New answers tagged

0

Let $r = x^2 + y^2$. Note that if the values are $a = \frac{x}{r}$ and $b = \frac{y}{r}$, then we will have: $w = \frac{x}{r}$ $dx$ + $\frac{y}{r}$ $dy$. Then, the following follows: $ w(X) = \frac{1}{r} (-xy +xy)$ = $0$,$ w(Y) = \frac{1}{r} (x^2+y^2)$ = $\frac{1}{r} r = 1.$ This contradicts the conditions given ($w(X) =1$, $w(Y) =0$). Thus, the solution in ...


0

Once you know the Frobenius Theorem, you know that the integral manifolds of $\omega = 0$ are (locally) given by level surfaces of some function $f$. This means that $df = \mu\omega$ for some nonzero function $\mu$.


1

We first note that $\omega(\gamma(t))$ (the $1$-form $\omega$ evaluated at $\gamma(t)$) is simply $\omega(\gamma(t))\gamma'(t)\,dt=\dfrac{x(t)dy-y(t)dx}{x(t)^2+y(t)^2}$. But $dx,dy$ map $\gamma'(t)=(x'(t),y'(t))$ to its components, so $\omega(\gamma(t))\gamma'(t)=\dfrac{x(t)y'(t)-y(t)x'(t)}{x(t)^2+y(t)^2}.$ Hence one regains the 'calc 3' notion of $(x,y)$ ...


3

When you have a binary operation $\newcommand{\bop}{\mathop{\scriptstyle\top}}\bop \colon S\times S \to S$, that induces a corresponding operation on the space of functions $D \to S$ by applying the operation pointwise, $$(f \bop g)(x) := f(x) \bop g(x).$$ The pointwise sum or product of real-valued functions are very familiar examples. The same ...


1

This is actually just calc 3, substitute $x=\cos t$ $y=\sin t$ and use $dx=-\sin t dt $ etc.It all simplifys to $\int dt$.


3

If $$\omega = \sum_{i=1}^n\frac{(-1)^{i-1}x_i}{\|x\|^n}\,dx_1 \wedge\cdots \wedge \widehat{dx_i}\wedge \cdots \wedge dx_n,$$then: $$d\omega = \sum_{i=1}^n\sum_{j=1}^n\frac{\partial}{\partial x_j}\left(\frac{(-1)^{i-1}x_i}{\|x\|^n}\right) dx_j \wedge dx_1 \wedge\cdots \wedge \widehat{dx_i}\wedge \cdots \wedge dx_n.$$Now, the only surviving term is when $j = ...


2

The mistake is subtle. While - for sufficiently regular $g$ - we have $$\int_{Q^k} g(u_1,u_2,u_3,\dotsc, u_k)\,d\mathbf{u} = \int_{Q^k} g(u_2, u_1,u_3,\dotsc,u_k)\,d\mathbf{u},\tag{$\ast$}$$ the $u_j$ in the "denominator" of $\dfrac{\partial (\sigma_{i_1},\dotsc, \sigma_{i_k})}{\partial (u_1,\dotsc, u_k)}$ are not arguments of the integrand, they just ...


0

Intuitively they are same and can be taken as simple. I state about how it appears to me. We consider small scalars and worry more about what is to be done among them in group operations rather than be bogged down what it deeply and definitively is.I say this as you are comfortable at advanced level and not so at the fundamental level. The Pythagoras ...


1

Say you have an nth order differential equation $$x^{(n)} = F(x, x', x'', ... , x^{(n-1)}) \tag{1}$$ This can be transformed into a system of $n$ first order differential equations by introducing the new variables: \begin{align} x_1 &:= x \\ x_2 &:= x' \\ &\vdots \\ x_n &:= x^{(n-1)} \end{align} Using these, equation $(1)$ becomes ...


0

This is indeed a transformation into the following first order differential system $x_1' = x_2 $ $x_2' = -x_1 + \alpha x_1^3 $ Does the released solution offer the resolution of this system afterwards?


1

Well, I think you want to find nonzero $G$ as $0$ is exact. Another approach : De Rham cohomology of $\mathbb{R}^2$ vanishes, thus every closed form is exact. Hence it suffices to show $dG\omega=(G_x dx+G_y dy)\wedge (ydx+dy)-G(dx\wedge dy)=(G_x-yG_y-G)dx\wedge dy=0$. Hence you need to find a function $G$ with $G_x-yG_y-G=0$.


0

I think that if you look at the corresponding vector field <1,y>, it has non-zero curl, and so it is not a gradient field, not conservative, and hence the differential form you have is not exact.


