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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{...


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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}...


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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 $...


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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 ...


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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{(...


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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 $V$ $C\ell(Q)=$ 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 ...


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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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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 ...


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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)}\...


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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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Let $x\in M$, let $\gamma : [0, t] \to M$ be $\gamma(s) = \exp_x (sX)$ for some $X \in T_xM$, with $\|X\| =1$. Then $\ell (\gamma) = t$. Thus $$ \int_0^t \omega_{ \gamma(s)} (\gamma'(s)) ds = t$$ for all $t$. Differentiating gives $$1 = \omega_{\gamma(t)} (\gamma'(t))$$ for all $t$. Putting $t=0$ gives $1 = \omega_x (X)$. But $X \in T_pM$ is ...


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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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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 ...


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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( ...


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. $v = v^r e_r + v^\theta e_\theta + v^z e_z$,...


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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.


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$x$ 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 ...


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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+...


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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 ...


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$\def\L{\mathcal L}$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{...


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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.)



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