Square root of the determinant line

Question

Let $\Sigma$ be a compact Riemann surface equipped with a spin structure (a square root of $\Omega^1_\Sigma$, denoted $\Omega^{1/2}_\Sigma$).

Let $\Gamma(\Omega^1_\Sigma)$ be the space of holomorphic differentials on $\Sigma$, and let $\Gamma(\Omega^{1/2}_\Sigma)$ be the space of holomorphic $\frac12$-forms.

I believe that there is a canonical isomorphism

$$ \Lambda^{top}\big(\Gamma(\Omega^{1/2}_\Sigma)\big)^{\otimes 2} \cong \Lambda^{top}\big(\Gamma(\Omega^1_\Sigma)\big). $$

Did I get this right?  [I got it wrong! See the answers below.]
What is the isomorphism? Where can I read about it?

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PS: Both $\Lambda^{top}(\Gamma(\Omega^{1/2}_\Sigma))$ and $\Lambda^{top}(\Gamma(\Omega^1_\Sigma))$ are line bundles over the moduli stack of spin Riemann surfaces.

Answer 1

There is no such isomorphism (at least for $g \geq 9$).

In

O. Randal-Williams, The Picard group of the moduli space of r-Spin Riemann surfaces. Advances in Mathematics 231 (1) (2012) 482-515.

I computed the Picard groups of moduli spaces of Spin Riemann surfaces (for $g \geq 9$). Grothendieck--Riemann--Roch shows that, in the notation of that paper, the right hand side of your formula is the class $\lambda$, and the left-hand side is the class $\lambda^{1/2}=2\mu$. (See page 511 of the published version for the calculation of the latter; the preprint has some mistakes at this point.)

But the Picard group has presentation $\langle \lambda, \mu \,\vert\, 4(\lambda + 4\mu)\rangle$ as an abelian group, so these are not equal (even modulo torsion).

You should not really need my calculation to see this: you can calculate the rational first Chern class of both sides by GRR, and see that they are distinct multiplies of the Miller--Morita--Mumford class $\kappa_1$; all that remains to know is that $\kappa_1 \neq 0$, which was shown in

J. L. Harer, The rational Picard group of the moduli space of Riemann surfaces with spin structure. Mapping class groups and moduli spaces of Riemann surfaces (Göttingen, 1991/Seattle, WA, 1991), 107–136, Contemp. Math., 150, Amer. Math. Soc., Providence, RI, 1993.

EDIT: To answer the question in the comments.

Yes, I think that (in my notation) the relation $4(\lambda + 4\mu)=0$ holds for $g \geq 3$ (for $g < 3$ it can probably be checked by hand). To see this, let me shorten notation by writing $\mathcal{M}_g = \mathcal{M}_g^{1/2}[\epsilon]$ for the moduli space of Spin Riemann surfaces of Arf invatriant $\epsilon \in \{0,1\}$, $\pi : \mathcal{M}_g^1 \to \mathcal{M}_g$ for the universal family (i.e. $\mathcal{M}_g^1$ is the moduli space of Spin Riemann surfaces with one marked point), and $\mathcal{M}_{g,1}$ for the moduli space of Spin Riemann surfaces with one boundary component.

Firstly, the Serre spectral sequence for $\pi$ has $$E_2^{0,1} = H^0(\mathcal{M}_g ; H^1(\Sigma_g ; \mathbb{Z}))$$ and one can show that this is zero: the fundamental group of $\mathcal{M}_g$ acts on $H^1(\Sigma_g ; \mathbb{Z}) = \mathbb{Z}^{2g}$ via a surjection onto a finite-index subgroup of $\mathrm{Sp}_{2g}(\mathbb{Z})$, but this finite-index subgroup will still be Zariski-dense in $\mathrm{Sp}_{2g}(\mathbb{C})$, so the (complexified) invariants will be zero.

It follows from the Serre spectral sequence that $$\pi^* : H^2(\mathcal{M}_g; \mathbb{Z}) \to H^2(\mathcal{M}_g^1; \mathbb{Z})$$ is injective, so it is enough to prove the relation when there is a marked point. In fact, it even follows that $$H^2(\mathcal{M}_g; \mathbb{Z}) \oplus \mathbb{Z}\cdot e \to H^2(\mathcal{M}_g^1; \mathbb{Z})$$ is injective, where $e$ denotes the Euler class of the vertical tangent bundle of $\pi$.

Now there is a fibration sequence $$\mathcal{M}_{g,1} \to \mathcal{M}_g^1 \overset{\frac{e}{2}}\to BSpin(2)$$ and so, from the Serre spectral sequence, an exact sequence $$0 \to \mathbb{Z}\cdot \frac{e}{2} \to H^2(\mathcal{M}_g^1;\mathbb{Z}) \to H^2(\mathcal{M}_{g,1};\mathbb{Z}) \overset{d^2}\to H^1(\mathcal{M}_{g,1};\mathbb{Z}).$$ Now it follows from

A. Putman, A note on the abelianizations of finite-index subgroups of the mapping class group, Proc. Amer. Math. Soc. 138 (2010) 753-758.

that for $g \geq 3$ the fundamental group of $\mathcal{M}_{g,1}$ has torsion abelianisation, so its first cohomology is zero. Putting it all together, we get an injection $$H^2(\mathcal{M}_g; \mathbb{Z}) \to H^2(\mathcal{M}_{g,1}; \mathbb{Z}),$$ so it is enough to verify the relation $4(\lambda + 4\mu)=0$ on $\mathcal{M}_{g,1}$. But for $g \leq 9$ there is a map $$\mathcal{M}_{g,1} \to \mathcal{M}_{9,1} \to \mathcal{M}_9,$$ given by gluing on 2-holed tori then a disc, so the relation holds because it holds on $\mathcal{M}_9$.

Answer 2

I'm not sure what you mean by 'canonical'. For example, when $\Sigma$ has genus $1$, there are 4 distinct spin structures, one representing the trivial line bundle, whose space of holomorphic sections has dimension $1$, and three nontrivial, whose spaces of holomorphic sections have dimension $0$. If the above isomorphism were 'canonical' in the three nontrivial cases, since the left hand side is canonically isomorphic to $\mathbb{C}$, it would produce a 'canonical' choice of holomorphic $1$-form on $\Sigma$, i.e., a 'canonical' volume form. Is there such a thing? I don't think so.

You may find the paper arXiv:1201.2557 useful. It's possible that the answer depends on whether the chosen spin structure (aka 'theta characteristic') $\Omega^{1/2}_\Sigma$ is even or odd.