Exterior Derivatives via Linear Algebra

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

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

Solution

Let $𝜔=𝑃𝑑𝑥+𝑄𝑑𝑦+𝑅𝑑𝑧$, we deduce that

$$𝑑𝑃=𝜕{𝑃}_{𝑥}𝑑𝑥+𝜕{𝑃}_{𝑦}𝑑𝑦+𝜕{𝑃}_{𝑧}𝑑𝑧$$$$𝑑𝑄=𝜕{𝑄}_{𝑥}𝑑𝑥+𝜕{𝑄}_{𝑦}𝑑𝑦+𝜕{𝑄}_{𝑧}𝑑𝑧$$$$𝑑𝑅=𝜕{𝑅}_{𝑥}𝑑𝑥+𝜕{𝑅}_{𝑦}𝑑𝑦+𝜕{𝑅}_{𝑧}𝑑𝑧$$$$𝑑𝜔=𝑑𝑃\wedge 𝑑𝑥+𝑑𝑄\wedge 𝑑𝑦+𝑑𝑅\wedge 𝑑𝑧$$

. Expand item by item to obtain

$$𝑑𝑃\wedge 𝑑𝑥=𝜕{𝑃}_{𝑦}𝑑𝑦\wedge 𝑑𝑥+𝜕{𝑃}_{𝑧}𝑑𝑧\wedge 𝑑𝑥$$$$𝑑𝑄\wedge 𝑑𝑦=𝜕{𝑄}_{𝑥}𝑑𝑥\wedge 𝑑𝑦+𝜕{𝑄}_{𝑧}𝑑𝑧\wedge 𝑑𝑦$$$$𝑑𝑅\wedge 𝑑𝑧=𝜕{𝑅}_{𝑥}𝑑𝑥\wedge 𝑑𝑧+𝜕{𝑅}_{𝑦}𝑑𝑦\wedge 𝑑𝑧$$

. Substitute them into the expression of $𝑑𝜔$, we obtain that

$$𝑑𝜔=\left(𝜕{𝑅}_{𝑦}-𝜕{𝑄}_{𝑧}\right)𝑑𝑦\wedge 𝑑𝑧+\left(𝜕{𝑃}_{𝑧}-𝜕{𝑅}_{𝑥}\right)𝑑𝑧\wedge 𝑑𝑥+\left(𝜕{𝑄}_{𝑥}-𝜕{𝑃}_{𝑦}\right)𝑑𝑥\wedge 𝑑𝑦$$

. And we are done.

2. The Climax

But this solution is extremely ugly. We obtain that $𝜔=\left(\begin{array}{ccc}𝑃& 𝑄& 𝑅\end{array}\right)\left(\begin{array}{c}𝑑𝑥\\ 𝑑𝑦\\ 𝑑𝑧\end{array}\right)$. So is there a way to solve the problem through linear algebra? Let’s have a try.

Solution

We obtain that for any function $𝑓$, $𝑑𝑓=\left(\begin{array}{ccc}𝜕{𝑓}_{𝑥}& 𝜕{𝑓}_{𝑦}& 𝜕{𝑓}_{𝑧}\end{array}\right)\left(\begin{array}{c}𝑑𝑥\\ 𝑑𝑦\\ 𝑑𝑧\end{array}\right)$. Let $𝜔=\left(\begin{array}{ccc}𝑃& 𝑄& 𝑅\end{array}\right)\left(\begin{array}{c}𝑑𝑥\\ 𝑑𝑦\\ 𝑑𝑧\end{array}\right)$, then $𝑑𝜔=\left(\begin{array}{ccc}𝑑𝑃& 𝑑𝑄& 𝑑𝑅\end{array}\right)\left(\begin{array}{c}𝑑𝑥\\ 𝑑𝑦\\ 𝑑𝑧\end{array}\right)$.

