Exterior Derivatives via Linear Algebra

Mathematics/Geometry/All notes11 Aug 2026

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1. The Start

This note starts with the exterior derivative in Euclidean space, which is the following.

Example 1.1. Derive the 2-form of the external differential by defining it in Euclidean space.

Solution. Let 𝜔​=𝑃​​d𝑥​+𝑄​​d𝑦​+𝑅​​d𝑧, we deduce that

d𝑃=𝜕𝑃𝑥d𝑥+𝜕𝑃𝑦d𝑦+𝜕𝑃𝑧d𝑧d𝑃=𝜕𝑃𝑥d𝑥+𝜕𝑃𝑦d𝑦+𝜕𝑃𝑧d𝑧d𝑃=𝜕𝑃𝑥d𝑥+𝜕𝑃𝑦d𝑦+𝜕𝑃𝑧d𝑧d𝑄=𝜕𝑄𝑥d𝑥+𝜕𝑄𝑦d𝑦+𝜕𝑄𝑧d𝑧d𝑄=𝜕𝑄𝑥d𝑥+𝜕𝑄𝑦d𝑦+𝜕𝑄𝑧d𝑧d𝑄=𝜕𝑄𝑥d𝑥+𝜕𝑄𝑦d𝑦+𝜕𝑄𝑧d𝑧d𝑅=𝜕𝑅𝑥d𝑥+𝜕𝑅𝑦d𝑦+𝜕𝑅𝑧d𝑧d𝑅=𝜕𝑅𝑥d𝑥+𝜕𝑅𝑦d𝑦+𝜕𝑅𝑧d𝑧d𝑅=𝜕𝑅𝑥d𝑥+𝜕𝑅𝑦d𝑦+𝜕𝑅𝑧d𝑧d𝜔=d𝑃∧d𝑥+d𝑄∧d𝑦+d𝑅∧d𝑧d𝜔=d𝑃∧d𝑥+d𝑄∧d𝑦+d𝑅∧d𝑧d𝜔=d𝑃∧d𝑥+d𝑄∧d𝑦+d𝑅∧d𝑧

. Expand item by item to obtain

d𝑃∧d𝑥=𝜕𝑃𝑦d𝑦∧d𝑥+𝜕𝑃𝑧d𝑧∧d𝑥d𝑃∧d𝑥=𝜕𝑃𝑦d𝑦∧d𝑥+𝜕𝑃𝑧d𝑧∧d𝑥d𝑃∧d𝑥=𝜕𝑃𝑦d𝑦∧d𝑥+𝜕𝑃𝑧d𝑧∧d𝑥d𝑄∧d𝑦=𝜕𝑄𝑥d𝑥∧d𝑦+𝜕𝑄𝑧d𝑧∧d𝑦d𝑄∧d𝑦=𝜕𝑄𝑥d𝑥∧d𝑦+𝜕𝑄𝑧d𝑧∧d𝑦d𝑄∧d𝑦=𝜕𝑄𝑥d𝑥∧d𝑦+𝜕𝑄𝑧d𝑧∧d𝑦d𝑅∧d𝑧=𝜕𝑅𝑥d𝑥∧d𝑧+𝜕𝑅𝑦d𝑦∧d𝑧d𝑅∧d𝑧=𝜕𝑅𝑥d𝑥∧d𝑧+𝜕𝑅𝑦d𝑦∧d𝑧d𝑅∧d𝑧=𝜕𝑅𝑥d𝑥∧d𝑧+𝜕𝑅𝑦d𝑦∧d𝑧

. Substitute them into the expression of d𝜔d𝜔, we obtain that

d𝜔=(𝜕𝑅𝑦−𝜕𝑄𝑧)d𝑦∧d𝑧+(𝜕𝑃𝑧−𝜕𝑅𝑥)d𝑧∧d𝑥+(𝜕𝑄𝑥−𝜕𝑃𝑦)d𝑥∧d𝑦d𝜔=(𝜕𝑅𝑦−𝜕𝑄𝑧)d𝑦∧d𝑧+(𝜕𝑃𝑧−𝜕𝑅𝑥)d𝑧∧d𝑥+(𝜕𝑄𝑥−𝜕𝑃𝑦)d𝑥∧d𝑦d𝜔=(𝜕𝑅𝑦−𝜕𝑄𝑧)d𝑦∧d𝑧+(𝜕𝑃𝑧−𝜕𝑅𝑥)d𝑧∧d𝑥+(𝜕𝑄𝑥−𝜕𝑃𝑦)d𝑥∧d𝑦d𝜔=(𝜕𝑅𝑦−𝜕𝑄𝑧)d𝑦∧d𝑧+(𝜕𝑃𝑧−𝜕𝑅𝑥)d𝑧∧d𝑥+(𝜕𝑄𝑥−𝜕𝑃𝑦)d𝑥∧d𝑦

. And we are done.

