Differential Forms January 8, 2025
We express differential forms in terms of tensors and index notation. The
closest treatment seems to be Winitzki (2009)
It seems the usual old-fashioned index-based tensor formalism described in
Čertík, O. et al. (2025)
Any vector u \mathbf{u} u
u = u i e ⃗ i . \mathbf{u}=u^i \vec{e}_i\,. u = u i e i . Any form p \mathbf{p} p
p = p i e ~ i . \mathbf{p}=p_i \tilde{e}^i\,. p = p i e ~ i . The basis vectors are e ⃗ i \vec{e}_i e i e ~ i \tilde{e}^i e ~ i
Any rank 2 tensor A \mathbf{A} A
A = A i j e ⃗ i ⊗ e ⃗ j . \mathbf{A}=A^{ij} \vec{e}_i \otimes \vec{e}_j\,. A = A ij e i ⊗ e j . Any rank 3 tensor A \mathbf{A} A
A = A i j k e ⃗ i ⊗ e ⃗ j ⊗ e ⃗ k . \mathbf{A}=A^{ijk} \vec{e}_i \otimes \vec{e}_j \otimes \vec{e}_k\,. A = A ijk e i ⊗ e j ⊗ e k . The tensor basis is e ⃗ i ⊗ e ⃗ j \vec{e}_i \otimes \vec{e}_j e i ⊗ e j e ⃗ i ⊗ e ⃗ j ⊗ e ⃗ k \vec{e}_i \otimes \vec{e}_j \otimes \vec{e}_k e i ⊗ e j ⊗ e k
A = A i j e ⃗ i ⊗ e ~ j . \mathbf{A}=A^i{}_{j} \vec{e}_i \otimes \tilde{e}^j\,. A = A i j e i ⊗ e ~ j . The vector basis and the dual (form) basis is related by a dot product:
e ~ i ⋅ e ⃗ j = δ i j . \tilde{e}^i \cdot \vec{e}_j = \delta^i{}_j\,. e ~ i ⋅ e j = δ i j . If the vector basis is a coordinate basis, then it can be written as:
e ⃗ i = ∂ r ∂ x i ≡ ∂ ∂ x i , \vec{e}_i = {\partial\mathbf{r}\over\partial x^i} \equiv
{\partial\over\partial x^i}\,, e i = ∂ x i ∂ r ≡ ∂ x i ∂ , where the position vector r \mathbf{r} r
e ~ i ≡ d x i . \tilde{e}^i \equiv dx^i\,. e ~ i ≡ d x i . The form can be said to act on a vector to produce a number (definition of a
dual space) as follows using a dot product:
p ( u ) ≡ p ⋅ u = p i e ~ i ⋅ u j e ⃗ j = p i u j e ~ i ⋅ e ⃗ j = p i u j δ i j = p i u i . \mathbf{p}(\mathbf{u})\equiv\mathbf{p} \cdot \mathbf{u}
=p_i \tilde{e}^i \cdot u^j \vec{e}_j
=p_i u^j \tilde{e}^i \cdot \vec{e}_j
=p_i u^j \delta^i{}_j
=p_i u^i\,. p ( u ) ≡ p ⋅ u = p i e ~ i ⋅ u j e j = p i u j e ~ i ⋅ e j = p i u j δ i j = p i u i . We will not use the p ( u ) \mathbf{p}(\mathbf{u}) p ( u ) p i u i p_i u^i p i u i
Index Notation ¶ We omit writing the basis vectors, so equations from the previous sections
would be just u i u^i u i p i p_i p i A i j A^{ij} A ij A i j k A^{ijk} A ijk A i j A^i{}_j A i j p i u i p_i u^i p i u i
The upper and lower indices are treated equally, they are just contravariant
and covariant components of the same tensor.
Antisymmetric Tensors ¶ An antisymmetric part of a tensor can be written as follows for rank 2:
A [ i j ] = 1 2 ! ε i j ε k l A k l = 1 2 det ( δ i k δ i l δ j k δ j l ) A k l = 1 2 ( δ i k δ j l − δ i l δ j k ) A k l = A_{[ij]} = {1\over2!}\varepsilon_{ij}\varepsilon^{kl} A_{kl}
= {1\over2}\det\begin{pmatrix} \delta_i{}^k & \delta_i{}^l \\ \delta_j{}^k & \delta_j{}^l \end{pmatrix} A_{kl}
= {1\over2}(\delta_i{}^k\delta_j{}^l - \delta_i{}^l \delta_j{}^k) A_{kl}= A [ ij ] = 2 ! 1 ε ij ε k l A k l = 2 1 det ( δ i k δ j k δ i l δ j l ) A k l = 2 1 ( δ i k δ j l − δ i l δ j k ) A k l =
= 1 2 ( A i j − A j i ) . = {1\over2}(A_{ij}-A_{ji})\,. = 2 1 ( A ij − A ji ) . Rank 3:
A [ i j k ] = 1 3 ! ε i j k ε l m n A l m n = 1 3 ! det ( δ i l δ i m δ i n δ j l δ j m δ j n δ k l δ k m δ k n ) A l m n = A_{[ijk]} = {1\over3!}\varepsilon_{ijk}\varepsilon^{lmn} A_{lmn}
= {1\over3!}\det\begin{pmatrix}
\delta_i{}^l & \delta_i{}^m & \delta_i{}^n \\
\delta_j{}^l & \delta_j{}^m & \delta_j{}^n \\
