New answers tagged

0

I speculate your problem is that you didn't take advantage of the fact that on spheres centered on the origin, we have an identity $$ \sum_i x_i^2 + y_i^2 = r^2 $$ and consequently $$ \sum_i 2 x_i \mathrm{d} x_i + 2 y_i \mathrm{d} y_i = 0 $$ is a linear dependence between the differential forms involved.


4

Write $\omega = \alpha_1 + \dots + \alpha_n$ where $\alpha_i = dx_{2i - 1} \wedge dx_{2i}$. The important observation is that if you expand $$ \omega^n = (\alpha_1 + \dots + \alpha_n) \wedge \dots \wedge (\alpha_1 + \dots + \alpha_n) = \sum_{i_1, \dots, i_n} \alpha_{i_1} \wedge \dots \wedge \alpha_{i_n} = \sum_{I} \alpha_I$$ then if $I$ contains a repeated ...


1

We use that $${\rm d}\alpha(X,Y) = X(\alpha(Y)) - Y(\alpha(X)) -\alpha([X,Y]). \qquad (\ast)$$ Suppose that $\alpha \wedge{\rm d}\alpha = 0$ and let's check that $\xi$ is closed by the Lie bracket. Let $X,Y \in \xi$. Then $(\ast)$ becomes ${\rm d}\alpha(X,Y) = -\alpha([X,Y])$. Evaluating $\alpha \wedge {\rm d}\alpha$ at the triple $([X,Y],X,Y)$ and using ...


1

For a closed $k$-form defined in a star-shaped region relative to the origin, there is an algorithm for finding an antiderivative that I describe here. Applying this to $\alpha$: Step 1: substitute $x_k\to tx_k$ and $dx^k\to t\,dx^k+x_k\,dt$.$$\begin{align}x_2\,dx^1\land dx^2 &\to tx_2(t\,dx^1+x_1\,dt)\land(t\,dx^2+x_2\,dt)\\ &= tx_2(t^2\,dx^1\...


3

$\newcommand{\Reals}{\mathbf{R}}\newcommand{\Basis}{\mathbf{e}}\newcommand{\dd}{\partial}$First, $(dz + x\, dy)(-\dd_{z}) = -1$. Generally, working in $\Reals^{3}$ for concreteness, if $(x, y, z)$ denotes some coordinate system, $(\dd_{x}, \dd_{y}, \dd_{z})$ denotes the associated coordinate vector fields, and $(dx, dy, dz)$ denotes the coordinate $1$-forms,...


4

If $\beta$ and $\beta'$ are one-forms such that $d\beta = d\beta' = \alpha$, then $\beta' - \beta$ is closed, i.e. $d(\beta' - \beta) = 0$. It then follows from the Poincaré Lemma that $\beta' - \beta$ is exact, i.e. $\beta' - \beta = df$ for some function $f$. So we see that if $\beta$ and $\beta'$ satisfy $d\beta = d\beta' = \alpha$, then $\beta' = \beta + ...


6

In Leibniz notation, even in the beginning, you weren't integrating functions: you were integrating differential forms. This is true even back in introductory calculus, even though you didn't have words to put to the concept. e.g. you didn't compute $\int f(x)$; you computed $\int f(x) \, \mathrm{d}x$. And even then is was important not to forget that $\...


6

Differential forms are not introduced to answer the question "How do I integrate functions of manifolds?" Instead, they are introduced to answer a different question, namely "What are the correct objects to integrate on manifolds?" The answer to that question is: differential forms! More precisely, the correct objects to integrate on an orientable $m$-...


1

I think I see the issue. Let $\omega$ be an $m$-form on $M$, supported in $U \subset M$. Let $\tau \colon U \to \mathbb{R}^m$ be an oriented chart, with coordinates $(r_1, \ldots, r_m)$. On $U \subset M$, we can write $\omega = f\,dr_1 \wedge \cdots \wedge dr_m$. Let $\phi, \psi \colon U \to \mathbb{R}^m$ be oriented charts, also on $U \subset M$. It is ...


0

If I understand correctly, $A_0$ is a connection on a principal bundle $P \to X$, while $\nabla_0$ is the corresponding connection on some associated vector bundle $E \to X$. The connection $\nabla_0$ induces a connection (denoted, say, $\nabla'$) on the vector bundle $\text{End}(E) \to X$. Meanwhile, I'm imagining that the gauge transformation $g \in \...


