Morel and Voevodsky construct the motivic stable homotopy category, a category through which all cohomology theories factor and where they are representable, by starting with a category of schemes, Yoneda-embedding it into simplicial presheaves, endowing those with the $\mathbb{A}^1$-local model structure, and then passing to $S^1 \wedge \mathbb{G}_m$-spectra. The last step ensures that smashing with $S^1$ or with $\mathbb{G}_m$ induce functors with a quasi-inverse on the homotopy category.

Inverting $S^1$ leads to a triangulated structure on the homotopy category, which is very welcome, but I would like a motivation for inverting $\mathbb{G}_m$. Since $\mathbb{P}^1$ is $\mathbb{A}^1$-equivalent to $S^1 \wedge \mathbb{G}_m$ I would also be content with a motivation to invert $\mathbb{P}^1$.

I must admit I already know some answers which certainly are reason enough to invert $\mathbb{G}_m$, e.g. (from slides by Marc Levine, start at page 64):

  1. Inverting $\mathbb{G}_m$ is necessary to produce a Gysin sequence

  2. The algebraic K-theory spectrum appears naturally as a $\mathbb{P}^1$-spectrum

However, I am greedy and would like to hear a motivation like the one for inverting the Lefschetz motive in the construction of pure motives: There one could say that for all envisaged realization functors which should factor through the category of pure motives, the effect of tensoring with the Lefschetz motive can be undone (e.g. is just a change of Galois representation leaving the cohomology groups unchanged).

Or, related to this, as Emerton explained in his nice answer here one has to invert the Lefschetz motive in order to make the Pure Motives a rigid tensor category. Ideally one would like the triangulated category of motives to arise as derived category of some rigid tensor category - if this was true, would it be reflected in the fact that $\mathbb{P}^1$ or $\mathbb{G}_m$ are invertible? (in this case of course one should ensure iinvertibility when constructing a candidate for this derived category)


4 Answers 4


I guess there are "internal" and "external" motivations. External for instance -- most natural examples of functors we have from the stable motivic homotopy category to some other category invert G_m (e.g. any of the usual realizations, or K-theory). Internal for instance -- we want the suspension spectra of varieties to be dualizable under the smash product.

  • 5
    $\begingroup$ BTW the phenomenon of wanting to invert extra spheres in nonstandard stable settings is not special to motivic theory. For example in equivariant stable homotopy theory, you gotta invert those representation spheres, and if you want to do stable homotopy theory over a space it's probably handy to invert sphere bundles... $\endgroup$ Commented Feb 24, 2010 at 23:07
  • $\begingroup$ wait was i right about K-theory? is it monoidal for smash product? i dunno. $\endgroup$ Commented Feb 24, 2010 at 23:15
  • $\begingroup$ .. and these two motivations are exactly the same as those mentioned (for the Lefschetz motive) in the last two paragraphs of the question! $\endgroup$ Commented Feb 24, 2010 at 23:45
  • $\begingroup$ toly: true!! just in the stable instead of abelian context, but the motivation is the same. $\endgroup$ Commented Feb 24, 2010 at 23:51
  • 2
    $\begingroup$ Remember how duality works for suspension spectra of compact manifolds in the ordinary stable homotopy category: X is dual to X^(-T_X), i.e. you need Thom spectra of virtual bundles. Here the duality comes from the purity map X x X --> X^(T_x). Now, in the motivic case, there a notion of Thom space for which purity works, namely take your vector bundle mod the complement of its zero section. These are the things you need to invert, and they look like smash products of P^1's. (p.s. I learned this stuff form a lecture of Jacob Lurie at Harvard...) $\endgroup$ Commented Feb 25, 2010 at 16:46

$H^1({\mathbb G}_m)$ is the same motive as $H^2(\mathbb P^1)$, so I believe that inverting $\mathbb G_m$ is the same as inverting the Lefschetz motive. (Topologically, fundamental classes all ultimately arise from the $H^1$ of the circle, so we must invert this if fundamental classes are to be invertible.) In the pure context, one is not allowed to talk about ${\mathbb G}_m$, and so talks of $H^2(\mathbb P^1)$ instead; but I do think it is the same process.

