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There isn't much you need to do. The wedge product satisfies the relationship: $\alpha \wedge \beta = (-1)^{pq} \beta \wedge \alpha$ if $\alpha, \beta \in \Lambda^p, \Lambda^q$ respectively. In your case $$\omega \wedge \omega = (-1)^{(2q+1)(2q+1)} \omega \wedge \omega = -\omega \wedge \omega$$ That can only happen if $\omega \wedge \omega = 0$.

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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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A function $f : U \to \mathbb{R}$ is locally constant if for each $x \in U$, there is an open neighbourhood $V$ of $x$ such that $f|_V$ is constant. Note that for any $y \in \mathbb{R}$, $f^{-1}(y)$ is open, so for any $A\subseteq \mathbb{R}$, $f^{-1}(A) = \bigcup_{y\in A}f^{-1}(y)$ is open. In particular, $f^{-1}(A)$ is open for every open $A\subseteq ... 3 In general, if a finite group$G$acts on a manifold$M$without fixed points and we let$N=M/G$, which is also a manifold, and$p:M\to N$is the canonical projection map, then$p$induces a map$p^*:H^*(N)\to H^*(M)$which is injective. In fact, the action of$G$on$M$induces one on$H^*(M)$, and the image of$p^*$is precisely$H^*(M)^G$, the set of ... 3 In that sentence, I was referring to the tangent-cotangent isomorphism, defined on pages 341-343 of the book. It probably would have been a good idea to include a page reference. 2 Why it has to do with index lowering: Because it maps the vector of components$p^i$to the form of components$p_i$. Why it has to do with the Euclidean metric: One cannot just copy the components of a vector and write them into a differential form. They are different objects in different spaces. What really happens is that you get a linear map ... 2 Well, in order to make thing clear, you can write these equalities explicitly so that you has to see when the assumption : $$f^*\omega= \omega$$ is needed. Let$p\in M$and$u\in T_pM, one has : $$\begin{array}{rcl} \omega_p(X_{H\circ f}(p),u) & = & d_p(H\circ f)(u) \\ & = & d_{f(p)}H(df_p(u)) \\ & = & ... 2 The symbol \iota is a common notation for the insertion of a vector fields. It maps k-forms to k-1-forms having the eating the vector, that is $$\iota_v(\omega)(u_1, \ldots, u_{k-1})=\omega(v,u_1,\ldots,u_{k-1})$$ Different conventions are around and sometimes the vector is fed into the last argument. So in your case ... 2 To finish this off you need to write the d(y_j \circ f) in terms of dx_i, use linearity of the wedge product, and produce the definition of the determinant. You begin with:$$d(y_j \circ f) = \sum_{i=1}^n \frac{\partial(y_j \circ f)}{\partial x_i} dx_i = \sum_{i=1}^n \frac{\partial f_j}{\partial x_i} dx_i$$The minuses in the determinant formula come ... 2 Let V denote an n-dim vector space (over \mathbb{R}, say). Recall that a p-form \alpha \in \Lambda^p(V^*) is decomposable (or simple) iff it can be written as a wedge product of 1-forms -- i.e., \alpha = \omega_1 \wedge \cdots \wedge \omega_p, with each \omega_i \in \Lambda^1(V^*) = V^*. Definition: Let \alpha \in \Lambda^p(V^*) be a ... 2 First, a couple of terminological corrections. \int_{\gamma} \theta where \theta = p dx is the symplectic 1-form and \gamma a closed curve. "Symplectic 1-form" is a misnomer. A symplectic form is by definition a 2-form. The form p\, dx is called a symplectic potential, or in the case of a phase space, the tautological 1-form or canonical ... 