Calculate the  group cohomology classes $H^d[U(1)\rtimes Z_2, Z]$ and   $H^d[U(1)\rtimes Z_2, Z_T]$ I would like to know what are the group cohomology classes $H^d[U(1)\rtimes Z_2, Z]$ and   $H^d[U(1)\rtimes Z_2, Z_T]$, and/or how to calculate them.
It can be shown that $H^d[U(1), Z]$ is $Z$ for even $d$ and 0 for odd $d$.
Here $\rtimes$ is the semidirect product: $T U T = U^{-1}$ for $U \in U(1)$, where
$T$ is the generator of $Z_2$.
In $H^d[U(1)\rtimes Z_2, Z]$, the module $Z$ is the trivial module.
In $H^d[U(1)\rtimes Z_2, Z_T]$, the module $Z_T$ is still the Abelian group $Z$,
 but $U(1)\rtimes Z_2$ has a non-trivial action on  $Z_T$:
$(U,1) a = a$ and $(U,T) a = -a$, $a \in Z_T$.
Thanks!
 A: The group $U(1) \rtimes \mathbb{Z}/2$ you describe is nothing but the group $O(2)$ (as $U(1) = SO(2)$).
As such I think one can see the spectral sequence for the extension does collapse, and one obtains
$$H^*(BO(2);\mathbb{Z}) = \mathbb{Z}[x_2, x_3, x_4]/(2x_2, 2x_3, x_3^2-x_2x_4).$$
Here we can take $x_2 = \beta(w_1)$ and $x_3 = \beta(w_2)$, the Bocksteins on the Stiefel--Whitney classes, and $x_4 = p_1$ the Pontrjagin class.
If you allow me to write $\mathbb{Z}^-$ for your $\mathbb{Z}_T$, there is a short exact sequence of $\mathbb{Z}[\mathbb{Z}/2]$-modules
$$0 \to \mathbb{Z} \overset{1 \mapsto 1+T}\to \mathbb{Z}[\mathbb{Z}/2] \to \mathbb{Z}^- \to 0$$
and $H^*(BO(2); \mathbb{Z}[\mathbb{Z}/2]) = H^*(BSO(2);\mathbb{Z}) = \mathbb{Z}[e_2]$ is a polynomial algebra on the Euler class. The long exact sequence on cohomology gives the following short exact sequences, once you know that $x_4 \mapsto e_2^2$ under
$H^*(BO(2); \mathbb{Z}) \to H^*(BO(2); \mathbb{Z}[\mathbb{Z}/2])$ (since $p_1 = e^2 \in H^4(BSO(2);\mathbb{Z})$):
\begin{eqnarray*}
0 \to H^{4i}(BO(2); \mathbb{Z}^-) \to H^{4i+1}(BO(2); \mathbb{Z}) \to 0 \\
0 \to H^{4i+1}(BO(2); \mathbb{Z}^-) \to H^{4i+2}(BO(2); \mathbb{Z}) \to 0 \\
0 \to \mathbb{Z}\langle e_2^{2i+1} \rangle \to H^{4i+2}(BO(2); \mathbb{Z}^-) \to H^{4i+3}(BO(2); \mathbb{Z}) \to 0\\
0 \to H^{4i+3}(BO(2); \mathbb{Z}^-) \to H^{4i+4}(BO(2); \mathbb{Z}) \to \mathbb{Z}\langle e_2^{2i+2}\rangle \to 0\\
\end{eqnarray*}
One can read off the groups $H^*(BO(2);\mathbb{Z}^-)$ from these short exact sequences.
(It is probably more efficient to describe $H^*(BO(2);\mathbb{Z}^-)$ as a module over $R := H^*(BO(2);\mathbb{Z})$.)
A: I hope this question gets attention from an expert, since I'm curious about the answer.  Here is what I know:
The classifying space of $U(1)$ is $\mathbb{C}P^{\infty}$.  This can be seen by considering the action of $U(1)$ by scalar multiplication on the vector space $\mathbb{C}^{\infty}$  ; removing the origin this action becomes free and the space remains contractible.  The quotient deformation retracts to $\mathbb{C}P^{\infty}$.
The classifying space of $Z_2$ is $\mathbb{R}P^{\infty}$ for the same reason--this time we have $O(1)$ acting freely on an infinite-dimensional real vector space without the origin.
We seek a contractible space on which $U(1) \rtimes Z_2$ acts freely.  An obvious starting point is the product space $\mathbb{C}^{\infty} \times \mathbb{R}^{\infty}$.  Keep the usual action of $U(1)$ on the first coordinate, but allow the generator of $Z_2$ to act by conjugation on the first coordinate, in addition to acting by negation on the second coordinate.  Removing the origin, we obtain a free action since any element $(U,T^0)$ changes the first coordinate unless $U$ is trivial, and any element $(U,T^1)$ changes the second coordinate.  Also, the space is contractible since it still deformation retracts to the sphere $S^{\infty}$.
It seems like the quotient is a bundle over $\mathbb{R}P^{\infty}$ with fibers $\mathbb{C}P^{\infty}$.  Traversing the non-trivial loop in $\mathbb{R}P^{\infty}$ should result in monodromy--conjugation in $\mathbb{C}P^{\infty}$.
I think the Serre spectral sequence is well-suited to compute cohomology of this space.  We consider cohomology of the base with local coefficients from the cohomology of the fiber.  The sequence abuts to cohomology of the total space.  I imagine this computation is routine for semidirect products in general, and may even have a uniform answer.
--Edit--
The spectral sequence in the paper you linked is exactly right.  In fact, it seems to me that the sequence has already converged at the $E^{p,q}_2$ page; after all, in the exact couple all three maps are $0$ .  I think you were worried that some of the diagonal maps on later pages might be isomorphisms, but there is no way to obtain non-zero maps on the next page using only zero maps on the current page.  I believe the cohomology is exactly equal to the upper bound given in J88.
--Edit #2--
In fact, it is not true that all three maps in the exact couple are 0.  I forgot that the differentials in the spectral sequence are the composites of the lower two maps in the exact couple.  The reference I am using is http://www.math.cornell.edu/~hatcher/SSAT/SSATpage.html if you'd like to take a look as well.
Unfortunately, the base space is $\mathbb{R}P^{\infty}$, so the relative cohomology groups are never $0$; further, there may be subtlety with our non-trivial action of $\Pi_1$.
--Edit #3--

The rumor seems to be that a non-zero differential can pop up even after many pages of zero differentials.  Nevertheless, in a spectral sequence arising from an exact couple, it is easy to see that a single page of zeros gives zero on all subsequent pages.  Indeed, using Hatcher's notation for an exact couple we see that
$$ d = jk = 0$$
implies
$$ d'(e + \mbox{Im } d) = j'k'[e] = j'(ike) = jke + \mbox{Im } d = 0 + \mbox{Im } d $$.
This makes Somnath's comment precise.

