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Define the functions $$ F(x,y):= \frac{f(x,y)}{xf(x,y)+yg(x,y)}$$ and $$G(x,y):= \frac{g(x,y)}{xf(x,y)+yg(x,y)}.$$ Then $$ d\omega =\left(\frac{\partial F}{\partial x}(x,y)dx+\frac{\partial F}{\partial y}(x,y)dy\right)\wedge dx + \left(\frac{\partial G}{\partial x}(x,y)dx+\frac{\partial G}{\partial y}(x,y)dy\right)\wedge dy. $$


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This seems to be false, even if one assumes that $\Sigma$ is isometrically embedded. Counterexample: Let $M$ be the $2$-sphere and $\Sigma$ some great circle in $M$; the volume form $\Omega$ of $\Sigma$ is some parallel one-form on $\Sigma$. $\Omega$ can't possibly extend even locally to some neighborhood of $M$, because $M$ doesn't admit even locally ...


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There is. Let $\pi : \mathbb R^4\to \mathbb R^2$ be the projection to the $y_1, y_2$-plane and $i : \mathbb R^2 \to \mathbb R^4$ be $(y_1, y_2) \mapsto (0,0, y_1, y_2)$. Then your $P_{[dy_1,dy_2]}$ is $\pi^* \circ i^* = (i\circ\pi)^*$.


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Let $V$ denote an $n$-dim vector space (over $\mathbb{R}$, say). Recall that a $p$-form $\alpha \in \Lambda^p(V^*)$ is decomposable (or simple) iff it can be written as a wedge product of $1$-forms -- i.e., $\alpha = \omega_1 \wedge \cdots \wedge \omega_p$, with each $\omega_i \in \Lambda^1(V^*) = V^*$. Definition: Let $\alpha \in \Lambda^p(V^*)$ be a ...


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Advice: First, you need understand some basic concepts of differential forms. I recommend two excellent readings of about differential forms, in my point-view. 1: Differential Forms, by Henri Cartan. This book is ideal for understand differential forms in various contexts, for example, Cartan develops the theory of forms in space of finite and infinite ...


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Let's recall the definition of the module of differentials on a $\mathbb{C}$-algebra $R$: take the free $R$-module on the symbols $\{ dr \colon r \in R \}$ and mod out by the relations $d(r+s ) = dr + ds$, $d(rs) = sdr + r ds$, and $da = 0$ for $a \in \mathbb{C}$. Then, the sheaf $\Omega_{R/\mathbb{C}}^1$ on $\textrm{Spec}(R)$ is the "tilde" of this module. ...


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We need to use the Natural Transformation Law defined in Guillemin p.168 and p.176: If $f:Y\to X$ is an orientation-preserving diffeomorphism, then $\int_X{\omega}=\int_Y{f^*\omega}$ for every compactly supported, smooth $k$-form on $X$. We can now claim $\int_X{f^*_j\omega}=\int_Y{\omega}$, where $j=0$ or $1$, which would be enough, but we observe ...


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Following the hint, if $F: \partial W \to Y$ is a smooth map that extends to all $W$ and $\omega$ is a closed $k$-form on $Y$, then $d\omega=0$, thus Stokes Theorem gives $$ \int_{\partial W} F^*\omega = \int_{W} d(F^*\omega) =\int_{W} F^*(d\omega) = \int_W 0 = 0 \tag{1} $$ Now let $F: X \times I \to Y$ be an homotopy between $f_0$ and $f$. Since ...


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It's just componentwise. Fix a basis $e_i$ with dual basis $\theta^i$ and define $$\int \omega = \sum_i \left(\int \theta^i \circ \omega\right) e_i.$$ If you have another basis $v_i$ with dual $\pi^i$, then let $A$ be the change of basis matrix such that $v_j = \sum_i A_{ji} e_i$. A little bit of linear algebra tells us that the dual bases are then related ...


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Interesting. That exercise got me thinking for awhile, I had no choice but to open everything in coordinates, but I nailed it. I wonder if there's a better way to do it. What you did so far is correct. Since we want to compare these forms in ${\bf x}(D)$ (and you should really use ${\bf x}$ instead of $x$ here, for reasons that'll be clear shortly), it is ...


