observer1 said:
But could you try that explanation again.. but simpler... even if you have to exaggerate... I am looking for a picture of the idea/definition.
Well, I can give it a try although the explanation above has been already rather simplified.
exterior
An algebra is a vector space ##\mathcal{A}## which allows a (interior) multiplication, e.g. ##\mathcal{A} = \{##
analytic functions on ##ℂ\}## with ##f\circ g: z \mapsto f(g(z))##.
The Graßmann or exterior algebra ##\Lambda(\mathcal{A}) ## of ##\mathcal{A}## is a vector space of countably many copies of ##\mathcal{A}## together with certain rules (identifications). The exterior multiplication of ##\mathcal{A}## is now the (interior) multiplication in ##\Lambda(\mathcal{A})##. It is between, e.g. an element in ##n## copies of ##\mathcal{A}## with an element in ##m## copies of ##\mathcal{A}##. A differential form can be defined as a alternating, multilinear mapping from such an exterior algebra to the space of smooth functions. Let me quote an example from Wikipedia (
translated by me):
Consider ##ℝ^3## with cartesian coordinates ##(x,y,z)##, the 1-form ##ω = z^2 dx +2ydy+xzdz## and the 2-form ##ν=zdz ∧ dx##.
The
exterior product is thus
##ω ∧ ν = z^3 dx ∧ dz ∧ dx + 2yz dy ∧ dz ∧ dx + xz^2 dz ∧ dz∧ dx = 2yz dx ∧ dy ∧ dz##
where I applied some commutation rules mentioned above
and the
exterior derivation is
\begin{align*}
dω &= d(z^2 dx +2ydy+xzdz) \\
&= (\frac{\partial}{\partial x} z^2 dx + \frac{\partial}{\partial y} z^2 dy + \frac{\partial}{\partial z} z^2 dz) ∧ dx + (\frac{\partial}{\partial x} 2y dx + \frac{\partial}{\partial y} 2y dy + \frac{\partial}{\partial z} 2y dz) ∧ dy + (\frac{\partial}{\partial x} xz dx + \frac{\partial}{\partial y} xz dy + \frac{\partial}{\partial z} xz dz) ∧ dz \\
&= 2z dz ∧ dx + 2 dy ∧ dy + (z dx + xdz) ∧ dz \\
&= 2z dz ∧ dx - z dz ∧ dx \\ &= z dz ∧ dx \\& =ν
\end{align*}
##ν## is exact because it is the image of ##ω## under the boundary, resp. differential operator ##d##.
##ν## is also closed because ##dν=0##.
exact (##ν = dω##)
Boundary operators ##d## always satisfy ##d^2 = d \circ d = 0##. Therefore it's always ## \mathcal{im} \, d ⊆ \mathcal{ker}\, d## and ##\{0\} → \mathcal{im} \, d \stackrel{φ}{→} \mathcal{ker}\, d \stackrel{ψ}{→} \mathcal{ker}\, d / \mathcal{im} \, d → \{0\} ## is a short
exact sequence, i.e. ##\mathcal{im} \, φ = \mathcal{ker} \, ψ##. (I don't know why they first called such a sequence exact. Perhaps because it is not only ##\mathcal{im} \, φ ⊆ \mathcal{ker} \, ψ## but exactly ##\mathcal{im} \, φ = \mathcal{ker} \, ψ## which is an important property of such sequences.) You could say as well that a exact form is a boundary (of another form).
(... continued example. Only to show how those abstract concepts fit in situations we know.)
Let ##γ: [0,1] → ℝ^3## be a parametrization ##γ(t) = (t^2,2t,1)## of a curve in ##ℝ^3##.
Thus we have for ##x = t^2 \; , \; y=2t \; , \; z=1## for the
pullback ##γ^*##
\begin{align*}
γ^*ω &= ω(dγ)\\ &= (z^2 dx + 2y dy + xz dz)dγ\\
& = (1^2dx + 2\cdot 2t dy + t^2 \cdot 1 dz)dγ\\& =\left(1^2 \frac{d(t^2)}{dt} + 2(2t) \frac{d(2t)}{dt} + t^2 \cdot 1 \frac{d(1)}{dt}\right)\,dt \\&= 2t dt + 4t \cdot 2 dt + 0 \\&= 10t \, dt
\end{align*}
For the integral of ##ω## along the curve ##Γ = γ([0,1]) ⊆ ℝ^3## we thus get ##\int_Γ ω = \int_{[0,1]} \gamma^*ω = \int_0^1 10t \, dt = 5##.
closed (##dν = 0##)
Let ##M## be a compact differential manifold. Let further ##M## be closed, that is its boundary is empty. (Not to be confused with the eventual boundaries which ##M## might have embedded in an ambient Euclidean space. A closed ball or sphere alone has no boundary. Think of the universe when people say it might be closed without a boundary.)
Then we have ##\int_M ν = \int_M dω \stackrel{Stokes}{=} \int_{\delta M} ω = \int_∅ ω = 0##.
I added this because it's the most elegant statement of Stokes' theorem which is about boundaries. Remember that you've been told you that Stokes' theorem and the chain rule are the most important facts in calculus!
You could say as well that a closed form is a cycle, a
closed path. Going once around it and you will end up at the start. (But this is a rather stretched picture. However, I guess it is the origin of it.)
I used the term boundary operator for ##d## occasionally to show where the geometrical term 'closed' comes from.
