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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 ...


3

You might look at Ted Frankel's book, as well as Abraham-Marsden-Ratiu.


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If $\omega$ is a differential $k$-form and $g$ is a smooth real-valued function then $g\cdot\omega$ (or just $g\omega$) is a differential $k$-form with $$(g\cdot\omega)(p)(v_1, \dots, v_k) = g(p)\omega(p)(v_1, \dots, v_k).$$ What you wrote doesn't make sense as $p \in \mathbb{R}^m$ while $v_1, \dots, v_k \in T_p\mathbb{R}^m$, so $g(v_1), \dots, g(v_k)$ are ...


2

You need to determine a function $f : \mathbb{R}^3 \to \mathbb{R}$ such that $\omega = df$.


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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 $dt$ ...


2

Hint: The conditions $d(f\alpha) = 0$ and $f$ nowhere zero allow you to obtain an expression for $d\alpha$. Once you have this expression, use it to simplify $\alpha\wedge d\alpha$.


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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 = ...


1

Your idea for $\omega_1$ is correct. To make it rigorous, recall that at each $p \in \mathbb S^2$, we have $T_p \mathbb S^2 = \{ v\in \mathbb R^3 : v\cdot p = 0\}$. So for all $v\in T_p\mathbb S^2$ we have $$\omega_1 (v) = x (v\cdot p) = 0$$ and so $\omega_1 |_{\mathbb S^2} = 0$. In particular it is both exact and closed. For $\omega_2$, using the idea ...


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, ...


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Note that $$d\hat{f} = \frac{\partial\hat{f}}{\partial x}dx + \frac{\partial\hat{f}}{\partial y}dy + \frac{\partial\hat{f}}{\partial z}dz.$$ First we have \begin{align*} \frac{\partial\hat{f}}{\partial x} &= \frac{\partial}{\partial x}\int_{0}^{1}(f(tx,ty,tz)x+g(tx,ty,tz)y+h(tx,ty,tz)z)dt\\ &= \int_{0}^{1}\frac{\partial}{\partial ...


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

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 ...


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That's a standard arguemnt using Stokes' Theorem. Let $\gamma (t)$ be a parametrization of $\gamma$. Then consider $$F : \mathbb S^1 \times [0,1] \to \mathbb R^2, $$ where $F(t, s) = s \alpha (t) + (1-s) \gamma (t)$. Then $F(\cdot, 0) = \gamma$ and $F(\cdot, 1) = \alpha$. Using Stokes' Theorem, as $d\omega = 0$, $$0=\int_{\mathbb S^1 \times [0,1]} F^* ...


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I believe I am correct. It doesn't matter all that much that $P$ is actually parametrized by $a$. The point is that for each $a$ this provides a homotopy of chain maps between $\text{id}^\#$ and $\pi^\# i_a^\#$, which is the essential lemma in this context.


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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

Notation/definitions: if $V$ is a vector space, and $V^*$ its dual, in the following write $$\langle \omega,v \rangle = \omega (v),$$ where $\omega \in V^*$, and $v\in V$. Suppose $g\colon M \to \mathbb R$ is a (nice) function: the $1$-form $dg$ is the unique element in the cotangent-space, (i.e., the dual vector space to the tangent space), such that ...


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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\\ ...


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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 ...



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