\delta_k{}^l & \delta_k{}^m & \delta_k{}^n \end{pmatrix} A_{lmn}= A [ ijk ] = 3 ! 1 ε ijk ε l mn A l mn = 3 ! 1 det ⎝ ⎛ δ i l δ j l δ k l δ i m δ j m δ k m δ i n δ j n δ k n ⎠ ⎞ A l mn =
= 1 3 ! ( δ i l δ j m δ k n − δ i m δ j l δ k n + ⋯ ) A l m n = = {1\over3!}(\delta_i{}^l\delta_j{}^m\delta_k{}^n - \delta_i{}^m\delta_j{}^l\delta_k{}^n + \cdots) A_{lmn}= = 3 ! 1 ( δ i l δ j m δ k n − δ i m δ j l δ k n + ⋯ ) A l mn =
= 1 6 ( A i j k − A j i k + ⋯ − ⋯ + ⋯ ) . = {1\over6}(A_{ijk}-A_{jik}+\cdots - \cdots + \cdots)\,. = 6 1 ( A ijk − A jik + ⋯ − ⋯ + ⋯ ) . Rank 4:
A [ i j k l ] = 1 4 ! ε i j k l ε m n o p A m n o p = 1 4 ! det ( δ i m δ i n δ i o δ i p δ j m δ j n δ j o δ j p δ k m δ k n δ k o δ k p δ l m δ l n δ l o δ l p ) A m n o p = A_{[ijkl]} = {1\over4!}\varepsilon_{ijkl}\varepsilon^{mnop} A_{mnop}
= {1\over4!}\det\begin{pmatrix}
\delta_i{}^m & \delta_i{}^n & \delta_i{}^o & \delta_i{}^p \\
\delta_j{}^m & \delta_j{}^n & \delta_j{}^o & \delta_j{}^p \\
\delta_k{}^m & \delta_k{}^n & \delta_k{}^o & \delta_k{}^p \\
\delta_l{}^m & \delta_l{}^n & \delta_l{}^o & \delta_l{}^p \end{pmatrix} A_{mnop}= A [ ijk l ] = 4 ! 1 ε ijk l ε mn o p A mn o p = 4 ! 1 det ⎝ ⎛ δ i m δ j m δ k m δ l m δ i n δ j n δ k n δ l n δ i o δ j o δ k o δ l o δ i p δ j p δ k p δ l p ⎠ ⎞ A mn o p =
= 1 4 ! ( δ i m δ j n δ k o δ l p − δ i n δ j m δ k o δ l p + ⋯ ) A m n o p = = {1\over4!}(\delta_i{}^m\delta_j{}^n\delta_k{}^o\delta_l{}^p - \delta_i{}^n\delta_j{}^m\delta_k{}^o\delta_l{}^p + \cdots) A_{mnop}= = 4 ! 1 ( δ i m δ j n δ k o δ l p − δ i n δ j m δ k o δ l p + ⋯ ) A mn o p =
= 1 24 ( A i j k l − A j i k l + ⋯ ) . = {1\over24}(A_{ijkl}-A_{jikl}+\cdots)\,. = 24 1 ( A ijk l − A jik l + ⋯ ) . The fraction at the beginning tells us how many terms there are in the final
answer: 2, 6, 24 for ranks 2, 3, 4.
The above expressions work for any dimension, except the Levi-Civita expression
that only works for rank equal to dimension (e.g., rank 2, dimension 2). For
dimension higher than the rank, one can skip the Levi-Civita expression and go
straight to the determinant, which is correct in any dimension.
There is a dimension-dependent expression using Levi-Civita symbols that is
equal to the determinant. For example for rank 3 and dimension 4, the
expression is ε i j k s ε s l m n \varepsilon_{ijks}\varepsilon^{slmn} ε ijk s ε s l mn
Exterior (Wedge) Product ¶ The exterior (wedge) product of two forms u = u i e ~ i \mathbf{u}=u_i \tilde{e}^i u = u i e ~ i v = v i e ~ i \mathbf{v}=v_i \tilde{e}^i v = v i e ~ i
u ∧ v ≡ u ⊗ v − v ⊗ u . \mathbf{u} \wedge \mathbf{v} \equiv
\mathbf{u} \otimes \mathbf{v} - \mathbf{v} \otimes \mathbf{u}\,. u ∧ v ≡ u ⊗ v − v ⊗ u . Expressing this in components in terms of the basis:
u ∧ v = u i e ~ i ∧ v j e ~ j = u ⊗ v − v ⊗ u = u i e ~ i ⊗ v j e ~ j − v i e ~ i ⊗ u j e ~ j = \mathbf{u} \wedge \mathbf{v}
= u_i \tilde{e}^i \wedge v_j \tilde{e}^j
=\mathbf{u} \otimes \mathbf{v} - \mathbf{v} \otimes \mathbf{u}
= u_i \tilde{e}^i \otimes v_j \tilde{e}^j
- v_i \tilde{e}^i \otimes u_j \tilde{e}^j = u ∧ v = u i e ~ i ∧ v j e ~ j = u ⊗ v − v ⊗ u = u i e ~ i ⊗ v j e ~ j − v i e ~ i ⊗ u j e ~ j = = u i v j e ~ i ⊗ e ~ j − v i u j e ~ i ⊗ e ~ j = = u_i v_j \tilde{e}^i \otimes \tilde{e}^j
- v_i u_j \tilde{e}^i \otimes \tilde{e}^j = = u i v j e ~ i ⊗ e ~ j − v i u j e ~ i ⊗ e ~ j = = ( u i v j − u j v i ) e ~ i ⊗ e ~ j = = (u_i v_j-u_j v_i) \tilde{e}^i \otimes \tilde{e}^j = = ( u i v j − u j v i ) e ~ i ⊗ e ~ j = = det ( u i u j v i v j ) e ~ i ⊗ e ~ j = A i j e ~ i ⊗ e ~ j . = \det\begin{pmatrix} u_i & u_j \\ v_i & v_j \end{pmatrix}
