[proofplan] We prove both directions. For the "if" direction ($n$ odd), we exhibit an explicit nowhere-zero tangent vector field on $S^{2k-1} \subset \mathbb{R}^{2k}$ by pairing coordinates: $v(x_1, y_1, \ldots, x_k, y_k) = (-y_1, x_1, \ldots, -y_k, x_k)$. For the "only if" direction ($n$ even), we show that a nowhere-zero tangent vector field would imply $\deg(a) = +1$ for the antipodal map $a$ on $S^n$, contradicting the [Degree of the Antipodal Map](/theorems/2247), which gives $\deg(a) = (-1)^{n+1} = -1$ when $n$ is even. The key step is constructing a homotopy from the identity to the antipodal map using the normalised vector field as a "rotation axis." [/proofplan] [step:Construct an explicit nowhere-zero tangent vector field when $n$ is odd] Let $n = 2k - 1$, so $S^n \subset \mathbb{R}^{2k}$. Write points of $\mathbb{R}^{2k}$ as $(x_1, y_1, x_2, y_2, \ldots, x_k, y_k)$. Define \begin{align*} v: S^{2k-1} &\to \mathbb{R}^{2k} \\ (x_1, y_1, \ldots, x_k, y_k) &\mapsto (-y_1, x_1, -y_2, x_2, \ldots, -y_k, x_k). \end{align*} **$v(x) \neq 0$ for all $x \in S^{2k-1}$:** We have $|v(x)|^2 = \sum_{j=1}^k (y_j^2 + x_j^2) = |x|^2 = 1$ for all $x \in S^{2k-1}$, so $v(x)$ is a unit vector and in particular never zero. **$v(x) \perp x$ for all $x \in S^{2k-1}$:** The standard inner product gives \begin{align*} v(x) \cdot x = \sum_{j=1}^k (-y_j x_j + x_j y_j) = 0. \end{align*} Since $v(x) \cdot x = 0$ and $x \in S^{2k-1}$ (so $v(x) \perp x$), the vector $v(x)$ lies in the tangent space $T_x S^{2k-1} = \{w \in \mathbb{R}^{2k} : w \cdot x = 0\}$. Therefore $v$ is a nowhere-zero tangent vector field on $S^{2k-1}$. [/step] [step:Show that a nowhere-zero tangent vector field on $S^n$ produces a homotopy from $\operatorname{id}$ to the antipodal map] Suppose $v: S^n \to \mathbb{R}^{n+1} \setminus \{0\}$ is a continuous map with $v(x) \cdot x = 0$ for all $x \in S^n$. Normalise $v$ to obtain \begin{align*} w: S^n &\to S^n \\ x &\mapsto \frac{v(x)}{|v(x)|}. \end{align*} This is well-defined and continuous since $v(x) \neq 0$ for all $x$, and $w(x) \cdot x = 0$ since $v(x) \cdot x = 0$. Define \begin{align*} F: [0, 1] \times S^n &\to S^n \\ (t, x) &\mapsto \cos(\pi t) \, x + \sin(\pi t) \, w(x). \end{align*} **$F$ is well-defined on $S^n$:** For each $(t, x)$, we verify $|F(t, x)| = 1$: \begin{align*} |F(t, x)|^2 &= \cos^2(\pi t) |x|^2 + 2 \cos(\pi t) \sin(\pi t) (x \cdot w(x)) + \sin^2(\pi t) |w(x)|^2 \\ &= \cos^2(\pi t) \cdot 1 + 2 \cos(\pi t) \sin(\pi t) \cdot 0 + \sin^2(\pi t) \cdot 1 = 1. \end{align*} Here we used $|x| = 1$, $x \cdot w(x) = 0$, and $|w(x)| = 1$. **$F$ is a homotopy from $\operatorname{id}_{S^n}$ to $a$:** At $t = 0$: $F(0, x) = \cos(0) \, x + \sin(0) \, w(x) = x$. At $t = 1$: $F(1, x) = \cos(\pi) \, x + \sin(\pi) \, w(x) = -x = a(x)$. Therefore $\operatorname{id}_{S^n} \simeq a$. [guided] The homotopy $F$ rotates each point $x$ through an angle $\pi t$ in the $2$-plane spanned by $x$ and $w(x)$. Since $x$ and $w(x)$ are orthonormal vectors in $\mathbb{R}^{n+1}$, the map $t \mapsto \cos(\pi t) \, x + \sin(\pi t) \, w(x)$ traces out a great semicircle from $x$ to $-x$ in $S^n$. At $t = 0$ we are at $x$; at $t = 1$ we have rotated by $\pi$ and arrived at $-x$. Why do we need $v(x) \neq 0$ everywhere? Because the normalisation $w(x) = v(x)/|v(x)|$ requires $v(x) \neq 0$. If $v$ vanished at some point $x_0$, then $w(x_0)$ would be undefined, and the homotopy $F$ would fail to be continuous (or well-defined) at $x_0$. Why do we need $v(x) \perp x$? The orthogonality $x \cdot w(x) = 0$ ensures $|F(t, x)| = 1$, so that $F$ maps into $S^n$. Without orthogonality, the Pythagorean identity $\cos^2 + \sin^2 = 1$ would not produce $|F|^2 = 1$. The geometric intuition: a nowhere-zero tangent vector field provides, at each point $x$, a "direction to rotate toward $-x$." Without such a field, there is no continuous way to choose a rotation axis at every point simultaneously. [/guided] [/step] [step:Derive a contradiction when $n$ is even] Suppose $n$ is even and $S^n$ admits a nowhere-zero tangent vector field. By the previous step, $\operatorname{id}_{S^n} \simeq a$. By part (4) of the [Elementary Properties of Degree](/theorems/2245), homotopic maps have equal degree, so \begin{align*} \deg(\operatorname{id}_{S^n}) = \deg(a). \end{align*} By part (1) of the [Elementary Properties of Degree](/theorems/2245), $\deg(\operatorname{id}_{S^n}) = 1$. By the [Degree of the Antipodal Map](/theorems/2247), $\deg(a) = (-1)^{n+1}$. When $n$ is even, $(-1)^{n+1} = -1$, so \begin{align*} 1 = \deg(\operatorname{id}_{S^n}) = \deg(a) = -1, \end{align*} a contradiction. Therefore $S^n$ admits no nowhere-zero tangent vector field when $n$ is even. [guided] The chain of reasoning is: nowhere-zero tangent vector field $\implies$ $\operatorname{id} \simeq a$ $\implies$ $\deg(\operatorname{id}) = \deg(a)$ $\implies$ $1 = (-1)^{n+1}$. For even $n$, $(-1)^{n+1} = -1 \neq 1$, and we have a contradiction. For odd $n$, the same argument gives $1 = (-1)^{n+1} = 1$, which is consistent. Indeed, the first step of this proof exhibits a concrete nowhere-zero tangent vector field on $S^{2k-1}$. The argument does not prove the existence statement for odd $n$ via degree theory; it only rules out existence for even $n$. The existence is established by explicit construction. The name "hairy ball theorem" comes from the intuitive formulation: you cannot comb the hair on a ball (sphere) flat without creating a cowlick (a zero of the vector field). This applies to $S^2$ (and all even-dimensional spheres), but not to $S^1$ (where you can comb tangentially around the circle) or $S^3$ (where the quaternionic structure provides a vector field). [/guided] [/step]