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Let $D_{4}$ be the dihedral group of a square, $r$ and $s$ denote rotation by $90^{\circ}$ in the clockwise direction and reflection respectively. Suppose through a series of reflections and rotations you attained a reflection prove that there are odd number of reflections you have done.

It can be proved by using the relation $rs=sr^{-1}$ and that every element of the dihedral group can be written uniquely as $ r^is^j,0\leq i\leq3,0\leq j\leq1$.

I'm looking for a proof by the action of dihedral group on a polynomial say $f(x_1,x_2,x_3,x_4) \in \mathbb{Z}[x_1,x_2,x_3,x_4]$ (I label the vertices from 1 to 4 cyclically) such that sign of the polynomial changes when we apply a reflection and its unchanged by a rotation.

I tried using the polynomial $f(x_1,x_2,x_3,x_4) = \prod_{1\leq i < j \leq 4} (x_i-x_j)$ but it gives a negative sign for a rotation.

Thanks in advance

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    $\begingroup$ You can proof this geometrically. $D_4$ is the set of isometries of a square in the plane. Rotations preserve the orientation of the plane, whereas reflections reverse it. There may be a clever way to phrase this as a polynomial. $\endgroup$ – quarague Aug 6 at 13:22
  • $\begingroup$ @quarague I tried taking determinant of the matrix whose $ith$ row is $[x_i ... x_4 x_1..x_{i-1}]$ but still rotations change the sign of the determinant. Could this be altered a bit to get the proof? $\endgroup$ – nrynn Aug 6 at 14:20
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Might this be the kind of thing you're after? Take $$ f(x_1,x_2,x_3,x_4)=x_1x_2x_3^2+x_2x_3x_4^2+x_3x_4x_1^2+x_4x_1x_2^2 $$ For any point in $\Bbb Z^4$, cyclically permuting the coordinates clearly doesn't change the value of the function, but flipping them does: $$ f(0,1,2,3)=18\neq 6=f(3,2,1,0) $$ Or, if you want a polynomial such that cyclic permutation leaves it unchanged while flipping just changes the sign, then $f(x_1,x_2,x_3,x_4)-f(x_4,x_3,x_2,x_1)$ is the one you're looking for.

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  • $\begingroup$ That works.Thanks. Could you please give some motivation on how you got the function? $\endgroup$ – nrynn Aug 6 at 16:58
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    $\begingroup$ @nrynn I knew that whatever I did, as long as I added together all the four corresponding cyclicly permuted terms, it would be unchanged by cyclic permutation. But whatever I did, each term still had to convey "direction". I started with $\frac{x_1x_3}{x_2}$ (one can always multiply away denominators afterward), but that didn't change when flipped; the middle term was the special one. So I changed to $\frac{x_1x_2}{x_3}$, which works. Add all four cyclic variations, multiply by $x_1x_2x_3x_4$ to make a polynomial, and you basically have my $f$. I did some simplifications, and that was it. $\endgroup$ – Arthur Aug 6 at 17:05
  • $\begingroup$ Looking at it now, $x_1x_2^2$ would've been enough, but what's done is done. $\endgroup$ – Arthur Aug 6 at 17:06
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I had asked for a polynomial in 4 variables and Arthur has answered it but one can also consider this polynomial with 8 variables $$ \begin{vmatrix} 1 & x_1 & y_1 \\ 1 & x_2 & y_2 \\ 1 & x_3 & y_3 \\ \end{vmatrix} + \begin{vmatrix} 1 & x_2 & y_2 \\ 1 & x_3 & y_3 \\ 1 & x_4 & y_4 \\ \end{vmatrix} + \begin{vmatrix} 1 & x_3 & y_3 \\ 1 & x_4 & y_4 \\ 1 & x_1 & y_1 \\ \end{vmatrix} + \begin{vmatrix} 1 & x_4 & y_4 \\ 1 & x_1 & y_1 \\ 1 & x_2 & y_2 \\ \end{vmatrix} $$ One can check that rotations don't change the polynomial but reflections change the sign of the polynomial (by the property of the determinants). Motivation for this polynomial is consider the square in a cartesian plane with vertices $(x_i,y_j)$ cyclically. Then the above polynomial is twice the area of the square but when we reflect, the area remains the same but the orientation of the square is changed.

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