\tilde{e}^i \otimes \tilde{e}^j
= A_{ij}\tilde{e}^i \otimes \tilde{e}^j\,. = det ( u i v i u j v j ) e ~ i ⊗ e ~ j = A ij e ~ i ⊗ e ~ j . The tensor A i j A_{ij} A ij
A i j e ~ i ⊗ e ~ j = A i j 1 2 ( e ~ i ⊗ e ~ j + e ~ j ⊗ e ~ i ) + A i j 1 2 ( e ~ i ⊗ e ~ j − e ~ j ⊗ e ~ i ) = A_{ij}\tilde{e}^i \otimes \tilde{e}^j
=A_{ij}{1\over2}(\tilde{e}^i \otimes \tilde{e}^j+\tilde{e}^j \otimes \tilde{e}^i)
+A_{ij}{1\over2}(\tilde{e}^i \otimes \tilde{e}^j-\tilde{e}^j \otimes \tilde{e}^i)= A ij e ~ i ⊗ e ~ j = A ij 2 1 ( e ~ i ⊗ e ~ j + e ~ j ⊗ e ~ i ) + A ij 2 1 ( e ~ i ⊗ e ~ j − e ~ j ⊗ e ~ i ) =
= A i j 1 2 ( e ~ i ⊗ e ~ j + e ~ j ⊗ e ~ i ) + A i j 1 2 e ~ i ∧ e ~ j = 1 2 A i j e ~ i ∧ e ~ j . =A_{ij}{1\over2}(\tilde{e}^i \otimes \tilde{e}^j+\tilde{e}^j \otimes \tilde{e}^i)
+A_{ij}{1\over2} \tilde{e}^i \wedge \tilde{e}^j
={1\over2} A_{ij} \tilde{e}^i \wedge \tilde{e}^j\,. = A ij 2 1 ( e ~ i ⊗ e ~ j + e ~ j ⊗ e ~ i ) + A ij 2 1 e ~ i ∧ e ~ j = 2 1 A ij e ~ i ∧ e ~ j . The left term contains an antisymmetric tensor
A i j = u i v j − u j v i A_{ij}=u_i v_j-u_j v_i A ij = u i v j − u j v i
u i e ~ i ∧ v j e ~ j = ( u i v j − u j v i ) e ~ i ⊗ e ~ j = 1 2 ( u i v j − u j v i ) e ~ i ∧ e ~ j . u_i \tilde{e}^i \wedge v_j \tilde{e}^j
=(u_i v_j-u_j v_i) \tilde{e}^i \otimes \tilde{e}^j
= {1\over2} (u_i v_j-u_j v_i) \tilde{e}^i \wedge \tilde{e}^j\,. u i e ~ i ∧ v j e ~ j = ( u i v j − u j v i ) e ~ i ⊗ e ~ j = 2 1 ( u i v j − u j v i ) e ~ i ∧ e ~ j . The equation
A i j e ~ i ⊗ e ~ j = 1 2 A i j e ~ i ∧ e ~ j A_{ij}\tilde{e}^i \otimes \tilde{e}^j
={1\over2} A_{ij} \tilde{e}^i \wedge \tilde{e}^j A ij e ~ i ⊗ e ~ j = 2 1 A ij e ~ i ∧ e ~ j shows that any rank 2 antisymmetric tensor can be written in terms of a wedge
product as a differential form. And the definition (18)
Thus differential forms are simply antisymmetric tensors.
The 1 2 {1\over2} 2 1 i = 1 , 2 i=1,2 i = 1 , 2
A i j e ~ i ⊗ e ~ j = A 12 e ~ 1 ⊗ e ~ 2 + A 21 e ~ 2 ⊗ e ~ 1 + A 11 e ~ 1 ⊗ e ~ 1 + A 22 e ~ 2 ⊗ e ~ 2 = A_{ij}\tilde{e}^i \otimes \tilde{e}^j
=A_{12}\tilde{e}^1 \otimes \tilde{e}^2
+A_{21}\tilde{e}^2 \otimes \tilde{e}^1
+A_{11}\tilde{e}^1 \otimes \tilde{e}^1
+A_{22}\tilde{e}^2 \otimes \tilde{e}^2= A ij e ~ i ⊗ e ~ j = A 12 e ~ 1 ⊗ e ~ 2 + A 21 e ~ 2 ⊗ e ~ 1 + A 11 e ~ 1 ⊗ e ~ 1 + A 22 e ~ 2 ⊗ e ~ 2 =
= A 12 e ~ 1 ⊗ e ~ 2 − A 12 e ~ 2 ⊗ e ~ 1 = A 12 ( e ~ 1 ⊗ e ~ 2 − e ~ 2 ⊗ e ~ 1 ) = A 12 ( e ~ 1 ∧ e ~ 2 ) . =A_{12}\tilde{e}^1 \otimes \tilde{e}^2
-A_{12}\tilde{e}^2 \otimes \tilde{e}^1
=A_{12}(\tilde{e}^1 \otimes \tilde{e}^2 -\tilde{e}^2 \otimes \tilde{e}^1)
=A_{12}(\tilde{e}^1 \wedge \tilde{e}^2)\,. = A 12 e ~ 1 ⊗ e ~ 2 − A 12 e ~ 2 ⊗ e ~ 1 = A 12 ( e ~ 1 ⊗ e ~ 2 − e ~ 2 ⊗ e ~ 1 ) = A 12 ( e ~ 1 ∧ e ~ 2 ) . Above we used the fact that A 11 = A 22 = 0 A_{11}=A_{22}=0 A 11 = A 22 = 0 A 21 = − A 12 A_{21}=-A_{12} A 21 = − A 12
1 2 A i j e ~ i ∧ e ~ j = 1 2 ( A 12 e ~ 1 ∧ e ~ 2 + A 21 e ~ 2 ∧ e ~ 1 + A 11 e ~ 1 ∧ e ~ 1 + A 22 e ~ 2 ∧ e ~ 2 ) = {1\over2}A_{ij}\tilde{e}^i \wedge \tilde{e}^j
={1\over2}\left(A_{12}\tilde{e}^1 \wedge \tilde{e}^2
+A_{21}\tilde{e}^2 \wedge \tilde{e}^1
+A_{11}\tilde{e}^1 \wedge \tilde{e}^1
+A_{22}\tilde{e}^2 \wedge \tilde{e}^2\right)= 2 1 A ij e ~ i ∧ e ~ j = 2 1 ( A 12 e ~ 1 ∧ e ~ 2 + A 21 e ~ 2 ∧ e ~ 1 + A 11 e ~ 1 ∧ e ~ 1 + A 22 e ~ 2 ∧ e ~ 2 ) =
= 1 2 ( A 12 e ~ 1 ∧ e ~ 2 − A 12 e ~ 2 ∧ e ~ 1 ) = 1 2 A 12 ( e ~ 1 ∧ e ~ 2 − e ~ 2 ∧ e ~ 1 ) = ={1\over2}\left(A_{12}\tilde{e}^1 \wedge \tilde{e}^2
