Reflection symmetry of operations in 3D
How does a field with reflectional symmetry change when it undergoes algebraic / differential operations?
Suppose a field defined in the 3D Euclidean space has reflectional symmetry with respect to a plane. How does the symmetry change if it undergoes algebraic / differential operations? This note collects practical conclusions for use in this context.
Definitions
I start by introduce some definitions of reflection symmetry of the fields. Apart from one or two notations, this part is essentially the same to that listed in Parity of spherical harmonics.
Given transformation
\[\mathcal{R}: \begin{pmatrix} a_x \\ a_y \\ a_z \end{pmatrix} \mapsto \begin{pmatrix} a_x \\ a_y \\ -a_z \end{pmatrix}\]we can define \(\mathcal{S}\) (and \(A\)) as the spaces of scalar fields that are symmetric (anti-symmetric) w.r.t. the \(z=0\) plane, i.e.
\[\begin{aligned} \mathcal{S} &= \{f:\ f(\mathbf{r}) = f(\mathcal{R}\mathbf{r})\quad \forall \mathbf{r}\in \mathbb{R}^3\}, \\ \mathcal{A} &= \{f:\ f(\mathbf{r}) = -f(\mathcal{R}\mathbf{r})\quad \forall \mathbf{r}\in \mathbb{R}^3\} \end{aligned}\]where \(f: \mathbb{R}^3\mapsto \mathbb{R}\) is a scalar function defined on 3D Euclidean space.
Similarly, let us define \(\mathcal{S}_1\) (and \(A_1\)) as the spaces of vector fields (\(1\)-subscript for rank-\(1\) tensors) that are symmetric (anti-symmetric) w.r.t. the \(z=0\) plane, i.e.
\[\begin{aligned} \mathcal{S}_1 &= \{\mathbf{f}:\ \mathbf{f}(\mathbf{r}) = \mathcal{R} \mathbf{f}(\mathcal{R}\mathbf{r}) \quad \forall \mathbf{r}\in \mathbb{R}^3 \}, \\ \mathcal{A}_1 &= \{\mathbf{f}:\ \mathbf{f}(\mathbf{r}) = -\mathcal{R} \mathbf{f}(\mathcal{R}\mathbf{r})\quad \forall \mathbf{r}\in \mathbb{R}^3\} \end{aligned}\]where \(\mathbf{f}: \mathbb{R}^3\mapsto \mathbb{R}^3\) is a vector function defined on 3D Euclidean space. A spatially uniform field perpendicular to the plane of symmetry, \(\hat{\mathbf{z}}\), for instance, is a anti-symmetric vector field.
For convenience (which will become obvious later), let us introduce an index for reflectional symmetry, \(\mu\), which outputs \(1\) for symmetric fields, and \(-1\) for antisymmetric fields. In other words,
\[\mu(\mathbf{f}) = \left\{\begin{aligned} & +1, \quad \mathbf{f} \in \mathcal{S}_r \\ & -1, \quad \mathbf{f} \in \mathcal{A}_r \end{aligned}\right.\]regardless of the rank of the tensor \(r\) (if \(r=0\), these are the scalar symmetry sets \(\mathcal{S}\) and \(\mathcal{A}\); if \(r=1\), these are the vector symmetry sets \(\mathcal{S}_1\), \(\mathcal{A}_1\)).
As a remark, these sets are linear spaces as they are defined by linear operations, hence they are close to addition and scalar products, i.e. if \(\mu(f) = \mu(g)\) and \(a \in \mathbb{R}\),
\[\mu(f + g) = \mu(f), \quad \mu(af) = \mu(f).\]where \(f\) and \(g\) in the equation above can be either scalar or vector fields.
Properties
Here I list the symmetry transformation under some algebraic and differential operations.
