New answers tagged

0

If $f$ is a diffeomorphism, $f^*$ is an isomorphism of vector-spaces of $(k+1)$-forms, hence $0=df^*\omega = f^*d\omega$ implies $d\omega=0$.


1

There is a general method for finding an antiderivative of a closed differential form defined on a star-shaped region (more generally, on some region $D$ that’s the image of a star-shaped region) that’s particularly simple in the case of a one-form. It’s basically the same as what’s described in Tsemo Aristide’s answer, elsewhere. Step 1: Replace $x^i$ ...


1

Unfortunately, you really need to compute, either integrating (you know that integrals along any path with same initial and final points are equal), or solving: $$ \frac{\partial f}{\partial x_1}=1, \quad \frac{\partial f}{\partial x_2}=\sin(x_3),\quad \frac{\partial f}{\partial x_3}=x_2\cos{x_3},. $$ The computational difficulty is the same with both ...


2

I believe you want to calculate $f$ locally. So take a chart whose image is a ball which contains $0$ and identify it with a neighborhood of $x\in M$ you have: $f(x)=\int_0^t\omega_{tx}(x)dt$.


1

To carry out the computation even though we know the answer will be zero: We have $df_1 = dx_1 + dx_2$, $df_2 = -dx_1$ and $df_3 = -dx_2$. Thus \begin{align*} f^*(dx_1 \wedge dx_2 \wedge dx_3) &= df_1 \wedge df_2 \wedge df_3 \\ &= (dx_1 + dx_2) \wedge (-dx_1) \wedge (-dx_2) \\&= dx_1 \wedge dx_1 \wedge dx_2 + dx_2 \wedge dx_1 \wedge dx_2 \\ ...


1

They do cancel, if you keep track of the signs correctly. For example, the second term produces $(Yf) \omega(Z,X)$, while the fourth produces $-\omega( -(Yf)X,Z) = (Yf)\omega(X,Z)$. These are negatives of each other because of the antisymmetry of $\omega$.


1

The idea makes more sense if you think about surfaces. For example take $U \subseteq \mathbb R^2$ open and some sufficiently nice function $f: U \to \mathbb R$. Then, the graph $G_f = \{(x,y,f(x,y))\}$ of $f$ is a surface in $\mathbb R^3$ and one can consider the tangential vector space to a point $p \in G_f$ which would then be defined to $$T_p(G_f) = \{ ...


1

If $T$ is a two dimensional vector space, an euclidian structure is given by a symmetric bilinear form, positive, non degenerate. In a given coordinate system $(x,y)$ it is given by $ Ex^2+2Fxy+Gy^2$. $E,F,G$ depends on the coordinate system, but the euclidian structure do not. This is what happen in your case, just you have a two parameter family of ...


1

Check out Kock's Synthetic Geometry of Manifolds for a nice geometric description of connection forms, curvature forms, and torsion in terms of parallel transport. My mathoverflow answer here gives an explanation of torsion from this point of view. I'll give a run down for how to intuitively think of the connection form. Let $\nabla$ be a choice of ...


0

Here is about the simplest example you can take, which is what you were trying to accomplish (I think!). Let $dx$ and $dy$ be the standard $1$-forms on $\mathbb{R}^2$, and let $\partial_x$, $\partial_y$ be the corresponding dual vector fields (i.e. $\partial_x$ is the unit vector field in the $x$ direction). Then $$i_{\partial_x}(dx \wedge dy)(\partial_x) = ...


1

I find your first question pretty unclear as it is stated, so I will concentrate on the second half. Consider the basis of $\Lambda^kV^*$ given by $$\{v_{i_1}\wedge\ldots v_{i_k}|1\le i_1<\ldots<i_k\le n\}.$$ We can write $\alpha$ as $$\alpha = \alpha^{i_1\ldots i_k}v_{i_1}\wedge\ldots v_{i_k}.$$ We have $$v_j\wedge\alpha = ...


0

This seems false. Consider $R^6$ with coordinates $a,b,c,x,y,z$. Then $$ da \wedge db + dc \wedge dx + dy \wedge dz $$ which is a 2-form, can't be written (I think!) as a product of 1-forms $$ s \wedge \phi $$ where $s$. At the very least, you can generalize the number of coordinates to something quite large; if you're restricting to just two $f_j$s, ...


0

If $f\in C^{\infty}(M)$ and $X$ is a vector field on $M^n$, then $$X(f)=df(X),$$ where in a coordinate chart $(U,x^i)$, $df$ is given by $$df=\sum \frac{df}{dx^i}dx^i.$$ Note that $\{dx^1,\cdots,dx^n\}$ is the dual basis to the basis of coordinate vectors $\{\frac{\partial}{\partial x^1},\cdots,\frac{\partial}{\partial x^n}\}.$ Advice: I recommend two ...


1

Well, $df$ doesn't make any sense outside exterior derivatives (in this context). You can see that if $(U,x)$ is a local chart, and $\partial/\partial x^\mu$ are the coordinate basis vectors, then the $dx^\mu$ one-forms are their dual basis. After all, the definition of $d$ on functions is that for an arbitrary vector field $X$, $$ X(f)=df(X). $$ Then, ...


1

Indeed yes. A standard example is the time-dependent Schrödinger equation% \begin{equation*} \partial _{t}\psi (\mathbf{x},t)=-iH\psi (\mathbf{x},t) \end{equation*} With \begin{equation*} \psi (\mathbf{x},t)=f(t)\varphi (\mathbf{x)} \end{equation*} one has \begin{eqnarray*} \partial _{t}f(t)\varphi (\mathbf{x}) &=&-if(t)\varphi (\mathbf{x)} \\ ...


1

Note that \begin{align*} T'^{a} &= de'^a + \omega'^a_b\wedge e'^b\\ &= d((\Lambda^{-1})^a_b e^b) + ((\Lambda^{-1})^a_cd\Lambda^c_b + (\Lambda^{-1})^a_c\omega^c_d\Lambda^d_b)\wedge e'^b\\ &= (\Lambda^{-1})^a_bde^b + d(\Lambda^{-1})^a_b\wedge e^b + (\Lambda^{-1})^a_cd\Lambda^c_b\wedge e'^b + (\Lambda^{-1})^a_c\omega^c_d\Lambda^d_b\wedge e'^b\\ ...


2

Yes, I think so: let $\omega$ be the (complex-valued) $1$-form $f(z)dz$. In this case, the line integral of $\omega$ along a curve $\gamma:[a,b]\to \mathbb{C}$ agrees with the line integral of the function $f$ along $\gamma$ (by the relevant definitions). You can compute the pullback $\gamma^{\ast}\omega$ as $f(\gamma(t))\gamma'(t)dt$ on $[a,b]$, where $dt$ ...


3

First extend $J$ complex linearly so that they are defined on $TM\otimes_{\mathbb{R}}\mathbb{C}$. Note that $TM\otimes_{\mathbb{R}}\mathbb{C}$ decomposes as the direct sum of the two eigenspaces of $J$, namely $T^{1, 0}M$ and $T^{0,1}M$, with eigenvalues $i$ and $-i$ respectively. Also extending the metric and $\nabla$ complex linearly, $A$ becomes a real ...



Top 50 recent answers are included