Let $G$ be a compact [Lie group](https://en.wikipedia.org/wiki/Lie_group) with invariant measure $\mu$. An odd function is a continuous function, $\phi:G\to \mathbb{C}$, such that $\int\_{G} \phi d\mu=0$. An odd map is a continuous map, $f:G\to G$, such that $\phi \circ f$ is an odd function for all odd functions $\phi:G\to \mathbb{C}$.


For example, $f\_{g}(h)=gh$ is an odd map.


Is it true to say that every odd map is not null homotopic?


**A weak and indirect motivation:** "Every odd map on sphere is not homotopic to a constant map"


**Another motivation:** The above $f\_{g},s$ are not null homotopic.



Can an odd map be null homotopic?
Here's a counterexample: take $G=S^1=\{z\in\mathbb{C}:|z|=1\}$,
$$
f(z)=\begin{cases}z^2:\mathrm{Im}(z)\geq 0,\\
\overline{z}^2:\mathrm{Im}(z)\leq 0.
\end{cases}
$$
This $f$ is nullhomotopic, but is an odd map because $\int\_G \varphi\circ f=\int\_G \varphi$ for all $\varphi:G\to\mathbb{C}$.




---


In the example above, $f$ is continuous, but not differentiable at $z=\pm 1$. The result is true for $C^1$ functions.


Suppose $f:G\to G$ is a continuously differentiable odd map, and for simplicity assume $G$ is connected. We have $\int\_G \varphi\circ f\,d\mu=\int\_G \varphi\,df\_\*(\mu)$, where $f\_\*\mu$ is the pushforward measure. Since $\int\_G \varphi\,d\mu=0\Rightarrow\int\_G\varphi\,df\_\*\mu=0$, we must have that $f\_\*\mu$ is a scalar multiple of $\mu$, and therefore $f\_\*\mu=\mu$ because $\mu$ has finite total measure.


For $g\in G$, let's define the Jacobian determinant of $f$ at $g$ to be the determinant of the map induced on $T\_g(G)$ by $L\_{gf(g)^{-1}}\circ f$, and let's write $J\_f(g)$ for this determinant. If $g$ is a regular value of $f$, then there is a neighborhood $U$ of $g$ which is evenly covered by $f$, say $f^{-1}(U)=V\_1\sqcup\ldots\sqcup V\_k$, and on $U$
$$
f\_\*\mu = \sum\_{i=1}^k\frac{1}{|J\_f\circ (f|\_{V\_i})^{-1}|}\mu,
$$
so that $f\_\*\mu=\mu$ implies that on $U$:
$$
\sum\_{i=1}^k\frac{1}{|J\_f\circ (f|\_{V\_i})^{-1}|}=1
$$
on $U$.


This should imply that $J\_f$ doesn't vanish anywhere, and since we assume $G$ is connected, $J\_f$ must have constant sign. For any $g\in G$, we can compute the action of $f$ on the top homology group of $G$ locally, and is is multiplication by $\pm(\#f^{-1}(g))$ (where the $\pm$ is the sign of $J\_f$). In particular this value is non-zero for some (hence all) $g\in G$. Since $f$ induces a non-zero map on the top homology of $G$, $f$ is not null-homotopic.


