function $S^1 \to (0, 2\pi)$, but this isn't true for two reasons: First, it is not defined at the point $(1, 0) \in S^1$. More seriously, being closed but not exact, $d\theta$ is not actually the differential of any real-valued function on ...


3

You write that the function $f$ is on a form $f(x,y)=1,$ so I assume that $S^1$ lies on a plane and $x$ in $dx$ stands for first coordinate on a plane. To calculate your integral, you have to specify orientation of $S^1$. Hence I additionaly assume that the orientation of the circle is standard, i.e. counterclockwise. Now take an arbitrary orientation ...


2

$dx$ is exact by definition, provided that $x$ is a globally well-defined function on your space. So if I understood your question correctly you're asking about an abstract torus that is considered to be the quotient space of $R^2$, on which the function $x$ is not globally defined. In this case, you'll need to compute your integral like ...


1

On a smooth manifold with a smooth chart $(U, \phi)$, the $1$-forms $dx^1, \ldots, dx^n$ form a local coframe of $TM$ defined on $U$, that is, for any $1$-form $\alpha$ on $M$, there are smooth functions $\alpha_i$ such that $\alpha\vert_U = \alpha_1 dx^1 + \cdots + \alpha_n dx^n$; this only requires knowing that for an $n$-manifold the rank of $TM$ is $n$, ...


2

You can define two forms which give you a basis of the cohomology, and show by hand thta every other 1-form is cohomologous to a linear combination of them. This is not hard to do.


0

Here are two ways of seeing that $H^1_{dR}(S^1\times S^1)$ is two-dimensional without using the Künneth Theorem. For any smooth connected manifold $X$, $H^1_{dR}(X) \cong \operatorname{Hom}(\pi_1(X), \mathbb{R})$ (see Lee's Introduction to Smooth Manifolds (second edition) Theorem $17.17$ and Problem $18$-$2$), so $$H^1_{\text{dR}}(S^1\times S^1) \cong ...


1

Let $f = \sum_{i=1}^n x_ix_{i+n}$ and let $\mathcal{V} = x_1 \wedge \dots \wedge x_{2n}$ denote the volume form. Then$$d\Omega = {{d(\omega \wedge \theta)}\over{f^n}} - n \sum_{i=1}^n x_{i+n}dx_i \wedge {{\omega \wedge \theta}\over{f^{n+1}}} = {n\over{f^n}}\left( \mathcal{V} - \sum_{i=1}^n x_{i+n}dx_i \wedge {{\omega \wedge ...


4

I think the explanation comes from what is hidden, namely the range of the summation indices. Because of the anticommutativity $$ \alpha=\sum_{i<j}a_{ij}\,dx^i\wedge dx^j. $$ When you take the exterior derivative it comes out as (I switch temporarily to indices $r,s,t$ - bear with me for a moment) $$ \begin{aligned} ...


2

This may not be the answer, you were looking for, but a similar thing puzzled me some years ago, namely why the technique separation of variables can proceed through the meaningless intermediate equation which seems to be an abuse of notation: $$ \begin{align} \frac{dy}{dx}&=g(x)h(y)\\ &\iff\\ \frac{1}{h(y)}\ dy&=g(x)\ dx \end{align} $$ But later ...


0

In the chart $x,y,z$, the cosets $G/H$ are the lines $x=\mathrm{const.}, z=\mathrm{const.}$ Hence a well-defined two-form on the coset space cannot depend on $y$ or $dy$ and must be on the form $f(x,z)dx\wedge dz$. The two steps taken is to ensure that this is indeed the case. 1) $i^*\omega=0$ says that $\omega$ does not depend on $dy$. 2) ...


2

Yes, there is an expression for $d\omega$ without reference to coordinate systems. If $\omega$ is a $k$-form, it “eats” $k$ vector fields at once and produces a function on $M$. The derivative $d\omega$ can be determined by what it does to a $(k+1)$-tuple of vectors. Remember also that vector fields on $M$ are derivations of functions on $M$. The ...


1

I don't know if this is not just placing the solution in and determining that it satisfies..however I will proceed! $$ y_1'' + f(x)y_1' = r(x)\\ y_2'' + f(x)y_2' = r(x). $$ adding the two equations together we obtain $$ y_1'' + y_2'' + f(x)y_1' + f(x)y_2' = 2r(x)\\ \left(y_1 + y_2\right)'' + f(x)\left(y_1'+y_2'\right) = \\ \left(y_1 + y_2\right)'' + ...