Scalar product
Product of two symmetric (or anti-symmetric) scalar fields is symmetric; product of one symmetric and one anti-symmetric scalar fields is anti-symmetric. In summary,
\[\mu(fg) = \mu(f) \, \mu(g)\]Proof: Under the premise that \(\mu\) is defined for \(f\) and \(g\) (i.e. \(f,g\) are either symmetric or anti-symmetric),
\[f(\mathcal{R}\mathbf{r}) = \mu(f) \, f(\mathbf{r}), \quad g(\mathcal{R}\mathbf{r}) = \mu(g) \, g(\mathbf{r}).\]Then their product \(w(\mathbf{r}) = f(\mathbf{r}) g(\mathbf{r})\) satisfies
\[\begin{aligned} w(\mathcal{R}\mathbf{r}) &= f(\mathcal{R}\mathbf{r})\, g(\mathcal{R}\mathbf{r}) = [\mu(f) f(\mathbf{r})][\mu(g) g(\mathbf{r})] \\ &= [\mu(f)\mu(g)][f(\mathbf{r})g(\mathbf{r})] = [\mu(f)\mu(g)] \, w(\mathbf{r}) \end{aligned}\]Therefore
\[\mu(w) = \mu(fg) = \mu(f) \, \mu(g). \qquad \square\]Scalar product with vector
Product of two symmetric (or anti-symmetric) scalar & vector fields is symmetric; product of one symmetric and one anti-symmetric scalar / vector fields is anti-symmetric. In summary,
\[\mu(f\mathbf{g}) = \mu(f) \, \mu(\mathbf{g})\]Proof: One can either prove this by showing the symmetry component by component, or show that the product \(\mathbf{w} = f\mathbf{g}\) satisfies
\[\begin{aligned} \mathbf{w}(\mathcal{R}\mathbf{r}) &= f(\mathcal{R}\mathbf{r})\, \mathbf{g} (\mathcal{R}\mathbf{r}) \\ &= [\mu(f) f][\mu(g) \mathcal{R} \mathbf{g}(\mathbf{r})] \\ &= [\mu(f)\mu(g)][\mathcal{R}f(\mathbf{r})g(\mathbf{r})] \\ &= [\mu(f)\mu(g)] \, \mathcal{R} \mathbf{w}(\mathbf{r}). \qquad \square \end{aligned}\]Vector inner product
Inner product of two symmetric (or anti-symmetric) vector fields is symmetric; inner product of one symmetric and one anti-symmetric vector fields is anti-symmetric. In summary,
\[\mu(\mathbf{f}\cdot \mathbf{g}) = \mu(\mathbf{f})\, \mu(\mathbf{g})\]Proof: Under the premise that \(\mu\) is defined for \(\mathbf{f}\) and \(\mathbf{g}\), we have
\[\mathbf{f}(\mathcal{R}\mathbf{r}) = \mu(\mathbf{f}) \, \mathcal{R}\mathbf{f}(\mathbf{r}), \quad \mathbf{g}(\mathcal{R}\mathbf{r}) = \mu(\mathbf{g}) \, \mathcal{R}\mathbf{g}(\mathbf{r}).\]Note here we used the definition and the property \(\mathcal{R}^2 = \mathcal{I}\), with \(\mathcal{I}\) being identity, or \(\mathcal{R}=\mathcal{R}^{-1}\). The inner product \(w(\mathbf{r}) = \mathbf{f}(\mathbf{r}) \cdot \mathbf{g}(\mathbf{r})\) satisfies
\[\begin{aligned} w(\mathcal{R}\mathbf{r}) &= \mathbf{f}(\mathcal{R}\mathbf{r})\cdot \mathbf{g}(\mathcal{R}\mathbf{r}) \\ &= [\mu(\mathbf{f}) \mathcal{R}\mathbf{f}(\mathbf{r})]\cdot [\mu(\mathbf{g}) \mathcal{R}\mathbf{g}(\mathbf{r})] \\ &= \mu(\mathbf{f})\mu(\mathbf{g}) \mathcal{R}^2 \, [\mathbf{f}(\mathbf{r})\cdot \mathbf{g}(\mathbf{r})] \\ &= \mu(\mathbf{f})\mu(\mathbf{g}) \, w(\mathbf{r}).\quad \square \end{aligned}\]Vector cross product
Cross product of two symmetric (or anti-symmetric) vector fields is anti-symmetric; inner product of one symmetric and one anti-symmetric vector fields is symmetric. In summary,
\[\mu(\mathbf{f}\times \mathbf{g}) = - \mu(\mathbf{f}) \, \mu(\mathbf{g})\]Proof: Under the premise that \(\mu\) is defined for \(\mathbf{f}\) and \(\mathbf{g}\), we have
