Let $G$ be a compact Lie group. Before defining $G$-prespectra, we have to define a $G$-universe $\mathcal U$.

Question: Why do we need a $G$-universe?

A $G$-universe is defined to be a countably infinite-dimensional (real) representation of $G$ with an inner product such that

  1. $\mathcal U$ contains the trivial representation.
  2. $\mathcal U$ contains countably many copies of each finite-dimensional subrepresentation.

After fixing a $G$-universe, we can define a $G$-prespectrum (indexed on $\mathcal U$): a collection $\lbrace EV \rbrace_V$ of $G$-spaces indexed by finite-dimensional subrepresentations $V$ of $\mathcal U$ together with $G$-maps $\sigma_{V,W}:\Sigma^{W-V}EV \to EW$ for pairs $V \subset W$. (For details and relevant papers, see "Basic notions in equivariant stable homotopy theory".)

A $G$-universe seems to be used only for taking (co)limits and to be regarded as a source of finite-dimensional (orthogonal) representations of $G$. I feel that it is very artificial to use a $G$-universe for the purposes.

For example, what is wrong if we deal with the category of (equivalent classes of) finite dimensional (orthogonal) representations of $G$? This is a small category, so we do not have any problem to consider (co)limits.

  • $\begingroup$ I asked my question on SE, but I deleted it, because MO seems to be more suitable for the question. $\endgroup$
    – H. Shindoh
    Oct 14, 2013 at 8:47
  • 7
    $\begingroup$ This is not really an answer: it's just a comment relating to the first parentheses in the last paragraph. Equivalence classes of [insert anything you want here] do NOT form a category. Indeed: you can't define morphisms between equivalence classes (and even if you think that you can define morphisms, then you can't compose them). $\endgroup$ Oct 14, 2013 at 10:27

3 Answers 3


Tyler, you are too fast: didn't give me a chance to answer first! Of course, I agree with everything you say. I wrote the following before seeing your answer (except for the last paragraph).

Since I introduced this choice, let me explain. But first, echoing André, taking equivalence classes would be a wrong choice even if it gave a category, which it doesn't. In fact, the word “representation” in this context is a convenient lie: we are not doing representation theory here, and we must not think at all in terms of equivalence classes. For example, isomorphisms between “representations” control signs in equivariant cohomology theory.

One point is to obviate set theoretic nonsense. It has become unfashionable, perhaps, to pay attention to this, but of course the collection of all finite dimensional representations is not a set, and for many purposes, such as taking colimits as you say, one does want a set.

A mathematical point is that different universes give different categories of G-spectra, and that matters enormously: change of universe plays an essential role in equivariant stable homotopy theory. This could be dealt with in other ways, but use of universes is convenient.

Actually, how essential a universe is, depends on which choice of a category of $G$-spectra one has in mind. For all choices, it is very convenient to work with $G$-vector spaces with a fixed given $G$-inner product. For orthogonal $G$-spectra, the fact that the category $\mathcal I$ of such $G$-inner product spaces is essentially small (equivalent to a small category) allows us to use it without actually specifying a universe, although one does obtain a different $\mathcal I$ for each choice of a set of irreducible representations (the complete universe, allowing all, being the most important).

For $G$-spectra in the sense of Gaunce Lewis and myself, and therefore for the $S$-modules of EKMM (Elmendorf–Kriz–Mandell–May) use of a universe is truly essential: $G$-spectra are obtained from $G$-prespectra as colimits over inclusions of sub $G$-inner product spaces of a universe. Such colimits make no sense without use of some device to ensure smallness. In this line of development, use of a universe seems truly essential. The linear isometries $G$-operad $\mathcal L$ is central to the construction of the smash product (and to lots of work in equivariant infinite loop space theory), and $\mathcal L(j)$ is the $G$-space of linear isometries $U^j\to U$, where $U$ is the universe in which one is working. It would be ludicrous to try to make sense of that without working in a universe.

As a philosophical point, it is essential to be eclectic in this area and to allow use of different categories of $G$-spectra, such as orthogonal and Lewis–May or EKMM, since there are many things that one can readily prove with one and not the other. For a comparison of these two and discussion of change of universe, see for example Mandell–May, Equivariant orthogonal spectra and $S$-modules.

That source explains how, in orthogonal $G$-spectra, one can actually work with one fixed universe, even the trivial one, and obtain equivalent categories as Tyler says. Hill–Hopkins–Ravenel took that observation from Mandell–May and ran with it. To be honest, I sometimes regret we made that observation; as Tyler notes, it can be a source of confusion, and it can sometimes obscure the mathematics. Here again it is wise to be eclectic and think in terms of both physical change of universe and “phantom” change of universe in terms of that observation.


I suspect that Prof. May will have the definitive answer here. However, here are some of the more mundane problems that the choice of universe addresses.

  • It cuts down on set-theoretic issues, by describing an equivariant spectrum as a functor on a small category.

  • Back when Lewis-May-Steinberger put equivariant stable theory on a firm footing, one should note that prespectra and spectra were not functorial in representations: we only had structure relating $EV$ and $EW$ when $V \subset W$. The functoriality in isomorphisms only came later.

  • Getting a little confused about this last point was the source of some mistakes, and if you want to read a scathing criticism of this (such as people casually writing $S^{V-W}$ when $V$ and $W$ are unrelated $G$-representations) then you can look at Adams' "Prerequisites for Carlsson's lecture", section 6.

  • It helps to understand the smash product. The smash product of a spectrum $E$ indexed on ${\cal U}$ and a spectrum $F$ indexed on ${\cal U'}$ is most naturally a spectrum $E \overline\wedge F$ indexed on ${\cal U \oplus U'}$, because it is built out of $EV \wedge FV'$ for $V \subset {\cal U}$ and $V' \subset \cal U'$. This exterior wedge is nicely associative and commutative, and so this concentrates the "issues" with the smash product into the problem of internalizing the smash power.

  • It conveniently describes which type of equivariant spectrum you have by describing which $G$-representations you are allowing formal deloopings by. Different isomorphism classes of universe lead to quite different theories.

  • Finally: Just as André says, you don't technically need a universe anymore. "Orthogonal spectra" are essentially some kind of functor defined on representations, and you may find this version palatable. In fact, the Hill-Hopkins-Ravenel construction of the norm critically used that $G$-equivariant orthogonal spectra form a category equivalent to a category of plain-old ordinary spectra equipped with a $G$-action, and so representation spheres didn't even appear. (I can't speak for anyone else, but this caught me by surprise. My confusion turned out to be that you have several categories that are equivalent and which look basically the same, but have quite different homotopy categories and homotopy theories because these equivalences of categories do not respect the notion of "weak equivalence" at all.)


The short answer is, I think, that one doesn't really need a $G$-universe, and that one could also index a $G$-spectrum over all finite dimensional representations of $G$.

  • $\begingroup$ I'd add to your short answer the comment that most if not all of what people do with representations of compact Lie groups has a classical flavor, so the added terminology relative to group actions is just a convenience. $\endgroup$ Oct 14, 2013 at 12:53

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