In his paper "Categories and cohomology theories" Graeme Segal gives examples how to construct a Gamma category and therefore also a Gamma space from a strict monoidal category like finite chain complexes of complex fin.-dim. vector bundles with the alternating sum over the dimensions equal to 1 and chain homotopy equivalences as morphisms. This yields a nice model for $BU_{\otimes}$ and proves that this is an infinite loop space. My question is

Can you still construct a Gamma category if the monoidal structure is not strict, for example if there is a non-trivial associator? If such a Gamma category $C$ exists, what would $C(3)$ and the induced morphisms look like?

  • $\begingroup$ You can, as long as you can strictify associators... $\endgroup$ Sep 29, 2011 at 13:51
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    $\begingroup$ My hope was that one can somehow avoid strictifications of the category and see directly what happens, but maybe that is too naive. $\endgroup$ Sep 29, 2011 at 14:26
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    $\begingroup$ On the contrary I'd say it would be very complicated. Gamma spaces are too strict. Either you strictify associators or weaken the notion of Gamma space by means of a "homotopy coherent" version. $\endgroup$ Sep 29, 2011 at 17:31

1 Answer 1


If your monoidal category is not strict you can first form a multicategory out of it. This process involves some choices (how to bracket higher tensor products) but they are not essential (see e.g. "Tom Leinster - Higher Categories, Higher Operads", chapter 3.3 for discussion). In nature symmetric monoidal categories very often come from multicategoires, like the category of vector bundles etc. So in some sense the underlying multicategory is the even more natural objects.

As a next step you can extract a gamma category out of a multicategory, see "Permutative categories, multicategories, and algebraic K-theory" by Elmendorf and Mandell. This is a direct generalization of the Gamma category of Segal associated to a permutative category. In particular if your category from the start was permutative it gives back the old construction of Segal.

  • $\begingroup$ That sounds interesting. I will look into it. Thanks, Thomas! $\endgroup$ Mar 1, 2012 at 9:28

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