0

Here's a sketch of a solution along the lines you're looking for, @KlaasVanderBroeck. In some neighborhood $U$ of each point $p\in M$, we can choose local coordinates $(x^1,\dots,x^n)$ so that $Z = \partial/\partial x^1$. (This is sometimes called the Flowbox Theorem.) Cover $M$ with such neighborhoods $U_\alpha$, with $x_\alpha\colon U_\alpha\to \Bbb R^n$ ...


1

First note that $\langle Z\rangle$ defines a subbundle of $TM$. By choosing a Riemannian metric on $M$, we can write $TM = \langle Z\rangle \oplus E$, so any $V \in \Gamma(TM)$ can be written uniquely as $fZ + e$ where $f \in C^{\infty}(M)$ and $e \in \Gamma(E)$. Now define $\gamma(V) = \gamma(fZ+e) = f$. Then $\gamma$ is a smooth one-form on $M$ and ...


2

There's a mistake in the last line in the calculation of $i_{\hat{X}_t}\beta$. The $t$'s in the numerators should be in the denominators: $$i_{\hat{X}_t}\beta = \frac{-zx}{\color{red}{t}r^3}\, dy + \frac{zy}{\color{red}{t}r^3}\, dx = \frac{z}{\color{red}{t}r^3}(y\, dx - x\, dy).$$ Then $$\Phi_t^*i_{\hat{X}_t}\beta = \frac{z}{(x^2 + y^2 + t^2 ...


2

Since $\partial_0P$ is a product of $n$ circles, you can just think of this, indeed, as an $n$-fold iterated integral such as you do with the usual one-dimensional Cauchy Integral Formula. That is, parametrizing each of the circles in the usual way as $\zeta_j = z_j+r_je^{i\theta_j}$, $j=1,\dots,n$, you can rewrite the integral as the iterated integral ...


1

By definition of $\Bbb X_t$, $\Bbb X_t(\Phi_t(x,y,z)) = \frac{\partial}{\partial t}(\Phi_t(x,y,z))$, i.e., $\Bbb X_t(tx,y,z) = \frac{\partial}{\partial t}(tx,y,z)$. Thus $\Bbb X_t(tx,y,z) = (x,0,0)$, and consequently $\Bbb X_t(x,y,z) = (\frac{x}{t},0,0)$. Since $$\Phi_t^*\beta = \frac{t^2x^2 - y^2}{(t^2x^2 + y^2)^2}\, dy\wedge dz - \frac{2txy}{(t^2x^2 + ...


1

We have \begin{align}i_{\Bbb X}(dz\wedge dx\wedge dy) &= i_{\Bbb X}(dz\wedge dx) \wedge dy + (-1)^2 dz \wedge dx \wedge i_{\Bbb X} dy\\ &= i_{\Bbb X}(dz \wedge dx) \wedge dy + dz\wedge dx \wedge (-x)\\ &= [i_{\Bbb X}dz \wedge dx + (-1)^1 dz \wedge i_{\Bbb X} dx]\wedge dy - x\, dz\wedge dx\\ &= [(-1)\wedge dx - dz \wedge 0]\wedge dy - x\, ...


1

You can further simplify by using the fact that $i_{\mathbb{X}}$ is linear over smooth functions, i.e. $i_{\mathbb{X}}(f\omega) = fi_{\mathbb{X}}\omega$. To see this, note that $f$ is a $0$-form, so by the Leibniz property $$i_{\mathbb{X}}(f\omega) = i_{\mathbb{X}}f\wedge\omega + (-1)^0fi_{\mathbb{X}}\omega = fi_{\mathbb{X}}\omega$$ where we have used ...


1

I don't see how you arrive at your conclusion. In $x,y$ coordinates, $$ \begin{align}\frac{\partial f}{\partial x} \, \mathrm{d}x + \frac{\partial f}{\partial y} \, \mathrm{d}y &= \frac{\partial f}{\partial x} (1,0) + \frac{\partial f}{\partial y} (0,1) \\&= \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \end{align}$$ ...


