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I'm trying to prove that the linear transformation in an affine mapping always is an isometry, when the affine mapping is an element of a space group.

I have a proof of this from one paper, but I don't quite understand it. (Paper here: https://www.math.ru.nl/~souvi/krist_09/cryst.pdf)

The affine mapping {g | t} acts on the vectors v of the vector space Rn by {g | t}(v) := g · v + t.

g is the linear transformation and t is the translation

The proof says: Let o be the (chosen) origin of Rn and let φ be an isometry in a space group, then we denote by t the translation by the vector φ(o) − o. Since a translation is an isometry, the mapping φ − t is also an isometry and by construction it fixes the origin o: (φ − t)(o) = φ(o) − (φ(o) − o) = o.

We do not need to assume that φ − t is a linear mapping, since an elementary (but not so wellknown) fact just states that every isometry fixing the origin automatically has to be an invertible linear mapping g. We thus conclude that the isometry φ is the composition of the invertible linear mapping g = φ − t and the translation t, i.e. an affine mapping {g | t}. Moreover, the linear part g has to be an isometry.

My problem is: I want to show WHY the linear part g is an isometry, and not just that it is an invertible linear mapping. I do not quite understand what makes it an isometry.

Is every invertible linear mapping that fixes the origin an isometry?

I am very grateful for any help.

Carl Tudén
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2 Answers2

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the isometry φ is the composition of the invertible linear mapping g = φ − t and the translation t, i.e. an affine mapping {g | t}. Moreover, the linear part g has to be an isometry.

You have $\varphi=g \circ t$, where $\varphi$ and $t$ are isometries, and you ask why $g$ is also an isometry, right?

Compose with $t^{-1}$ on the right: $$\varphi \circ t^{-1}=g\circ t\circ t^{-1}=g.$$ That is, $g$ is a composition of two isometries $g=\varphi \circ t^{-1}$, hence an isometry on its own.

If you're still unsure, here we used:

  1. invertibility of $t$ to compose with $t^{-1}$
  2. associativity of function composition: $(g\circ t)\circ t^{-1}=g\circ (t\circ t^{-1})$
  3. "transitivity" of isometries under composition: if $\varphi$ and $t^{-1}$ are isometries, then their composition is one, too.

You can make a very crude analogy with parity of natural numbers and ask: given $a,b\in\mathbb N$ even and $k+a=b$, why is $k$ also even? Well, it's because you can subtract $a$ on the right and get $k=b+(-a)$, where $(-a)$ is also even, and the sum of two even numbers is even, too.

Al.G.
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  • Thank you so much!! I have one question. To my understanding: this proof only holds if we initially assume the function to be an isometry? As in, we start with the assumption that g*v + t = φ

    Is there a way of just showing that g is an isometry, using the properties that g is an invertible linear mapping fixing the origin?

    Kind regards

     – Carl Tudén Apr 19 '23 at 20:36
  • No, because it is not even true - take the example from @Randall under your question, e.g. $g=2I$ where $I$ is the identity linear operator. Then $g$ is a linear isomorphism (its inverse being $g^{-1}=1/2 I$), but it's not an isometry: $\left\Vert Iv\right\Vert =\left\Vert 2v\right\Vert =2\left\Vert v\right\Vert$, so $v$'s length is not preserved under $g=2I$ if $v\neq 0$. – Al.G. Apr 19 '23 at 20:55
  • You may find it illuminating that my answer does not even mention linearity of $g$. $g$ being an isometry is completely independent of its linearity in this argument. – Al.G. Apr 19 '23 at 20:58
  • And to your first question: To my understanding: this proof only holds if we initially assume the function to be an isometry? - yes, this proof heavily relies on $\varphi$ and $t$ being isometries. – Al.G. Apr 19 '23 at 21:05
  • Glad it helped! Please note that you can indicate which answer helped you most by clicking the green checkmark next to it :) – Al.G. Apr 21 '23 at 07:45
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Firstly, all linear maps fix the origin since $\varphi(0)=0$. Secondly many linear maps are not isometries, even if they're bijective. You can see this easily in $\mathbb{R}$ since for $c \neq 1$ then $f(x)=cx$ is linear but $\vert f(1) \vert = \vert c \vert \neq 1$ so it isn't an isometry.

However if we choose a basis and represent a linear transformation $\varphi$ by the matrix $A$ then we can show the transformation is an isometry if and only if $A$ is orthogonal, which is to say $A^TA=I$ with $A^T$ the transpose of $A$. Towards this end we note that $v \cdot u = v^Tu$ so that $\vert u \vert ^2 = u^Tu$. This means that $\vert Au \vert ^2 = (Au)^TAu = u^TA^TAu = u^Tu = \vert u \vert^2$ and so $\vert Au \vert = \vert u \vert$.

Orthogonal matrices will have determinant $\pm 1$. This makes sense since the determinant can be interpreted as a signed volume and isometries preserve triangles, thus volumes. Geometrically these are rotations and reflections. You can think of translations as changing the origin and the linear part as changing the viewing direction without scaling.

CyclotomicField
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