Obtain that $\left(\begin{array}{ccc}𝑑𝑃& 𝑑𝑄& 𝑑𝑅\end{array}\right)=\left(\begin{array}{ccc}𝑑𝑥& 𝑑𝑦& 𝑑𝑧\end{array}\right)\left(\begin{array}{ccc}𝜕{𝑃}_{𝑥}& 𝜕{𝑃}_{𝑦}& 𝜕{𝑃}_{𝑧}\\ 𝜕{𝑄}_{𝑥}& 𝜕{𝑄}_{𝑦}& 𝜕{𝑄}_{𝑧}\\ 𝜕{𝑅}_{𝑥}& 𝜕{𝑅}_{𝑦}& 𝜕{𝑅}_{𝑧}\end{array}\right)$. Substitute this in, thus, $𝑑𝜔=\left(\begin{array}{ccc}𝑑𝑥& 𝑑𝑦& 𝑑𝑧\end{array}\right)\left(\begin{array}{ccc}𝜕{𝑃}_{𝑥}& 𝜕{𝑃}_{𝑦}& 𝜕{𝑃}_{𝑧}\\ 𝜕{𝑄}_{𝑥}& 𝜕{𝑄}_{𝑦}& 𝜕{𝑄}_{𝑧}\\ 𝜕{𝑅}_{𝑥}& 𝜕{𝑅}_{𝑦}& 𝜕{𝑅}_{𝑧}\end{array}\right)\left(\begin{array}{c}𝑑𝑥\\ 𝑑𝑦\\ 𝑑𝑧\end{array}\right)$.

Let $\overrightarrow{𝑣}=\left(\begin{array}{c}𝑑𝑥\\ 𝑑𝑦\\ 𝑑𝑧\end{array}\right)$, $𝑨=\left(\begin{array}{ccc}𝜕{𝑃}_{𝑥}& 𝜕{𝑃}_{𝑦}& 𝜕{𝑃}_{𝑧}\\ 𝜕{𝑄}_{𝑥}& 𝜕{𝑄}_{𝑦}& 𝜕{𝑄}_{𝑧}\\ 𝜕{𝑅}_{𝑥}& 𝜕{𝑅}_{𝑦}& 𝜕{𝑅}_{𝑧}\end{array}\right)$, then $𝑑𝜔={\overrightarrow{𝑣}}^{𝑇}𝑨\overrightarrow{𝑣}$.

Here, we may need the following lemma.

Lemma 2.1

Let $𝑅$ be an exchange ring, $𝐴$ is combination algebra on $𝑅$. Let $\overrightarrow{𝑣}\in {𝐴}^{𝑛}$, for any matrix $𝑺=\left({𝑠}_{𝑖𝑗}\right)\in {𝑀}_{𝑛}\left(𝑅\right)$, If $𝑄={\overrightarrow{𝑣}}^{𝑇}𝑺\overrightarrow{𝑣}$, then $𝑄={\sum }_{𝑖=1}^{𝑛}{\sum }_{𝑗=1}^{𝑛}{𝑠}_{𝑖𝑗}{𝑣}_{𝑖}{𝑣}_{𝑗}$.

Since $\forall 𝑑𝑥,𝑑𝑦,𝑑𝑥\wedge 𝑑𝑥=0,𝑑𝑥\wedge 𝑑𝑦=-𝑑𝑦\wedge 𝑑𝑥$, after applying Lemma 2.1, we obtain that:

$$𝑑𝜔=\left(𝜕{𝑅}_{𝑦}-𝜕{𝑄}_{𝑧}\right)𝑑𝑦\wedge 𝑑𝑧+\left(𝜕{𝑃}_{𝑧}-𝜕{𝑅}_{𝑥}\right)𝑑𝑧\wedge 𝑑𝑥+\left(𝜕{𝑄}_{𝑥}-𝜕{𝑃}_{𝑦}\right)𝑑𝑥\wedge 𝑑𝑦$$

. 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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#import "/themes/site/notes.typ": *

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

= The Climax

But this solution is extremely ugly.
We obtain that $omega = mat(P, Q, R) vec(d x, d y, d 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 $f$,
  $d f = mat(partial f_x, partial f_y, partial f_z) vec(d x, d y, d z)$.
  Let $omega = mat(P, Q, R) vec(d x, d y, d z)$,
  then $d omega = mat(d P, d Q, d R) vec(d x, d y, d z)$.

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

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

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