Are reflections in higher dimensions commutative?

Question

If $~r_1, r_2 ,\cdots, r_t~$ are a sequence of reflections about $~x_1=\frac{1}{2},\cdots,x_n=\frac{1}{2}~$ respectively in n dimensions does the sequence of reflections matter?

Answer 1

First answering the original version of the question.

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Absolutely! The order matters.

In $\Bbb{R}^n$ the orthogonal reflection $s_{ij}$ with respect to the hyperplane $x_i=x_j$ interchanges the two involved coordinates: $$s_{ij}:(x_1,\ldots,x_i,\ldots,x_j,\ldots,x_n)\mapsto (x_1,\ldots,x_{i-1},x_j,x_{i+1},\ldots,x_{j-1},x_i,x_{j+1},\ldots,x_n). $$ In other words, it permutes the coordinates according to the 2-cycle $(ij)$. But from an encounter with permutation groups you probably remember that 2-cycles don't always commute. For they generate the entire symmetric group $S_n$, which is non-commutative when $n\ge3$.

• As Lee Mosher explained, an even simpler case of non-commuting reflections is with respect to two parallel hyperplanes. In that case the commutator of the two reflections is a translation.
• But, if two hyperplanes, $H_1$ and $H_2$, in $\Bbb{R}^n$ are orthogonal to each other (= if their respective normal vectors $\vec{n}_1,\vec{n}_2$ are orthogonal), then they do commute.

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Editing the answer to reflect the specification made to the question, if only a bit late :-)

So the hyperplanes $H_i$ determined by the equations $x_i=1/2$, with respective normals $\vec{n}_i=\vec{e}_i$, are all orthogonal to each other, the corresponding reflections do commute. In fact, reflection w.r.t. $H_i$ maps the point $$(x_1,\cdots,x_n)\mapsto (x_1,x_2,\ldots,x_{i-1},1-x_i,x_{i+1},\ldots,x_n).$$ Affecting only the $i$th coordinate, two such mappings commute.

Answer 2

Presumably you already know this is true in dimension $1$: for example, reflection of the real line across $x=0$ and across $x=1$ do not commute.

This is easily convertible into an example in any number of dimensions. Consider $\mathbb R^n$ with coordinates $x_1,...,x_n$: reflection across the plane $x_1=0$ and across the plane $x_1=1$ do not commute.