2. The Climax

But this solution is extremely ugly. We obtain that 𝜔​=(𝑃𝑄𝑅)(d𝑥d𝑦d𝑧). So is there a way to solve the problem through linear algebra? Let’s have a try.

Solution. We obtain that for any function 𝑓𝑓, d𝑓​=(𝜕𝑓𝑥𝜕𝑓𝑦𝜕𝑓𝑧)(d𝑥d𝑦d𝑧). Let 𝜔​=(𝑃𝑄𝑅)(d𝑥d𝑦d𝑧), then d𝜔​=(d𝑃d𝑄d𝑅)(d𝑥d𝑦d𝑧).

Obtain that (d𝑃d𝑄d𝑅)​=(d𝑥d𝑦d𝑧)(𝜕𝑃𝑥𝜕𝑃𝑦𝜕𝑃𝑧𝜕𝑄𝑥𝜕𝑄𝑦𝜕𝑄𝑧𝜕𝑅𝑥𝜕𝑅𝑦𝜕𝑅𝑧). Substitute this in, thus, d𝜔​=(d𝑥d𝑦d𝑧)(𝜕𝑃𝑥𝜕𝑃𝑦𝜕𝑃𝑧𝜕𝑄𝑥𝜕𝑄𝑦𝜕𝑄𝑧𝜕𝑅𝑥𝜕𝑅𝑦𝜕𝑅𝑧)(d𝑥d𝑦d𝑧).

Let ⃗𝑣​=(d𝑥d𝑦d𝑧), 𝑨​=(𝜕𝑃𝑥𝜕𝑃𝑦𝜕𝑃𝑧𝜕𝑄𝑥𝜕𝑄𝑦𝜕𝑄𝑧𝜕𝑅𝑥𝜕𝑅𝑦𝜕𝑅𝑧), then d𝜔​=⃗𝑣T𝑨⃗𝑣.

Here, we may need the following lemma.

Lemma 2.1. Let 𝑅𝑅 be an exchange ring, 𝐴𝐴 is combination algebra on 𝑅𝑅. Let ⃗𝑣​∈𝐴𝑛, for any matrix 𝑺​=(𝑠𝑖𝑗)​∈𝑀𝑛(𝑅), If 𝑄​=⃗𝑣T𝑺⃗𝑣, then 𝑄​=∑𝑛𝑖=1∑𝑛𝑗=1𝑠𝑖𝑗𝑣𝑖𝑣𝑗.

Since ∀​​d𝑥,​​d𝑦,​​d𝑥​∧​​d𝑥​=0,​​d𝑥​∧​​d𝑦​=​−​​d𝑦​∧​​d𝑥, after applying Lemma 2.1, we obtain that:

d𝜔=(𝜕𝑅𝑦−𝜕𝑄𝑧)d𝑦∧d𝑧+(𝜕𝑃𝑧−𝜕𝑅𝑥)d𝑧∧d𝑥+(𝜕𝑄𝑥−𝜕𝑃𝑦)d𝑥∧d𝑦d𝜔=(𝜕𝑅𝑦−𝜕𝑄𝑧)d𝑦∧d𝑧+(𝜕𝑃𝑧−𝜕𝑅𝑥)d𝑧∧d𝑥+(𝜕𝑄𝑥−𝜕𝑃𝑦)d𝑥∧d𝑦d𝜔=(𝜕𝑅𝑦−𝜕𝑄𝑧)d𝑦∧d𝑧+(𝜕𝑃𝑧−𝜕𝑅𝑥)d𝑧∧d𝑥+(𝜕𝑄𝑥−𝜕𝑃𝑦)d𝑥∧d𝑦d𝜔=(𝜕𝑅𝑦−𝜕𝑄𝑧)d𝑦∧d𝑧+(𝜕𝑃𝑧−𝜕𝑅𝑥)d𝑧∧d𝑥+(𝜕𝑄𝑥−𝜕𝑃𝑦)d𝑥∧d𝑦

. And we are done.