\[\mathbf{f}(\mathcal{R}\mathbf{r}) = \mu(\mathbf{f}) \, \mathcal{R}\mathbf{f}(\mathbf{r}), \quad \mathbf{g}(\mathcal{R}\mathbf{r}) = \mu(\mathbf{g}) \, \mathcal{R}\mathbf{g}(\mathbf{r}).\]The cross product \(\mathbf{w}(\mathbf{r}) = \mathbf{f}(\mathbf{r}) \times \mathbf{g}(\mathbf{r})\) satisfies
\[\begin{aligned} w_i(\mathcal{R}\mathbf{r}) &= \sum_{j,k} \epsilon_{ijk} \, f_j(\mathcal{R}\mathbf{r}) \, g_k (\mathcal{R}\mathbf{r}) \\ &= \sum_{j,k} \epsilon_{ijk} \, [\mu(\mathbf{f}) \mathcal{R}_j f_j] [\mu(\mathbf{g}) \mathcal{R}_k g_k] \\ &= [\mu(\mathbf{f}) \mu(\mathbf{g})] \, \sum_{j,k} \epsilon_{ijk} \, [ \mathcal{R}_j f_j \mathcal{R}_k g_k] \end{aligned}\]Here we note that \(\mathcal{R}_1 \mathcal{R}_2 = 1 = - \mathcal{R}_3\), \(\mathcal{R}_2 \mathcal{R}_3 = \mathcal{R}_1 \mathcal{R}_3 = -1 = - \mathcal{R}_1 = - \mathcal{R}_2\). This allows the latter part of the equation to be written in a simple form:
\[\sum_{j,k} \epsilon_{ijk} \mathcal{R}_j f_j \mathcal{R}_k g_k = - \mathcal{R}_i \sum_{j,k} \epsilon_{ijk} f_j g_k\]and hence
\[\begin{aligned} w_i(\mathcal{R} \mathbf{r}) &= - [\mu(\mathbf{f}) \mu(\mathbf{g})] \mathcal{R}_i \, [\mathbf{f}\times\mathbf{g}]_i = [- \mu(\mathbf{f}) \mu(\mathbf{g})] [\mathcal{R} \mathbf{w}]_i \\ \mathbf{w}(\mathcal{R} \mathbf{r}) &= [- \mu(\mathbf{f}) \mu(\mathbf{g})] \, \mathcal{R} \mathbf{w}(\mathbf{r}). \quad \square \end{aligned}\]Scalar gradient
Gradient of a(n) symmetric (resp. anti-symmetric) scalar field produces a(n) symmetric (resp. anti-symmetric) vector field. In summary,
\[\mu(\nabla f) = \mu(f)\]Proof: The gradient satisfies
\[\begin{aligned} (\nabla f)(\mathcal{R}\mathbf{r}) &= [\nabla (f(\mathbf{r}'))]_{\mathbf{r}'=\mathcal{R}\mathbf{r}} \\ &= [\nabla (\mu(f)f(\mathcal{R}\mathbf{r}'))]_{\mathbf{r}'=\mathcal{R}\mathbf{r}} \\ &= \mu(f) \, [\nabla (f(\mathcal{R}\mathbf{r}'))]_{\mathbf{r}'=\mathcal{R}\mathbf{r}} \\ &= \mu(f) \, \mathcal{R} [(\nabla f)(\mathcal{R}\mathbf{r}')]_{\mathbf{r}'=\mathcal{R}\mathbf{r}} \\ &= \mu(f) \, \mathcal{R} [(\nabla f)(\mathcal{R}^2\mathbf{r})] \\ &= \mu(f) \, \mathcal{R} (\nabla f)(\mathbf{r}), \quad \square \end{aligned}\]Alternatively, one can also first derive the symmetry of the scalar differential operators \(\partial_i\):
\[\begin{aligned} \mu(\partial_1 f) = \mu(\partial_x f) &= \mu(f), \\ \mu(\partial_2 f) = \mu(\partial_y f) &= \mu(f), \\ \mu(\partial_3 f) = \mu(\partial_z f) &= -\mu(f) \end{aligned}\]These operators can then be formally treated as normal scalars and assigned their own symmetry, i.e. \(\partial_1,\partial_2 \in \mathcal{S}\), hence \(\mu(\partial_1) = \mu(\partial_2) = 1\) , and \(\partial_3 \in \mathcal{A}\), hence \(\mu(\partial_3)=-1\). The operator \(\nabla\) can be formally treated as a normal vector with \(\nabla\in \mathcal{S}_1\) or \(\mu(\nabla) = 1\). Using these formal treatments, \(\nabla f\) is just a scalar-vector product. Applying known properties, we have
\[\mu(\nabla f) = \mu(\nabla) \, \mu(f) = \mu(f).\]Vector divergence
Divergence of a(n) symmetric (resp. anti-symmetric) vector field produces a(n) symmetric (resp. anti-symmetric) scalar field. In summary,
\[\mu(\nabla\cdot \mathbf{f}) = \mu(\mathbf{f})\]A proof can be given by combining the formal treatment \(\mu(\nabla) = 1\) and the vector inner product property \(\mu(\mathbf{f}\cdot \mathbf{g}) = \mu(\mathbf{f})\, \mu(\mathbf{g})\). Alternatively, this can be proven using vector identities and chain rules.
