Rotations in Euclidean space

by Stéphane Haussler

On this page, I systematically explore the representation of infinitesimal rotations in 3-dimensional Euclidean space using the language of differential forms. We establish the relation to matrix multiplication rules, as well as the direct relation to the cross product. In preparation for the analysis of rotations in 4-dimensional Minkowski spacetime, we determine the symmetries of the exterior product in mixed form, which, in this case, are found to be trivial.

Free matrix representation of rotations

The free matrix representation is intuitive when using a bivector basis, since the elements can be organized and re-ordered at will. With three dimensions, rotations are possible on the three planes and can be expressed as a linear combinations the three basis bivectors:

\[R^{♯♯} = R^x \; ∂_y ∧ ∂_z + R^y \; ∂_z ∧ ∂_x + R^z \; ∂_x ∧ ∂_y\]

We can rewrite as a single column:

Representation of rotations as a single column of bivectors

\[\begin{split}R^{♯♯} = \begin{bmatrix} R^x \; ∂_y ∧ ∂_z \\ R^y \; ∂_z ∧ ∂_x \\ R^z \; ∂_x ∧ ∂_y \\ \end{bmatrix}\end{split}\]

We could also represent the rotation with a row/column notation:

\[\begin{split}R^{♯♯} = \left[ \begin{alignedat}{2} & R^z \; ∂_x ∧ ∂_y & \\ & & R^x \; ∂_y ∧ ∂_z \\ R^y \; ∂_z ∧ ∂_x & & \\ \end{alignedat} \right]\end{split}\]

However there is a more natural representation. The exterior product is anti-symmetric \(∂_i ∧ ∂_j = - ∂_j ∧ ∂_i\) and strictly equivalent to \(∂_i ∧ ∂_j = \frac{1}{2} (∂_i ∧ ∂_j - ∂_j ∧ ∂_i)\), which permits to rewrite:

\[\begin{split}R^{♯♯} = \begin{bmatrix} R^x \; ∂_y ∧ ∂_z \\ R^y \; ∂_z ∧ ∂_x \\ R^z \; ∂_x ∧ ∂_y \\ \end{bmatrix} = \frac{1}{2} \begin{bmatrix} R^x \; ∂_y ∧ ∂_z - R^x \; ∂_z ∧ ∂_y \\ R^y \; ∂_z ∧ ∂_x - R^y \; ∂_x ∧ ∂_z \\ R^z \; ∂_x ∧ ∂_y - R^z \; ∂_y ∧ ∂_x \\ \end{bmatrix}\end{split}\]

With a row/column representation, we obtain a fully anti-symmetric and doubly contravariant representation:

The doubly contravariant row/column representation of rotations

\[\begin{split}R^{♯♯} = \frac{1}{2} \begin{bmatrix} & - R^z \; ∂_y ∧ ∂_x & + R^y \; ∂_z ∧ ∂_x \\ + R^z \; ∂_x ∧ ∂_y & & - R^x \; ∂_z ∧ ∂_y \\ - R^y \; ∂_x ∧ ∂_z & + R^x \; ∂_y ∧ ∂_z & \\ \end{bmatrix}\end{split}\]

The doubly contravariant rotation obtained exclusively operates on covectors. Falling back to matrix multiplication rules requires a mixed tensor that takes a vector as input, and output a vector as output. Specifically, we need to flatten the first component in oder to obtain the \(♯♭\) tensor representation, which corresponds exactly to the matrix representation commonly encountered in linear algebra.

\(♭♯\) representation

Flattening the first index of of the doubly contravariant form of the rotation \(R\), we obtain:

\[\begin{split}R^{♭♯} = \begin{bmatrix} & - R^z \; dy ∧ ∂_x & + R^y \; dz ∧ ∂_x \\ + R^z \; dx ∧ ∂_y & & - R^x \; dz ∧ ∂_y \\ - R^y \; dx ∧ ∂_z & + R^x \; dy ∧ ∂_z & \\ \end{bmatrix}\end{split}\]

Calculations

Flatten the first index, keeping the second sharp

\[R^{♭♯} = (R^{♯♯})^{♭♯}\]