1

Note that $$d\omega = \sum_{i=1}^n(-1)^{i-1}dF_i\wedge dx_1\wedge\dots\wedge dx_{i-1}\wedge dx_{i+1}\wedge\dots\wedge dx_n.$$ and $dF_i = \sum_{j=1}^n\frac{\partial F_i}{\partial x_j}dx_j$. As wedge product is skew-symmetric, if some $dx_k$ appears twice in a term, that term must be zero. So we see that \begin{align*} &\ (-1)^{i-1} dF_i\wedge dx_1\...


2

HINT: Generalized Laplace expansion by cofactors. The way I like to think of it is this. Reduce to the case of an oriented orthonormal basis $v_1,\dots,v_n$ for $\Bbb R^n$ (in your case, some will be tangent to $M$ and the rest normal to $M$). Consider the Hodge star operator on both forms and multivectors. Then it boils down to $$(dx^i\wedge dx^j)(v_1\wedge ...


0

The form $d\theta$ is not defined at $0$ (and, indeed, can't be continued across $0$), so Stokes' theorem doesn't apply. For example, Stokes' theorem would give \begin{align*} \int_{\partial \Delta} d\theta = \int_{\Delta} d(d\theta) = 0,\;\;\;\;\;(!) \end{align*} although the first integral is clearly $2\pi$. The best you can do in this situation removing a ...


0

The problem is the differential form $d\theta$. In Cartesian coordinates this so-called "form" is $$ d\theta = \frac{xdy-ydx}{x^2+y^2} $$ This is not a well-defined form as it is not well-defined at $x=y=0$. Another way of seeing this is if $d\theta$ was a well-defined form then $\oint d\theta = 0$ but it is not! This is because $\theta$ is not a continuous ...


1

Let $x=(x_1,...,x_n)$ be (global) coordinates for $\mathbb{R}^n$. Then $p$ lives in the tangent space, $T\mathbb{R}^n|_x$, which has corresponding basis $\dfrac{\partial}{\partial x_1},...,\dfrac{\partial}{\partial x_n}$ The dual space to the tangent space, $T^{\star}\mathbb{R}^n$ is the cotangent space, which is the space of $1$-forms. It has the ...


2

There is no need for orientations or bases to show that the two given (linear!) maps $$\nu_1:\ A\ \longmapsto\ \omega_A^1\qquad\text{ and }\qquad \nu_2:\ A\ \longmapsto\ \omega_A^2,$$ are isomorphisms. The spaces of $1$-forms and $2$-forms are of dimensions $\tbinom{3}{1}=3$ and $\tbinom{3}{2}=3$, respectively. So to show that $\nu_1$ and $\nu_2$ are ...


3

You are effectively asked to show that the $2$-forms $x_i\wedge x_j$ with $1\leq i<j\leq n$ form a basis for the vector space of all $2$-forms. It is instructive to verify that the set of all $2$-forms is indeed a (real) vector space. I'll first prove the fact without the hint, and later with the hint. Let $\omega^2$ be a $2$-form. Then $\omega^2=\...


0

So is this correct? First, let us assume that we talk about $\mathbb{R}^n$. By taking the standard basis $B={e_1,...,e_n}$, I then calculate the exterior product of $x_i, x_j$, the basic forms, element of the dual of $\mathbb{R}^n$ on this basis. I get: \begin{equation} (x_i \wedge x_j)(e_i,e_j)= \begin{vmatrix} x_i(e_i) & x_j(e_i) \\ x_i(e_j) & ...


5

By Stokes, our condition is necessary. The converse direction follows from Serre duality. Consider $$\omega_k=g(z)\,d\bar z\in\Gamma(\mathbb{P}_\mathbb{C}^1,\bar K\otimes \mathcal O(-k))$$ as well-defined forms with values in $\mathcal O(-k)$, by using the standard trivialization of $\mathcal O(-k)$ and the assumption that $g$ has compact support. By ...


