synthetic differential geometry
Introductions
from point-set topology to differentiable manifolds
geometry of physics: coordinate systems, smooth spaces, manifolds, smooth homotopy types, supergeometry
Differentials
Tangency
The magic algebraic facts
Theorems
Axiomatics
(shape modality $\dashv$ flat modality $\dashv$ sharp modality)
$(ʃ \dashv \flat \dashv \sharp )$
dR-shape modality$\dashv$ dR-flat modality
$ʃ_{dR} \dashv \flat_{dR}$
(reduction modality $\dashv$ infinitesimal shape modality $\dashv$ infinitesimal flat modality)
$(\Re \dashv \Im \dashv \&)$
fermionic modality$\dashv$ bosonic modality $\dashv$ rheonomy modality
$(\rightrightarrows \dashv \rightsquigarrow \dashv Rh)$
Models
Models for Smooth Infinitesimal Analysis
smooth algebra ($C^\infty$-ring)
differential equations, variational calculus
Euler-Lagrange equation, de Donder-Weyl formalism?,
Chern-Weil theory, ∞-Chern-Weil theory
Cartan geometry (super, higher)
In Riemannian geometry, the curl or rotation of a vector field $v$ on an oriented $3$-dimensional Riemannian manifold $(X,g)$ is the vector field denoted $curl(v)$ (or $rot(v)$ or $\Del \times v$) defined by
where
$\Gamma(T X) \underoverset{\underset{g}{\longrightarrow}}{\overset{g^{-1}}{\longleftarrow}}{\phantom{AA}\simeq\phantom{AA}} \Omega^1(X)$ is the linear isomorphism between vector fields and differential 1-forms given by the metric tensor $g$;
$d_{dR} \;\colon\; \Omega^n(X) \longrightarrow \Omega^{n+1}(X)$ is the de Rham differential
$\star_g \;\colon\; \Omega^n(X) \to \Omega^{dim(X)-n}(X)$ is the Hodge star operator (which uses the orientation of $X$).
Notice that for this to make sense it is crucial that the dimension of $X$ is $3$, for only then is the Hodge dual of the de Rham differential of a $1$-form again a $1$-form; that is, $n = 3$ is the unique solution of $n - (1 + 1) = 1$.
Alternatively, the curl/rotation of a vector field $\vec v$ at some point $x\in X$ may be defined as the limit integral formula
where $D$ runs over the smooth oriented surfaces (submanifolds of dimension $2$) containing the point $x$ and with smooth boundary $\partial D$, $\vec{n}$ is the unit vector through the surface $S$, and $\vec{t}$ is the unit vector tangent to the curve $\partial S$. (We use the orientation of $X$ to determine the direction of $\vec{n}$ from the orientation of $S$.)
This formula does not depend on the shape of boundaries taken in limiting process, so one can typically take a coordinate chart and discs with decreasing radius in this particular coordinate chart. One can even use $\pi r^2$ in place of the actual area of the disc around $x$ with coordinate-radius $r$, to save on calculating this area, as long as the coordinate chart assigns the standard coordinates to the metric at $x$.
The proof that this definition is coherent and agrees with the previous one is essentially the Kelvin–Stokes Theorem; see below for discussion.
More generally, if $(X,g)$ is a Riemannian manifold whose cotangent spaces (equivalently, tangent spaces) are smoothly equipped with a binary cross product $⨉\colon \Omega^2(X;\mathbb{R}) \to \Omega^1(X;\mathbb{R})$, then the curl of any vector field $v$ is
However, this is not as general as it may appear:
That said, there are also cross products of other arity? in other dimensions; using essentially the same formula, we can take the curl of a $k$-multivector field if we have a smooth $(k+1)$-ary cross product. Or if the cross product is other than vector-valued, then we can obtain a curl that is other than a vector field.
In particular, in $2$ dimensions, we have the scalar curl
where $⨉\colon \Omega^2(X;\mathbb{R}) \to \Omega^0(X;\mathbb{R})$ is the volume form (or area form) on the $2$-dimensional Riemannian manifold $X$.
If $(X,g)$ is $\mathbb{R}^3$ endowed with the canonical Euclidean metric, then the curl of a vector field $(v^1,v^2,v^3) = v^1\partial_1 + v^2\partial_2 + v^3\partial_3$ is
This is the classical curl from vector analysis?.
If $(X,g)$ is $\mathbb{R}^2$ endowed with the canonical Euclidean metric, then the curl of a vector field $(v^1,v^2) = v^1\partial_1 + v^2\partial_2$ is
Recall that if $X$ is an $n$-dimensional differentiable manifold, $D$ is a $p$-dimensional submanifold with boundary, and $\alpha$ is a differentiable $(p-1)$-rank exterior differential form on a neighbourhood of $D$ in $X$, then the generalized Stokes Theorem says that the integral of $\alpha$ on the boundary $\partial{D}$ equals the integral on $S$ of the de Rham differential $\mathrm{d}_{DR}\alpha$.
When $n=3$, $p=2$, and $X$ is equipped with an orientation and a metric, then this is equivalent to saying that the integral of a vector field $v$ along the boundary of an oriented surface $D$ in $X$ is equal to the integral of the vector field's curl across the surface:
(In particular, when $X$ is $\mathbb{R}^3$ with its standard orientation and metric, then this is equivalent to the classical Kelvin–Stokes Theorem.) At least, this is what it says if the curl is defined in terms of differential forms; if the curl is defined via integration instead, then (1) is immediate, and the Kelvin–Stokes Theorem says that this definition matches the other one.
When $n=2$, $p=2$, and $X$ is equipped with an orientation and a metric, then the Stokes Theorem is equivalent to saying that the integral of a vector field $v$ along a simple closed curve in $X$ is equal to the integral of the vector field's scalar curl on the region $D$ enclosed by the curve:
(In particular, when $X$ is $\mathbb{R}^2$ with its standard orientation and metric, then this is equivalent to the curl-circulation form of Green's Theorem.)
In many classical applications of the curl in vector analysis?, the Riemannian structure is actually irrelevant, and the gradient can be replaced with the deRham differential $d_{dR}$. That is, $X$ is treated as the $1$-form $g(X)$, its curl is treated as the $2$-form $d_{dR} g(X)$, and once these identifications are made there is no need to involve $g$ or $X$ directly at all.
See also
Last revised on September 17, 2018 at 02:14:29. See the history of this page for a list of all contributions to it.