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0

You have to find a parametrization of $C$ first: $$ \gamma(t)=\begin{cases} (t,0) & \mbox{ if } 0\le t \le 1\\ (1,t-1)&\mbox{ if } 1<t\le 2\\ (3-t,1)&\mbox{ if } 2<t\le 3\\ (0,4-t)&\mbox{ if } 3<t\le 4 \end{cases}. $$ With the above parametrization we have: \begin{eqnarray} ...


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

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 ...


0

It is precisely an infinitesimal change. But an infinitesimal difference in what? A differential 1 form measures the flux across an infinitesimal line situated at some point. By extension a differerential 0 form evaluated at a point in space measures the flux across an infinitesimal point (an infinitiseimAl point is the same as a finite point is the same ...


0

As has been mentioned in the comments, you are correct. The beginning of the Mayer-Vietoris sequence is $$0 \to H^0(M) \xrightarrow{k^*\oplus l^*} H^0(U)\oplus H^0(V) \xrightarrow{i^*-j^*} H^0(U\cap V) \xrightarrow{\delta} H^1(M) \to \dots$$ and the end is $$\dots \to H^{n-1}(U\cap V) \xrightarrow{\delta} H^n(M) \xrightarrow{k^*\oplus l^*} H^n(U)\oplus ...


0

Following Spivak Notation: So we know by definition that $\omega:=\frac{1}{2}\sum_{i_1<i_2}\omega_{i_1,i_2}\space dx^{i_1} \wedge dx^{i_2}$ so $\omega$ is a 2-form. Thus if we compute $d\omega$ we attain $$d\omega = d\left[ \frac{1}{2}\sum_{i_1<i_2}\omega_{i_1,i_2}\space dx^{i_1} \wedge dx^{i_2} \right]$$ ...


0

$d\omega$ is $1/2 \sum \partial \omega_{i,j}/\partial x_k dx_k\wedge dx_i\wedge dx_j$.


0

To calculate $d\omega$ use $$d\omega_{i,j}=\sum_s\frac{\partial \omega_{i,j}}{\partial x_s}dx^s,$$ for the components and then ensamble $$d\omega = \frac{1}{2} \sum_{i,j} d\omega_{i,j}\wedge dx_i \wedge dx_j.$$


0

Hint: Suppose $X$ is a non-empty connected open set, and that $f:X \to \mathbf{R}$ is locally constant. Pick an arbitrary point $x_{0}$ in $X$, and let $U = \{x \in X : f(x) = f(x_{0})\}$. Use local constancy of $f$ to prove $U$ is open and closed in $X$.


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 ...


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 ...


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 ...


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

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 ...


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 ...


0

Write $\theta = \sum f_i \,{\rm d}x_i$, so: $$F^\ast\theta = F^\ast\left(\sum_{i=1}^n f_i\,{\rm d}x_i\right) = \sum_{i=1}^n \left(f_i \circ F\right) F^\ast {\rm d}x_i.$$ Now we apply a point $p$: $$(F^\ast\theta)_p = \sum_{i=1}^n f_i(F(p)) (F^\ast{\rm d}x_i)_p.$$ Now take a tangent vector $X$: $$(F^\ast\theta)_p(X) = \sum_{i=1}^n f_i(F(p)) (F^\ast{\rm ...


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 $\omega \wedge \omega = 0$.


0

OP might find the following rephrasing useful. Consider a graded associative algebra $({\cal A},\circ,+)$ containing the exterior derivative $d$ of form degree $|d|:=+1$, and the contraction $i_X$ of form degree $|i_X|:=-1$, where $X$ is a vector field. The graded Lie superbracket is the supercommutator $$[a,b]~:=~a\circ b -(-1)^{|a||b|}b\circ ...


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.


0

Consider the following equivalent formulation of the first fundamental theorem: given a function $f$ defined on an interval $I = [a,b]$. Let $G$ be the 1d Green's function for differentiation: $G(s) = \text{sgn } s/2$. Then an antiderivative $F$ for the function $f$ can be found by, for $x \in I$, $$F(x) G(x-t) |_{t=a}^b = \int_a^b f(t) G(x-t) \, dt + ...


0

Hint: Prove this via induction on $k + \ell$. You'll want to use the fact that any $k$-form can be written as a sum $\alpha_1\wedge \mu_1 + \cdots +\alpha_s \wedge \mu_s$, where $\alpha_1,\ldots,\alpha_s$ are $1$-forms and $\mu_1,\ldots,\mu_s$ are $(k-1)$-forms.


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 ...


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 ...


0

This one is exact, unfortunately: it is just $$\mathrm d\biggl(\frac{-1}{(x^2+y^2+z^2)^{1/2}}\biggr).$$


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

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 ...


0

A more general result can be found in Conditions on a $1$-form in $\mathbb{R}^3$ for there to exist a function such that the form is closed. For your special case (1-form in $\mathbb{R}^2$) a direct argument can be given as follows. (I am going to liberally mix 1-forms and vector fields in what follows, since we have Euclidean structure.) You want a ...


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 $x\in U$, ...


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.


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 ...


0

Yes, we can! If a Riemannian manifold has a metric in the form you have written, then the Levi-Civita connection is split into the two directions, which implies that vectors (and differential forms, just as well) decompose exactly. It is explained in detail in D.Joyce, Riemannian Holonomy Groups and Calibrated Geometry, if I'm not mistaken.


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 ...


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 ...


0

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 $$ d\omega =\left(\frac{\partial F}{\partial x}(x,y)dx+\frac{\partial F}{\partial y}(x,y)dy\right)\wedge dx + \left(\frac{\partial G}{\partial x}(x,y)dx+\frac{\partial G}{\partial y}(x,y)dy\right)\wedge dy. $$


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 ...


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 $\pi^* \circ i^* = (i\circ\pi)^*$.


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 ...



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