Take the doubly contravariant row/column representation

\[\begin{split}R^{♭♯} = \frac{1}{2} \begin{bmatrix} & - R^z \; ∂_y ∧ ∂_x & + R^y \; ∂_z ∧ ∂_x \\ + R^z \; ∂_x ∧ ∂_y & & - R^x \; ∂_z ∧ ∂_y \\ - R^y \; ∂_x ∧ ∂_z & + R^x \; ∂_y ∧ ∂_z & \\ \end{bmatrix}^{♭♯}\end{split}\]

Distribute the musical operators

\[\begin{split}R^{♭♯} = \frac{1}{2} \begin{bmatrix} & - R^z \; (∂_y ∧ ∂_x)^{♭♯} & + R^y \; (∂_z ∧ ∂_x)^{♭♯} \\ + R^z \; (∂_x ∧ ∂_y)^{♭♯} & & - R^x \; (∂_z ∧ ∂_y)^{♭♯} \\ - R^y \; (∂_x ∧ ∂_z)^{♭♯} & + R^x \; (∂_y ∧ ∂_z)^{♭♯} & \\ \end{bmatrix}\end{split}\]

Apply the musical operators, using the euclidean metric

\[\begin{split}R^{♭♯} = \frac{1}{2} \begin{bmatrix} & - R^z \; δ_{yi} dx^i ∧ ∂_x & + R^y \; δ_{zi} dx^i ∧ ∂_x \\ + R^z \; δ_{xi} dx^i ∧ ∂_y & & - R^x \; δ_{zi} dx^i ∧ ∂_y \\ - R^y \; δ_{xi} dx^i ∧ ∂_z & + R^x \; δ_{yi} dx^i ∧ ∂_z & \\ \end{bmatrix}\end{split}\]

Identify the non-zero components and conclude

\[\begin{split}R^{♭♯} = \frac{1}{2} \begin{bmatrix} & - R^z \; dy ∧ ∂_x & + R^y \; dz ∧ ∂_x \\ + R^z \; dx ∧ ∂_y & & - R^x \; dz ∧ ∂_y \\ - R^y \; dx ∧ ∂_z & + R^x \; dy ∧ ∂_z & \\ \end{bmatrix}\end{split}\]

Expanding the exterior product to its tensor form and simplifying, we find the explicit expression in terms of tensor products \(⊗\):

\[\begin{split}(∂_x ∧ ∂_y)^{♭♯} &= dx ⊗ ∂_y - dy ⊗ ∂_x \\ (∂_y ∧ ∂_z)^{♭♯} &= dy ⊗ ∂_z - dz ⊗ ∂_y \\ (∂_z ∧ ∂_x)^{♭♯} &= dz ⊗ ∂_x - dx ⊗ ∂_z \\\end{split}\]

Calculations

Expand the exterior product into tensor products

\[\begin{split}(∂_x ∧ ∂_y)^{♭♯} &= (∂_x ⊗ ∂_y - ∂_y ⊗ ∂_x)^{♭♯} \\ (∂_y ∧ ∂_z)^{♭♯} &= (∂_y ⊗ ∂_z - ∂_z ⊗ ∂_y)^{♭♯} \\ (∂_z ∧ ∂_x)^{♭♯} &= (∂_z ⊗ ∂_x - ∂_x ⊗ ∂_z)^{♭♯} \\\end{split}\]

Distribute the musical operators

\[\begin{split}(∂_x ∧ ∂_y)^{♭♯} &= ∂_x^♭ ⊗ ∂_y^♯ - ∂_y^♭ ⊗ ∂_x^♯ \\ (∂_y ∧ ∂_z)^{♭♯} &= ∂_y^♭ ⊗ ∂_z^♯ - ∂_z^♭ ⊗ ∂_y^♯ \\ (∂_z ∧ ∂_x)^{♭♯} &= ∂_z^♭ ⊗ ∂_x^♯ - ∂_x^♭ ⊗ ∂_z^♯ \\\end{split}\]

