Rotations have degree $1$

Question

In p.233 of Hatcher's Algebraic Topology, there is a statement that a rotation of a sphere $S^n$ has degree, homotopic to the identity. I have some questions about these.

1. How do we define a rotation in $\Bbb R^n$?

2. In the above definition, how can we show that it is homotopic to the identity?

Answer 1

1. As @CyclotomicField said, I'll interpret a rotation of $\mathbb{R}^n$ to be an element of $\text{SO}(n)$, i.e. an orientation-preserving Euclidean isometry fixing the origin. Equivalently, they are $n\times n$ real matrices $R$ with $\det R = 1$ and $R R^T = 1$.

2. There is a nice factorization of a rotation $A \in \text{SO}(n)$ into $A = QRQ^T$, where

   • $Q$ is orthogonal,
   • $R$ is block-diagonal, with blocks being $2 \times 2$ rotation matrices $R_{\theta}$, or possibly a 1 if $n$ is odd.

   See Corollary 0.2 in this explanation by J. Maurice Rojas, for example. This factorization recovers the familiar result that a 3-D rotation is described uniquely by an axis and an angle of rotation about that axis.

   Now, to deform $R$ to the identity, simply "dial down" the parameters $\theta_i$ in the block decomposition of $R$ until each $\theta_i = 0$.

Answer 2

1. Define an $n$-dimensional axis and a rotation angle $0 \leq \theta < 2 \pi$ around it.
2. We can always continuously apply a rotation to an $n$-sphere to bring it back to its original orientation (identity). That defines the homotopy.