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For all those who are unlikely to have answers to my questions, I provide some

Background:

In some sense, pure motives are generalisations of smooth projective varieties. Every Weil cohomology theory factors through the embedding of smooth projective varieties into the category of pure Chow motives.

Pure effective motives

In the definition of pure motives (say over a field k), the last step is to take the category of pure effective motives and formally invert the Lefschetz motive L.

The category of pure effective motives is the pseudo-abelian envelope of a category of correspondence classes, which has as objects smooth projective varieties over k and as morphisms X → Y cycle classes in X×Y of dimension dim X (think of it as a generalisation of morphisms, where morphisms are included as their graphs), where an adequate equivalence relation is imposed, to have a well-defined composition (therefore the word "classes"). When the adequate relation is rational equivalence, the resulting category is called the category of pure effective Chow motives.
In each step of the construction, the monoidal structure of one step defines a monoidal structure on the next step.

For more background see Ilya's question about the yoga of motives.

Definition of the Lefschetz motive

The Lefschetz motive L is defined as follows:

For each point p in P¹ (1-dimensional projective space over k), there is the embedding morphism Spec k → P¹, which can be composed with the structural morphism P¹ → Spec k to yield an endomorphism of P¹. This is an idempotent, since the other composition Spec k → Spec k is the identity.
The category of effective pure motives is pseudo-abelian, so every idempotent has a kernel and thus [P¹] = [Spec k] + [something] =: 1+L, where the summand [something] is now named Lefschetz motive L.

Properties

The definition of L doesn't depend on the choice of the point p.

From nLab and Kahn's leçons I learned that the inversion of the Lefschetz motive is what makes the resulting monoidal category a rigid monoidal category - while the category of pure effective motives is not necessarily rigid.

In the category of pure motives, the inverse $L^{-1}$ is called T, the Tate motive.

Questions:

These questions are somehow related to each other:

  1. Why is this motive L called "Lefschetz"?
  2. Why is its inverse $L^{-1}$ called "Tate"?
  3. Why is it exactly this construction that "rigidifies" the category?
  4. Would another construction work, too - or is this somehow universal?
  5. How should I think of L geometrically?

I have almost no background in number theory, so even if you have good answers, it may remain totally unclear to me, why the name "Tate" intervenes. I expect however, that the name "Lefschetz" has something to do with the Lefschetz trace formula. I guess that the procedure of inverting L is the only one which makes the category rigid, in some universal way, but I have no idea, why. In addition, I guess there is no "geometric" picture of L.

If I made any mistakes in the background section, feel free to edit. As I'm currently taking a first course on motives, I may now have asked something completely stupid. If this is the case, please point me politely to some document which will then enlighten me or, at least, let me ascend to a higher level of confusion.

UPDATE: Thanks so far for the answers, questions 1-4 are now clear to me. It remains, if the "rigidification" could be accomplished by another construction - maybe some universal way to turn a monoidal category into a rigid one? Then one could later identify the Lefschetz motive as some kind of a generator of the kernel of the rigidification functor.
The geometric intuition, to think of L as a curve and of $L^{\otimes d}$ as a d-dimensional manifold, remains fuzzy, but I have the hope that this becomes clear when I have worked a little bit on the classical Lefschetz/Poincare theorems and the proof of Weil conjectures for Betti cohomology (is this hope justified?).

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Very nice question! –  Mariano Suárez-Alvarez Feb 8 '10 at 2:26
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Not good enough to be an answer (to your VERY good question) but I've been told, when asking how to think about motives, to think of the equation $[\mathbb{P}^1]=[pt]+[\mathbb{A}^1]$, which suggests that the Lefschetz motive should be thought of as adjoining a non-projective variety to the category to allow some decompositions. I have no real way to make any of what I've been told rigorous, though. –  Charles Siegel Feb 8 '10 at 3:07
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While I'm sure that everyone is correct who says that L stands for Lefschetz, I have always found it a helpful mnemonic to think that P^1 minus a point is a LINE. –  David Speyer Feb 9 '10 at 12:06
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4 Answers 4

up vote 20 down vote accepted

The motive $L$ is called Lefschetz because it is the cycle class of the point in ${\mathbb P}^1$, and so underlies (in a certain sense) the Lefschetz theorems about the cohomology of projective varieties. To understand this better, you may want to read about how the hard Lefschetz theorem for varieties over finite fields follows from the Riemann hypothesis, as well as a discussion of Grothendieck's standard conjectures and how they relate to the Weil conjectures.