1

Following my comment, observe that each $t_{i_j}$ is a function so that $dt_{i_1}\wedge\cdots \wedge dt_{i_k}$ is not of the form $\sum_J b_J(x)dx_J$. Let's take an example to understand that point with $T:\mathbb R^2\rightarrow \mathbb R^3$ given by $$T(x)=(x_1,x_2)=(x_1x_2,x_1+x_2,x_1^3)$$ so that $$t_1(x)=x_1x_2, \ t_2(x)=x_1+x_2, \ t_3(x)=x_1^3.$$ ...


1

Actually, if $f\in C'$ and satisfies $$df=xdy$$ it means that $$\dfrac{\partial f}{\partial x}(x,y)=0 \text{ and } \dfrac{\partial f}{\partial y}(x,y)=x.$$ And as you can see each partial derivative is again of class $C'$ and thus $f$ has to be of class $C''$. Now, stating that $d^2f\neq 0$ is just an other way to notice that the Schwarz theorem doesn't ...


0

As in your previous question you can make the curvature form anything you want, provided it's closed and lives in the right cohomology class. Most 2-forms are not nondegenerate. In particular, there are manifolds that support no symplectic form (take $S^4$, say...) or cohomology classes $c_1$ such that $c_1^n$ is zero in cohomology, hence cannot be ...


2

This post just consolidates the comments. If $V$ is a real vector bundle, a fiberwise inner product on $V$ determines an isomorphism $V \to V^*$ by $v \mapsto \langle v, -\rangle$. So anything else you could possibly do to $V$ and $V^*$ are isomorphic: in particular, it determines an isomorphism $\Lambda^i T^*M \otimes V \to \Lambda^i T^*M \otimes V^*$. ...


0

Since it might be of interest in the future, here's my solution only using Warner's definition $\star(e_1 \wedge \dots \wedge e_k) = e_{k+1} \wedge \dots \wedge e_n$ on a ONB $e_1,\dots,e_n$ with orientation determined by $e_1 \wedge \dots \wedge e_n$: We can exclude the easy cases $k=0$ and $k=n$. We have $e_1 \wedge \dots \wedge e_n = ...


1

Given an arbitrary $k$-element ordered list $i_1,i_2,\dots,i_k$ (consisting of distinct elements), choose complementary numbers $j_{k+1},\dots,j_n$ so that $\{i_1,\dots,i_k,j_{k+1},\dots,j_n\} = \{1,2,\dots,n\}$. If you order the $j$'s so that $$e_{i_1}\wedge e_{i_2} \wedge \dots \wedge e_{i_k}\wedge e_{j_{k+1}}\wedge \dots \wedge e_{j_n} = e_1\wedge ...


2

The expression $dx_1 \wedge cdx_2$ does make sense. It's not in the form given in definition 10.11, but if you skip ahead and take a look at section 10.17 where the (exterior/wedge) product of differential forms is defined, you'll notice that for the $1$-forms $\omega = dx_1$ and $\lambda = cdx_2$, both of which are in the canonical form of definition 10.11, ...


0

First of all, since $\pi = \pi\circ\sigma$, if $\beta$ is a $k$-form on $\Bbb RP^n$, then $\alpha = \pi^*\beta = (\pi\circ\sigma)^*\beta = \sigma^*(\pi^*\beta) = \sigma^*\alpha$. Conversely, we know that $\pi$ is a local diffeomorphism. So, given a point $[x]\in\Bbb RP^n$, choose a neighborhood $U$ of $x\in S^n$ so that $\pi|_U\colon U\to \pi(U)$ is a ...


0

S.Kobayashi, "Transformation groups in differential geometry", discusses G-structures (2-tensors are special cases) and when their automorphism groups are finite dimensional Lie groups. The key is the "elliptic" condition, Theorem 4.1 (Chapter I). This theorem deals with the case of compact manifolds, but, I think, this condition is irrelevant in the case ...


3

Note that $$(D\theta)(X, Y) = (\nabla\theta)(X, Y) = (\nabla_X\theta)(Y) = X(\theta(Y)) - \theta(\nabla_XY).$$ So the skew-symmetric part of $(D\theta)(X, Y)$ is \begin{align*} \frac{1}{2}[(D\theta)(X, Y) - (D\theta)(Y, X)] &= \frac{1}{2}[X(\theta(Y)) - \theta(\nabla_XY) - Y(\theta(X)) + \theta(\nabla_YX)]\\ &= \frac{1}{2}[X(\theta(Y)) - ...


7

Say $\alpha$ is your exact degree $n$ form. Since it is exact, there exists $\tau$ such that $\alpha = d\tau$. Suppose now $M$ is orientable. Then, since it's also compact, we can do integration, in which case Stokes' Theorem tells us that $\int_M\alpha = \int_Md\tau = \int_{\partial M}\tau = 0$ since the boundary $\partial M$ of $M$ is empty. From this we ...