2

Remember that $$\int_c d\omega=\int_{I^2}c^*d\omega.$$ I supppose $I=[0,1]$, now its just a easy calculation.


1

You showed that $[\mathbb{X},\mathbb{Y}]=0 \implies \frac{dq(x,y)}{dx}=\frac{dp(x,y)}{dy} , r(z) \neq 0$. All that you need to show now is that the converse holds: $[\mathbb{X},\mathbb{Y}]=0 \Leftarrow \frac{dq(x,y)}{dx}=\frac{dp(x,y)}{dy} , r(z) \neq 0$.


1

If $M$ is a manifold and $(U, \varphi)$ is a coordinate chart with $\varphi = (x^1, \dots, x^n)$ then $dx^i$ is a one-form on $U$ and is exact as $dx^i = d(x^i)$; note, this one-form is only defined on $U$, not $M$. In $M$ has a global coordinate chart, which occurs if and only if $M$ is diffeomorphic to $\mathbb{R}^n$, then $dx^1, \dots, dx^n$ are exact ...


3

What's wrong with $\mathbb{R}^n$? The notation $dx$ isn't a coincidence; $dx$ is the differential of the coordinate function (zero-form) $x$ which takes a point $p=(p_1, p_2,...,p_n)$ to the number $p_1$. $dx$ is exact on $S^1$ because the restriction to a subspace of an exact form is exact: If $\omega=df$ on $X$, than $\omega|_S = d(f|_S)$ and is exact ...


3

There are fancier ways to do this, which you'll learn eventually: If a compact group $G$ acts on $M$, then any closed $k$-form $\omega$ on $M$ is cohomologous to a $G$-invariant closed $k$-form $\tilde\omega$ on $M$ (i.e., $\omega-\tilde\omega = d\eta$ for some $k-1$-form $\eta$). But let's just do this bare-hands here. Consider $$\frac ...


2

I prefer to avoid evaluating on basis vectors if I can, so here we go: Your hypothesis that $|dr|=1$ tells us that $dr\wedge\star dr = 1 dx\wedge dy = \dfrac i2 dz\wedge d\bar z$. This in turn, by the way, tells us that $\left|\dfrac{\partial r}{\partial z}\right| = \dfrac12$ (since $r$ is real-valued). (Simply write out $dr = \dfrac{\partial r}{\partial ...


1

This is not a very satisfying answer, but the defining function for the disk has not been normalized. In fact, its gradient has length two. Also, I see that James Cook already answered this question.


1

The solution is to set $r = \frac{1}{2}(z\bar{z}-1)$. This gives $\nabla r = x\frac{\partial}{\partial x} +y\frac{\partial}{\partial y}$ for which $|\nabla r|=1$ for $|z|^2=x^2+y^2=1$. Then the sanity check works out just fine since $\frac{\partial r}{\partial \bar{z}}=\frac{z}{2}$ and: $$ \begin{align*} \int_{b\Omega} \frac{1}{z} \mathrm{d}z &= ...


2

I addressed how to calculate an orientation form on $S^n$ in this answer. I have reproduced it below. Consider $S^n$ as the unit sphere centred at zero in $\mathbb{R}^{n+1}$. First we will write down an orientation form $\alpha$ on $S^n$. On $\mathbb{R}^{n+1}$, we have the orientation form $\omega = dx^1\wedge\dots\wedge dx^{n+1}$. By Proposition ...


1

If you're already using exterior algebra, clifford algebra isn't that much of a stretch to use. It just relies on an associative product called the geometric product. In an orthonormal basis, its properties are $$e_i e_j = \begin{cases} 1 & i = j \\ -e_j e_i & i \neq j \end{cases}$$ Elements of the clifford algebra are members of a $2^n$ ...


0

Because the similarity between symmetric tensor product and wedge product, I will discuss only the wedge product here. It is common to see both two definitions of wedge product in different textbooks. Given two differential forms $\alpha\in\bigwedge^p(V)$ and $\beta\in\bigwedge^q(V)$, we can define the wedge product as $\displaystyle ...