-A_{12}\tilde{e}^2 \wedge \tilde{e}^1\right)
={1\over2}A_{12}(\tilde{e}^1 \wedge \tilde{e}^2 -\tilde{e}^2 \wedge \tilde{e}^1)= = 2 1 ( A 12 e ~ 1 ∧ e ~ 2 − A 12 e ~ 2 ∧ e ~ 1 ) = 2 1 A 12 ( e ~ 1 ∧ e ~ 2 − e ~ 2 ∧ e ~ 1 ) =
= 1 2 A 12 ( e ~ 1 ∧ e ~ 2 + e ~ 1 ∧ e ~ 2 ) = A 12 ( e ~ 1 ∧ e ~ 2 ) . ={1\over2}A_{12}(\tilde{e}^1 \wedge \tilde{e}^2 +\tilde{e}^1 \wedge \tilde{e}^2)
=A_{12}(\tilde{e}^1 \wedge \tilde{e}^2)\,. = 2 1 A 12 ( e ~ 1 ∧ e ~ 2 + e ~ 1 ∧ e ~ 2 ) = A 12 ( e ~ 1 ∧ e ~ 2 ) . The final answer is exactly the same in both cases:
A 12 ( e ~ 1 ∧ e ~ 2 ) A_{12}(\tilde{e}^1 \wedge \tilde{e}^2) A 12 ( e ~ 1 ∧ e ~ 2 ) e ~ 1 ∧ e ~ 2 \tilde{e}^1 \wedge \tilde{e}^2 e ~ 1 ∧ e ~ 2 − e ~ 2 ∧ e ~ 1 -\tilde{e}^2
\wedge \tilde{e}^1 − e ~ 2 ∧ e ~ 1 1 2 {1\over2} 2 1 e ~ i ⊗ e ~ j \tilde{e}^i \otimes \tilde{e}^j e ~ i ⊗ e ~ j
e ~ 1 ∧ e ~ 2 = 1 2 ( e ~ 1 ∧ e ~ 2 − e ~ 2 ∧ e ~ 1 ) = ( e ~ 1 ⊗ e ~ 2 − e ~ 2 ⊗ e ~ 1 ) . \tilde{e}^1 \wedge \tilde{e}^2
={1\over2}(\tilde{e}^1 \wedge \tilde{e}^2 -\tilde{e}^2 \wedge \tilde{e}^1)
=(\tilde{e}^1 \otimes \tilde{e}^2 -\tilde{e}^2 \otimes \tilde{e}^1)\,. e ~ 1 ∧ e ~ 2 = 2 1 ( e ~ 1 ∧ e ~ 2 − e ~ 2 ∧ e ~ 1 ) = ( e ~ 1 ⊗ e ~ 2 − e ~ 2 ⊗ e ~ 1 ) . Each wedge product contains two tensor products, thus requiring the
1 2 {1\over2} 2 1 1 2 {1\over2} 2 1
A i j = 1 2 ( A i j − A j i ) A_{ij} = {1\over2}(A_{ij}-A_{ji}) A ij = 2 1 ( A ij − A ji ) For an antisymmetric tensor A i j A_{ij} A ij
Thus we get:
u i e ~ i ∧ v j e ~ j = det ( u i u j v i v j ) e ~ i ⊗ e ~ j = A i j e ~ i ⊗ e ~ j = 1 2 A i j e ~ i ∧ e ~ j . u_i \tilde{e}^i \wedge v_j \tilde{e}^j =
\det\begin{pmatrix} u_i & u_j \\ v_i & v_j \end{pmatrix}
\tilde{e}^i \otimes \tilde{e}^j
= A_{ij}\tilde{e}^i \otimes \tilde{e}^j
= {1\over2}A_{ij}\tilde{e}^i \wedge \tilde{e}^j\,. u i e ~ i ∧ v j e ~ j = det ( u i v i u j v j ) e ~ i ⊗ e ~ j = A ij e ~ i ⊗ e ~ j = 2 1 A ij e ~ i ∧ e ~ j . The wedge product of three 1-forms is defined using:
u ∧ v ∧ w ≡ u ⊗ v ⊗ w − v ⊗ u ⊗ w + w ⊗ u ⊗ v − w ⊗ v ⊗ u + \mathbf{u} \wedge \mathbf{v} \wedge \mathbf{w} \equiv
\mathbf{u} \otimes \mathbf{v} \otimes \mathbf{w}
-\mathbf{v} \otimes \mathbf{u} \otimes \mathbf{w}
+\mathbf{w} \otimes \mathbf{u} \otimes \mathbf{v}
-\mathbf{w} \otimes \mathbf{v} \otimes \mathbf{u}+ u ∧ v ∧ w ≡ u ⊗ v ⊗ w − v ⊗ u ⊗ w + w ⊗ u ⊗ v − w ⊗ v ⊗ u +
+ v ⊗ w ⊗ u − u ⊗ w ⊗ v . +\mathbf{v} \otimes \mathbf{w} \otimes \mathbf{u}
-\mathbf{u} \otimes \mathbf{w} \otimes \mathbf{v}
\,. + v ⊗ w ⊗ u − u ⊗ w ⊗ v . Expressing this in components we get a rank 3 antisymmetric tensor:
u i e ~ i ∧ v j e ~ j ∧ w k e ~ k = det ( u i u j u k v i v j v k w i w j w k ) e ~ i ⊗ e ~ j ⊗ e ~ k = A i j k e ~ i ⊗ e ~ j ⊗ e ~ k . u_i \tilde{e}^i \wedge v_j \tilde{e}^j \wedge w_k\tilde{e}^k =
\det\begin{pmatrix} u_i & u_j & u_k \\ v_i & v_j & v_k \\ w_i & w_j & w_k \end{pmatrix}
\tilde{e}^i \otimes \tilde{e}^j \otimes \tilde{e}^k
=A_{ijk}\tilde{e}^i \otimes \tilde{e}^j \otimes \tilde{e}^k\,. u i e ~ i ∧ v j e ~ j ∧ w k e ~ k = det ⎝ ⎛ u i v i w i u j v j w j u k v k w k ⎠ ⎞ e ~ i ⊗ e ~ j ⊗ e ~ k = A ijk e ~ i ⊗ e ~ j ⊗ e ~ k . Likely this is true:
A i j k e ~ i ⊗ e ~ j ⊗ e ~ k = 1 3 ! A i j k e ~ i ∧ e ~ j ∧ e ~ k . A_{ijk}\tilde{e}^i \otimes \tilde{e}^j \otimes \tilde{e}^k
={1\over 3!}A_{ijk}\tilde{e}^i \wedge \tilde{e}^j \wedge \tilde{e}^k\,. A ijk e ~ i ⊗ e ~ j ⊗ e ~ k = 3 ! 1 A ijk e ~ i ∧ e ~ j ∧ e ~ k . But we need to prove it.
