# Tag Info

3

The condition of being a lagrangian submanifold $L\subset M$ means that for every point $p\in L$ and every pair of tangent vectors $X,Y\in T_pL$, $$\omega_p(X,Y)=0\,.$$ This does not violate the non-degeneracy of $\omega$, since $T_pL$ is merely a subspace of $T_pM$. However, non-degeneracy implies that any submanifold satisfying this condition is of ...

2

In particular, I read that $L\subset M$ is lagrangian iff the symplectic form field $\omega(x)$ evaluated on every point $p\in L$ gives zero. More precisely, the restriction of $\omega$ to $T_p L$ is zero. As an element of $\wedge^2 T_p M$, $\omega(p)$ is nonzero and nondegenerate. For an example, consider $\mathbb{R}^2$ with the symplectic form ...

2

Define the characteristic vector (fields) of a form as $\iota_v\omega=0$. By the properties of the interior product if both $v_1$ and $v_2$ are c.v., then so are their linear combinations, $\text{span}(v_1,v_2)$. Furthermore, if the rank of the form $r$, the number of linearly independent characteristic vectors is $n-r$. One can describe a tangent plane to ...

3

These identities can be found using the formulae $(i_Y\alpha)(V_1, \dots, V_k) = \alpha(Y, V_1, \dots, V_k)$ and $$(\mathcal{L}_X\alpha)(V_1, \dots, V_k) = X(\alpha(V_1, \dots, V_k)) - \sum_{i=1}^k\alpha(V_1, \dots, V_{i-1}, [X, V_i], V_{i+1}, \dots, V_k).$$ First of all, we have \begin{align*} & (\mathcal{L}_Xi_Y\alpha)(V_1, \dots, V_k)\\ =&\ ...

1

I've done this question before. I can't remember how to do part (b) right now, but I have included an argument for part (a). Fix $p \in M$ and let $\hat{x}^1, \dots, \hat{x}^n$ be local coordinates in a neighbourhood $U$ of $p$. Then we have $$\phi|_U = f(\hat{x}^1, \dots, \hat{x}^n)d\hat{x}^1\wedge\dots\wedge d\hat{x}^n$$ for some smooth $f$. Let ...

0

You might try Sternberg's Curvature in Mathematics and Physics (Sternberg) or Chern's Lectures on Differential Geometry (Chern). Both are very reasonably priced and make extensive use of moving frames.

2

But, being a beginner in differential forms on smooth manifolds: can we extend to a basis globally? In general, no. On parallelizable manifolds, you can extend to a global basis in the sense that $(\omega(p),\omega_2(p),\dotsc,\omega_k(p))$ are a basis of the cotangent space at every point, but not on other manifolds. Or does this suffices locally? ...

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No: $M=\mathbb R^2,\: \omega=xd y$ Edit In the first version of the question "non vanishing" was in brackets and I interpreted the question as "not identically zero but maybe having zeros", as is indeed the case in my example..

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math.n00b's answer is very good and detailed, I just wanted to add a general observation that pulling back differential forms expressed in the basis involving $dy_i$ is done simply by substituting $y_i=f_i(x_j)$ under differentials and using the chain rule to arrive at a form expressed in $dx_i$. For example, $f^*{dy_i}=\dfrac{\partial f_i}{\partial ... 0 Either$\omega\wedge \omega=0$or$d\omega\wedge d\omega=0$, because firstly, the exterior product is graded anticommutative:$\alpha\wedge\beta=(-1)^{|\alpha||\beta|}\beta\wedge\alpha$, where$|\alpha|$denotes the degree of$\alpha$. It is 0 for odd-degree forms. And secondly, the exterior derivative increments the degree. 1 You just made a silly mistake in the last line of computing$i_{\mathbb{X}_t}\beta$and grabbed the wrong coefficient. You should have$2xy$in the numerator of the first factor instead of$x^2-y^2$. Then what you get after the pullback is easily integrated. 2 Now apply the magic formula again: $$L_X(\eta)=i_Xd\eta+di_X\eta$$ where$\eta=i_Yd\omega+di_Y\omega$and get $$L_X(L_Y\omega)=i_Xd(i_Yd\omega+di_Y\omega)+di_X(i_Yd\omega+di_Y\omega)$$ use that$d$is linear $$i_Xd(i_Yd\omega)+i_X(di_Y\omega)+di_X(i_Yd\omega)+di_X(di_Y\omega)$$ Subtract$L_Y L_X\omegaand compare with $$... 1 I don't know where your error is, but using the exterior derivative is simpler:$$ \begin{align} d\beta &= d\left(\sum_{i=1}^{n}(-1)^{i-1}\frac{x^i dx^1 \wedge dx^2 \wedge \dots \wedge \hat{dx^i} \wedge \dots \wedge dx^n}{((x^1)^2+\dots+(x^n)^2)^p}\right) \\ % &= \sum_{i=1}^{n}(-1)^{i-1} \sum_{k=1}^n\frac{\partial}{\partial x^k}\frac{x^i ... 2 If\omega$and$\tau$are closed forms (which you have not used), then $$d(\omega \wedge \tau) = d\omega \wedge \tau + (-1)^{p-1}\omega \wedge d\tau = 0,$$ hence$\omega \wedge \tau$is also closed, hence it corresponds to a class in cohomology. Let us see that this class depends only on the classes of$\omega$and$\tau$. Let us suppose that$\omega = ...

