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How is the interior derivative a derivative? I wouldn't say it is. My background is in Clifford algebra, and that discipline's equivalent of this operation is universally referred to as a product operation, not a derivative operation. What is the geometric content of Hodge duality? Short version: you're finding the orthogonal complement of whatever ...

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$\DeclareMathOperator{sgn}{sgn}$ For the sake of practice I've done normed version. Normalization means that I use factorials and sum over all permutations in definition of wedge product, i.e.: $$(\alpha\wedge\beta)(v_1,\dots,v_{k+l})=\frac{(k+l)!}{k!l!}\sum_{\sigma\in S_{k+l}}\sgn(\sigma)\alpha(v_{\sigma(1)},\dots,v_{\sigma(k)})\beta(v_{\sigma(k+1)},\dots,... 4 A main possibly non-intuitive usage of "form" is as a somewhat particular type of map/function. Traditionally, the word function was used in a more restrained way and it was mainly used for real and complex functions, only. For example, classically in (real) functional analysis one would have: A function would be a map from \mathbb{R} to \mathbb{R}... 1 I'm not a Mathistorian, but... Likely it originally meant its English meaning of "appearance", and it still does in most usages. Quadratic forms have a very specific appearance, namely a homogenous quadratic polynomial. Modular forms are functions satisfying a certain form of equation and some other conditions. Conjunctive/disjunctive/Skolem normal forms are ... 0 Computing the integral is easy. Pick the following natural parametrization of T:$$(x, y, z, w) = \frac 1 {\sqrt[4] 2}(\cos s, \sin s, \cos t, \sin t) $$with s,t \in [0, 2 \pi). Notice that$$\begin{align} \sqrt[4] 2 \omega &= \Bbb d (\cos s) \wedge \Bbb d (\sin s) + \Bbb d (\cos t) \wedge \Bbb d (\sin t) \\ &= (- \sin s \ \Bbb d s) \wedge (\...

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Because $f_i$ is homotopic to the identity, the induced map on cohomology is the identity. So $[f_i^* \omega] = [\omega]$. To say that two forms are cohomologous means precisely that they differ by an exact form, so that $\omega = f_i^*\omega + d\theta$.