Vector curl
Curl of a(n) symmetric (resp. anti-symmetric) vector field produces an(a) anti-symmetric (resp. symmetric) vector field. In summary,
\[\mu(\nabla\times \mathbf{f}) = - \mu (\mathbf{f})\]A proof can be given by combining the formal treatment \(\mu(\nabla) = 1\) and the vector cross product property \(\mu(\mathbf{f}\times \mathbf{g}) = - \mu(\mathbf{f})\, \mu(\mathbf{g})\). Alternatively, this can be proven using vector identities and chain rules.
Vector contraction with vector gradient
Contraction of a(n) symmetric (anti-symmetric) vector field with the gradient of a(n) symmetric (anti-symmetric) vector field produces a symmetric vector field. Vice versa, contraction of a(n) symmetric (anti-symmetric) vector field with the gradient of an(a) anti-symmetric (symmetric) vector field produces an anti-symmetric vector field. In summary,
\[\mu(\mathbf{f}\cdot \nabla \mathbf{g}) = \mu(\mathbf{f}) \, \mu(\mathbf{g})\]A proof can be given by combining the formal treatment \(\mu(\nabla) = 1\) and
\[\mu((\mathbf{f}\cdot \mathbf{g})\mathbf{h}) = \mu(\mathbf{f})\, \mu(\mathbf{g})\, \mu(\mathbf{h}).\]Alternatively, this can be proven using vector identities and chain rules.
Dyadic / tensor product
The tensor product of two vectors also have certain properties of reflectional symmetry, if the two vectors themselves have reflectional symmetry. However, this also relies on a definition for reflectional symmetry of tensors. For rank-2 tensor \(\mathbf{T}\), we can further introduce definition
\[\begin{aligned} \mathcal{S}_2 &= \{\mathbf{T}:\ \mathbf{T}(\mathbf{r}) = \mathcal{R} \, \mathbf{T}(\mathcal{R}\mathbf{r}) \, \mathcal{R} \quad \forall \mathbf{r}\in \mathbb{R}^3 \}, \\ \mathcal{A}_2 &= \{\mathbf{T}:\ \mathbf{T}(\mathbf{r}) = -\mathcal{R} \, \mathbf{T}(\mathcal{R}\mathbf{r})\, \mathcal{R} \quad \forall \mathbf{r}\in \mathbb{R}^3\} \end{aligned}\]where \(\mathcal{R} \mathbf{T} \mathcal{R}\) indicates applying the reflection operator both to the left and to the right of the tensor. In this case, the tensor product of two symmetric (anti-symmetric) vectors is a symmetric dyad, whereas the tensor product of one symmetric and one anti-symmetric vector is an anti-symmetric dyad. In summary,
\[\mu(\mathbf{f}\otimes \mathbf{g}) = \mu(\mathbf{f}) \, \mu(\mathbf{g})\]Proof: using component form, if \(\mathbf{T} = \mathbf{f}\otimes \mathbf{g}\), then
\[\begin{aligned} T_{ij}(\mathcal{R}\mathbf{r}) &= f_i(\mathcal{R}\mathbf{r}) \, g_j(\mathcal{R}\mathbf{r}) \\ &= [\mu(f_i) f_i(\mathbf{r})] \, [\mu(g_j) g_j(\mathbf{r})] \\ &= [\mu(\mathbf{f}) \mathcal{R}_i f_i(\mathbf{r})] \, [\mu(\mathbf{g}) \mathcal{R}_j g_j(\mathbf{r})] \\ &= [\mu(\mathbf{f}) \mu(\mathbf{g})] \mathcal{R}_i \mathcal{R}_j f_i(\mathbf{r}) g_j(\mathbf{r}) \\ &= [\mu(\mathbf{f}) \mu(\mathbf{g})] \mathcal{R}_i \mathcal{R}_j T_{ij}(\mathbf{r}) \\ \mathbf{T}(\mathcal{R}\mathbf{r}) &= [\mu(\mathbf{f}) \mu(\mathbf{g})] \, \mathcal{R}\mathbf{T}(\mathbf{r})\mathcal{R}. \quad \square \end{aligned}\]Vector gradient
The gradient of a(n) symmetric (resp. anti-symmetric) vector is a(n) symmetric (resp. anti-symmetric) rank-2 tensor. In summary,
\[\mu(\nabla \mathbf{f}) = \mu (\mathbf{f})\]This can be similar shown using the tensor product rule and the formal treatment \(\mu(\nabla) = 1\).