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I wonder if the intersection of two planes of symmetry for some three dimensional object, is a symmetry axis of that object (i.e. an axis for which there exists an angle (smaller than 360°), so that rotation around this angle maps the object in itself).

I also wonder (if the above statement is true), if I can generalize this:

Consider a function $f: \mathbb{R}^3 \rightarrow\mathbb{R}$. Also consider a plane, we can define then the reflection about this plane ($R : \mathbb{R}^3 \rightarrow \mathbb{R}^3 $). I then would like to call this plane by definition a plane of symmetry for $f$, if $f(P) = f(R(P))$ (where $P \in \mathbb{R}^3$). I define an axis of symmetry for $f$ in the same way: if there is a rotation $R_2$ around the axis so that $f$ takes,for every point $P$,the same values in $P$ as in $R_2(P)$. I wonder here too if the intersection of two planes of symmetry for $f$, is an axis of symmetry for $f$.

Rayman
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1 Answers1

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The answer is definitely "yes". If $R$ and $R'$ are your two reflections, then the product (composition) $R_2=RR'$ is a rotation about the line of intersection. Why is $R_2$ a rotation? Because each of $R$ and $R'$ reverse the orientation, so the product is orientation preserving. Moreover, if $x$ belongs to the line of intersection then $R_2x=R(R'x)=Rx=x$, so $R_2$ fixes every point in that line.

Similarly, $f(R_2x)=f(R(R'x))=f(R'x)=f(x)$ so $f$ has symmetry $R_2$ as well.

Hope that helps.

ps. This argument shows that the set of symmetries of an object or a function forms a group (you also need to show that inverses of symmetries are also symmetries, which is also straightforward).

JamesM
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