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It's common in subjects that have some version of the "yoga of weights" that you have a functor called "Tate twist" and that the most natural version of it seems like it should be the square of something which obviously doesn't exist. For example, in the category of mixed Hodge structures, there's the natural Tate Hodge structure $\mathbb{Q}(k)$ (which is, for example, the Hodge structure on the top degree of a smooth $k$-dimensional projective variety). You'd like there to be a Hodge structure $\mathbb{Q}(\frac{1}{2})$ such that $\mathbb{Q}(\frac{1}{2})^{\otimes 2k}=\mathbb{Q}(k)$, but there's no way that such a thing could exist in the standard definition of mixed Hodge structures, because Hodge structures of pure odd weight must be even dimensional.

A way of fixing this is suggested by Beilinson, Ginzburg and Soergel - Koszul Duality Patterns in Representation Theory on page 514: in essence, you consider two copies of the category of mixed Hodge structures with one of them formally Tate twisted by $(\frac{1}{2})$.

I was writing a paper that used this idea, and thought of an explanation of this I like quite a bit better, and now am wondering if I'm really the first to have thought of it.

The idea of this: you think of the two vector spaces in MHS as the even and odd parts of a super vector space (i.e. a $\mathbb{Z}/2$-graded vector space). The definition of a Hodge structure on an even vector space is the usual one; the definition of a Hodge structure on a purely odd vector space is the usual one, but now the Hodge filtration is indexed by elements of $\mathbb{Z}+\frac{1}{2}$.

So, for example, $\mathbb{Q}(\frac{1}{2})$ is pure of weight 1, with $F_{-1/2}=\mathbb{C}$ and $F_{1/2}=0$ (with the purity condition being exactly that these guys are transverse).

Has anyone encountered this approach before? Once I saw it, it seemed like obviously the right thing to do.

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    $\begingroup$ From some points of view (gamma motives and hypergeometric motives) it makes sense to extend even further: to allow Hodge structures to have a "phase" in $[0,1)$, so that usual Hodge structures have phase $0$, your odd Hodge structures have phase $1/2$, but for general phase $s$ the Hodge filtratiion is indexed by elements of $\mathbb{Z}+s$. It's like take some differential equation. Then the space of solutions around the origin is filtered by their growth: if a solution starts with $t^s$ you put it in $F_s$. $\endgroup$
    – Anton Mellit
    Commented Dec 13, 2021 at 13:13
  • $\begingroup$ @AntonMellit Maybe you had this in mind already, but the ODE story is quite precisely true if you take the corresponding twistor: you take $z^sV[z,z^{-1}]$ as a bundle with flat connection on $\mathbb{C}^*$, and then the two Hodge filtrations tell you how to extend over $0$ and $\infty$. I think you should exactly get what you said about growth rates of flat sections recovering the Hodge filtrations (by construction). Being an honest Hodge structure means you get actual flat sections, rather than multi-valued ones. $\endgroup$
    – Ben Webster ♦
    Commented Dec 13, 2021 at 13:48

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I am not sure this is relevant to you, and you may know this already, but in the theory of monodromic ("exponential") mixed Hodge structures, there is a very natural (even) square root of the Tate motive. See Kontsevich-Soibelman's Cohomological Hall algebra, exponential Hodge structures and motivic Donaldson–Thomas invariants, Section 3.4: Square root of Tate motive. Most papers on DT theory will use this, see e.g. Davison-Meinhardt, Cohomological Donaldson–Thomas theory of a quiver with potential and quantum enveloping algebras, Section 2.1 for a summary.

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    $\begingroup$ More generally, you may look at the values of the gamma function at rational arguments, and they are not motivic, i.e. they are not periods, but any product $\Gamma(s_1) \cdots \Gamma(s_k)$ for $s_1,\ldots,s_k\in \mathbb{Q}$ satisfying $\sum_i s_i\in \mathbb{Z}$ is motivic. So you want to say that individual gamma-values are motivic of fractional weight. For instance $\Gamma(1/2)=\sqrt{\pi}$ wants to be the square root of the Tate motive. $\endgroup$
    – Anton Mellit
    Commented Dec 13, 2021 at 10:20
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    $\begingroup$ Do these live in MMHS, or some over-category of that? $\endgroup$
    – Balazs
    Commented Dec 13, 2021 at 10:22
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    $\begingroup$ @Balazs I presume the $\Gamma$ correspond to twisted de Rham cohomology classeson $\mathbb{A}^{1}$ (evaluated along rapid decay cycles etc) for some potential $t^{n}$ (maybe after étale pullback) so yes this should be encodable in EMHS $\endgroup$
    – user108998
    Commented Dec 13, 2021 at 11:29
  • $\begingroup$ @Balasz I don't know exact definitions, but what EBz wrote makes sense. By the way, the Hodge decomposition for $\Gamma(s)$ should be $H^{s,1-s}=\mathbb{C}$, so the weight is in fact always integer, but the Hodge filtration has jumps at fractional indices, similar to how the original post suggests. $\endgroup$
    – Anton Mellit
    Commented Dec 13, 2021 at 12:45
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If you introduce super-structure, it doesn't matter whether $\mathbb{C}(1/2)$ is even or odd. The reason is that your category of motives has a symmetric monoidal automorphism which acts by $-1$ on $\mathbb{C}(1/2);$ this implies that after introducing super-structure, there is a symmetric monoidal auto-equivalence which takes $\mathbb{C}(1/2)$ to the supershift $\Pi \mathbb{C}(1/2).$

More informally, I would suspect that's not the right thing to do. The reason is that there's nothing "super" about cohomology of varieties: the tensor product on motives compatible with cohomology is the usual one, modified by cohomological degree in the usual way. It sounds like Beilinson's $\mathbb{C}(1/2)$ is a cohomological degree-zero object, so it should behave commutatively. Another reason I think this isn't the right thing to do is that in the world of super-schemes, while motives will tend to "kill" the super-structure, you might still want to look at things like the Hodge-to-de Rham degeneration which will have genuinely "super" pieces. The fact that $$\mathbb{A}^{0\mid 1}\times \mathbb{A}^{0\mid 1} \cong \mathbb{A}^{0\mid 2}$$ should intuitively mean that these super pieces correspond to a new "purely odd" piece of the cohomology rather than a square root of the Tate motive.

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