3. The End

Now, the only work is to prove Lemma 2.1, and here it is.

Proof. What are you looking at? It’s just some dumb, straightforward calculation — do it.

□

View Typst source

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#show: tylenotes.with(
  title: "Exterior Derivatives via Linear Algebra",
  date: "2026-08-11",
  tags: ("geometry", "lean"),
  summary: "Deriving the exterior derivative of a 1-form as a 2-form in Euclidean space.",
)

= The Start

This note starts with the exterior derivative in Euclidean space,
which is the following.

#example[
  Derive the 2-form of the external differential by defining it in Euclidean space.
]

#solution[
  Let $omega = P dif x + Q dif y + R dif z$,
  we deduce that
  $ dif P = partial P_x dif x + partial P_y dif y + partial P_z dif z $
  $ dif Q = partial Q_x dif x + partial Q_y dif y + partial Q_z dif z $
  $ dif R = partial R_x dif x + partial R_y dif y + partial R_z dif z $
  $ dif omega = dif P and dif x + dif Q and dif y + dif R and dif z $.
  Expand item by item to obtain
  $ dif P and dif x = partial P_y dif y and dif x + partial P_z dif z and dif x $
  $ dif Q and dif y = partial Q_x dif x and dif y + partial Q_z dif z and dif y $
  $ dif R and dif z = partial R_x dif x and dif z + partial R_y dif y and dif z $.
  Substitute them into the expression of $dif omega$, we obtain that
  $ dif omega = 
  (partial R_y - partial Q_z) dif y and dif z +  
  (partial P_z - partial R_x) dif z and dif x +
  (partial Q_x - partial P_y) dif x and dif y $.
  And we are done.
]

= The Climax

But this solution is extremely ugly.
We obtain that $omega = mat(P, Q, R) vec(dif x, dif y, dif z) $.
So is there a way to solve the problem through linear algebra?
Let's have a try.

#solution[
  We obtain that for any function $f$,
  $dif f = mat(partial f_x, partial f_y, partial f_z) vec(dif x, dif y, dif z)$.
  Let $omega = mat(P, Q, R) vec(dif x, dif y, dif z)$,
  then $dif omega = mat(dif P, dif Q, dif R) vec(dif x, dif y, dif z)$.

  Obtain that $mat(dif P, dif Q, dif R) = mat(dif x, dif y, dif z)
  mat(
    partial P_x, partial P_y, partial P_z;
    partial Q_x, partial Q_y, partial Q_z;
    partial R_x, partial R_y, partial R_z;
  )$.
  Substitute this in, thus,
  $dif omega = mat(dif x, dif y, dif z) mat(
    partial P_x, partial P_y, partial P_z;
    partial Q_x, partial Q_y, partial Q_z;
    partial R_x, partial R_y, partial R_z;
  ) vec(dif x, dif y, dif z)$.

  Let $arrow(v) = vec(dif x, dif y, dif z)$, $bold(A) = mat(
    partial P_x, partial P_y, partial P_z;
    partial Q_x, partial Q_y, partial Q_z;
    partial R_x, partial R_y, partial R_z;
  )$, then $dif omega = arrow(v)^"T" bold(A) arrow(v)$.

  Here, we may need the following lemma.

  #lemma[
    Let $R$ be an exchange ring, $A$ is combination algebra on $R$.
    Let $arrow(v) in A^n$, for any matrix $bold(S) = (s_(i j)) in M_n (R)$,
    If $Q = arrow(v)^"T" bold(S) arrow(v)$, then
    $Q = sum^n_(i=1) sum^n_(j=1) s_(i j) v_i v_j$.
  ]

  Since $forall dif x, dif y, dif x and dif x = 0, dif x and dif y = - dif y and dif x$,
  after applying *Lemma 2.1*, we obtain that:
  $ dif omega = 
  (partial R_y - partial Q_z) dif y and dif z +  
  (partial P_z - partial R_x) dif z and dif x +
  (partial Q_x - partial P_y) dif x and dif y $.
  And we are done.
]

= The End

Now, the only work is to prove *Lemma 2.1*, and here it is.

#proof[
  What are you looking at? It's just some dumb, straightforward calculation — do it.
]