Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Is it possible to define a derivative for rigid transformations eg. reflection and translation?

I am especially interested on reflections shortly $\sigma$.

Because I am trying to relate $\int_{A}f(\sigma x)dx$ with $\int_{\sigma A}f(x)dx$, where ,say, f is an integrable function $f:X\to [0,1]$ and $A\subset X$ Borel set.

So if we try change of variables, then we need $det(D \sigma)$. Feel free to go as advanced and technical as you like.

Does $lim_{h\to 0} \frac{\sigma(x+h)-\sigma(x)}{h}$ exist? I think it equals 1 because

$|x-y|=|\sigma(x)-\sigma(y)|\Rightarrow lim_{h\to 0} |\frac{\sigma(x+h)-\sigma(x)}{h}|= lim_{h\to 0} \frac{|x+h-x|}{h}=1$


share|improve this question
As a linear transformation (it may actually be an affine mapping, but the translation part does not change things here) a reflection typically has a single eigenvalue $-1$ and several $+1$:s. So on a line I would expect to see $-1$. In higher dimensions a matrix with appropriate eigenvalues. In an infinite dimensional space...? –  Jyrki Lahtonen May 22 '14 at 6:17

1 Answer 1

Rigid motions are just multiplications with a matrix (plus a translation), i.e. $x\mapsto Ax+b$. Then the derivative is simply $A$. (In fact, to be "rigid", i.e. an isometry, wwe require some conditions about $A$, but even without these conditions the derivative of $x\mapsto Ax+b$ is simply $A$).

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.