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A spectrum is a sequence $X_0,X_1,...$ of spaces together with structure morphisms $\Sigma X_n\to X_{n+1}$. To get the usual model for the stable homotopy category based on the category of spectra, one "inverts" the suspension functor (or the shift functors) which is not an isomorphism in the category of spectra. It seems to me that this is a kind of allowing the objects to be indexed over the integers. So why does one not define a spectrum to be indexed over the integers or at least bounded below?

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3 Answers 3

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There are certainly lots of people who think of spectra as being indexed over the integers. For example, if you ask a topologist to define the Bott spectrum representing (complex, say) topological K-theory, I think most would say that it has a space for each integer, and in even dimensions that space is $\mathbb{Z}\times BU$, while in odd dimensions it's $U$. On the other hand, I'm not sure where you can look to see a careful construction of a category of spectra indexed over the integers, and as Peter explained in his answer, there's no real need for this.

There are actually many essentially equivalent ways to index spectra. One popular one is "coordinate-free" spectra, which are indexed over the finite-dimensional linear subspaces of a countably infinite dimensional inner product space over the reals ($\mathbb{R}^\infty$, say). This is the perspective taken by May and his collaborators in many papers and books (lots of them available on May's homepage). In this setting, the "bonding maps" are defined by viewing the one-point compactification of a subspace $V$ as a sphere $S_V$, and smashing it with the associated space to form the (iterated) suspension. That is, if $V\subset W$ are finite dimensional subspaces to which we've associated topological spaces $X_V$ and $X_W$, then the bonding map associated to this inclusion is a map $S_{W\setminus V} \wedge X_V \to X_W$, where $W\setminus V$ is the orthogonal complement of $V$ in $W$ (surely there's better notation for this?). Note that if you consider an increasing sequence of subspaces in which the dimensions increase by 1, you'll see your notion of a spectrum pop out.

There are various other indexing schemes, including the (maximal?) choice in which a spectrum consists of a space associated to every pointed space homeomorphic to a finite CW complex! See the paper of Mandell, May, Schwede, and Shipley for a discussion of various indexing categories for spectra. It's #96 on May's webpage: http://www.math.uchicago.edu/~may/PAPERSMaster.html

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    $\begingroup$ The practice of doing things coordinate free is very important. This helps to set up the appropriate foundations for equivariant stable homotopy theory. And it feels a bit more natural. $\endgroup$ Sep 9, 2010 at 2:27
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The extra indices would be of no use - natural numbers already make the shift functor invertible and if you add more data to a spectrum you have to identify more again in the passage to the homotopy category. The shift down functor, which, applied to suspension spectra, is the suspension functor, has an opposite which is the shift up functor which puts a one point space at the zero level (where you have to "invent" a new space). In the homotopy category the two become inverse: If you shift first up, then down, you have the identity anyway. If you shift first down then up you have deleted the zeroth space. In the homotopy category this makes no difference because you pass to the limit along the whole sequence of spaces, so there it is also the identity. This passing to the limit only needs one direction ("positive infinity"), so negative indices would have no effect...

Maybe it is also helpful to think of the passage to spectra as of allowing homotopy groups in negative degrees also: If you go along the sequence of spaces in an $\Omega$-spectrum (and every spectrum is equivalent to such an $\Omega$-spectrum) you find the same homotopy groups at every stage, but shifted down by one at each step. So potentially you get non-trivial homotopy groups in every degree, also negative degrees.

In this text on page 21, section 6.5, Bjørn Dundas explains it very nicely with an analogy to chain complexes: If you had only non-negative chain-complexes, you could model arbitrary ones by sequences of non-negative ones which you shift down at each step.

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    $\begingroup$ Thank you Peter. I was a bit sloppy and of course meant "shift down functor" instead of "shift functor". Ah, I can acquire a taste for your argument that it doesn't matter since one considers "the limit" only anyway. But I cannot follow Dundas on the bottom of page 22 where he says that the homotopy category of Spt is equivalent to the "linearized" one via the linearization functor. Aren't there spaces with the same stable homotopy type but different homology? $\endgroup$
    – roger123
    Sep 9, 2010 at 9:10
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    $\begingroup$ Whoops, that's a typo on the bottom of page 22! All the adjunctions right of Spt(A) induce equivalences of homotopy categories - the one on the left doesn't. Also the Z[-] should be below the arrow pair not on top. The fact that Ho(spectra of positive chain complexes) is equivalent to chain complexes the analogy that I thought would be helpful. $\endgroup$ Sep 9, 2010 at 13:28
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    $\begingroup$ But the relation between stable homotopy type and homology type is the other way round (where "homology" means ordinary, e.g. singular, homology): The homology functor comes from Top, goes via suspension spectrum into Spt, then from left to right through the whole chain on page 22, and in the end you take homology of the chain complex. At the first step, the linearizing, you lose information - so there are spaces with different stable homotopy type (i.e. non-isomorphic in Ho(Spt)) which become isomorphic in Ho(Spt(A)) and thus have the same homology. $\endgroup$ Sep 9, 2010 at 13:37
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The reason may be slightly historical, the first examples of spectra were most easily seen to be indexed over the natural numbers: $MU$, $HR$, and $\mathbb{S}$. Now though, we know we ought to be using finite dimensional inner product subspaces of $\mathbb{R}^\infty$! Also, we want there to be an underlying "point set" category of spectra that we do something to do in order to get the stable homotopy category. What we really care about is some spectrum, not the image of that spectrum in the stable homotopy category. I guess I am trying to say that even though we will end up inverting something later in order to work on it does not mean we should throw out the original and only remember the localization.

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