Higher-order products are defined in analogous manner (n n n n n n
A i j = det ( u i u j v i v j ) = u i v j − u j v i A_{ij}= \det\begin{pmatrix} u_i & u_j \\ v_i & v_j \end{pmatrix}
= u_i v_j - u_j v_i A ij = det ( u i v i u j v j ) = u i v j − u j v i and
A i j k = det ( u i u j u k v i v j v k w i w j w k ) . A_{ijk} = \det\begin{pmatrix} u_i & u_j & u_k \\ v_i & v_j & v_k \\ w_i & w_j & w_k \end{pmatrix}\,. A ijk = det ⎝ ⎛ u i v i w i u j v j w j u k v k w k ⎠ ⎞ . The wedge product is equal to the tensor written
using the tensor basis in equations (34) (37) (39) (40)
Using properties of a determinant, we can see that the wedge product is just a
fully antisymmetric tensor.
Show that in 2D and 3D the wedge product is the oriented area and volume.
Ideally using tensors first, since we only want to use that, and then show that
the wedge product when written using tensors gives the area and volume exactly.
Examples ¶ Example: d x ∧ d z dx \wedge dz d x ∧ d z ¶ We can now use the definitions to compute various special cases. For example if
we have the exterior product of two forms d x dx d x d z dz d z d x i = δ 1 i dx_i=\delta_{1i} d x i = δ 1 i d z i = δ 3 i dz_i=\delta_{3i} d z i = δ 3 i
d x ∧ d z = det ( δ 1 i δ 1 j δ 3 i δ 3 j ) e ~ i ⊗ e ~ j = e ~ 1 ⊗ e ~ 3 − e ~ 3 ⊗ e ~ 1 , dx \wedge dz =
\det\begin{pmatrix} \delta_{1i} & \delta_{1j} \\ \delta_{3i} & \delta_{3j} \end{pmatrix}
\tilde{e}^i \otimes \tilde{e}^j
= \tilde{e}^1 \otimes \tilde{e}^3 - \tilde{e}^3 \otimes \tilde{e}^1 \,, d x ∧ d z = det ( δ 1 i δ 3 i δ 1 j δ 3 j ) e ~ i ⊗ e ~ j = e ~ 1 ⊗ e ~ 3 − e ~ 3 ⊗ e ~ 1 , thus the components of this rank 2 tensor are:
A i j = det ( δ 1 i δ 1 j δ 3 i δ 3 j ) = δ 1 i δ 3 j − δ 1 j δ 3 i = ( 0 0 1 0 0 0 − 1 0 0 ) . A_{ij}=\det\begin{pmatrix} \delta_{1i} & \delta_{1j} \\ \delta_{3i} & \delta_{3j} \end{pmatrix}=\delta_{1i}\delta_{3j}-\delta_{1j}\delta_{3i}
=\begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ -1 & 0 & 0 \end{pmatrix} \,. A ij = det ( δ 1 i δ 3 i δ 1 j δ 3 j ) = δ 1 i δ 3 j − δ 1 j δ 3 i = ⎝ ⎛ 0 0 − 1 0 0 0 1 0 0 ⎠ ⎞ . Example: e ~ 1 ∧ e ~ 3 \tilde{e}^1 \wedge \tilde{e}^3 e ~ 1 ∧ e ~ 3 ¶ The exterior product of two basis forms e ~ 1 \tilde{e}^1 e ~ 1 e ~ 3 \tilde{e}^3 e ~ 3
e ~ 1 ∧ e ~ 3 = det ( δ 1 i δ 1 j δ 3 i δ 3 j ) e ~ i ⊗ e ~ j = e ~ 1 ⊗ e ~ 3 − e ~ 3 ⊗ e ~ 1 , \tilde{e}^1 \wedge \tilde{e}^3 =
\det\begin{pmatrix} \delta_{1i} & \delta_{1j} \\ \delta_{3i} & \delta_{3j} \end{pmatrix}
\tilde{e}^i \otimes \tilde{e}^j
= \tilde{e}^1 \otimes \tilde{e}^3 - \tilde{e}^3 \otimes \tilde{e}^1 \,, e ~ 1 ∧ e ~ 3 = det ( δ 1 i δ 3 i δ 1 j δ 3 j ) e ~ i ⊗ e ~ j = e ~ 1 ⊗ e ~ 3 − e ~ 3 ⊗ e ~ 1 , thus the components of this rank 2 tensor are:
A i j = det ( δ 1 i δ 1 j δ 3 i δ 3 j ) = δ 1 i δ 3 j − δ 1 j δ 3 i = ( 0 0 1 0 0 0 − 1 0 0 ) . A_{ij}=\det\begin{pmatrix} \delta_{1i} & \delta_{1j} \\ \delta_{3i} & \delta_{3j} \end{pmatrix}=\delta_{1i}\delta_{3j}-\delta_{1j}\delta_{3i}
=\begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ -1 & 0 & 0 \end{pmatrix} \,. A ij = det ( δ 1 i δ 3 i δ 1 j δ 3 j ) = δ 1 i δ 3 j − δ 1 j δ 3 i = ⎝ ⎛ 0 0 − 1 0 0 0 1 0 0 ⎠ ⎞ . We can see that e ~ 1 ∧ e ~ 3 = e ~ 1 ⊗ e ~ 3 − e ~ 3 ⊗ e ~ 1 \tilde{e}^1 \wedge \tilde{e}^3
= \tilde{e}^1 \otimes \tilde{e}^3 - \tilde{e}^3 \otimes \tilde{e}^1 e ~ 1 ∧ e ~ 3 = e ~ 1 ⊗ e ~ 3 − e ~ 3 ⊗ e ~ 1
Example: d x 1 ∧ d x 2 ∧ d x 4 dx^1 \wedge dx^2 \wedge dx^4 d x 1 ∧ d x 2 ∧ d x 4 ¶ When evaluating the form (composed of basis forms) on vectors u \mathbf{u} u v \mathbf{v} v w \mathbf{w} w
( d x 1 ∧ d x 2 ∧ d x 4 ) ⋅ ( u ⊗ v ⊗ w ) = (dx^1 \wedge dx^2 \wedge dx^4) \cdot
