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4

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 $\omega \wedge \omega = 0$.


3

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 ...


3

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 \begin{equation} \iota_v(\omega)(u_1, \ldots, u_{k-1})=\omega(v,u_1,\ldots,u_{k-1}) \end{equation} 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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