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It seems pretty intuitive, and I have seen many sources make this claim. From what I've seen, congruence in geometry is defined as a direct isometry between two shapes, that is, an isometry that preservers handedness.

Here is what Wikipedia says: "The direct isometries comprise a subgroup of E(n), called the special Euclidean group. They include the translations and rotations, and combinations thereof; including the identity transformation, but excluding any reflections."

However, I haven't found any proof that the set of all direct isometries is composed solely of rotations and translations, or that these transformations do preserve handedness.

This might be a well-known result, but I'm new to geometry, so if anyone could suggest some books related to this topic, that would be greatly appreciated.

Jambon Au Chocolat
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I don't have a book on hand for this, but it's a pleasant exercise to verify that isometries of Euclidean space are of the form $x \mapsto Mx + b$ where $b$ is the translation vector and $M$ is an orthogonal matrix (I'm ignoring orientation for now).

It requires a bit of vector algebra (not only to prove, but even to formulate). Let $x \mapsto f(x)$ be an isometry on $\mathbb{R}^n$. Write $g(x) = f(x) - f(0)$; then $g$ is an isometry such that $g(0) = 0$. It remains to show that $g$ is represented by an orthogonal matrix. Notice

$$(g(x) - g(y)) \cdot (g(x) - g(y)) = (x - y) \cdot (x - y) \qquad (1)$$ (both sides represent squared distances); in particular, using $g(0) = 0$,

$$g(x) \cdot g(x) = x \cdot x, \qquad g(y) \cdot g(y) = y \cdot y. \qquad (2)$$ Combining (1) and (2) appropriately, and using distributivity and symmetry properties of the dot product, derive

$$g(x) \cdot g(y) = x \cdot y \qquad (3)$$ for all $x, y \in \mathbb{R}^n$. Then

$$g(x + x') \cdot g(y) = (x + x') \cdot y = x \cdot y + x' \cdot y = g(x) \cdot g(y) + g(x') \cdot g(y) = (g(x) + g(x')) \cdot g(y)$$ and so

$$[g(x) + g(x') - g(x + x')] \cdot g(y) = 0.$$ Considering the cases $y = x, x', x+x'$ and combining them, one easily derives

$$\|g(x) + g(x') - g(x+x')\|^2 = 0$$ whence $g(x) + g(x') = g(x+x')$ for all $x, x' \in \mathbb{R}^n$. Similar sorts of arguments show that $g(rx) = r g(x)$ for all scalars $r \in \mathbb{R}$. So $g$ is a linear transformation, represented by a matrix $M$ which by (3) satisfies $Mx \cdot My = x \cdot y$, and therefore $M$ is an orthogonal matrix, completing the argument.

user43208
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  • Thank you for your response! I'm not very familiar with analytic geometry, so I'm not seeing how being an affine function implies that it must be the composition of translations, rotations and reflections. – Jambon Au Chocolat Feb 28 '24 at 19:28
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    Referring to the Wikipedia article on rigid transformations, https://en.wikipedia.org/wiki/Rigid_transformation, a rotation is represented by a rotation matrix, which (see https://en.wikipedia.org/wiki/Rotation_matrix) is an orthogonal matrix $M$ with determinant 1 (orthogonal matrices always have determinant $\pm 1$; the positive determinant requirement relates to handedness). The transformation $x \mapsto Mx + b$ is the composition of the rotation $x \mapsto Mx$ followed by the translation $x \mapsto x + b$. – user43208 Feb 28 '24 at 19:35
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    (Or rather, $M$, being an orthogonal matrix, is either a rotation matrix = orthogonal matrix with det 1, or, if its determinant is $-1$, it is a the composition of a reflection $R$ with a rotation matrix, where $R$ is for example a matrix whose first column is $-e_1$, and the other columns $e_i$, the standard basis elements. This $R$ has det $-1$, so $RM$ in this case will have det $1$: a rotation matrix $O$, whereby $M$ is a composite $RO$.) – user43208 Feb 28 '24 at 19:42
  • So, @JambonAuChocolat: does this make sense to you? Or do you have any questions? – user43208 Feb 29 '24 at 12:14
  • It was a very clear explanation. I don't have any more questions, thank you. – Jambon Au Chocolat Mar 01 '24 at 09:06