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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!

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It seems that you are asking about the cohomology of the classifying space $BG$ (where $G$ is your semi-direct product)? This would not normally be called "group cohomology", since that name seems to be reserved for the case of discrete groups, see here mathoverflow.net/questions/72810/… Could you clarify? –  Mark Grant Sep 14 '11 at 10:45
    
Dear Mark, I just know the algebraic definition of group cohomology and I am not familiar with classifying space (I am a physicist). But I thought for integer coefficient, we do have $H^d[G,Z]=H^d[BG,Z]$ even for Lie group. So maybe what I asked is really $H^d[BG,Z]$. –  Xiao-Gang Wen Sep 14 '11 at 18:31

2 Answers 2

up vote 7 down vote accepted

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 $R := H^*(BO(2);\mathbb{Z})$.)

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Thanks a lot, Oscar. –  Xiao-Gang Wen Sep 17 '11 at 1:46
    
@ Oscar - your result suggests that $H^2(BO(2),Z^-)=Z\oplus Z_2$. But in a spectral sequence calculation (see eqn. J91 in arxiv.org/pdf/1106.4772v2 ) it appears that $H^2(BO(2),Z^-)=Z$. Also, $H^2(BO(2),Z^-)=H^1[BO(2),U_T(1)]$ corresponds to the 1D (anti unitary) representations of $O(2)$ where $T$ is anti unitary. (See eqn. C6 the above paper.) This consideration also suggest that $H^2(BO(2),Z^-)=Z$. So do you really mean $H^2(BO(2),Z^-)=Z\oplus Z_2$. Also does your result imply $H^3(BO(2),Z^-)=Z_2$ –  Xiao-Gang Wen Sep 17 '11 at 3:25
    
The sequence I wrote above is consistent with this: it gives $0 \to \mathbb{Z} \to H^2(BO(2);\mathbb{Z}^-) \to \mathbb{Z}/2 \to 0$, and what you have said shows that this is not split: it is isomoprhic to $0 \to \mathbb{Z} \overset{2}\to \mathbb{Z} \to \mathbb{Z}/2 \to 0$. –  Oscar Randal-Williams Sep 17 '11 at 9:02
    
And yes, it implies $H^3(BO(2);\mathbb{Z}^-) = \mathbb{Z}/2$. –  Oscar Randal-Williams Sep 17 '11 at 9:03
    
@ Oscar - Thanks! –  Xiao-Gang Wen Sep 17 '11 at 14:42

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.

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I have tried to use spectral sequence. But I cannot use easy arguments to show that the $E_2$ page stabilizes. See eqn. J87 in a physics paper arxiv.org/pdf/1106.4772v2 –  Xiao-Gang Wen Sep 15 '11 at 1:35
    
All the maps on the $E_2$ page are zero, so the sequence has already converged. I think the upper bound you give in J88 is actually the correct answer. –  John Wiltshire-Gordon Sep 15 '11 at 2:19
    
Indeed that all the maps on the $E_2$ page are zero. But I do not know how to argue that all the map on the $E_3$ page are zero. So I only have $E_2^{p,q}=E_3^{p,q}$, but I cannot show that $E_3^{p,q}=E_\infty^{p,q}$. It appears that you are suggesting that if the maps in one page are all zero, then the maps on all the later page will automatically be zero. I would love to have such a result. But some one told me that I cannot assume that. –  Xiao-Gang Wen Sep 15 '11 at 2:35
    
Sorry! I think I was wrong that having zero differentials means the next page of differentials is zero. We need more data to compute the next page of differentials, and that that extra data comes from the relative cohomology groups of the skeleta of the base space. –  John Wiltshire-Gordon Sep 15 '11 at 3:19
    
@ John - I believe that if the differential $d_r$ (which is a composition of two maps in the exact triple) is zero then even in the exact couple business, the resulting differential $d_{r+1}$ will be zero, as it essentially the composition of the two derived maps and is also zero when composed. –  Somnath Basu Sep 15 '11 at 3:49

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