The motvie $L^{-1}$, when converted into an $\ell$-adic Galois representation, is precisely the $\ell$-adic Tate module of the roots of unity. Tensoring by this Galois representation is traditionally called Tate twisting, and so the motive underlying this Galois representation is called the Tate motive.

One needs to have $L^{-1}$ at hand in order for the category to admit duals.

If one were working with just usual singular cohomology, this wouldn't be necessary; Poincare duality would pair $H^i$ with $H^{\text{top}-i}$ into $H^{\text{top}}$, which would be isomorphic with ${\mathbb Q}$ via the fundamental class.

But motivically, if $X$ (smooth, connected, projective) has dimension $d$, so that the top dimension is $2d$, then $H^{\text{top}} = L^{\otimes d}$, and so $H^i$ and $H^{2d-i}$ pair into $L^{\otimes d}$. To get a pairing into $\mathbb Q$ (the trivial 1-dim'l motive) we need to be able to tensor by powers of $L^{-1}$. Traditionally tensoring by the $n$th tensor power of $L^{-1}$ is denoted $(n)$; so we find e.g. that $H^i$ pairs with $H^{2d -i}(d)$ into ${\mathbb Q}$, and we have our duality.

You can see from the fact that cup product pairs cohomology into powers of $L$ that inverting $L$ is precisely what is needed in order to obtain duals.

Finally, one should think of $L$ as the fundamental class of a curve, think of $L^{\otimes d}$ as the fundamental class of a smooth projective $d$-dimensional manifold, and also become comfortable with Poincare duality and the Lefschetz theorems; these are the basic ideas which will help give solid geometric sense to motivic constructions.

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Thanks! I don't understand why L is the cycle class of the point in P¹ - I thought the cycle class of any point would be 1 = [Spec k]. The Poincaré duality explanation / geometric picture for L and it's tensor powers seems to go in the direction I was looking for, but I don't understand why $H^{top}=L^{\otimes d}$. Can you name some standard reference(s) for "how the hard Lefschetz theorem for varieties over finite fields follows from the Riemann hypothesis"? –  Konrad Voelkel Feb 8 '10 at 3:14
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For $H^{2d} = L^{\otimes d}$, the corresponding result after passing to $\ell$-adic etale cohomology is a standard theorem in that theory. To see it directly, and motivically (i.e. geometrically), I wonder if one could use Noether normalization (i.e. finite projection to a linear subspace of the ambient projective space) to reduce to the case of projective space, and then inductively and/or via Kunneth, reduce to the case $H^1({\mathbb P}^1) = L$. Maybe someone else can confirm/refute this approach? –  Emerton Feb 8 '10 at 3:42
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You mean the analogue of RH for finite fields? –  Hailong Dao Feb 8 '10 at 4:20
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Hard Lefschetz is deduced from RH in Weil II, if I remember correctly. As for "${\rm{H}}^{2d} = L^{\otimes d}$" in the $\ell$-adic etale realization (so to speak), I assume one means that if $X$ is sept'd, finite type, & irred. of dim $d$ over alg (or sep) closed field then ${\rm{H}}^{2d}_ c(X,\mathbf{Q}_ {\ell}(d))$ is canonically isomorphic to $\mathbf{Q}_ {\ell}$ via a certain "trace" map. The proof I've read is really hard, involving delicate curve fibrations (so hard part is proving canonicity of trace map) to reduce to relative curve case. One does not reduce to projective spaces. –  BCnrd Jul 17 '10 at 3:38
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Just to clarify, Emerton's suggestion with Noether normalization is a useful technique in proofs if one wants to reduce some problems to either a projective space or a power of the projective line. But to actually set up the theory of the trace is an entirely different matter, and one has to confront huge well-definedness problems. Much work needed. It's like $\ell$-adic Kunneth formula: one can do 2-line isomorphism proof with derived category stuff, but just shifts the burden of work to relating the isom. to cup product (a major undertaking). As always in geometry, no royal road. :) –  BCnrd Jul 17 '10 at 3:42
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I think the name L is chosen because it corresponds to the 'L operator' of 'old algebraic geometry' over the complex numbers (and that in turn is named L, I would agree, probably to honour Lefschetz). I don't know what notation Lefschetz used himself, but for example Griffiths-Harris book "Principles of Algebraic Geometry" introduces this L operator on page 111 in the deRham setting. Also as the key player for the Lefschetz decomposition/Hard Lefschetz.

