I think the Freudenthal suspension theorem tells you that an element of $\pi_kS$ will act on $\pi_*gl_1S$ in the same way as it acts on $\pi_*S$, for $*>k$. So on the one hand the construction won't be symmetric in the two manifolds but on the other hand will "almost always" agree with direct product (up to framed bordism of course.) And in the specific case of $\eta$, the action only differs from the direct product at $\pi_1$ where it acts by zero I think...so as @EricPeterson suggested, it will be a strange operation indeed

@RobertoLadu Good question! You go by induction. First you do the case $n=1$ by hand. Then you use the fiber bundle $U(n-1)\rightarrow U(n) \rightarrow S^{2n-1}$ find, by induction, that the $n$-th chern class is $ax_1y_{2n-1}$ for some integer $a$. To show that that integers is 1, you just need to exhibit a bundle constructed using your method whose $n$-th chern class is not divisible. For that, take the rank $n$ bundle over $(S^1)^{2n}$ which is the $n$-fold exterior direct sum of the line bundle over $(S^1)^2$ whose first chern class is a generator of $H^2$.

@YCor I am truly, deeply sorry for any offense I may have caused you or anyone else.

Thanks alot for this! Some questions: what is the smallest ring over which $G/\Gamma$ is defined (I think that's $A'$ in the above?) and how does it relate to $A$ and $\mathfrak{o}$? How can I show that $G/\Gamma$ does not even inject into any $\phi^*G$? ($\phi^*$ is any map $Spf\mathfrak{o}\rightarrow SpfA$? Also, where can I learn Newtonian magic (preferably without selling my soul)? I looked through your paper and couldn't find "Newton" anywhere.

@Lubin So $u\cdot f$ "is" $T$ - it is not quite an endomorphism of $F$ but rather sends $F$ to $\phi^*F$. And instead of $G/\text{ker}T\cong G$ all I have is $G/\text{ker} T\hookrightarrow \phi^*G$.

@QiaochuYuan It's a closed formal subscheme that's finite and free and is also a group. That's equivalent to the data of a monic polynomial $f(x)$ with nilpotent lower order coefficients such that $f(F(x,y))=0$ mod $(f(x),f(y))$.