2

(1) $M,\ N$ has charts $x,\ y$ Then if $\omega$ is a pull back then $\omega= (f\circ \pi) (x,y) dy_{1}\cdots dy_p$ where $f: N\rightarrow {\bf R}$ (For convenience, we can write it) So $$ \partial_{x_i} (f\circ \pi )= df\ d\pi \partial_{x_i} =0 $$ If $d\pi X=0$, then $X=g_i(x,y)\partial_{x_i}$ So $i_X\omega =0$ In further since $L_X=i_Xd +di_X$ so that ...


0

Notice with your definition of $g$: $$ \partial_i g_j = x_j \partial_i f( ||x| |) = x_j f'( ||x|| ) \partial_i ||x||= \frac{ x_j x_i }{ ||x||} f'(||x||)$$ (remember $f:\mathbb{R} \to \mathbb{R}$)Thus $$ \partial_i g_j - \partial_j g_i = 0$$ thats the only step I'd say you missed to justify in a). For b), define $$ \eta = \int_{x_0}^x \omega $$ where the ...


1

Green’s theorem is a good place to start. It says that $\oint_C P\,dx+Q\,dy=\iint_D\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\,dx\,dy$. What do those partial derivatives equal in this case? If you work that out before trying to convert to polar coordinates or other things, you might find that the double integral is particularly simple.


1

Integrate directly, by parametrizing. I.e., go to a single-variable Riemann integral (from your line integral). Compute your $dy$ and $dx$ terms like in Calculus I. And you should know how to proceed from here.


1

The Sturm-Liouville trick goes like this: start out with the differential equations for eigenfunctions belonging to different eigenvalues $$\frac d{dx}\left[p(x)y_n^{\prime}\right]+\lambda_nq(x)y_n=0$$ $$\frac d{dx}\left[p(x)y_m^{\prime}\right]+\lambda_mq(x)y_m=0$$ Multiply the first equation by $y_m(x)$ and the second by $y_n(x)$ and subtract to get ...


0

The only case I am aware of when you can integrate a smooth $1$-form $\omega$ along a continuous path with no differentiability property is when $\omega$ is closed: $d\omega=0$. You then have the Recipe Consider the universal covering $p:\tilde X\to X$ of your manifold, lift your path $\gamma:[0,1]\to X$ to a path $\tilde {\gamma}:[0,1]\to \tilde X$, find a ...


0

You might look at Chapter 11 in my Introduction to Smooth Manifolds (2nd ed.), which has an extensive discussion of integration of $1$-forms along curves.


0

I can't think of a reference for this in particular, but (assuming basic familiarity with manifolds) I feel like it's a simple enough topic to just explain. Given a differentiable manifold $M$, a curve in $M$ is a differentiable function from some interval $[a,b]$ to $M$, and comes equipped with a tangent vector $\dot\gamma(t) = d\gamma(t)/dt$. A one-form ...


1

I) You can't find such $D_3$. You have correctly proved that if you can, the integral is zero. If you could find such a $D_3$, the surface would be null-homologous; but there are plenty of non-null surfaces (eg $\Bbb{CP}^1 \subset \Bbb{CP}^2$). II) Your proof only works if $A$ is defined on both sides of the curve, so that $F=dA$ globally. But if this were ...


1

Try starting from the other direction and use $(F^*\omega) \otimes (F^*\eta)(v_{\sigma(1)},\ldots, v_{\sigma(p+q)}) = (F^*\omega)(v_{\sigma(1)},\ldots,v_{\sigma(p)})\cdot (F^*\eta)(v_{\sigma(p+1)},\ldots, v_{\sigma(p+q)}) = \omega(F(v_{\sigma(1)}),\ldots,F(v_{\sigma(p)}))\cdot\eta(F(v_{\sigma(p+1)}),\ldots,F(v_{\sigma(p+q)})) = \omega \otimes \eta ...


1

For the second integral, $\int_C \omega$, note that the integrand $\omega=\frac{x dy - y dx}{x^2+y^2}$ is defined on the interior of the circle. This means that we can apply the Poincaré lemma, which says that all forms on a contractible manifold ar exact. Hence there is some 2-form $\omega'$ such that $d\omega'=\omega$. Let's call the closure of the ...


0

On ${\bf R}^3$, consider $\alpha =xdy + ydz$ so that $d\alpha =dxdy+dydz$ That is by Stokes, we have $$ \int_S d\alpha =\int_{\partial S} \alpha $$ Boundary of $S$ is $x=0,\ z^2+y^2=1$ so that $$ \alpha =ydz,\ \int_{\partial S} ydz =\int_0^{2\pi} (\cos\ t) d\sin\ t = \int_0^{2\pi}\frac{1}{2} dt= \pi $$


1

Proof Let $\varphi : M\to N$ and $f\in C^{\infty}(N)$: The relation: $$(\psi^{*}f)\circ \phi^{-1}=(f\circ \chi^{-1})\circ(\chi \circ \psi \circ \phi^{-1})$$ It follows: $$\psi^{*}f\in C^{\infty}(M)$$


0

If $X \in K_M$, then $\iota_X\sigma = 0$, so for any $Z \in V(M)$, $0 = (\iota_X\sigma)(Z) = \sigma(X, Z)$. Also note that $\sigma(Z, X) = -\sigma(X, Z) = 0$. So if $X \in K_M$ appears as either argument of $\sigma$, then that term is zero. Note that $$d\sigma(X, Y, Z) = X\sigma(Y, Z) - Y\sigma(X, Z) + Z\sigma(X, Y) - \sigma([X, Y], Z) + \sigma([X, Z], Y) ...