The whole book makes no reference to motives or Weil conjecture stuff, just down to earth complex manifold stuff. So, just as a reading suggestion, if you want to read about it on an elementary level.

Maybe this is an easier starting point than things over finite fields etc., even though admittedly maybe not very sexy.

The L operator here wedges with the Kähler form, and that would be just an explicit (1,1)-form representing the point in P1.

(by representing I mean that it represents the cycle class Chow -> deRham of the point via an explicitly given differential form - and that should come down to representing the 'motive' of it)

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Thank you! That explains the "Lefschetz" name issue pretty well :-) –  Konrad Voelkel Feb 8 '10 at 21:56
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Let me make a comment on the question number 3: why do we have to invert $\mathbb L$ in order to get a rigid category?

Note that the category formed by direct sums of powers of $\mathbb L$ is equivalent to the category of nonnegatively graded vector spaces over $\mathbb Q$ (if we work with $\mathbb Q$ - coefficients), with $\mathbb L$ corresponding to $\mathbf Q[1]$ a 1-dimensional vector space in degree 1. The equivalence is provided by $CH^*$ (Chow groups) functor.

The fact that it is an equivalence follows from the fact that $Hom(\mathbb L^i, \mathbb L^j) = \mathbb Q$ if $i=j$ and 0 otherwise.

Now to make a rigid category out of nonegatively graded vector spaces we must include all $\mathbb Z$-graded spaces, so that $\mathbb Q[1]$ is dual to $\mathbb Q[-1]$. This corresponds to inverting $\mathbb L$.

And once we invert $\mathbb L$ we indeed get a rigid category, as Matt Emerton explained: the dual to $M(X)$ is $M(X)(-dim X)$ (I believe that whether you have a minus sign or not depends on definitions).

As for the question how do we think of $\mathbb L$, the picture is that the decomposition $M(\mathbb P^1) = 1 \oplus \mathbb L$ corresponds to cellular decomposition of $\mathbb P^1$ into disjoint union of a point and a line. This generalizes to any variety admitting cellular decomposition (Grassmannians and quadrics for example): their motives are sums of Lefschetz motives, a summand $L^d$ stands for each cell of dimensions d.

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A little more on questions 3 and 4.

The basic property of the Lefschetz motif (in this aspect) is the Cancellation Theorem: $Hom(X,Y)\cong Hom(X\otimes L,Y\otimes L)$ for any effective motives X and Y. If you don't have this property for an object, then inverting it is a 'bad' operation: the functor from the 'old' category to the 'new' one is very far from being a full embedding.

So, one should not invert any L if it does not satisfy the cancellation theorem. If it does, then its image in the ('usual') Chow motives is invertible (with respect to the tensor product). Conjecturally, any invertible Chow motif is of the form $M(n)$ where $M$ is an (effective) zero-dimensional motif which becomes isomorphic to 1 after a finite base field extension.

We obtain: there IS some flexibility in the choice of a motif which you want to invert, but they yield the same result. I do not know how to define 'universal rigidifying' for a tensor category; probably there does not exist such a construction that is nice enough.

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I'm really curious about your statement on motives which become 1 (you mean Z, right?) under a finite field extension. At some point I was trying to prove that there's no such Chow motives, but of course I didn't succeed. Are there any constructions of such motives? Also do you have a reference to the conjectural statement on invertible Chow motives? –  Evgeny Shinder Feb 10 '10 at 17:47
    
Well, zero-dimensional (effective) Chow motives are just Artin ones i.e. direct summands of motives of finite field extensions of the base field. I am not sure that there exist any non-trivial direct summands of this sort if you take integral coefficients, but there are plenty of those if you consider motives with rational coefficients. The conjecture I wrote about comes from: Kahn, motivic zeta functions of motives, arxiv.org/abs/math/0606424 –  Mikhail Bondarko Feb 10 '10 at 19:12
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