## 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

• Dear Matt, good question. The answer is, it’s not used anywhere, so the argument applies to all motives with coefficients in $\mathbf{Q}$. As a sanity check, if $\chi$ is a character of a finite group with values in $\mathbf{Q}$, then the dual character is $\overline{\chi} = \chi$, so $\chi$ is self-dual. Indeed, this is basically the same argument (in weight $w = 0$.)

• Dear GR,