5

The coordinates $x$ and $y$ are not global. If they were, the torus would be diffeomorphic to $\mathbb{R}^2$. If you are thinking of $S^1$ as $[0, 2\pi]$ where the endpoints are identified, then $x$ and $y$ are coordinates on $(0, 2\pi)\times(0, 2\pi)$ which is the complement of the red and purple circles in the image below. $\hspace{7mm}$ The question ...


1

This is very similar to a part of John's answer to your previous question, so I will leave you to fill in the details. Suppose $dx$ and $dy$ differed by an exact form. Then $dx - dy = df$ for some $f \in C^{\infty}(T)$. Now let $\gamma$ be a closed loop and consider $\int_{\gamma}dx - dy$; you should be able to conclude that this must be zero. Now find a ...


3

Since $$X\times Y=(X^2Y^3-X^3Y^2)e_1+(X^3Y^1-X^1Y^3)e_2+(X^1Y^2-X^2Y^1)e_3$$ then $$(X\times Y)^i=\eta_{ijk}X^jY^k,$$ is the correct formula for components. In the other hand if we agree the volume form be $e^{*1}\wedge e^{*2}\wedge e^{*3}$ and defined by $$e^{*1}\wedge e^{*2}\wedge e^{*3}=\sum_{\sigma\in S_3}(-1)^{\sigma} e^{*\sigma(1)}\otimes ...


1

John's answer is good. I would just like to add that randomly creating differential forms and testing to see if they are closed and exact is not the best way to build a relationship with cohomology. As a grad student, I tried that approach and it was not very fruitful. Since we're in a low degree, you might understand what's going on by dualizing the ...


1

Just realised the answer, I was looking for the imaginary part of the right hand side, when it was actually $i$ times $m$. Carrying on from my last line, I have Show: $(xi-y)m(x+iy)^{m-1}=im(x+iy)^m$ RHS $=im(x+iy)^m=im(x+iy)(x+iy)^{m-1}=m(ix-y)(x+iy)^{m-1}=$ LHS That $im$ notation is annoying, for all of my working I thought of it as the imaginary part.


1

So with some guidance from @TedShifrin I think I got it. First since $\omega$ is decomposable and closed taking the exterior derivative, $0=d\phi \wedge \theta -(1) \phi \wedge d\theta$. Hence, $d\phi \wedge \theta = \phi \wedge d\theta$. Wedging the expression with $\phi$ gives $d\phi \wedge \theta \wedge \phi=0$, since the right side has $\phi \wedge ...


2

By definition of $D$, we get $$\begin{align} D^2 \omega=D(D(\omega) &= D(d\omega + \alpha \wedge \omega) \\ &= d(\alpha \wedge \omega)+ \alpha \wedge d\omega + \alpha \wedge \alpha \wedge \omega \\ &= d\alpha \wedge \omega-\alpha \wedge d\omega + \alpha \wedge d\omega + \alpha \wedge \alpha \wedge \omega\\ &= d\alpha \wedge \omega , ...


0

That function doesn't have a pole at infinity. The form does. The reason why you look at the form is because that (the form) is what is inside the integral. For example. Take the function $g(z)=1$. It doesn't have a pole anywhere. It is constant. But $1\text{d}z=-\frac{1}{w^2}\text{d}w$ has a pole at infinity. An important observation is that the residue ...


3

Your computation of $i_{\mathbb{X}}\alpha$ is correct. Let me explain some general facts that work on any manifold (at least locally), but keep it restricted to $\mathbb{R}^3$. First note that a one-form $\alpha$ is a function on vector fields. Any one-form can be written as a linear combination of $dx$, $dy$, and $dz$ where the coefficients are functions ...


1

$\beta$'s irrelevance notwithstanding, your computation of $\iota_{\mathbb X}\alpha$ is correct. I'm not sure what your final question is supposed to mean.


3

Take T to be the quotient space of $\pi:\ \mathbb{R}^2\rightarrow \mathbb{R}^2 / \mathbb{Z}^2$, where $\mathbb{Z}^2$ acts on $\mathbb{R}^2$ by the translations $t_{mn}: (x,y)\mapsto(x+m,y+n)$ for arbitrary integers $m$ and $n$. Those actions obviously leaves $dx$ and $dy$ unchanged: $$t_{mn}^*(dx)=d(x+m)=dx,\ t_{mn}^*(dy)=d(y+n)=dy$$ And if $\tilde X$ is ...



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