(\mathbf{u} \otimes \mathbf{v} \otimes \mathbf{w})= ( d x 1 ∧ d x 2 ∧ d x 4 ) ⋅ ( u ⊗ v ⊗ w ) =
= det ( δ 1 i δ 1 j δ 1 k δ 2 i δ 2 j δ 2 k δ 4 i δ 4 j δ 4 k ) e ~ i ⊗ e ~ j ⊗ e ~ k ⋅ ( u l v m w n e ⃗ l ⊗ e ⃗ m ⊗ e ⃗ n ) = = \det\begin{pmatrix}
\delta_{1i} & \delta_{1j} & \delta_{1k} \\
\delta_{2i} & \delta_{2j} & \delta_{2k} \\
\delta_{4i} & \delta_{4j} & \delta_{4k} \\
\end{pmatrix}
\tilde{e}^i \otimes \tilde{e}^j \otimes \tilde{e}^k
\cdot (u^l v^m w^n \vec{e}_l \otimes \vec{e}_m \otimes \vec{e}_n)= = det ⎝ ⎛ δ 1 i δ 2 i δ 4 i δ 1 j δ 2 j δ 4 j δ 1 k δ 2 k δ 4 k ⎠ ⎞ e ~ i ⊗ e ~ j ⊗ e ~ k ⋅ ( u l v m w n e l ⊗ e m ⊗ e n ) =
= det ( δ 1 i δ 1 j δ 1 k δ 2 i δ 2 j δ 2 k δ 4 i δ 4 j δ 4 k ) u i v j w k = det ( u 1 v 1 w 1 u 2 v 2 w 2 u 4 v 4 w 4 ) . = \det\begin{pmatrix}
\delta_{1i} & \delta_{1j} & \delta_{1k} \\
\delta_{2i} & \delta_{2j} & \delta_{2k} \\
\delta_{4i} & \delta_{4j} & \delta_{4k} \\
\end{pmatrix}
u^i v^j w^k
= \det\begin{pmatrix}
u_1 & v_1 & w_1 \\
u_2 & v_2 & w_2 \\
u_4 & v_4 & w_4 \\
\end{pmatrix}\,. = det ⎝ ⎛ δ 1 i δ 2 i δ 4 i δ 1 j δ 2 j δ 4 j δ 1 k δ 2 k δ 4 k ⎠ ⎞ u i v j w k = det ⎝ ⎛ u 1 u 2 u 4 v 1 v 2 v 4 w 1 w 2 w 4 ⎠ ⎞ . The form has thus selected the first, second and fourth component row of the
vectors. The result is just a scalar.
Exterior Derivative ¶ The exterior derivative d d d
0-form f f f d f ≡ ∂ f ∂ x i d x i = ∂ i f d x i df \equiv {\partial f\over\partial x^i} dx^i
=\partial_i f \, dx^i df ≡ ∂ x i ∂ f d x i = ∂ i f d x i
k-forms α \alpha α β \beta β a a a b b b d ( a α + b β ) = a d α + b d β d(a\alpha+b\beta) = a d\alpha + b d\beta d ( a α + b β ) = a d α + b d β
k-form α \alpha α β \beta β d ( α ∧ β ) = d α ∧ β + ( − 1 ) k α ∧ d β d(\alpha \wedge \beta) = d\alpha\wedge\beta + (-1)^k \alpha\wedge d\beta d ( α ∧ β ) = d α ∧ β + ( − 1 ) k α ∧ d β
k-form α \alpha α d ( d α ) = 0 d(d\alpha) = 0 d ( d α ) = 0
It turns out in terms of tensors, the exterior derivative is simply a regular
derivative that is made antisymmetric:
0-form: ∂ i f \partial_i f ∂ i f
1-form: ∂ [ i f j ] \partial_{[i} f_{j]} ∂ [ i f j ]
2-form: ∂ [ i A j k ] \partial_{[i} A_{jk]} ∂ [ i A jk ]
3-form: ∂ [ i A j k l ] = 0 \partial_{[i} A_{jkl]} = 0 ∂ [ i A jk l ] = 0
One can use the formulas from the section “Antisymmetric Tensors” above to
evaluate these antisymmetric derivatives. We show examples of this below. We
first compute the exterior derivative using differential forms and the above
rules, and then we do exactly the same operations using tensors.
Input:
f ( x , y ) . f(x, y)\,. f ( x , y ) . Derivative:
d f = ∂ x f d x + ∂ y f d y . df = \partial_x f\, dx + \partial_y f\, dy\,. df = ∂ x f d x + ∂ y f d y . In general, we get:
d f = ∂ i f d x i . df = \partial_i f \, dx^i\,. df = ∂ i f d x i . Input:
ω = f 1 ( x 1 , x 2 ) d x 1 + f 2 ( x 1 , y 2 ) d x 2 . \omega = f_1(x^1,x^2) dx^1 + f_2(x^1,y^2) dx^2\,. ω = f 1 ( x 1 , x 2 ) d x 1 + f 2 ( x 1 , y 2 ) d x 2 . Derivative:
d ω = d f 1 ∧ d x 1 + d f 2 ∧ d x 2 = ∂ i f 1 d x i ∧ d x 1 + ∂ i f 2 d x i ∧ d x 2 = d\omega = d f1 \wedge dx^1 + df_2 \wedge dx^2
= \partial_i f_1 \, dx^i \wedge dx^1 + \partial_i f_2 \, dx^i \wedge dx^2= d ω = df 1 ∧ d x 1 + d f 2 ∧ d x 2 = ∂ i f 1 d x i ∧ d x 1 + ∂ i f 2 d x i ∧ d x 2 =
= ∂ 2 f 1 d x 2 ∧ d x 1 + ∂ 1 f 2 d x 1 ∧ d x 2 = ( ∂ 1 f 2 − ∂ 2 f 1 ) d x 1 ∧ d x 2 . = \partial_2 f_1 \, dx^2 \wedge dx^1 + \partial_1 f_2 \, dx^1 \wedge dx^2
= (\partial_1 f_2 - \partial_2 f_1) dx^1 \wedge dx^2\,. = ∂ 2 f 1 d x 2 ∧ d x 1 + ∂ 1 f 2 d x 1 ∧ d x 2 = ( ∂ 1 f 2 − ∂ 2 f 1 ) d x 1 ∧ d x 2 . In general, the derivative of a 1-form f ~ = f i d x i \tilde f=f_i dx^i f ~ = f i d x i