Apply musical operators using the Euclidean metric:

\[\begin{split}(∂_x ∧ ∂_y)^{♭♯} &= δ_{xi} dx^i ⊗ ∂_y - δ_{yi} dx^i ⊗ ∂_x \\ (∂_y ∧ ∂_z)^{♭♯} &= δ_{yi} dx^i ⊗ ∂_z - δ_{zi} dx^i ⊗ ∂_y \\ (∂_z ∧ ∂_x)^{♭♯} &= δ_{zi} dx^i ⊗ ∂_x - δ_{xi} dx^i ⊗ ∂_z \\\end{split}\]

Identify the non-zero terms

\[\begin{split}(∂_x ∧ ∂_y)^{♭♯} &= δ_{xx} dx ⊗ ∂_y - δ_{yy} dy ⊗ ∂_x \\ (∂_y ∧ ∂_z)^{♭♯} &= δ_{yy} dy ⊗ ∂_z - δ_{zz} dz ⊗ ∂_y \\ (∂_z ∧ ∂_x)^{♭♯} &= δ_{zz} dz ⊗ ∂_x - δ_{xx} dx ⊗ ∂_z \\\end{split}\]

Conclude

\[\begin{split}(∂_x ∧ ∂_y)^{♭♯} &= dx ⊗ ∂_y - dy ⊗ ∂_x \\ (∂_y ∧ ∂_z)^{♭♯} &= dy ⊗ ∂_z - dz ⊗ ∂_y \\ (∂_z ∧ ∂_x)^{♭♯} &= dz ⊗ ∂_x - dx ⊗ ∂_z \\\end{split}\]

\(♯♭\) representation

Flattening the second index of of the doubly contravariant form of the rotation \(R\), we obtain:

\[\begin{split}R^{♯♭} = \frac{1}{2} \begin{bmatrix} & + R^z \; ∂_x ∧ dy & - R^y \; ∂_x ∧ dz \\ - R^z \; ∂_y ∧ dx & & + R^x \; ∂_y ∧ dz \\ + R^y \; ∂_z ∧ dx & - R^x \; ∂_z ∧ dy & \\ \end{bmatrix}\end{split}\]

Calculations

Flatten the second index, keeping the first sharp:

\[R^{♯♭} = (R^{♯♯})^{♯♭}\]

Take the rotation in its matrix form:

\[\begin{split}R^{♯♭} = \frac{1}{2} \begin{bmatrix} & - R^z \; ∂_y ∧ ∂_x & + R^y \; ∂_z ∧ ∂_x \\ + R^z \; ∂_x ∧ ∂_y & & - R^x \; ∂_z ∧ ∂_y \\ - R^y \; ∂_x ∧ ∂_z & + R^x \; ∂_y ∧ ∂_z & \\ \end{bmatrix}^{♯♭}\end{split}\]

Distribute the musical operators:

\[\begin{split}R^{♯♭} = \frac{1}{2} \begin{bmatrix} & - R^z \; (∂_y ∧ ∂_x)^{♯♭} & + R^y \; (∂_z ∧ ∂_x)^{♯♭} \\ + R^z \; (∂_x ∧ ∂_y)^{♯♭} & & - R^x \; (∂_z ∧ ∂_y)^{♯♭} \\ - R^y \; (∂_x ∧ ∂_z)^{♯♭} & + R^x \; (∂_y ∧ ∂_z)^{♯♭} & \\ \end{bmatrix}\end{split}\]

Apply the musical operators using the euclidean metric:

\[\begin{split}R^{♯♭} = \frac{1}{2} \begin{bmatrix} & - R^z \; ∂_y ∧ δ_{xi} dx^i & + R^y \; ∂_z ∧ δ_{xi} dx^i \\ + R^z \; ∂_x ∧ δ_{yi} dx^i & & - R^x \; ∂_z ∧ δ_{yi} dx^i \\ - R^y \; ∂_x ∧ δ_{zi} dx^i & + R^x \; ∂_y ∧ δ_{zi} dx^i & \\ \end{bmatrix}\end{split}\]

Identify the non-zero components:

\[\begin{split}R^{♯♭} = \frac{1}{2} \begin{bmatrix} & - R^z \; ∂_y ∧ dx^x & + R^y \; ∂_z ∧ dx^x \\ + R^z \; ∂_x ∧ dx^y & & - R^x \; ∂_z ∧ dx^y \\ - R^y \; ∂_x ∧ dx^z & + R^x \; ∂_y ∧ dx^z & \\ \end{bmatrix}\end{split}\]

