# Non invertibility of certain integral arising from group action

Edit 1: According to the comment of Andreas Cap I revise the integral formula in the question.

Edit 2: I understand from the following post that some part of the previos version of my question has negative answer, so I completely revise the question:

https://math.stackexchange.com/questions/1506133/is-there-an-odd-continuous-map-fs2-to-gl2-mathbbc

Let a compact topological group $G$ with invariant measure $\mu,$ acts on $S^{2}$ and $\rho$ is a $n$-dimensional unitary irreducible representation of $G$. Let $f:S^{2}\to M_{n}(\mathbb{C})$ be a continuous function. is it true to say that:

There exist $x_{0}\in S^{2}$ such that the following integral is non invertible:

$$\int_{G} \rho(g)f(g^{-1}.x_{0})d\mu$$

As another question we ask: Is the following statement, true?

Assume that $\phi$ is a order $n$ homeomorphism on $S^{4}$ and $\lambda \neq 1$ is a quaternion with $\lambda^{n}=1$.Assume that $f:S^{4}\to \mathbb{R}^{4} \simeq \mathbb{H}$ is a continuous function. Then there is a point $x_{0}\in S^{4}$ such that $\sum_{i=0}^{n-1} \lambda^{n-i}f(\phi^{i}(x_{0}))=0$.

Note that the particular case $\lambda=-1$ and $\phi=$ The antipodal map is the classical Borsuk-Ulam theorem.

The motivation for the first question is that the statement is true for $1$ dimensional representation.

The motivation for the last part of the question is that the above statement is true if we replace $S^{4}$ and $\mathbb{R}^{4}$ by $S^{2}$ and $\mathbb{R}^{2}\simeq \mathbb{C}$, respectively and choose $\lambda \in \mathbb{C}$.

• Algebraic topology? – Fernando Muro Oct 8 '14 at 20:07
• The fact that the fundamental group shows up doesn't make the question an algebraic topology one. – Fernando Muro Oct 8 '14 at 20:16
• @AliTaghavi No, the complex general linear group has higher homotopy groups, so it cannot have a contractible covering space. – Fernando Muro Oct 9 '14 at 9:55
• I don't quite understand the motivation for looking at that integral. If you would put $\int_G\rho(g)f(g^{-1}x)$ then there is a simple interpretation: This would mean that you view $f$ as a function $X\to V^n$, where $V$ is the space on which $G$ is represented and average for the natural action of $G$ on the space of such maps. Thus the result would be $G$-equivariant for that action. One would get an analogous statement for the natural action on maps $X\to L(V,V)$ when considering $\int_G\rho(g)f(g^{-1}x)\rho(g)^{-1}$. – Andreas Cap Oct 30 '15 at 8:09
• I mean $V^n$ and not $V$, and the analog of invertibility is that all values consist of linearly independent vectors. For the other part, the question is whether you can interpret the statement as the value at $x$ of $\int_G g\cdot f$ for some action on $G$ on a space of functions or not. (I don't claim that such an interpretation would readily answer your question, but it provides a way how to think about the quesiton.) Concerning motivation, I rather mean motivation why the claimed property should hold, rather than what it could be used for. – Andreas Cap Nov 1 '15 at 10:01