If we think of reflection in $\mathbb{R}^1$ (multiplication by $-1$), this can be represented as $180$ degree rotation in euclidean plane (assuming a "natural embedding" notion of $\mathbb{R}^1$ into $\mathbb{R}^2$).

Similarly all reflections of the square in the euclidean plane can be represented as rotations in $\mathbb{R}^3$ (around the axis of reflection).

I am wondering if there is a general principle here that any reflection in $\mathbb{R}^n$ can be represented as rotation in a higher dimension "in a natural way"?


3 Answers 3


Yes! A reflection $r$ (across a hyperplane through the origin) in $\mathbb{R}^n$ is a linear transformation with $n-1$ linearly independent eigenvectors of eigenvalue $1$ (these span the hyperplane), together with $1$ eigenvector of eigenvalue of $-1$ (orthogonal to the hyperplane). So, in this eigenbasis, a reflection looks like the block diagonal matrix, $$ \left( \begin{array}{c|c} I_{n-1} & 0 \\ \hline 0 & -1 \end{array}\right) $$ where $I_k$ is the identity matrix of size $k$.

Now consider the space sitting inside euclidean space of one dimension greater, via the embedding $\iota: \mathbb{R}^n \hookrightarrow \mathbb{R}^{n+1}$, mapping $(x_1, \dots, x_n) \mapsto (x_1, \dots, x_n, 0)$, as well as the projection $\pi: \mathbb{R}^{n+1} \to \mathbb{R}^n$, mapping $(y_1, \dots, y_n, y_{n+1}) \mapsto (y_1, \dots, y_n)$.

In this larger space, the proper rotation $\rho$ (i.e. in the special orthogonal group) $$ \left( \begin{array}{c|c} I_{n-1} & 0 \\ \hline 0 & -I_2 \end{array}\right) $$ has the property that it acts identically to the reflection on the $n$-dimensional subspace with final coordinate $0$. In other words, the following diagram commutes: $\require{AMScd}$ \begin{CD} \mathbb{R}^n @>{r}>> \mathbb{R}^n \\ @V{\iota}VV @AA{\pi}A \\ \mathbb{R}^{n+1} @>{\smash[t]{\rho}}>> \mathbb{R}^{n+1} \end{CD}

In other words, $r = \pi \circ \rho \circ \iota$.


Yes. If $L$ is a reflection in $\mathbb R^n$, $M: \mathbb R^{n+1}\to\mathbb R^{n+1},(\vec x,x_{n+1})\mapsto(L\vec x,-x_{n+1})$ is a rotation in $\mathbb R^{n+1}$. You can see this from the matrix representations of these linear maps: In the canonical basis, the matrix for $M$ has a block structure, with the matrix for $L$ in the top left and a $-1$ in the bottom right. By Laplace expansion, $\det M=-\det L=-(-1)=1$, and you can check that since $L$ is orthogonal, so is $M$. So $M$ is an orthogonal matrix with positive determinant, and thus a rotation matrix.


Yes. Reflection through the origin in even dimensions has determinant positive one, and is always rotation by 180 degrees. You can just pair off the dimensions and rotate in each plane. Like so:

$$ (x_1\cos\theta + x_2\sin\theta,x_2\cos\theta -x_1\sin\theta,\cdots,x_{2n-1}\cos\theta + x_{2n}\sin\theta,x_{2n}\cos\theta -x_{2n-1}\sin\theta). $$

Then as $\theta$ grows from $0$ to $\pi$ radians, you have a rotation from $(x_1,x_2,\cdots, x_{2n-1},x_{2n})$ to the antipodal point $(-x_1,-x_2,\cdots, -x_{2n-1},-x_{2n}).$

And in odd dimensions, reflection through the origin is orientation reversing (determinant is negative one) and so can never be a rotation. But just like you noted, you can always pass to higher dimension. If you view your odd tuple $(x_1,x_2,\cdots, x_{2n-2},x_{2n-1})$ as lying in the first $2n-1$ coordinates of $2n$ dimensional space, so it's actually $(x_1,x_2,\cdots, x_{2n-2},x_{2n-1},0)$, then the rotation considered above looks like

$$ (x_1\cos\theta + x_2\sin\theta,x_2\cos\theta -x_1\sin\theta,\cdots,x_{2n-1}\cos\theta , -x_{2n-1}\sin\theta) $$

which is exactly the higher dimensional analogue of viewing reflection in the line as a rotation in the plane into the negative axis.

  • $\begingroup$ Have you made a typo here because I think reflections always have determinant negative? $\endgroup$
    – nsimplex
    Commented Jan 17 at 7:38
  • $\begingroup$ @nsimplex the reflection through the origin can be represented as a composition of $n$ reflections through $n$ coordinate hyperplanes. Or alternately as an $n\times n$ matrix with all $-1$ on the diagonal. Either representation will show that the determinant is $(-1)^n$ which is $+1$ when $n$ is even. Reflection through the origin in even dimensions is a rotation. That's the whole point of your question and my answer. And rotations have determinant 1. $\endgroup$
    – ziggurism
    Commented Jan 17 at 14:21

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