Replace the covectors by their expressions

\[\begin{split}dx^x = dx \\ dx^y = dy \\ dx^z = dz \\\end{split}\]

\[\begin{split}R^{♯♭} = \frac{1}{2} \begin{bmatrix} & - R^z \; ∂_y ∧ dx & + R^y \; ∂_z ∧ dx \\ + R^z \; ∂_x ∧ dy & & - R^x \; ∂_z ∧ dy \\ - R^y \; ∂_x ∧ dz & + R^x \; ∂_y ∧ dz & \\ \end{bmatrix}\end{split}\]

Reorder and conclude

\[\begin{split}R^{♯♭} = \frac{1}{2} \begin{bmatrix} & + R^z \; ∂_x ∧ dy & - R^y \; ∂_x ∧ dz \\ - R^z \; ∂_y ∧ dx & & + R^x \; ∂_y ∧ dz \\ + R^y \; ∂_z ∧ dx & - R^x \; ∂_z ∧ dy & \\ \end{bmatrix}\end{split}\]

Expanding the wedge product to its tensor form and simplifying, we find the explicit expression of the mixed wedge products.

\[\begin{split}(∂_x ∧ ∂_y)^{♯♭} &= ∂_x ⊗ dy - ∂_y ⊗ dx \\ (∂_y ∧ ∂_z)^{♯♭} &= ∂_y ⊗ dz - ∂_z ⊗ dy \\ (∂_z ∧ ∂_x)^{♯♭} &= ∂_z ⊗ dx - ∂_x ⊗ dz \\\end{split}\]

Calculations

Expand exterior products into tensor products

\[\begin{split}(∂_x ∧ ∂_y)^{♯♭} &= (∂_x ⊗ ∂_y - ∂_y ⊗ ∂_x)^{♯♭} \\ (∂_y ∧ ∂_z)^{♯♭} &= (∂_y ⊗ ∂_z - ∂_z ⊗ ∂_y)^{♯♭} \\ (∂_z ∧ ∂_x)^{♯♭} &= (∂_z ⊗ ∂_x - ∂_x ⊗ ∂_z)^{♯♭} \\\end{split}\]

Distribute the musical operators

\[\begin{split}(∂_x ∧ ∂_y)^{♯♭} &= ∂_x^♯ ⊗ ∂_y^♭ - ∂_y^♯ ⊗ ∂_x^♭ \\ (∂_y ∧ ∂_z)^{♯♭} &= ∂_y^♯ ⊗ ∂_z^♭ - ∂_z^♯ ⊗ ∂_y^♭ \\ (∂_z ∧ ∂_x)^{♯♭} &= ∂_z^♯ ⊗ ∂_x^♭ - ∂_x^♯ ⊗ ∂_z^♭ \\\end{split}\]

Apply the musical operators using the euclidean metric

\[\begin{split}(∂_x ∧ ∂_y)^{♯♭} &= ∂_x ⊗ δ_{yi} dx^i - ∂_y ⊗ δ_{xi} dx^i \\ (∂_y ∧ ∂_z)^{♯♭} &= ∂_y ⊗ δ_{zi} dx^i - ∂_z ⊗ δ_{yi} dx^i \\ (∂_z ∧ ∂_x)^{♯♭} &= ∂_z ⊗ δ_{xi} dx^i - ∂_x ⊗ δ_{zi} dx^i \\\end{split}\]

Identify the non-zero terms

\[\begin{split}(∂_x ∧ ∂_y)^{♯♭} &= ∂_x ⊗ δ_{yy} dx^y - ∂_y ⊗ δ_{xx} dx^x \\ (∂_y ∧ ∂_z)^{♯♭} &= ∂_y ⊗ δ_{zz} dx^z - ∂_z ⊗ δ_{yy} dx^y \\ (∂_z ∧ ∂_x)^{♯♭} &= ∂_z ⊗ δ_{xx} dx^x - ∂_x ⊗ δ_{zz} dx^z \\\end{split}\]