2 First, a little bit of algebraic-topological intuition. Your set M is a hyperboloid of one sheet, which means it deformation retracts to a circle. A closed, nonexact one-form is a nontrivial cohomology class: It should tell you that a loop around the "waist" of the hyperboloid is not contractible. So look at the circle. Do you know a closed, nonexact ... 1 Hint A standard example of a 1-form that is closed but not exact is the form$$\frac{-z \,dy + y \,dz}{y^2 + z^2}$$on the punctured plane \Bbb R^2 - \{0\} (with standard coordinates suggestively named); suggestively (but not quite precisely) this is often denoted d \theta, where \theta is an angular polar coordinate. Remark Incidentally, if one ... 1 To be explicit, suppose F : U \rightarrow V is a smooth map for U \subset \mathbb{R}^m, V \subset \mathbb{R}^n open sets with coordinates (\tilde{x}_1,\ldots,\tilde{x}_m) and (x_1,\ldots,x_n) respectively. A 1-form \theta \in \Omega^1(V) can be written as \theta = \sum_if_i(x_1,\ldots,x_m)dx_i for functions f_i : V\rightarrow \mathbb{R}. The ... 1 There is. Let \pi : \mathbb R^4\to \mathbb R^2 be the projection to the y_1, y_2-plane and i : \mathbb R^2 \to \mathbb R^4 be (y_1, y_2) \mapsto (0,0, y_1, y_2). Then your P_{[dy_1,dy_2]} is \pi^* \circ i^* = (i\circ\pi)^*. 1 In general, we parameterize a smooth curve C with \vec r(t)=\hat xx(t)+\hat yy(t), t\in[0,1], such that$$\int_C\,\left(f(x,y)dx+g(x,y)dy\right)=\int_0^1 \left(f(x(t),y(t))\,\frac{dx(t)}{dt}+g(x(t),y(t))\,\frac{dy(t)}{dt}\right)\,dt$$In the example at hand, we parameterize each line segment of C separately. To that end, we have$$\begin{align} ... 1 Just do it explicitly for\omega = f dx^{i_1}\wedge\dots\wedge dx^{i_k}$and$\eta = g dx^{j_1}\wedge\dots\wedge dx^{j_\ell}$. (It's just the usual product rule for functions.) Then the general result follows by distributing$d$and wedge over sums. 1 If$\omega$is a closed$1$-form on a smooth manifold$M$, then in any contractible open set$U$, you can define a function$f$such that$df=\omega$. This is a special case of a the Poincaré lemma, which says that every closed form is locally exact. To define$f$, choose a point$x_0\in U$and let$f(x) = \int_{\gamma_x}\omega$, where for each$x\in U$, ... 1 Let$\{e_1, \dots ,e_n,f_1,\dots, f_n \}$be a basis for$T_xM$with$\omega (e_i,f_i)=1$and all other pairs vanish. Such a basis exists by the standard diagonalization theorem for symplectic matrices. Then$\omega ^n(e_1,f_1, \dots , e_n,f_n)=\prod_i \omega(e_i,f_i)=1$, so$\omega^n$is non-zero at$x$. Since this is true for each$x$,$\omega^n$is a ... 1 Metric can be thought of as a mapping that takes two vector fields as arguments and produces scalar field. By fixing one vector field (call it$X$) you get a map from vector fields to scalars, that is a diferential form$gX$. This differential form will map vector field$Y$to its scalar product with$X$. If you use Cartesian coordinates in Euclidean space ... 1 Note that, if $$A(x,y)=\frac{f(x,y)}{xf(x,y)+yg(x,y)}$$ and $$B(x,y)=\frac{f(x,y)}{xf(x,y)+yg(x,y)},$$ you can derive and find that $$A_y=\frac{yf_yg-fg-yfg_y}{(xf+yg)^2}$$ and $$B_x=\frac{xfg_x-fg-xf_xg}{(xf+yg)^2}.$$ Theorem: If$\omega=Adx+Bdy$a 1-form is$C^1$such that$A_y=B_x$, then$\omega$is closed. Hint: Use that$f$and$g$are ... 1 Based on Mariano's suggestion (and terminology), I believe I have a solution for this problem: First, we can see that$p^*\omega$must be$G$-invariant because for each$g \in G$we have$p\circ g = p$, which means at the level of cohomology we have$g^*\circ p^* = p^*$. Hence$g^*(p^*\omega) = p^*\omega$and so$p^*\omega$is$G\$-invariant. In fact we can ...

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