Hot answers tagged

4

For $k < 2$ and $k > \dim V - 2$ it's always true that a $k$-form $\omega$ both is decomposable and satisfies $\omega \wedge \omega = 0$. For $k = 2$ (and provided the field underlying $V$ does not have characteristic $2$), the condition $\omega \wedge \omega = 0$ is both necessary and sufficient for decomposability. (Proving this is a nice exercise.) ...


3

I think the function you are looking for is $f(x)=\int_{\gamma_x} \alpha$ where $\gamma_x$ is a path from some chosen base point $x_0$ to $x$. But then for this to be well defined you have to say that all paths between $x$ and $x_0$ are homotopic, i.e. that your sphere is simply connected, so the whole argument gets a bit circular... P.S. Traying to write ...


3

$\omega=g^\ast\omega$ means that for all $p\in M$, $\omega_p=g^\ast(\omega_{g(p)})$. Now, to consider your statement that $\omega(X)=g^\ast\omega(X)$, we consider the statement pointwise. Suppose for now that the action is transitive and fix $p\in M$. Now consider $(\omega(X))(q)$ for any $q\in M$. By transitivity, there is some $g\in G$ such that ...


3

A differential is a $1$-form. At each point a $1$-form gives a linear functional in a tangent space, and the level sets of a linear functional are parallel affines subspaces. Now $df$ in $\mathbb R^3$ gives a sort of a ruling in each tangent. For a given $p$, the level sets of $df_p$ are what we can see if we zoom in on the level sets of $f$ in a ...


3

Meditating on the p.d.e. for a minute, we can see that any function of the form $a(x, y) = b(xy)$ satisfies $x a_x = y a_y$, and this is (I think) the general solution, but I don't see offhand how to prove this. Instead, here's another strategy for finding the glide-invariant $2$-forms that avoids solving a p.d.e.: Computing gives $\mathcal L_V (dx \wedge ...


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 ...


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 ...


2

Your De Rham cohomology algebra is $H^*(\mathbb P^3_\mathbb C)=\mathbb R[w]/(w^4)$, where $w=[\omega]$ is the class of some closed $2$-form $\omega$. We then have $0\neq w^2=[\omega \wedge \omega]$, so that $\omega\wedge \omega$ is not exact and a fortiori $\neq 0$.


1

Write $S^n$ as the union of two open sets $U= S^n-N$, $V=S^n-S $where $N,S$ are the north and south pole. Note that the intersection $U\cap V$ is connected if $n\geq 2$. $U,V$ are contracile (diffeomorphic to $R^,$, and on each of these sets the form $\alpha$ is exact and andmits primitive $f_U, f_V$. On the intersection $d(f_U-f_V)=0$ and by connexity ...


1

Looking at this a different way, $\omega=g^\ast\omega$ means that pointwise: $$ (g^\ast\omega)_p=g^\ast(\omega_{g\cdot p})=\omega_p. $$ For a vector field $X$, $$ (\omega(X))(p)=\omega_p(X_p) $$ On the other hand, $$ (\omega(X))(p)=((g^\ast\omega)(X))(p)=(g^\ast\omega)_p(X_p)=(g^\ast(\omega_{g\cdot p}))(X_p)=\omega_{g\cdot p}(g_\ast X_p)=\omega_{g\cdot ...


1

The exterior derivative operator $d$ is real (i.e., real-valued on real-valued forms) and complex-linear (i.e., if $\alpha$ and $\beta$ are real $p$-forms, then $d(\alpha + i\beta) = d\alpha + i\, d\beta$, so yes, your calculation is correct. To deduce the Cauchy-Riemann equations from Green's theorem, it's convenient to differentiate the holomorphic ...


1

Hint: Use a local chart. Let $U\subset X$ be a trivialization of $T^*X$, $T^*U=U\times R^n$ with local coordinates $(x_1,...,x_n,y_1,..,y_n)$. You can write $d\alpha_X=\sum_{i=1}^{i=n}dx_i\wedge dy_j$. You have $\pi^*\mu= \sum u_{jk}dx_j\wedge dx_k$. This implies that $(d\alpha_X+\pi^*\mu)(\partial x_i,\partial y_j)=d\alpha_X(\partial x_i,\partial y_j)$.


1

In terms of the multilinear algebra, it looks like you're on the right track. We extend $\omega$ to a basis $\omega, v_2, \dots, v_n$ and then write $$\alpha = \sum a_I \omega \wedge v_{i_2} \wedge \dots \wedge v_{i_k} + $$ $$ \sum b_J v_{j_1} \wedge \dots \wedge v_{j_k} $$ Then as you've noted, wedging with $\omega$ kills everything in the first term. ...


1

Let's start with equation $(120)$. We have two assumptions, first that $\omega$ is closed, $d\omega = 0$, and second that $\omega \in Y_p$, that is, in $$\omega = \sum_I f_I(x)\,dx_I$$ all $k$-indices $I$ only involve $dx_i$ for $i \leqslant p$. We compute $$d\omega = \sum_I (d f_I)(x)\wedge dx_I = \sum_I \sum_{r = 1}^n (D_r f_I)(x)\, dx_r \wedge dx_I = ...


1

Note that $$d\left(\frac{dz}{z}\right) = d\left(\frac{1}{z}dz\right) = \frac{\partial}{\partial z}\left(\frac{1}{z}\right)dz\wedge dz + \frac{\partial}{\partial\bar{z}}\left(\frac{1}{z}\right) d\bar{z}\wedge dz = 0$$ as $dz\wedge dz = 0$ and $\dfrac{1}{z}$ is holomorphic. Therefore $\dfrac{dz}{z}$ is closed. In general, a complex one-form on an open ...


1

If $M$ is a manifold of dimension $n$, then for all $p > n$, $\Omega^p(M) = \{0\}$. In particular, as a point is a zero-dimensional manifold, $\Omega^p(\ast) = \{0\}$ for all $p > 0$.


1

I think this becomes easier by putting less emphasis on coordinates. View your forms a defining for each $x\in\mathbb R^n$ a linear map $\omega_x:\mathbb R^n\to\mathbb R^k$. By the assumption on linear independence, each of the maps $\omega_x$ is surjective. Now take a point $p$, choose a basis $\tilde X_1,\dots,\tilde X_{n-k}$ for $ker(\omega_p)$ and ...


1

Cargo's answer doesn't seem to give any geometric intuition for the statements he/she claims; let me try and remedy that. Let's just talk about $\mathbb{CP}^n$. There are ways of calculating $H^{\bullet}(\mathbb{CP}^n;\mathbb{Z}) = \mathbb{Z}[x]/x^{n+1}$, where $x$ has degree 2 (cellular homology + universal coefficients gives you the module structure; ...


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 $(x,y)$ ...


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 $\int dt$.


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 $G_x-yG_y-G=0$.



Only top voted, non community-wiki answers of a minimum length are eligible