Tag Info

New answers tagged

1

Another approach, ultimately equivalent to Jyrki's but from a different point of view, proceeds by first choosing arbitrarily a non-zero 2-form at the identity element (i.e., a vector in $\bigwedge^2T_e^*$) and then translating that vector to all other points by means of left-translations, so as to enforce left-invariance. (A different initial choice would ...


1

The element, call it $g(a,b)$, of your Lie group can be viewed as the matrix $$ g(a,b)=\left(\begin{array}{rr}a&b\\0&1\end{array}\right) $$ acting on column vectors $(x,1)^T$. We see that the matrix of $1$-forms $$ ...


1

Every differential form $\theta$ of degree n-1 has a corresponding 1 form $\eta$ with relation $\theta=i_\eta\omega$, it is called Hodge dual of $\theta$. To see this, assume $\theta\wedge\rho=\lambda\omega$, then $\theta$ can be seen as a linear functional $\theta:\Omega^1(M)\to\mathbb R$ by $\rho\mapsto \lambda$. By Rietz theorem, there exists a 1-form ...


1

It is the case. One way to see it is dimension count, spaces of vectors and $n-1$ exterior forms have the same dimension, and $v\mapsto i_v\omega$ is linear and injective, hence surjective, on every tangent space. Inverses to a smooth family of invertible linear maps also form a smooth family.


3

You can't differentiate a pointwise expression. In particular, $\lambda$ is not the pullback of a form on $M$.


0

Since $v^{2}:= \iota_{\vec{v}}vol^{3}$ the association explicitly depends on a choice of the volume form. Frankel probably doesn't add "pseudo" to forms in $\mathbb{R}^3$ because he thinks of it as having a fixed coordinate system, which determines the standard volume form. Riemannian metric plus orientation will also determine a volume form, as will a ...


1

Just write down the parametric form: $$ p=(x,y,z)=(\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta) $$ Now the two derivations (tangent vectors) can be computed by: $$ \begin{split} \partial/\partial\theta&=(\cos\theta\cos\phi,\cos\theta\sin\phi,-\sin\theta)\\ \partial/\partial\phi&=(-\sin\theta\sin\phi,\sin\theta\cos\phi,0) \end{split} $$ The only ...


0

A $1$-form is just a smooth choice of a linear functional on the tangent space at every point in the manifold. Note that every linear functional is alternating, but it doesn't really mean anything, since all permutations on a set of one vector are trivial. Likewise, the derivative of a function is a linear functional on every tangent space. For $v\in T_pM$, ...


2

Well, if those sets are pairwais disjoint, then everything splits as a direct sum: the $k$-forms $$\Omega^k(\cup_i U_i)\simeq\oplus_i\Omega^k(U_i)$$ is given by $\omega\mapsto (\omega|_{U_i})_i$ and the De Rham differential $d$ respects this splitting. So, both the kernel and image of $d$ respect this splitting as well.. Note, however, that for an infinite ...


4

As a definition, you should take $\Omega^k (U)$ to be zero for all $k$ not in $\{0, 1, \dots, n\}$, where $n$ is the dimension of $U$. So, for example, "$d^{-1}$" is the zero map, and $\text{im} ~ d^{-1} = \{0\} \subset \Omega^0(U)$. Thus $H^0(U) = \text{ker} ~ d_0$ (which consists of locally constant functions). Similarly, $d_n = 0$, so $\text{ker} ~ d_n = ...


2

Yes, your calculations are correct. Two things to remember: the wedge product is $C^\infty(M)$-bilinear rather than just $\Bbb R$-bilinear, and $du \wedge dv = -dv\wedge du$ (or, more generally, if $\omega$ is a k-form and $\mu$ an $l$-form, then $\omega \wedge \mu = (-1)^{kl}\mu \wedge \omega$). Using these, you should be able to simplify your last ...


1

Thank you for your answers and Phillip Andreae has the best answer so far. But I want to answer the question myself because I found an easier and more intuitive way to look at this problem. I just think of differential forms as alternating multilinear functions, for example the determinant ! Let's take a simpler example: $\omega = 3x dx\wedge dy$ $X = ...


2

Interior product is defined like this: if $\omega$ is a $k$-form and $X$ is a vector field, then $\iota_X \omega$ is a $(k-1)$-form defined by (remember, $\iota_X \omega$ "eats" $k-1$ vectors and returns a function): $$ [\iota_X \omega](V_1, V_2, \dots, V_{k-1}) := \omega(X,V_1, V_2, \dots, V_{k-1}) ,$$ i.e., we just put $X$ in the first argument of ...


0

I've found the answer. The point is that the n-form on the n-dimensional hypersurface is said to be a Lerya residue of a (n+1)-form, that is singular on the hypersurface. If we say that on each patch of the projective plane the n-form should be given by the Lerya residue, then the globality condition gives the power of the polynomial, that defines the ...


0

A 2-form eats 2 vectors, not 1. So if $\omega$ is a 2-form and $X$ is a vector field, $i_X\omega$ is not a function, but a 1-form. Alternatively, if $X,Y$ are vector fields, then $\omega(X,Y)$ is a function. More generally, a $k$-form eats $k$ vector fields and returns a function. The most important thing about differential forms is that they are linear and ...


0

If you need to calculate in the local coordinates with a metric, then you need to use the musical isomorphism. You can treat it purely algebraically if you like, but if you want to do calculation in an actual manifold this is needed.


2

Let $M$ be an n-manifold with boundary. Then $\partial M$ is the boundary of the manifold. The interior of $M$ is the set of points in $M$ such that they have a neighborhood which is homeomorphic to an open subset in $\mathbb{R}^n$. $M - IntM = \partial M$ If $M$ is a manifold without boundary, then $\partial M = \emptyset$ For example, consider the unit ...


1

I think one reason is a lot of basic differential geometry constructions are anti-symmetric by nature. For example, no matter which connection you choose, the curvature is always defined by $$ K(X,Y)Z=\nabla_{X}\nabla_{Y}(Z)-\nabla_{Y}\nabla_{X}(Z)-\nabla_{[X,Y]}(Z) $$ and it is clear from the definition that $K(X,Y)(Z)=-K(Y,X)Z$. So the curvature is a so ...



Top 50 recent answers are included