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I found this affirmation in a lot of books and articles, but I don't know how like to prove. Really, I very confused with this affirmation. I understand by a rigid motion a map $f:(X,d_1)\to (X,d_2)$ between two metric spaces $(X,d_1)$ and $(X,d_2)$, such that

$$d_1(x,y)=d_2(f(x_1),f(x_2)),$$

i.e, $f$ is a isometry. Is it ok?

Because, I know other definition to isometric maps, where I have a similar condition to $d_1(x,y)=d_2(f(x_1),f(x_2))$ over the differential map of $f$.

In the second, can I to define the Laplacian Operator in a metric space? I need at least of a normed vectorial space, or not? If yes, my rigid motion must be a linear isometry?

Finally, someone has some hint to prove this affirmation?

Thanks you.

Irddo
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2 Answers2

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That the Laplacian in ${\mathbb R}^n$ is invariant with respect to rigid motions follows from its "geometric" description in this answer:

Intuitive interpretation of the Laplacian

Community
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Christian Blatter
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This assertion is specific to rigid motions in $\mathbb{R}^n$. (And maybe on a Riemannian manifold, though I don't know too much about that situation yet.) You need at least a manifold with some Hermitian inner product structure to have a meaningful concept of the (classical) Laplacian.

Edit: As John notes, there are other contexts where the Laplacian makes sense, but just a metric isn't enough.

As for how to prove this in $\mathbb{R}^n$, this works because the origin-preserving rigid motions of $\mathbb{R}^n$ are given by the orthogonal group. You can express the Laplacian of a function $u$ in Cartesian coordinates, then apply an orthogonal change of variable and fiddle around with indices until you see (via orthogonality) that you get the same value.

Gyu Eun Lee
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