I am currently working through the book "Differential Forms in Algebraic Topology" by Bott and Tu, though I am confused with one of the problems.
Problem 6.43 on page 75 asks:
"... Find the class u on M such that
$\Phi^2 = \Phi\wedge\pi^\star u$ in $H^\star_{cv}(E)$"
with $\Phi$ the Thom class of a vector bundle $\pi:E\to M$.
My confusion is with the $\Phi^2$, I am not quite sure what it should mean. The first thing that springs to mind is that it is shorthand for $\Phi\wedge\Phi$, but this would mean $\Phi^2 = 0$, which leaves me wondering why they wouldn't just ask for a form $u$ with $\Phi\wedge\pi^\star u = 0$. Is this what I should be looking for, is there something I am missing about $\Phi\wedge\Phi$, or is $\Phi^2$ something entirely different?
As discussed in the comments, $\Phi^2$ is notation for $\Phi\wedge\Phi$. If $\Phi$ has odd degree, then $\Phi^2 = 0$, but this is not necessarily the case if $\Phi$ has even degree.