d f ~ = d ( f i d x i ) = d f i ∧ d x i = ( ∂ j f i d x j ) ∧ d x i = 1 2 ( ∂ j f i − ∂ i f j ) d x j ∧ d x i = d\tilde f
=d(f_i dx^i)
=df_i \wedge dx^i
=(\partial_j f_i dx^j) \wedge dx^i
={1\over2}(\partial_j f_i - \partial_i f_j) dx^j \wedge dx^i= d f ~ = d ( f i d x i ) = d f i ∧ d x i = ( ∂ j f i d x j ) ∧ d x i = 2 1 ( ∂ j f i − ∂ i f j ) d x j ∧ d x i =
= ( ∂ j f i − ∂ i f j ) d x j ⊗ d x i . =(\partial_j f_i - \partial_i f_j) dx^j \otimes dx^i\,. = ( ∂ j f i − ∂ i f j ) d x j ⊗ d x i . So we get:
d f ~ = A i j d x i ⊗ d x j , d\tilde f = A_{ij} dx^i \otimes dx^j\,, d f ~ = A ij d x i ⊗ d x j , where A i j = ∂ i f j − ∂ j f i A_{ij}=\partial_i f_j - \partial_j f_i A ij = ∂ i f j − ∂ j f i
We take a 1-form f j f_j f j ∂ i f j \partial_i f_j ∂ i f j d x i ⊗ d x j − d x j ⊗ d x i = 1 2 d x i ∧ d x j dx^i \otimes dx^j -
dx^j \otimes dx^i={1\over2}dx^i\wedge dx^j d x i ⊗ d x j − d x j ⊗ d x i = 2 1 d x i ∧ d x j ∂ i f j − ∂ j f i \partial_i f_j - \partial_j f_i ∂ i f j − ∂ j f i A i j = ∂ i f j − ∂ j f i A_{ij}=\partial_i f_j - \partial_j f_i A ij = ∂ i f j − ∂ j f i
Let’s now do the same computation using regular tensors:
A i j = ∂ [ i f j ] = 1 2 ε i j ε k l ∂ k f l = 1 2 ( δ i k δ j l − δ i l δ j k ) ∂ k f l = 1 2 ( ∂ i f j − ∂ j f i ) . A_{ij} = \partial_{[i} f_{j]}
= {1\over 2}\varepsilon_{ij}\varepsilon^{kl} \partial_{k} f_{l}
= {1\over 2}(\delta_i{}^k\delta_j{}^l-\delta_i{}^l\delta_j{}^k) \partial_{k} f_{l}
= {1\over 2}(\partial_i f_j - \partial_j f_i)\,. A ij = ∂ [ i f j ] = 2 1 ε ij ε k l ∂ k f l = 2 1 ( δ i k δ j l − δ i l δ j k ) ∂ k f l = 2 1 ( ∂ i f j − ∂ j f i ) . The exterior derivative is simply a regular (not antisymmetric) derivative
∂ i f j \partial_{i} f_{j} ∂ i f j ∂ [ i f j ] \partial_{[i} f_{j]} ∂ [ i f j ]
Input:
ω = F 1 d x 2 ∧ d x 3 + F 2 d x 3 ∧ d x 1 + F 3 d x 1 ∧ d x 2 . \omega = F^1 \,dx^2 \wedge dx^3 + F^2 \,dx^3 \wedge dx^1 + F^3 \,dx^1 \wedge dx^2\,. ω = F 1 d x 2 ∧ d x 3 + F 2 d x 3 ∧ d x 1 + F 3 d x 1 ∧ d x 2 . Output:
d ω = d F 1 ∧ d x 2 ∧ d x 3 + d F 2 ∧ d x 3 ∧ d x 1 + d F 3 ∧ d x 1 ∧ d x 2 = d\omega
= dF^1 \wedge dx^2 \wedge dx^3 + dF^2 \wedge dx^3 \wedge dx^1 + dF^3 \wedge dx^1 \wedge dx^2= d ω = d F 1 ∧ d x 2 ∧ d x 3 + d F 2 ∧ d x 3 ∧ d x 1 + d F 3 ∧ d x 1 ∧ d x 2 =
= ∂ 1 F 1 d x 1 ∧ d x 2 ∧ d x 3 + ∂ 2 F 2 d x 2 ∧ d x 3 ∧ d x 1 + ∂ 3 F 3 d x 3 ∧ d x 1 ∧ d x 2 = = \partial_1 F^1 dx^1 \wedge dx^2 \wedge dx^3 + \partial_2 F^2 dx^2 \wedge dx^3 \wedge dx^1 + \partial_3 F^3 dx^3 \wedge dx^1 \wedge dx^2= = ∂ 1 F 1 d x 1 ∧ d x 2 ∧ d x 3 + ∂ 2 F 2 d x 2 ∧ d x 3 ∧ d x 1 + ∂ 3 F 3 d x 3 ∧ d x 1 ∧ d x 2 =
= ( ∂ 1 F 1 + ∂ 2 F 2 + ∂ 3 F 3 ) d x 1 ∧ d x 2 ∧ d x 3 . = (\partial_1 F^1 + \partial_2 F^2 + \partial_3 F^3) \,dx^1 \wedge dx^2 \wedge dx^3\,. = ( ∂ 1 F 1 + ∂ 2 F 2 + ∂ 3 F 3 ) d x 1 ∧ d x 2 ∧ d x 3 . In general for antisymmetric A i j A_{ij} A ij
ω = A i j d x i ⊗ d x j = 1 2 A i j d x i ∧ d x j . \omega = A_{ij}\, dx^i \otimes dx^j = {1\over2} A_{ij}\, dx^i \wedge dx^j\,. ω = A ij d x i ⊗ d x j = 2 1 A ij d x i ∧ d x j . We differentiate the tensor:
d ω = 1 2 d A i j d x i ∧ d x j = 1 2 ∂ k A i j d x k ∧ d x i ∧ d x j = d\omega = {1\over2} d A_{ij} \, dx^i \wedge dx^j
= {1\over2} \partial_k A_{ij} \, dx^k \wedge dx^i \wedge dx^j= d ω = 2 1 d A ij d x i ∧ d x j = 2 1 ∂ k A ij d x k ∧ d x i ∧ d x j =
= 1 2 ∂ [ k A i j ] d x i ∧ d x j ∧ d x k = = {1\over2} \partial_{[k} A_{ij]} \, dx^i \wedge dx^j \wedge dx^k= = 2 1 ∂ [ k A ij ] d x i ∧ d x j ∧ d x k =
= 1 2 1 6 ε k i j ε l m n ∂ l A m n d x i ∧ d x j ∧ d x k = = {1\over2} {1\over 6}\varepsilon_{kij}\varepsilon^{lmn} \partial_{l} A_{mn} \, dx^i \wedge dx^j \wedge dx^k= = 2 1 6 1 ε kij ε l mn ∂ l A mn d x i ∧ d x j ∧ d x k =