2

If you look at a neighbourhood of a point $(t,0)$, and you project it on the $x$-axis, you see that the projection is $2$-to-$1$. This means that $(x-t)$ is not a uniformizer for the local ring at that point. The projection map $x : E \to \Bbb P^1(k)$ locally looks just like the map $z \mapsto z^2$ around $0$ (if you work over $\Bbb C$ there are actual ...


2

Let me remark that there is no need to convert to spherical coordinates. The coordinates $x,y,z$ are well defined smooth functions on $S^2$, so their exterior derivatives are well defined one-forms on $S^2$. Naturality of the exterior derivative and the operations on differential forms shows that the pullback of your form to $S^2$ is still given by ...


1

If $$T(\rho,\theta,\phi) = (\rho\cos\theta\cos\phi, \rho\sin\theta\cos\phi, \rho\sin \phi), $$ we wish to compute $\Omega = T^\ast\omega$, where $\omega$ is our initial $2-$form. This is the formal way to do it - the pull-back by the change of coordinates. We do this as follows: \begin{align} T^\ast\omega &= T^\ast(x\, dy \wedge dz + y\, dz \wedge dx + ...


4

First of all, there's just a point on the unit sphere where $\phi=0$ — the north pole (but there's also the south pole, where $\phi=\pi$, to worry about). But remember that spherical coordinates actually fail to give a coordinate system at these points (and we can debate what happens when $\theta = 0$ or $2\pi$). In the original cartesian coordinates, you ...


2

You are somehow right: the proof would work if you only knew that $\omega$ were closed, provided that $\omega$ were defined not just on $\partial C$, but on all $C$. But in that case, all closed forms on $C$ are exact. That is not the case on $\mathbb{R}^2 \setminus \{ 0\}$. If you pick $C_1$ and $C_2$ to bound a region $C$ that does not contain $0$, then ...


0

Denote by $\pi \colon \mathbb{R}^2 \rightarrow \mathbb{T}^2$ the universal cover map. Since $S^2$ is simply connected, the map $f \colon S^2 \rightarrow \mathbb{T}^2$ factors through $\pi$ and we can find a smooth $g \colon S^2 \rightarrow \mathbb{R}^2$ such that $f = \pi \circ g$. If $\omega$ is a two-form on $\mathbb{T}^2$, then we have $$ \int_{S^2} ...


1

I would like to make a remark. In a torsionless manifold, the link between these derivatives may be found in the (very good) reference mentionned by Yuri Vyatkin (book of Yano, 1955). Another point is very interesting for practical use of Lie derivative in the same reference : the index convention for the covariant derivative may lead to some errors when ...


1

If you ever want to get anywhere in geometry, you are going to have to learn to be more flexible with your notational qualms. I wish this were not the case, but this is the culture of the field. Here, in order to make sense of $dx_i(v)$, you must first realize that $v$ is supposed to be a vector, in order to make sense of projection. Since it is standing in ...


3

Consider the map $ u : H^n_c(M) \to \mathbb{R}$ defined setting $$u: \ [ \omega ] \mapsto \int_{M} \omega \ .$$ This is well defined because of Stokes theorem, and is injective because of your assumption. So (under your assumption) either $H^n_c(M)=0$ or $\mathbb{R}$. On the other hand $H^n_c(M)\not=0$ since $u$ is non-zero (working in local coordinates, ...


1

(1) If $\omega =fdx$, then define $g(t):=\int_{-\infty}^t\omega$ Then $ g'(t)=f(t)$ so that $dg=\omega$ (2) $H^1_c({\bf R})={\bf R}$ : If $\omega$ is $1$-form on compact support on ${\bf R}$, then $d\omega =0$ And let $\int_{\bf R}\omega =C$ If $\omega'$ is another form of compact support with $\int_{\bf R} \omega'=C$, then $$ \int_{\bf R} \omega ...


0

Neither $\theta_{1}$ nor $\theta_{2}$ is a (global, continuous) function on the torus, only on the universal cover. Integrating "$d\theta_{i}$" over a closed path parallel to the $\theta_{i}$ axis gives $1$, while the integral of an exact form would be $0$.



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