Apply numerical values

\[\begin{split}(∂_x ∧ ∂_y)^{♯♭} &= ∂_x ⊗ dx^y - ∂_y ⊗ dx^x \\ (∂_y ∧ ∂_z)^{♯♭} &= ∂_y ⊗ dx^z - ∂_z ⊗ dx^y \\ (∂_z ∧ ∂_x)^{♯♭} &= ∂_z ⊗ dx^x - ∂_x ⊗ dx^z \\\end{split}\]

Replace the covectors by their expressions:

\[\begin{split}dx^x = dx \\ dx^y = dy \\ dx^z = dz \\\end{split}\]

\[\begin{split}(∂_x ∧ ∂_y)^{♯♭} &= ∂_x ⊗ dy - ∂_y ⊗ dx \\ (∂_y ∧ ∂_z)^{♯♭} &= ∂_y ⊗ dz - ∂_z ⊗ dy \\ (∂_z ∧ ∂_x)^{♯♭} &= ∂_z ⊗ dx - ∂_x ⊗ dz \\\end{split}\]

Symmetries of the mixed exterior product

From the explicit calculation of the basis elements, we observe the following properties:

Basis element	Expression	Row/column matrix symmetry
\(∂_x ∧ dy\)	\(∂_x ⊗ dy - ∂_y ⊗ dx\)	Antisymetric
\(∂_y ∧ dz\)	\(∂_x ⊗ dz - ∂_z ⊗ dy\)	Antisymetric
\(∂_z ∧ dx\)	\(∂_x ⊗ dx - ∂_x ⊗ dz\)	Antisymetric

The \(\mathfrak{so}(3)\) rotation group

Whether as a transpose or not, we identify the \(\mathfrak{so}(3)\) matrices as well as get a first hint that we are about to identify the electromagnetic tensor. Choosing the implicit basis \(\mathbf{e}_i \wedge \mathbf{e}_j\) in a row major representation, we obtain:

\begin{align} R &= \frac{1}{2} \begin{bmatrix} & - R^z & + R^y \\ + R^z & & - R^x \\ - R^y & + R^x & \\ \end{bmatrix} \\ &= R^x \left[ \begin{alignedat}{4} \; 0 & \; & 0 & \; & 0 \\ \; 0 & \; & 0 & \;- & 1 \\ \; 0 & \; + & 1 & \; & 0 \\ \end{alignedat} \right] + R^y \left[ \begin{alignedat}{4} & 0 & \quad 0 & \; + & 1 \\ & 0 & \quad 0 & \; & 0 \\ - & 1 & \quad 0 & \; & 0 \\ \end{alignedat} \right] + R^z \left[ \begin{alignedat}{4} & 0 & - & 1 & \quad 0 \\ + & 1 & & 0 & \quad 0 \\ & 0 & & 0 & \quad 0 \\ \end{alignedat} \right] \end{align}

Which is a regular choice for the basis of the \(\mathfrak{so}(3)\) group.

The cross product

Rotations in three dimensions have a dual. We can either express a rotation along the three planes, or we can express a rotation along the three directions of space. Indeed, through the use of the Hodge star \(⋆\), we fall back to the description of rotations expressed as a cross product \(⨯\):

Apply the Hodge star:

\[⋆R = ⋆(R^x \; ∂_y ∧ ∂_z + R^y \; ∂_z ∧ ∂_x + R^z \; ∂_x ∧ ∂_y)\]

Distribute the Hodge star:

\[⋆R = R^x ⋆(∂_y ∧ ∂_z) + R^y ⋆(∂_z ∧ ∂_x) + R^z ⋆(∂_x ∧ ∂_y)\]

Identify the cross product:

\[⋆R = R^x \; ∂_x + R^y \; ∂_y + R^z \; ∂_z\]

That is, the Hodge star of the rotation ∂_xpressed as a linear comibination of bivectors is exactly a rotation in terms of cross products in the Hodge dual space:

\[⋆R = R^x \; ∂_y ⨯ ∂_z + R^y \; ∂_z ⨯ ∂_x + R^z \; ∂_x ⨯ ∂_y\]

We could have written a covector in the same explicit manner. This notation is very conveniant when performing calculations in Cartan’s framework as it permits to identify and organize terms for practical calculations by falling back to regular matrix multiplication.