# Do symmetric 3d solid figures always have a plane of symmetry?

Chapter 3 of Martin Gardner's The New Ambidextrous Universe begins as shown below. As you can see (highlighted), on page 13 he writes that not all solid symmetric objects have a plane of symmetry, and gives an example. On the next page, however, he then writes "to be symmetric a solid must have at least one plane of symmetry."

Can someone explain this apparent contradiction?

• Although I hvae much respect for Martin Gardner, there is a lot in this passage that I don't agree with. Already the first sentence: a propeller is neither mirror-symmetric (since then it wouldn't propel) nor is it asymmetric (it has $n$-fold rotational symmetry fo some $n$, usually $n\in\{2,3,4\}$). It is not clear what he means with the term "asymmetric" if he does talk about rotational symmetry. – Marc van Leeuwen Feb 27 '13 at 8:29

The text is formulated confusingly, and indeed as you note the second highlighted passage (which is false) contradicts the first; nevertheless there is some sound mathematical content that it wants to transmit. First of all it should be noted that the text uses "asymmetric" to mean "chiral" (object that are distinguishable from their mirror images, even when they may be rotated freely), which is a property that does not exclude rotational symmetry (a propeller in 3D, or in the 2D case a letter Z or S or a swastika, are examples of chiral but rotationally symmetric figures).

An linear isometry has determinant either $+1$ or $-1$, and is accordingly called orientation-preserving or orientation-reversing. Its (real-valued) matrix is also diagonalisable over the complex numbers, with all its eigenvalues on the unit circle, of which the non-real ones come in complex conjugate pairs (these correspond to a rotation component in the isometry). Since the determinant is the product of the eigenvalues, and the product of a pair of complex-conjugates on the unit circle is always $+1$, a linear isometry is orientation-reversing if and only if it has an eigenvalue $-1$ with odd multiplicity. In the plane this means that it has one eigenvalue $-1$ and one eigenvalue $+1$, in it therefore is a reflection in a line (through the origin). Bounded plane figures that are achiral (similar to their mirror image) must have a line of symmetry; there just aren't any other orientation-reversing linear isometries in dimension $2$.

In higher dimensions one still has the possibility of a single eigenvalue $-1$ and all other eigenvalues $+1$, and such isometries are called reflections, giving a direct mirror symmetry. (Some people confusingly allow for more than one eigenvalue $-1$ for a reflection, talking for instance about reflection in a line in $3$-space; I will however require reflections in $3$-space to fix an entire plane, and in higher dimension reflections must fix an entire hyperplane). However there are now other possibilities, notably in dimension $3$ one can combine a simple eigenvalue $-1$ with a pair of complex conjugate eigenvalues (giving a rotary reflection), or one can have a triple eigenvalue $-1$, the case of a central symmetry.

In both cases one can arrange that the isometry generates a finite subgroup of the orthogonal group that does not contain any reflections, and find a solid that has exactly that group of symmetries: an achiral solid that does not have any plane of mirror symmetry. This is easiest to see for the central symmetry which generates a $2$-element subgroup, and the "rectangular card with opposite corners folded" is intended to to illustrate this case. This shape does look the same as its mirror image: its mirror image can be brought back to the original shape by a $180^\circ$ rotation about an axis perpendicular to the plane of the mirror (which incidentally shows that a central symmetry is a limiting case of a rotary reflection, with the pair of conjugate eigenvalues going to $-1$). Another way to obtain this symmetry group is to start with a cube, and colour its corners by diametrically opposite pairs using $4$ different colours; then a colour-preserving isometry must stabilise each of the four diagonals, and the only non-trivial isometry that does this is the central symmetry. One may replace the colouring by cutting off different sized caps fom the corners, so as to obtain a pure (convex) solid with only central symmetry, and therefore no plane of reflection.

For the case of a rotary reflection one can make an example by starting with an $n$-blade propeller, add its mirror image by a plane orthogonal to the axis, and then rotate the mirror image by one $2n$-th turn (half the angle between the blades) so as to destroy the mirror symmetry. For a more classical (and convex) kind of solid, one could start with a uniform anti-prism and cut off an irregular cap from each corner so as to leave as symmetries only the subgroup generated by the rotary reflection that stabilises the anti-prism.

As a final remark, if one allows affine isometries, which have a translation component and do not fix the origin, then there are non-reflection orientation-reversing isometries already in dimension $2$: the glide reflections. They cannot be symmetries of a bounded figure (which has to fix say the center of the figure in some appropriate sense), but they can be symmetries of unbounded figures, as is illustrated by the following frieze pattern:

For one thing, the claim about Figure 10 is debatable: it does not "look the same in a mirror". I checked:

Figure 10 is centrally symmetric, that is, it's invariant under point reflection. In the plane, point reflection is the same as rotation by 180 degrees. So one can also say that it has rotational symmetry (of order 2, because we rotate by $1/2$ of full turn).

Of course, it's up to the author which sets to call symmetric. Maybe he contemplated the central symmetry before deciding that in this book symmetric should mean "invariant under reflection in a line/plane". Or maybe the second highlighted sentence is really meant to emphasize the possibility of having several planes of symmetry, not to formalize the concept of symmetric. At least this is how I read it.

• It looks the same as a flipped-over copy of itself in the mirror. I think that is what the author meant: It is not invariant under any nontrivial rotation, or any mirror reflection, but it is invariant under a mirror reflection followed by a rotation (i.e. it is point symmetric in 3D). – Rahul Feb 26 '13 at 22:26
• @ℝⁿ. After reading "such an object is said to have rotational symmetry" I concluded that the figure must be a planar object with 2-fold rotational symmetry. Now I see it was really 3-dimensional, with corners bent up/down. Well, this is not what I would call rotational symmetry. But it's the author's decision anyway. – user53153 Feb 26 '13 at 22:39
• I agree, I can't make much sense of the choice of terminology either. – Rahul Feb 26 '13 at 22:43