# Could someone explain chirality from a group theory point of view?

While answering this question my interest in the rotation/reflection group was piqued.

I personally know very basic group theory, not much more than what a group really is. I understand that the techniques I used in the answer to that question are similar to group theoretical techniques.

Could someone explain how chirality appears in $n$ dimensional objects living in $m>n$ dimensional space if and only if $n=m$ using group theory? Preferably (but not necessarily) giving some background on the techniques being used to better suit my current understanding of group theory.

Note thatwhen I say an $n$ dimensional object, I mean an object that can occupy a minimum of $n$ dimensions. A helix is thus a "three dimensional object", even if its topological dimension is 1.

Apologies if the question is too broad.

• I am confused by "$m\gt n$ dimensional space if and only if $n=m$".
– robjohn
Oct 20, 2013 at 13:50

When a dimension is added, one can rotate the axis of reflection and the added dimension by $180^\circ$ so that the reflection in the lower dimension is realizable as a rotation in the higher dimension. In the higher dimension, there is a rotation (determinant $1$) and a reflection (determinant $-1$) that affect the lower dimension in the same way.

To be precise (in the spirit of this answer), consider a reflection in $\mathbb{R}^n$ that negates $x_n$ $$A=\begin{bmatrix} 1&0&0&\dots&0\\ 0&1&0&\dots&0\\ 0&0&1&\dots&0\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ 0&0&0&\dots&-1 \end{bmatrix}$$ This can be lifted to an isometry in $\mathbb{R}^{n+1}$ in two ways, as a reflection (determinant $=-1$) $$B=\begin{bmatrix} 1&0&0&\dots&0&0\\ 0&1&0&\dots&0&0\\ 0&0&1&\dots&0&0\\ \vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\ 0&0&0&\dots&-1&0\\ 0&0&0&\dots&0&1 \end{bmatrix}$$ or as a rotation of $\theta=\pi$ (determinant $=1$) $$C=\begin{bmatrix} 1&0&0&\dots&0&0\\ 0&1&0&\dots&0&0\\ 0&0&1&\dots&0&0\\ \vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\ 0&0&0&\dots&\cos(\theta)&-\sin(\theta)\\ 0&0&0&\dots&\sin(\theta)&\cos(\theta) \end{bmatrix}$$ Since this determinant is a continuous function and the determinant of an isometry is $\pm1$, we cannot get from the identity to $A$ or $B$ via rotations. However, we can get to $C$ via rotation in $\mathbb{R}^{n+1}$, and the action of $C$ is indistinguishable from the action of $A$ as viewed within $\mathbb{R}^n$. Thus, chirality is lost when we add even one dimension.

Of course, you posted this answer, so perhaps I am misunderstanding your question.

• @Manishearth: I am sorry I can't help with algebraic understanding beyond linear algebra. I hope this helps, but let me know if I am totally missing the point.
– robjohn
Oct 20, 2013 at 16:35
• @Landscape: not uncomfortable at all. Your comments were constructive and my answer is better now. Thanks.
– robjohn
Oct 20, 2013 at 21:38
• You are welcome and thank you for your understanding.
– 23rd
Oct 20, 2013 at 21:42
• @robjohn I'd asked this question to get a better understanding of group theory. I already know the matrix approach :P Oct 21, 2013 at 5:50
• @Manishearth: I figured as much, considering you had written the answer I cited. I meant no offense, but this is what I could offer, and there weren't any other answers.
– robjohn
Oct 21, 2013 at 7:38