# The Antipodal Map is Orientation Preserving iff $n$ is Odd

The following result is well-known.

Theorem. The antipodal map on $S^n$ is orientation preserving if and only if $n$ is odd.

Below I provide a proof in which there must be an error since I reach to a wrong conclusion. Can somebody please point out the error?

Let $a:S^n\to S^n$ be the antipodal map on $S^n$ and $A$ be the antipodal map on $\mathbf R^{n+1}$. Consider the following diagram:

Let $\Omega=dx_1\wedge \cdots \wedge dx_{n+1}$ be an orientation form on $\mathbf R^{n+1}$. Let $X$ be the vector field on $\mathbf R^{n+1}$ defined as \begin{equation*} X = \sum_{i=1}^{n+1} x_i \frac{\partial}{\partial x_i} \end{equation*} This vector field is nowhere tangent to $S^n$. The standard orientation on $S^n$ is determined by the contracting $\Omega$ using $X$ and restricting it to $S^n$. Thus the $n$-form $\omega=i^*(X\lrcorner \Omega)$ determines the standard orientation on $S^n$. Now we have $$\begin{array}{rcl} a^*\omega &=& a^*(i^*(X\lrcorner \Omega))\\ &=& (i\circ a)^*(X\lrcorner \Omega)\\ &=& (A\circ i)^*(X\lrcorner \Omega)\\ &=& i^*(A^*(X\lrcorner \Omega)) \end{array}$$

Note that by the very definition of $A$, we have $A^*\eta=(-1)^{k} \eta$ for any $k$-form $\eta$ on $\mathbf R^{n+1}$.

Thus $A^*(X\lrcorner \Omega)=(-1)^{n}(X\lrcorner \Omega)$ and we get $a^*\omega=(-1)^n\omega$. Therefore $a$ is orientation preserving if $n$ is even and orientation reversing if $n$ is odd.

What is the mistake?

Your mistake is the following statement:

Note that by the very definition of $A$, we have $A^*\eta=(-1)^{k}\eta$ for any $k$-form $\eta$ on $\mathbf R^{n+1}$.

This is not true. Any $k$-form is a sum of terms like $f\,dx^{i_1}\wedge\dots\wedge dx^{i_k}$ where $f$ is some smooth function. If $\eta$ is such a form, then the pullback of $\eta$ by $A$ is $$A^*\eta = (f\circ A) (-1)^k dx^{i_1}\wedge\dots\wedge dx^{i_k}.$$ This is equal to $(-1)^k\eta$ if and only if the coefficient function $f$ is invariant under $A$.

For example, consider the $1$-form $\eta = x^1\, dx^1$. Then $A^* \eta = \eta\ne (-1)^1\eta$.

One way to see what happens when you pull back $X\lrcorner \Omega$ is to note that $A_*X=X$ and $A^*\Omega = (-1)^{n+1}\Omega$. Thus we can compute what the pullback form does to an $n$-tuple of vector fields: \begin{align*} (A^* ( X\lrcorner\Omega)) \left( V_1,\dots,V_n\right) &= (X\lrcorner\Omega) \left( A_*V_1,\dots, A_*V_n\right)\\ &= \Omega \left( X, A_*V_1,\dots, A_*V_n\right)\\ &= \Omega \left( A_*X, A_*V_1,\dots, A_*V_n\right)\\ &= (A^*\Omega) \left( X, V_1,\dots, V_n\right)\\ &= (-1)^{n+1}\Omega \left( X, V_1,\dots, V_n\right)\\ &= (-1)^{n+1}(X\lrcorner\Omega) \left( V_1,\dots, V_n\right). \end{align*} It follows that $A^*(X\lrcorner \Omega) = (-1)^{n+1}X\lrcorner \Omega$.

• Thank you Professor Lee. I was repeatedly making this mistake. – caffeinemachine Dec 7 '15 at 19:04