Recall that the Thom spectrum is an orthogonal spectrum $\operatorname{mO} \in \mathbf{Sp}$ defined via $\operatorname{mO}(V) = \operatorname{Th}(\operatorname{Gr}_{\dim{V}}(V \oplus \mathbb{R}^{\infty}))$ where this means that we take the Thom Space of the universal bundle.
Let $G$ be a finite group, $\rho_G = \mathbb{R}[G]$ be the regular representation and $\overline{\rho}_G = \left \{\sum_g a_g g : \sum_g a_g = 0 \right \}$ be the reduced regular representation. Then, $$ \operatorname{mO}(V \otimes \rho_G) = \operatorname{Th}(\operatorname{Gr}_{|G| \dim{V}}((V \otimes \rho_G) \oplus \mathbb{R}^{\infty})) \cong \operatorname{Th}(\operatorname{Gr}_{|G| \dim{V}}(V \oplus \mathbb{R}^{\infty} \oplus (V \otimes \overline{\rho}_G))). $$ Then, I want to understand the following (which is essentially Example 6.1.46 in Stefan Schwede's Global Homotopy Theory).
Claim. Let $V$ be an inner product space. An element $[x,U] \in \operatorname{mO}(V \otimes \rho_G)$ is $G$-fixed if and only if $x \in U^G$ and $U = U^G \oplus (U \cap (V \otimes \overline{\rho}_G))$.
This sounds like the following elementary group representation theory statement: Let $V \oplus W$ be a $G$-representation such that $G$ acts trivially on $V$ and $W^G = 0$. Let $U \subseteq V \oplus W$ be a linear subspace. Then, $U$ is $G$-fixed if and only if $U = U^G \oplus (U \cap W)$.
But the converse is wrong.
So I seem to be missing something. What is going on here?
Edited to respond to questions in the comments
As you observe, the claim is incorrect. Note though that this claim is not used in Example 6.1.46 of Stefan Schwede's book. What is used is only one direction: if $U \subseteq V \oplus W$ is $G$-invariant, with $V$ and $W$ as above, then $U = (U \cap V) \oplus (U \cap W)$ as subsets of $V \oplus W$.
I claim that it follows from this that there is a homeomorphism \begin{aligned} \mathbf{Gr}(V) \times \mathbf{Gr}(W)^G \cong \mathbf{Gr}(V \oplus W)^G. \end{aligned} There is a map $\mathbf{Gr}(V) \times \mathbf{Gr}(W)^G \to \mathbf{Gr}(V \oplus W)^G$ given by sending $(U',U'')$ to $U := U' \oplus U''$. Since $V$ has trivial action, any $U'$ is automatically $G$-invariant. As $U''$ is $G$-invariant by assumption, $U$ is also $G$-invariant, so the map is well-defined. There is also a map $\mathbf{Gr}(V \oplus W)^G \to \mathbf{Gr}(V) \times \mathbf{Gr}(W)^G$ given by sending $U$ to $(U^G,U \cap W)$, which is just another way of writing $(U \cap V, U \cap W)$. Both $U \cap V$ and $U \cap W$ are $G$-invariant. It is clear that you get back $(U',U'')$ from $U' \oplus U''$ by doing this. However, what is a priori not clear (which I overlooked in my previous post) is whether any $U$ is the same as $(U \cap V) \oplus (U \cap W)$. Since you say in the comments that you're already fine with this, we're done.