Compute the degree of $\phi:(x_1,x_2,x_3,x_4) \mapsto (x_1,-x_3,-x_2,x_4)$. Does $\phi$ preserve orientation?

Question

Consider the map $\phi:S^3 \rightarrow S^3$ given by $\phi(x_1,x_2,x_3,x_4)=(x_1,-x_3,-x_2,x_4)$.

Compute the degree of $\phi$. Does $\phi$ preserve orientation?

First I want to point everything that I made:

Some introduction: a regular value of a smooth map $\phi:M \rightarrow N$ is a point $a \in N$ such that for each $x \in \phi^{-1}(a)$ the derivative $D\phi_x$ is surjective.

If $a$ is a regular value of $\phi$ then

$$deg \ \phi=\sum_{x \in \phi^{-1}(a)} sgn \ (det \ D\phi_x)$$

Here in this particular case, $D\phi_x=\begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 0 & -1 & 0\\ 0 & -1 & 0 & 0\\ 0 & 0 & 0 & 1 \end{bmatrix}$, so every $a \in S^3$ is a regular value of $\phi$.

Moreover, $\phi$ is a bijection, so the above sum has only one component and $D\phi_x$ does not depend on $x \in S^3$.

Then $deg \ \phi = -1$.

For the second part of the question, I take the chart $(U,\varphi)$ of $S^3$, where $U=\{(x_1,x_2,x_3,x_4):x_1>0\}$ and $\varphi:U \rightarrow \mathbb{R}^3$ given by $\varphi(x_1,x_2,x_3,x_4)=(x_2,x_3,x_4)$.

So $\omega=-\frac{1}{x_1}dx_2 \wedge dx_3 \wedge dx_4$ is a well defined orientation on $U$.

If we can show that $\omega \circ \phi (=\phi^* \omega) = -\omega$, then we can say that $\phi$ does not preserve orientation (I have a feeling that $\phi$ does not preserve orientation, but if the point was to prove that it does preserve orientation, we have to check every chart in an atlas of $S^3$, which is not quick and interesting at all)

So $\omega \circ \phi = -\frac{1}{x_1}d(-x_3) \wedge d(-x_2) \wedge dx_4 = +\frac{1}{x_1}dx_2 \wedge dx_3 \wedge dx_4 = -\omega$.

Then $\phi$ does not preserve orientation.

Is this all correct? Something that I missed?

Thanks!

Answer 1

What you wrote for the degree is not correct because $D\phi_x : T_x\mathbb{R}^4 \to T_x\mathbb{R}^4$, you need to consider the map $D\phi_x : T_xS^3 \to T_xS^3$ which may have a different determinant. You can rectify this by considering a specific point $x$ where the tangent plane $T_xS^3$ is easy to identify. For example, choosing $x = (1, 0, 0, 0)$, then under the isomorphism $T_{(1,0,0,0)}\mathbb{R}^4 \cong \mathbb{R}^4$, the subspace $T_{(1,0,0,0)}S^3$ is given by $\{0\}\times\mathbb{R}^3$. Therefore, the map $D\phi_{(1,0,0,0)} : T_{(1,0,0,0)}S^3 \to T_{(1,0,0,0)}S^3$ is the linear map with standard matrix

$$\begin{bmatrix} 0 & -1 & 0\\ -1 & 0 & 0\\ 0 & 0 & 1 \end{bmatrix}$$

which indeed does have determinant $-1$, and hence the degree of $\phi : S^3 \to S^3$ is $-1$. The reason the two coincide is that $D\phi$ preserves the normal vector field to $S^3$, whenever this is true, the two determinants will be the same.

A diffeomorphism preserves orientation if it has degree $1$, while it reverses orientation if it has degree $-1$. Therefore $\phi$ reverses orientation, you don't need to do any extra work.

If you wanted to see this without passing to a chart, you could use a volume form on $S^3$. As I explain in this answer,

$$\omega = x^1dx^2\wedge dx^3\wedge dx^4 - x^2dx^1\wedge dx^3\wedge dx^4 + x^3dx^1\wedge dx^2\wedge dx^4 - x^4dx^1\wedge dx^2\wedge dx^3$$

is a volume form on $S^3$. It is easy to compute that $\phi^*\omega = -\omega$ (which again shows that $\deg(\phi) = -1$), and hence $\phi$ reverses orientation.