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For any $p \in \mathbb{R}^2$, $$\dfrac{\partial}{\partial x_1}\bigg|_p \in T_p\mathbb{R}^2$$ and $$(dx_1)\left(\dfrac{\partial}{\partial x_1}\bigg|_p\right) = 1 \neq 0$$ so $p \not\in Z_{\mathbb{R}^2}$. Therefore, $Z_{\mathbb{R}^2} = \emptyset$. Now let $i : S^1 \to \mathbb{R}^2$ be the inclusion map, then $dx_1|_{S^1} = i^*dx_1$. Note that $(i_*)_p ... 1 Just use the definition (with$c = (c_x, c_y, c_z)): \begin{align} \int \limits _c x \ \Bbb d x + y \ \Bbb d y + z \ \Bbb d z \\ &= \int \limits _0 ^{2\pi} c_x (t) c'_x (t) + c_y (t) c'_y (t) + c_z (t) c'_z (t) \ \Bbb d t \\ &= \int \limits _0 ^{2\pi} \frac 1 2 \left( c_x(t)^2 + c_y(t)^2 + c_z(t)^2 \right)' \ \Bbb d t \\ &= \frac 1 2 \left( ... 4 It is not true that (-(j_U)_* \omega , (j_V)_*\omega)=0 iff \omega=0. Note that (j_U)_*\omega and (j_V)_*\omega here are cohomology classes in H^1_c(U) and H^1_c(V), not just 1-forms. So we have to consider the possibility that they might be coboundaries. A 1-form on U is a coboundary iff its integral is 0, and similarly for V. So ... 1 You need to be careful about the distinction between points and vectors. While a point is described by cylindrical coordinates (r,\theta,z) and has distance \sqrt{r^2 + z^2} from the origin, a vector v based at a point is described by Cartesian coordinates locally aligned to the cylindrical coordinates - i.e. v = v^r e_r + v^\theta e_\theta + v^z e_z,... 1 Actually, I think I have it. On S^3, \gamma_k pulls back to i_E(dV). Then from Stoke's Theorem\int_{S^3} i_E (dV) = \int_{\bar{\Bbb B}_4} 4 dV = 4 \operatorname{Vol}(\bar{\Bbb B}_4)$$A form is exact if and only if its pullback is exact. And exactness would imply the above integral vanishes, so \gamma_k is never exact. 3 Suppose \begin{eqnarray} I=\int f'(x) f(x) dx. \end{eqnarray} Putting f(x)=t, we get f'(x)~dx=dt, and thus, we have \begin{eqnarray} I=\int t dt=\frac{t^{2}}{2}+c=\frac{(f(x))^{2}}{2}+c, \end{eqnarray} where c is a constant of integration. Using this formula, we get \begin{eqnarray} 0=\int f'(x) f(x) dx+\int g'(x) g(x) dx=\frac{(f(x))^{2}}{2}+\frac{(... 2 Use Laplace transform:$$ \begin{cases} f'(x)=g(x)\\ g'(x)=-f(x)\\ f(5)=f'(5)=2 \end{cases} $$Take the Laplace transform of both sides:$$\mathcal{L}_x\left[f'(x)\right]_{(s)}=\mathcal{L}_x\left[g(x)\right]_{(s)}\Longleftrightarrow s\text{F}(s)-f(0)=\text{G}(s)\mathcal{L}_x\left[g'(x)\right]_{(s)}=\mathcal{L}_x\left[-f(x)\right]_{(s)}\... 1x$is an element of$M$,$TM$the tangent bundle of$M$.$T_xM$is the fiber of the projection$p:TM\rightarrow M$,$T^*M$is the contangent bundle. If$\pi:T^*M\rightarrow M$is the projection, the fiber$T^*M_x$is the dual of$T_xM$. Thus an element$\epsilon_x\in T^*M_x$is a linear form$\epsilon_x:T^*M_x\rightarrow R$. The differential of$\pi$is a ... 3 This seems to be best answered by Lounesto's paper "Marcel Riesz's Work on Clifford Algebras" (see here or here). In what follows:$\bigwedge V=$the exterior algebra over$VC\ell(Q)=$the Clifford (geometric) algebra over$V$w.r.t. the quadratic form$Q$Note in particular that we always have$C\ell(0)=\bigwedge V$,$0$being the degenerate ... 1 Assuming you want to use$(x_1,\dots,x_{n+1},y_1,\dots,y_n)$as local coordinates on the sphere (where$y_{n+1}\ne 0$), here's a somewhat easier way to do your pullback calculation. Note that because$x_1^2+\dots+x_{n+1}^2+y_1^2+\dots+y_{n+1}^2=1$, we have $$\sum_{i=k}^{n+1} x_k\,dx_k + y_k\,dy_k = 0,$$ and so $$dy_{n+1} = -\frac1{y_{n+1}}\big(\sum_{k=1}^{n+... 3 EDIT: By multiplying by an appropriate polynomial, we may assume that \omega has poles (at most) at 0 and \infty. On \Bbb C-\{0\} you now have holomorphic functions f and g (your f_2) with$$z^2f(z)=-g(1/z).$$Since f and g have at worst poles at 0, this equation tells us that each of their Laurent series has only finitely many nonzero ... 4 Lemma. Every holomorphic function on a compact Riemann surface is constant. Proof. Let f:X \to Y be a nonconstant holomorphic mapping between (connected) Riemann surfaces, with X compact. Then f(X) is compact, therefore closed. But it is also open by the open mapping theorem. Therefore by connectedness Y = f(X), and Y is also compact. As \mathbb{... 1 As mentioned in the comments, you may simply differentiate componentwise. Alternatively, given any time-dependent k-form \omega_t on a manifold M, note that at each point p \in M, \omega_t(p) may be viewed as a curve \mathbb{R} \to \bigwedge^k(T_p^*M) in a finite-dimensional vector space, defined by t \mapsto \omega_t(p). There is a canonical ... 1 \def\L{\mathcal L}From what you've written there is no dependence of the Hamiltonian on time (indeed a time variable is not introduced at all), so the interpretation in the comments doesn't feel right. Without more context, I would assume that by (d/dt)\omega the author means the Lie derivative \L_X \omega where$$X = \sum_j \left(\nabla_{p_j} H \frac{... 1 You're wrong to say that$dx\,dy-dy\,dx=0$. You need to put tensor products into everything and then the definition of wedge product is precisely$dx\wedge dy = dx\otimes dy-dy\otimes dx$. (Some people will put a factor of$1/2$there, but I don't.) 2 For what follows, I will denote with$p$the points of$P$and with$u$the points of$U_i$, otherwise I will get confused. But be careful that what you have written in the question (which probably comes from the book) is the other way around! For example,$\sigma_i$is a function on$U_i$, so it takes$u$as an argument, and not$p\$...and so on! So, we ...

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I speculate your problem is that you didn't take advantage of the fact that on spheres centered on the origin, we have an identity $$\sum_i x_i^2 + y_i^2 = r^2$$ and consequently $$\sum_i 2 x_i \mathrm{d} x_i + 2 y_i \mathrm{d} y_i = 0$$ is a linear dependence between the differential forms involved.

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