# A Why the terms - exterior, closed, exact?

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1. May 15, 2016

### observer1

Hi all,

(Thank you for the continuing responses to my other questions...)

I am gaining more and more understanding of differential forms and differential geometry.

But now I must ask... Why the words?

I understand the exterior derivative, but why is it called "exterior?"
Ditto for CLOSED and EXACT forms: I understand the definition, but nothing in that definition informs me of why that word was chosen.

Could anyone elaborate, simply, on the reason for the words: exterior, closed, exact?

2. May 15, 2016

### Staff: Mentor

This comes from its topological, resp. geometrical origin. A closed differential form is a cycle $C$, a exact differential form a boundary $B$ which establishes the short exact sequence: $0 → B → C → C/B → 0$. (A exact form is always closed.) See also Wiki: "Loosely speaking, a cycle is a closed submanifold, a boundary is the boundary of a submanifold with boundary, and a homology class (which represents a hole) is an equivalence class of cycles modulo boundaries." Geometrically it is related to simplicial complexes where the language comes from.

Edit: I omitted the filtration for simplicity.

Edit 2: The Graßmann algebra $Λ V$ is also called the exterior algebra of $V$. It is the structure where exterior derivatives live. It is exterior because a inner product on $V$ would make it an algebra with a binary multiplication $V \times V → V$ whereas the exterior product is between the filtration of $Λ V$, e.g. $a ∧ b = (-1)^{ij} b ∧ a$ for $a \in Λ^i(V)$ and $b \in Λ^j(V)$.

Last edited: May 15, 2016
3. May 15, 2016

### observer1

I am still learning. Almost at the point of understanding some of what you wrote.

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.

In other words: these terms are often defined right at the start of an introduction to differential forms...
That means if they are using a word (exterior, closed, open) right there at the start, they should give a reason for the word, based on the minimal knowledge a first time reader has.

You have given me the reason one would give AFTER understanding it.

4. May 15, 2016

### lavinia

5. May 15, 2016

### Staff: Mentor

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
$\\ dω \\ = d(z^2 dx +2ydy+xzdz) \\ = (\frac{\partial}{\partial x} z^2 dx + \frac{\partial}{\partial y} z^2 dy + \frac{\partial}{\partial z} z^2 dz) ∧ dx + (\frac{\partial}{\partial x} 2y dx + \frac{\partial}{\partial y} 2y dy + \frac{\partial}{\partial z} 2y dz) ∧ dy + (\frac{\partial}{\partial x} xz dx + \frac{\partial}{\partial y} xz dy + \frac{\partial}{\partial z} xz dz) ∧ dz \\ = 2z dz ∧ dx + 2 dy ∧ dy + (z dx + xdz) ∧ dz \\ = 2z dz ∧ dx - z dz ∧ dx \\ = z dz ∧ dx \\ =ν$

$ν$ 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 $γ^*$

$γ^*ω = ω(dγ) = (z^2 dx + 2y dy + xz dz)dγ = (1^2dx + 2\cdot 2t dy + t^2 \cdot 1 dz)dγ =\left(1^2 \frac{d(t^2)}{dt} + 2(2t) \frac{d(2t)}{dt} + t^2 \cdot 1 \frac{d(1)}{dt}\right)\,dt = 2t dt + 4t \cdot 2 dt + 0 = 10t \, dt$
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.

Last edited: May 4, 2018