0

Let write the $1$-form as $$ \omega=A\:dx+B\:dy $$ with $$ A=\frac{y}{2y+x},\quad B=\log (2y+x)+\frac{2y}{2y+x}. $$ One has $$ \frac{\partial A}{\partial y}=-\frac{2 y}{(x+2 y)^2}+\frac1{x+2 y},\quad \frac{\partial B}{\partial x}=-\frac{2 y}{(x+2 y)^2}+\frac1{x+2 y} $$ giving $$ \begin{align} d\omega=&\left(\frac{\partial A}{\partial x}dx+ \frac{\...


2

No wonder you cannot prove that - your claim is false! Let $i : \Sigma \hookrightarrow T^*M$ be the natural embedding $i(x) = x$. Notice that $\theta = i^* \tilde \theta$. Let $\tilde \omega = \textrm d \tilde \theta .$ Notice that $$\omega = \textrm d \theta = \textrm d i^* \tilde \theta = i^* \textrm d \tilde \theta = i^* \tilde \omega ,$$ so $\omega$ ...


0

I believe the kernel of the contraction $\omega^\#$ at a point $p\in \Sigma$ is dependent on the vector field $X$. For example, if we consider $X$ such that $X(p) \in (T_p\Sigma)^{\omega}$ for each $p\in \Sigma$ then the one form is trivial. We can do this with a non-zero vector field anytime the submanifold is not symplectic so that we have a non trivial ...


1

Each cotangent space, i.e. the set of the 1-forms restricted to a point $x_0$ is a vector space. The set of the differential 1-forms over a manifold is more generally a module over the ring of the smooth functions: as you correctly pointed out, you can multiply differential forms by functions, not just by scalars. Nonetheless they form in particular also a ...


0

Agree with @user343900's answer. I would add that when one moves from $\Bbb{C}$ to $\Bbb{R}^2$, the term $\mathrm{i}u + v$ becomes the vector $(v,u)$ and and $\mathrm{d}x + \mathrm{i}\mathrm{d}y$ becomes the vector $\mathrm{d}s = (\mathrm{d}x,\mathrm{d}y)$, i.e. becomes a line element (as is required by the integral being along the boundary of the region). ...


1

The contour integral $\int_{\partial D} f(z) \, \mathrm{d}z$ when written out becomes $$\int_{\partial D} (u+iv) \cdot (\mathrm{d}x + i \mathrm{d}y) = \Big(\int_{\partial D} u \, \mathrm{d}x - v \, \mathrm{d}y\Big) + i \Big(\int_{\partial D} v \, \mathrm{d}x + u \, \mathrm{d}y\Big).$$ This is not heuristic - complex integrals are defined to make this true. ...


1

No, you've got it backwards. :) Let $i : S^2 \hookrightarrow \Bbb R^3$ be the usual inclusion. If $\omega$ is a form on $\Bbb R^3$, then its restriction to $S^2$ is defined as $\eta = i^* \omega$ (the pull-back of $\omega$). Therefore, the fact that $\eta$ is closed does not mean anything relevant for $\omega$ because $0 = \Bbb d \eta = \Bbb d (i^* \omega) =...


4

Expanding my comment: \begin{align*}(g^*\omega)(x)(v_x^1,\dots,v_x^{2n+1}) & =\omega(g(x))(d_xg(v_x^1),\dots,d_xg(v_x^{2n+1})) \\ & =\omega(\color{red}{-x})(-v_x^1,\dots,-v_x^{2n+1})\\ & =\sum_{j=1}^{n+1} (-x_j) (-v_x^{n+j}) - (-y_j) (-v_x^j)\\ &=\sum_{j=1}^{n+1} x_jv_x^{n+j} - y_jv_x^j\\ &=\omega(x)(v_x^1,\dots,v_x^{2n+1}).\end{align*} ...


2

Recall the product rule $d(\alpha \wedge \beta) = d\alpha \wedge \beta + (-1)^{|\alpha|} \alpha \wedge d\beta$. Hence, we have $$ d(\bar{z}_{\nu} d\bar{z}[\nu]) = d(\bar{z}_{\nu}) \wedge d\bar{z}[\nu] + \bar{z}_{\nu} \wedge d(d\bar{z}[\nu]). $$ Applying the product rule again, you see that $d(d\bar{z}[\nu])$ is a sum of a wedge of forms, each containing $d^...



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