Why it is good to be Pure

There do not exist any regular pure motives M over \mathbf{Q} which are not essentially self dual. Here is why. M gives rise to a compatible family of Galois representations for each rational prime v such that the characteristic polynomial R(X) of Frobenius is independent of this choice. By purity, the eigenvalues \alpha of R(X) are algebraic integers lying in a CM-field such that |\iota \alpha|^2 = p^{w} for some integral weight w and any complex embedding \iota. In particular, if \alpha is a root of R(X), then \alpha^c = p^w/\alpha is a root of X^n R(p^w/X). Since R(X) has coefficients in \mathbf{Z}, it follows that \alpha^c is also a root of R(X), from which one may deduce that R(X) = X^n R(p^w/X) (edit: up to the appropriate constant which makes the RHS monic – this doesn’t affect any of the arguments). Yet this implies that M^{\vee}(w) \simeq M, by the Cebotarev density theorem. (Caveat: it really says that the p-adic avatars of M are essentially self-dual. Perhaps deducing the result for M actually requires the standard conjectures.)

This argument no longer applies if one relaxes the conditions slightly; there do exist non-self dual motives of rank three with coefficients;  Bert van Geemen and Jaap Top found some explicit examples with coefficients in an imaginary quadratic extension of \mathbf{Q}. The point where the argument above fails is that it identifies the polynomial X^n R(p^w/X) with the complex conjugate polynomial R^c(X), which need not equal R(X) anymore.

Stefan Patrikis and Richard Taylor use a similar argument in their recent paper to prove a nice result. Start with a regular pure motive M over \mathbf{Q} (so by the above remarks, it is essentially self dual). Suppose that the corresponding v-adic Galois representation:

\displaystyle{r_v: G_{\mathbf{Q}} \rightarrow \mathrm{GL}_n(\mathbf{Q}_v)}

is not absolutely irreducible. One may ask: are the irreducible constituents s_v themselves essentially-self dual? They show that the answer is yes. Let S(X) denote the corresponding characteristic polynomials. If S(X) lies in \mathbf{Z}[X], then the same argument above applies to s_v. But it may be the case that the representation r_v only decomposes over an extension of \mathbf{Q}. By looking at the eigenvalues, it trivially follows that each of the S(X) may be defined over some CM field F/F^{+}. More importantly, by a technical argument which I will omit but which is not too difficult, one may find a fixed CM field M/M^{+} which contains all the polynomials S(X) (one may even do this [in some sense] independently of v, although we won’t use that here). Consider the Galois representation (s_v)^c, where c is acting on the coefficients. Let \alpha be a root of S(X). Then \alpha^c = p^w/\alpha is now a root of X^m S(p^w/X), and so S^c(X) and X^m S(p^w/X) coincide. Since X^n R(p^w/X) = R(X), we deduce that (s_v)^c is a sub-representation of (r_v)^c = r_v. In particular, (s_v)^c and s_v are both sub-representations of r_v. But the Hodge-Tate weights of s_v and (s_v)^c are the same! (Literally, the Hodge-Tate weights of (s_v)^c are the Hodge-Tate weights of {}^c (s_v) where {}^c(s_v)(g) = s_v(c g c^{-1}), but since s_v is a representation of \mathbf{Q}, conjugation by c is conjugation by a matrix, so there is an isomorphism s_v \simeq {}^c(s_v).) It follows (from the regularity assumption) that s_v = (s_v)^c, and then the argument above implies that s_v is self-dual.

One may use this argument as follows. As in BLGGT, one may find a prime v such that all of the s_v are residually irreducible, and so (if v is sufficiently large) are also potentially modular (by BLGGT again). In particular, either all of the r_v are reducible or they are irreducible for a set of density one set of primes. Moreover, any regular motive over \mathbf{Q} is potentially modular, which is only three adjectives away from the complete reciprocity conjecture!

Patrikis and Taylor do something slightly more general, instead of pure regular motives over \mathbf{Q}, they consider essentially self-conjugate regular compatible systems (with coefficients) of G_{F} for some CM field F/F^{+}. For reasons alluded to above, the coefficients live in some CM-field M. This extra generality (mostly) adds some notational complexity to the argument above. (To see the type of complications that arise, consider an elliptic curve E with CM and then restrict to the CM field F. Then any reducible constituent s_v = \chi_v is related not to its complex conjugate \chi^c_v acting on M, but the complex conjugate {}^c \chi^c_v of this where complex conjugation is now acting on the coefficients M and on the Galois group F.) As expected, one obtains (using BLGGT) some nice consequences, like potential automorphy of regular polarizable compatible systems, as well as irreducibility (for a density one set of primes) of Galois representations associated to RAESDC automorphic form \Pi.

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3 Responses to Why it is good to be Pure

  1. Dear GR,

    Probably a stupid question, but where is regularity used in the first result (the one that just involves M, without any s_v’s).

    Cheers,

    Matt

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