Consider the space $X=BSL(8,\mathbb{C})/(\mathbb{Z}/2)$. The topological Brauer group of $X$ is given by $Br_{top}(X)=Tor(H^{3}(X;\mathbb{Z}))=\mathbb{Z}/2$. I'm studying concepts of topological period and index of a class in $Br_{top}(X)$ and I would like to calculate $ind_{top}(\alpha)$ where $\alpha$ is the nontrivial class in the topological Brauer group of the space given above.

My question is: What should I know about this space that can help me to calculate the index?.

It's known that $per_{top}(\alpha)|ind_{top}(\alpha)$ and that they have the same prime divisors. Since $per_{top}(\alpha)=2$, then $ind_{top}(\alpha)=2^{m}$ for some positive integer $m>0$.

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    $\begingroup$ The index of $\alpha$ can be described in terms of differentials in the Atiyah-Hirzebruch spectral sequence for twisted $K$-theory with twist $\alpha$, according to this paper by Antieau and Williams: projecteuclid.org/euclid.gt/1513732766 In the same paper they also prove your statement about prime divisors. Note that these results require some assumptions on $X$ (such as that it is compact, or has the homotopy type of a finite complex) which may not be the case in your example. $\endgroup$ – Mark Grant Feb 20 '18 at 7:42

The index in this case is $8$. You can see that it divides $8$ as your space $X$ supports a tautological degree $8$ topological Azumaya algebra given by the map $B(SL_8/\mu_2) \rightarrow BPGL_8$. If the index were lower, it would be very strange: it would imply that if $Y$ is any space with $H^2(Y,\mathbb{Z})=0$ and a period $2$ Brauer class with a degree $8$ representative, then in fact the index would be less than $8$. Intuition suggests this is not the case.

To prove the claim above, you can argue as in Section $6$ of this paper: https://arxiv.org/abs/1208.4430. Basically, the $5$-skeleton of $B(SL_8/\mu_2)$ is $5$-equivalent to the $5$-skeleton of $K(\mathbb{Z}/2,2)\times K(\mathbb{Z},4)$. The paper above shows that the index of the tautological non-zero period $2$ Brauer class on the $5$-skeleton of $K(\mathbb{Z}/2,2)$ is $8$.

Another approach is to use the incompressibility of classifying spaces as explained in Jackowski-McClure-Oliver's work on maps between classifying spaces.


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