2

Let $\alpha \in E^p$, then we can write $\alpha = H(\alpha) + (\alpha - H(\alpha))$. As $G(\alpha) \in (H^p)^\perp$ satisfies $\Delta G(\alpha) = \alpha - H(\alpha)$, we have $$||G(\alpha)|| \leq C ||\Delta G(\alpha)|| = C ||\alpha - H(\alpha)|| \leq C||\alpha||,$$ thus $G$ is bounded. To show that it is self-adjoint, you need to restrict $G$ to ...

1

I'm not sure how you are getting the second equality and I don't think it is true. However, the Lie derivative of a function is just the directional derivative so that $L_X(x)=X(x)$, which looks silly. But you have $X=y\frac{\partial}{\partial x}+z\frac{\partial}{\partial z}$, so that $X(x)=y$. Anyway, a simpler way to solve this problem would be to use ...

1

Note, or recall, that $X(x) = \frac{d}{dt}|_{t=0}\Phi_t(x).$ That's a basic fact about the correspondence between vector fields and flows. Now, put $\Gamma_t := G \circ \Phi_t \circ G^{-1}.$ Then, according to part (a) of the exercise, $\Gamma_t$ is a flow. Denote the vector field corresponding to $\Gamma_t$ by $Y.$ From definitions and our constructions, we ...

3

Expanding on John's comment: the wedge product on differential forms is graded commutative (or skew commutative), meaning that for $\alpha, \beta \in \Omega^*(M)$, you have: $$\beta \wedge \alpha = (-1)^{|\alpha| |\beta|} \alpha \wedge \beta,$$ where $|\alpha|$, $|\beta|$ are the respective degrees of $\alpha$ and $\beta$. In particular if $\omega$ is a ...

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First, I will answer your question about the equality you quote from G&H; in the meanwhile, it could happen that I also answer your two general questions about currents. The setting is the following: we are considering currents in the product space $M\times M$, where $M$ is a $n$-dimensional complex manifold; $\Delta$ is the diagonal, i.e. $$\Delta=\{(x,x)\ ... 0 This isn't antisymmetric in u,v, the problem comes from the even terms. So this doesn't define a 2-form. You could however define the exponential of \omega using the standard formula (it's a finite sum in any case)$$\sum_{i=0}^\infty \frac{\omega^n}{n!}=1+\omega+\frac{\omega^2}{2}+\cdots+\frac{\omega^p}{p!}$$where p is defined so that 2p is the ... 4 This is basically a question of how one converts back and forth between vector calculus and differential forms notation. I suggest using an intermediary notation--the notation of clifford algebra and geometric calculus--which has the simplicity of vector calculus notation and the power of differential forms operations. Let's do it starting with ... 3 A differential 2-form can be understood in terms of integral operators, specifically what happens when you integrate them over (directed) 2-surfaces. Stoke's theorem, the higher-dimensional analog of the fundamental theorem of calculus, tells us that$$ \int_S d\omega = \int_{\partial S} \omega  where $\partial S$ is the boundary of $S$. (if $\omega$ ...

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To develop some intuition, you might find it helpful to go back to the physics that inspired the definitions of grad, div and curl: electromagnetism. There are nice physical reasons why we would expect $d^2$ to be zero on both $0$-forms/scalar fields, $\nabla \times \nabla \phi = \vec{0}$, and $1$-forms/vector fields, $\nabla\cdot\nabla\vec{A} = 0$. ...

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Henri Cartan, one of the greatest 20th century mathematicians wrote: ...utinam intelligere possim ratinationes pulcherrimae quae e propositione concisa DE QUADRATO NIHILO EXAEQUARI fluunt In the unlikely event that you don't speak Latin, o amice, this means: ...if I could only understand the beautiful consequences following from the concise proposition ...

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