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A point $\mathbf{c}$ is a center of symmetry of a set $S$ in Euclidean space if point reflection $\mathbf{x}\mapsto 2\mathbf{c}-\mathbf{x}$ maps $S$ onto itself. Every ellipse has its center of an ellipse as a center of symmetry. More generally, a nonempty bounded set has at most one: two such reflections compose to translation by twice the difference of their centers, and iterating a nonzero translation contradicts boundedness.

Molecular dipole cancellation occurs when nonzero contributions to a molecule's electric dipole moment cancel under vector addition. In a bond dipole model, use ${\mathbf{p}}_{\mathrm{mol}}={\sum }_b{\mathbf{p}}_b$; adding magnitudes instead discards the molecular geometry. A symmetric linear molecule with two equivalent opposed bond dipoles has ${\mathbf{p}}_1=p_0\hat{\mathbf{x}}$ and ${\mathbf{p}}_2=-p_0\hat{\mathbf{x}}$, hence zero total electric dipole moment even though each chemical bond is polarized. Carbon dioxide provides the familiar molecular example; water's bent molecular geometry permits a nonzero total.^[ref]

For an original geometry calculation, give a fictitious three-site model $A{B}_2$ a central effective electric charge $+2q$ at the origin and two terminal electric charges $-q$ atHere $r$ is the stipulated bond length, $\theta$ the angle between the two connections, and $q>0$ is independent of any electronegativity scale. Equivalently, split the central $+2q$ into two colocated $+q$ contributions, one for each neutral bond dipole. With $p_0=qr$, the physical electric dipole moments point from the terminal sites towards the central site:The sideways components cancel; the components along the angle bisector add. Thus $|{\mathbf{p}}_{\mathrm{mol}}|=2p_0\cos (\theta /2)$. At $\theta =180^{\circ }$ the result is zero; at the stipulated $\theta =120^{\circ }$ it is exactly $p_0$. This is an angle example for a model, not the measured angle or electric dipole moment of water.

Opposition alone is insufficient: a linear molecule with unequal opposed bond dipoles of magnitudes $p_1$ and $p_2$ retains magnitude $|p_1-p_2|$. Conversely, symmetry can enforce exact cancellation beyond a particular bond dipole partition. If the complete charge density invariant under point reflection about the origin satisfies $\rho (\mathbf{r})=\rho (-\mathbf{r})$, contributions to $\int \mathbf{r}\rho (\mathbf{r})d^{3}r$ cancel in pairs. That proof uses the whole electric charge distribution, including lone pairs and atomic nuclei. It explains why molecular symmetry is stronger evidence for zero permanent electric dipole moment than a rule based only on tabulated electronegativity differences.

A permanent molecular electric dipole moment is specified relative to the molecule's orientation. An ensemble with random orientations can have zero mean laboratory vector while each molecule is polar. That orientational average differs from internal molecular dipole cancellation. The Stark effect and molecular beam electric resonance probe molecular electric dipole moments through state-dependent responses to an electric field.

A point reflection about $c$ sends each point $x$ to $2c-x$. The point $c$ is the midpoint of $x$ and its image, and applying the map twice is the identity. It is a Euclidean isometry since $(2c-x)-(2c-y)=-(x-y)$ preserves the Euclidean norm. In the Euclidean plane it is a rotation through $\pi$; in dimension $n$ its determinant is $(-1{)}^n$, so it reverses orientation exactly in odd dimension. A figure invariant under this map has $c$ as a center of symmetry.

Let a simple closed regular planar curve be invariant under point reflection through a center and remain strictly on one side of a coplanar axis of rotation, at distance $d$ from its center. If $L$ is its arc length, its surface of revolution has surface area $2\pi dL$. To prove this, choose coordinates with the center at the origin and the axis of rotation at $y=-d$. The surface area formula is $2\pi \oint (d+y)ds$. Pair each point with its reflected point: the arc length element $ds$ is unchanged and $y$ changes sign. Thus $\oint yds=0$, proving the result. The one-side condition prevents the surface from passing through the axis of rotation or requiring an absolute value with changing sign.