Prove that $S^{2n}$ doesn't admit topological group structure only by hairy ball theorem It's not difficult to see that $S^{2n}$ doesn't admit a Lie group structure. Since if $S^{2n}$ admit a Lie group structure, then there exists a left invariant vector field. While the Hairy ball theorem says that there exists no continuous tangent vector field  on $S^{2n}$ non-vanishing at every point.
Now we want to prove that $S^{2n}$ admits no topological group structure. It follows from the well known theorem: a topological group G that is a topological manifold must be a Lie group.
However,  I want to give a direct proof without using this theorem.
Identify $S^{2n}$ the unit sphere in $R^{2n+1}$. For any two points $v_1,v_2\in R^{2n+1}$, denote $\langle v_1, v_2 \rangle$ the standard inner product. If $S^{2n}$ admits a topological group structure, let $e\in S^{2n}$ be the identity element, $\cdot$ the binary operation. Choose a unit vector $v$, s.t. $\langle e, v \rangle =0$. Then $v\in S^{2n}$. For any point $g\in S^{2n}$,  $g\cdot v \in S^{2n}$, however, we don't have that $\langle g\cdot v, g \rangle=0$. So we project $g\cdot v$ to the tangent space of $g$, define the vector field by
$$
X_g=\frac{g\cdot v-\langle g\cdot v,g \rangle g}{\|g\cdot v-\langle g\cdot v,g \rangle g\|}.
$$
As Will Sawin pointed out to me, if $g\cdot v=-g$, then $g\cdot v-\langle g\cdot v,g \rangle g =0$. So how to adjust the construction of $X_g$ to get a nonvanishing tangent vector field?
Thanks for Will Sawin's (1 2 3), Pierre PC's (1 2), and Moishe Kohan's (1 2) comments.
It's known that for any continous map $f: S^{2n} \to S^{2n}$, there exists some point $x\in S^{2n}$ s.t. either $f(x)=x$ or $f(x)=-x$. Because if no such $x$, then $f$ is homotopy to identity map and antipodal map (degree=-1), contradiction.
If $S^{2n}$ admit a topological structure, as Moishe Kohan comments, then for any $v\neq e \in G$,  the map $g \in G \to v\cdot g \in G$ is a continous map, and there is no $g$ s.t. $v\cdot g=g$. As Pierre PC comments, by the compactness of $S^2$, we can choose $v$ close to $e$, then $v\cdot g\neq -g$ for any $g\in G$.
Maybe this argument is known to experts.
 A: The function $g\mapsto\|(-g)\cdot g^{-1}-e\|$ is continuous and never zero on the sphere, so it admits a minimum $\varepsilon>0$. Let $v$ be an element of the sphere with $0<\|v-e\|<\varepsilon$. Because of this condition, we know that $v\cdot g$ is neither $g$ nor $-g$, for all $g\in S^{2n}$.
Denote by $X_g$ the projection of $v\cdot g$ on the plane orthogonal to $g$. $X_g$ is zero when $v\cdot g$ is parallel to $g$; since both of them belong to the sphere, this happens for $v\cdot g=\pm g$, which as discussed above never happens. So $X_g$ is a continuous nonvanishing section of the tangent bundle (by definition it is orthogonal to $g$), which is a contradiction.
If we really want to get back to the smooth unit version of the hairy ball theorem, we can proceed in the following way. By compactness, let $\delta>0$ be the smallest value of $g\mapsto\|X_g\|$. Then choose $Y/\|Y\|$, for $Y$ a smooth vector field $\delta$-close to $X$ (for instance using partitions of unity).
What this argument actually says is that a continuous map that sends $g$ outside of $\{-g,g\}$ (in our case $g\mapsto v\cdot g$) cannot exist.
A: Moishe Kohan's comments (1 2) can be translated to a "topological" hairy ball theorem with an identical proof as the classical case:
Recall, the classical hairy ball theorem states that $T(S^{2n})$ admits no nowhere vanishing sections. The proof is that the Euler class of the tangent bundle of a manifold evaluates on the fundamental class to the Euler characteristic. This implies for $S^{2n}$, the Euler class is nontrivial. In addition, the Euler class is an obstruction to a vector bundle having a nowhere vanishing section, so we are done.
Now to a topological manifold $M$, we can assign to it a topological tangent bundle. The type of object this is is a $\mathbb{R}^n$ fiber bundle (with a distinguished 0 section). Intuitively, the fiber over a point $x$ is a chart containing $x$. These objects have many of the classical properties of tangent bundles of smooth manifolds, in particular they have Euler classes, and these Euler classes also evaluate to the Euler characteristic on the fundamental class. Likewise, these Euler classes give obstructions to having nowhere vanishing sections.
Hence, the exact proof of the hairy ball theorem extends to this topological setting, and for the same reason it prohibits a topological group structure, since we may use the group operation to trivialize the tangent $\mathbb{R}^n$ bundle, implying it has a section.
