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This is a diagram of quadrilaterals and their duals. Within this diagram, there is a square (shown with 8 lines of symmetry) and right below that there is a "gyrational square" shown with no lines of symmetry but nonetheless given an orientation on its boundary.

What is a "gyrational square" in the context of Euclidean geometry?

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This is a description of the group of isometries that let the square invariant.

This group, called r8, has $8$ elements: $4$ axial symmetries, $3$ rotations (including the central symmetry), and identity. Rotations are composition of $2$ axial symmetries, by the way.

This group has subgroups, which have also a name (given by Conway, according to Wikipedia). For each proper subgroup, there are other quadrilaterals than the square, that exhibit the same symmetries. Except for g4 subgroup, which contains the $3$ rotations and identity: squares are the only quadrilaterals being invariant with these isometries.

What the diagram in Wikipedia and the name "gyrational square" mean, is that the g4 subgroup makes the square invariant but has one more property: it does not change the order of vertices. After rotation, the $4$ vertices are still in the same order clockwise, while an axial symmetry makes them anti-clockwise. So the object that has exactly this g4 group of isometries is the gyrational square ("oriented square" would perhaps have been clearer). In the same way as the non-square rectangle is the quadrilateral that has exactly the p4 group of isometries. But it is somewhat inhonomogeneous because we actually add a property (the vertices order orientation) that is not taken into account for other classes of quadrilaterals in the diagram.

So if we were in a context where orientation had to be taken into account, the group of isometries for a (gyrational) square would be g4, not r8. An example of such context is in algebraic topology, when using simplices: they are oriented. But of course squares are not simplices. And the situation with squares is more complex than with triangles: $4$ vertices actually have $6$ equivalence classes when considering rotational order: $(abcd), (abdc), (acbd), (acdb), (adbc), (adcb)$.
But only $2$ of them can be attained from $(abcd)$ by an isometry: $(abcd)$ and $(adcb)$. So g4 transforms the gyrational square $(abcd)$ into the same gyrational square $(abcd)$, while other isometries in r8 transform it into $(adcb)$, which we can call the opposite orientation gyrational square.

To find problems where square orientation is taken account, "oriented square" and "positively oriented square" seem to give much more Google answers than "gyrational square".

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  • $\begingroup$ What a remarkably clear answer. The best I've seen on math.stackexchange! Thank you. $\endgroup$
    – Fomalhaut
    Commented Oct 26, 2023 at 0:05

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