Note that this is compatible with Dustin Clausen's and Marty's answers: in all realizations (Galois representations, computing periods, ... ) of motives, we can and do invert the Lefschetz motive (since it just becomes the inverse cyclotomic character, or $2 \pi i$, or ... ). Incidentally, related to David Roberts's answer, smashing with $\mathbb G_m$ is what number theorists call a Tate twist, I believe, and on the level of Galois representations it it is just tensoring with the cyclotomic character (here I am thinking of $H_1(\mathbb G_m)$). So asking for $\mathbb G_m$ to be invertible is the same as asking that one can perform Tate twists (of arbitrary integer power) on the level of motives.

Now if $M$ is a motive over a number field, with $L$-function $L(M,s)$, and $M(n)$ is its $n$th Tate twist, then the $L$-function of $M(n)$ is simply $L(M,s + n)$. So the Tate twisting parameter is the same as the parameter $s$ in the $L$-function (restricted to integer values, of course). Thus the desire to have this parameter really be an integer (and not just a natural number) also has deep roots in the conjectured relations between special values of $L$-functions and the arithmetic geometry of motives. (If one looks at the simplest $L$-function, namely the Riemann zeta function, it has special values at positive and negative integers, and one certainly wants to consider all of these and relate all of them to motivic considerations.)


Though I'm not an expert on motives, by any measure, I think that an answer to your question can be given by considering periods. As Kontsevich and Zagier recall in their paper "Periods", publ. IHES, Section 4.2, one can form a square matrix of periods from a pair consisting of a smooth algebraic variety $X$ over $Q$ and a divisor with normal crossings $D \subset X$, also defined over $Q$. One simply pairs a basis for the relative singular homology of $(X,D)$ with coefficients in $Q$ with a basis of the relative de Rham cohomology of $(X,D)$ of appropriate degrees. This entries of this pairing matrix are the so-called "periods".

This matrix -- in general -- is almost invertible as a matrix over the ring of periods. However, to invert the matrix, one must also adjoin $1/2 \pi i$ to the ring of periods -- this corresponds precisely to inverting (the period of) $G_m$ as you mention.

So, as I understand it, one can not achieve comparison isomorphisms of Betti and de Rham realizations of motives (or at least not write down the isomorphism in both directions) without inverting the period of $G_m$.

Or, it also follows from Kontsevich-Zagier that one cannot define the triple-product on the ring of periods, without having $1/2 \pi i$. Defining this triple-product is necessary, if one wishes to endow $Spec$ of the ring of periods with the structure of a pro-algebraic torsor for the motivic Galois group.


As for the triangulated category perspective (and I'm not an expert) it comes down to having the shift functor invertible, and from what I understand, this is formally like suspension, i.e. smashing with $\mathbb{G}_m$ (the analogy is closest for $\mathbb{G}_m(\mathbb{C}) = \mathbb{C}^\times$ which is homotopic to the circle). There is a bit of abstract perspective on this at motivic cohomology at the nLab

  • $\begingroup$ The triangulated category structure comes only from inverting the simplicial circle, not G_m. As for the nLab page: I had a look and found it horribly misleading (there are just plainly wrong statements)!! I inserted two warnings there but have no time right now to edit it nicely... $\endgroup$ Commented Feb 25, 2010 at 16:33
  • $\begingroup$ Thanks, Peter, for these two warnings. The first of them I incorporated into the text. What else is "horribly misleading"? The first part at half is supposed to be pretty literally a summary of the lecture by Jardine linked to there. My apologies for that mistake I made which you pointed out. Good that you caught it. $\endgroup$ Commented Feb 25, 2010 at 20:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.