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First, we should define tensors. There are various types of tensors, and we focus on those which are relevant to your question. A tensor of type $(k,0)$ is a map$$T:\Gamma(TM)\times\ldots\times\Gamma(TM)\to C^\infty(M),$$which is $C^\infty(M)$-linear in every coordinate. In other words, $T$ eats $k$ vector fields and spits out a function. The linearity means ...


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Suppose the charts on $Y$ follow coordinates $w \in \mathbb{C}^n$. Then we may write $\omega$ locally in a chart as $$ \omega(w) = \frac{\partial^2 f}{\partial w \partial \bar w }(w) dw \wedge d\bar w $$ Now if we have a nice holo map $F: X \to Y$ (i.e. structure preserving), the pull back is just as you've written, $F^* \omega = \omega \circ F$. This ...


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You certainly don't need stereographic coordinates. By definition, you need to check that $\omega$ is a smooth 2-form, $d \omega=0$, and $\omega$ is non-degenerate, i.e. for all $v \in T_u S^2$, there exists $w \in T_u S^2$ such that $\omega_u(v,w)\neq 0$. You could try to verify (1) using stereographic coordinates, but that seems unnecessary. The ...


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By Sard's theorem, there is a dense subset $U \in N$ so that $Df$ is sujective on $f^{-1}(U)$. Then $d\omega$ is zero on $U$: this is a general fact in linear algebra: Let $L : V\to W$ be a surjective linear map. If $$L^*\alpha(X, Y, Z) = \alpha(LX, LY, LZ) = 0$$ for all $X, Y, Z\in V$, then $$\alpha (a, b, c) = 0$$ for all $a, b, c \in W$, as $L$ is ...


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Yes. In fact, this is true for any $k$-form, regardless of whether or not it is in the image of $d$. To see this, let $\omega$ be any smooth $k$-form on $N$. Let $U\subseteq N$ denote the open set consisting of all $p\in N$ for which $\omega_p$ is not identically $0$. We will assume $U\neq \emptyset$, and then prove that $f^\ast \omega$ is not the $0$ ...


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That's correct! We say a form $\omega$ is closed if $d\omega = 0$, and we say that $\omega$ is exact if $\omega = d\eta$ for some form $\eta$. Your remark says, in this terminology, that every exact form is closed. However, the converse is not true: not every closed form is exact. (Here, I am referring to forms defined on our whole manifold - the Poincare ...


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That's precisely correct! See This wikipedia page on closed and exact forms for more details.


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Notice that for $\Omega_i$ it is always possible to write: $$ \Omega_{i} = \sum_{j\neq i} a_j \Omega_j $$ for some constants $a_j$. For example: $\Omega_1 = \Omega_2 + \Omega_4 -\Omega_3 $. So if all $j\neq i$ forms are exact, so will be $\Omega_i$.


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I guess what I write here is wrong. But, I'm not sure exactly why, so I post it here in the hope someone points out the error of my ways. The two-form $\omega = e^{xy}dx \wedge dy$ may be written as $\omega = e^{r^2 \sin \theta \cos \theta} r \, dr \wedge d\theta$. Then, in polar coordinates, the unit circle is $r=1$ and $\partial / \partial r$ is an ...


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Let $\phi: S^k \setminus \{z_0\} \to \mathbb{R}^k$ be the stereographic projection, where $z_0$ denotes the north pole of $S^k$. Then for an arbitrary bump function $\rho: \mathbb{R}^k \to [0,1]$, define $\omega=\phi^* (\rho \, dx_1 \wedge\cdots \wedge dx_k)$ on $S^k\setminus \{z_0\}$. How should we define $\omega$ at $z_0$ so that $\omega$ is a smooth ...


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You've given a coordinate representation of the linear map $df_p:T_pM\to T_{f(p)}R$. One can work with it as a row-vector or introduce the "gradient vector" or indeed write it in the basis $\{dx_i$}. Generally, for any $df_p:T_pM\to T_{f(p)}N$, its coordinate representation is simply the Jacobian matrix which could too be written in the corresponding basis ...


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I have found a way to solve my problem : Instead of writing differential forms, it is also possible to simply differentiate the function $z = f(x, y) = f\left(x(u, v), y(u,v)\right)$ assuming that such a function exists. So I am looking for $$ \frac{\partial z}{\partial x} = \frac{\partial f}{\partial x} \quad{\rm and }\quad \frac{\partial z}{\partial y} ...



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