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I have to show that the length $l(\gamma)$ and the area $\mathcal{A}(\gamma)$ are unchanged by applying an isometry to $\gamma$.

Let $M$ be the isometry, then $M(v)=Pv+a$.

We define $\tilde\gamma = (\tilde x,\tilde y) = M(x,y) = M(\gamma)$. Then: $$\mathcal{A}(M(\gamma))=\mathcal{A}(\tilde\gamma)=\frac{1}{2}\int_0^T (\tilde x\dot{\tilde y}-\tilde y\dot{\tilde x})dt=\frac{1}{2} \int_0^T ||\tilde \gamma \times \tilde \gamma'||dt \\ =\frac{1}{2} \int_0^T ||M(\gamma) \times \frac{d}{dt}M(\gamma)|| dt \\ =\frac{1}{2} \int_0^T ||(P \gamma +a) \times \frac{d}{dt}(P \gamma +a)|| dt \\=\frac{1}{2} \int_0^T ||(P \gamma+a) \times (P \gamma ')|| dt$$

But it doesn't hold that $||(P \gamma+a) \times (P \gamma ')||=||\gamma \times \gamma'||$, that's why we have to take cases for $M$, if it is a translation or an orthogonal transformation, right?

  • $M(v)=v+a$

Then $A(M(\gamma))=\frac{1}{2}\int_0^T ||(\gamma+a) \times \gamma'||$.

Is this equal to $||\gamma \times \gamma'||$ ?

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It's easier to see directly from the second equality that $\mathcal A$ is invariant under $v\mapsto v + a$. Write $a = (a_1, a_2)$, then $$\begin{split} \mathcal A(\overline \gamma) &= \frac 12 \int_0^T (x+ a_1) \dot y - (y+ a_2) \dot x dt\\ &= \mathcal A(\gamma) +\frac 12\int_0^T a_1 \dot y - a_2 \dot xdt \\ &= \mathcal A(\gamma) +\frac 12\int_0^T (a_1 y - a_2 x)'dt \\ &= \mathcal A(\gamma), \end{split} $$ where the last equality used that $\gamma$ is a closed loop.

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  • $\begingroup$ I see... Then if we suppose that the isometry is an orthogonal transformation, we have that $Mv=Pv$. $A(\tilde \gamma)=\frac{1}{2} \int_0^T \tilde \gamma \times \tilde \gamma' dt= \frac{1}{2} \int_0^T \int_0^T P \gamma \times P \gamma'dt= \frac{1}{2}P det(P) \int_0^T \gamma \times \gamma ' dt= \pm \frac{1}{2}P A(\gamma)$ $$$$ Have I done something wrong? $\endgroup$ – evinda Nov 16 '15 at 21:30
  • $\begingroup$ You need to use that $\|P\gamma \times P\gamma'\| = \|\gamma \times \gamma'\|$. @evinda $\endgroup$ – user99914 Nov 17 '15 at 0:34
  • $\begingroup$ To we justify it just by saying that $P$ is orthogonal? Or could we explai it further? $\endgroup$ – evinda Nov 17 '15 at 0:38
  • $\begingroup$ You can use $\|a\times b\| = \|a\|\cdot \|b\|\sin\theta$, where $\theta$ is the angle between $a, b$. Note that $P$ preserves both the length and the angles. @evinda $\endgroup$ – user99914 Nov 17 '15 at 1:25

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