= 1 2 1 6 ε k i j ε l m n ∂ l ε m n q F q d x i ∧ d x j ∧ d x k = = {1\over2} {1\over 6}\varepsilon_{kij}\varepsilon^{lmn} \partial_{l} \varepsilon_{mnq} F^q \, dx^i \wedge dx^j \wedge dx^k= = 2 1 6 1 ε kij ε l mn ∂ l ε mn q F q d x i ∧ d x j ∧ d x k =
= 1 2 1 6 ε k i j 2 δ l q ∂ l F q d x i ∧ d x j ∧ d x k = = {1\over2} {1\over 6}\varepsilon_{kij} 2 \delta^l{}_q \partial_{l} F^q \, dx^i \wedge dx^j \wedge dx^k= = 2 1 6 1 ε kij 2 δ l q ∂ l F q d x i ∧ d x j ∧ d x k =
= 1 6 ε i j k ∂ q F q d x i ∧ d x j ∧ d x k = = {1\over 6}\varepsilon_{ijk} \partial_q F^q \, dx^i \wedge dx^j \wedge dx^k= = 6 1 ε ijk ∂ q F q d x i ∧ d x j ∧ d x k =
= ∂ q F q d x 1 ∧ d x 2 ∧ d x 3 . = \partial_q F^q \, dx^1 \wedge dx^2 \wedge dx^3\,. = ∂ q F q d x 1 ∧ d x 2 ∧ d x 3 . We used the fact that the basis is antisymmetric, so only the antisymmetric
part of the derivative survives. The basis becomes antisymmetric due to the
incorporation of the d x k dx^k d x k
The above is the correct result for differentiating any k k k n n n k = n − 1 k=n-1 k = n − 1
Let’s now do the same computation using regular tensors:
∂ [ k A i j ] = 1 6 ε k i j ε l m n ∂ l A m n = 1 6 ε k i j ε l m n ∂ l ε m n q F q = \partial_{[k} A_{ij]}
= {1\over 6}\varepsilon_{kij}\varepsilon^{lmn} \partial_{l} A_{mn}
= {1\over 6}\varepsilon_{kij}\varepsilon^{lmn} \partial_{l} \varepsilon_{mnq} F^q = ∂ [ k A ij ] = 6 1 ε kij ε l mn ∂ l A mn = 6 1 ε kij ε l mn ∂ l ε mn q F q =
= 1 6 ε k i j 2 δ l q ∂ l F q = 1 6 ε i j k ∂ q F q . = {1\over 6}\varepsilon_{kij} 2 \delta^l{}_q \partial_{l} F^q
= {1\over 6}\varepsilon_{ijk} \partial_q F^q\,. = 6 1 ε kij 2 δ l q ∂ l F q = 6 1 ε ijk ∂ q F q . The exterior derivative is simply a regular (not antisymmetric) derivative
∂ k A i j \partial_{k} A_{ij} ∂ k A ij ∂ [ k A i j ] \partial_{[k} A_{ij]} ∂ [ k A ij ]
The quantity F q F^q F q A i j = ε i j q F q A_{ij}=\varepsilon_{ijq}F^q A ij = ε ij q F q A i j A_{ij} A ij ε i j q \varepsilon_{ijq} ε ij q ∂ [ k A i j ] \partial_{[k} A_{ij]} ∂ [ k A ij ] ε i j k \varepsilon_{ijk} ε ijk
TODO ¶ Add examples of a 3D space, rank 0-, 1-, 2-, 3-forms that correspond to
scalars, vectors/forms, rank 2 antisymmetric tensor (which internally contains
a pseudo-vector I think, using A i j = ε i j k B k A_{ij}=\varepsilon_{ijk}B_k A ij = ε ijk B k ε i j k \varepsilon_{ijk} ε ijk B k B_k B k
Riemannian volume form (∣ g ∣ d x 1 ∧ . . . ∧ d x n \sqrt{|g|} dx^1 \wedge ... \wedge dx^n ∣ g ∣ d x 1 ∧ ... ∧ d x n
− g \sqrt{-g} − g
Show this:
a ∧ b = i a × b , \mathbf{a}\wedge \mathbf{b} = i \mathbf{a} \times \mathbf{b}\,, a ∧ b = i a × b , where i = e 1 ∧ e 2 ∧ e 3 i = e_1 \wedge e_2 \wedge e_3 i = e 1 ∧ e 2 ∧ e 3 i 2 = − 1 i^2=-1 i 2 = − 1
This is also related to the Hodge star operator that provides isomorphism
between axial vectors (pseudovectors) and bivectors (2-forms).
Give examples from physics for each. It should be possible to build this
structure fully using antisymmetric tensors.
Also add examples of differentiation, where derivative of a vector creates a
tensor, this must be related to F μ ν = ∂ μ A ν − ∂ ν A μ F_{\mu\nu}=\partial_\mu A_\nu - \partial_\nu A_\mu F μν = ∂ μ A ν − ∂ ν A μ
Knowing exactly where each physical entity belongs is crucial (give examples),
but that is possible already using tensors, we do not need differential forms
for that. Rewrite various equations that use differential forms to use tensors.
Winitzki, S. (2009). Linear algebra via exterior products . Sergei Winitzki. Čertík, O. et al. (2025). Theoretical Physics Reference: Differential Geometry . https://www.theoretical-physics.com/dev/math/differential-geometry.html