# SU(2) Lefschetz decomposition of Jacobian of Riemann surfaces

Start with a (closed and oriented) Riemann surface with $g$ handles $\Sigma_g.$ I'm interested in (the cohomology of) its Jacobian $Jac(\Sigma_g)=T^{2g},$ in particular how the $SU(2)$ or $SL(2,\mathbb{R})$ Lefschetz decomposition acts on it.

The picture I have in mind is, following Gopakumar-Vafa ideas (see http://arxiv.org/abs/hep--th/9809187 and http://arxiv.org/abs/hep-th/9812127 ) and Mirror Symmetry book (by Vafa et al., chapter 33), that $H^\bullet(T^{2g})$ is a representation of $SU(2),$ precisely it is $I_g=I_1^{\otimes g},$ where $I_1=(1/2)\oplus2(0).$

To see this, start from $g=1,$ i.e. $T^2:$ with a bit of hand-waving, the Jacobian story is equivalent to considering a ground state $| 0 \rangle$ and then act on it with the fermionic operators $\mathrm{d}z$ and $\mathrm{d}\overline{z}$ to get indeed the $I_1$ representation $$| 0 \rangle, \quad \mathrm{d}z| 0 \rangle, \quad \mathrm{d}\overline{z}| 0 \rangle, \quad \mathrm{d}z\wedge\mathrm{d}\overline{z}| 0 \rangle,$$ i.e. in terms of spin $$0,\pm \frac12, 0,$$ even if I don't see the precise arrangement or linear combinations involved.

A similar story can be repeated for higher $g$ and I guess this matches fine.

My QUESTIONS are the following:

i. is it possible to actually arrange the various states so obtained in $SU(2)$ multiplets? up to now, I can ony match the number of expected states, which brings me to

ii. how is this related to the $SU(2)$ Lefschetz decomposition, which has creation, annihilation and number operators given by $$J_+=\omega\wedge, \quad J_-=\omega\lrcorner,\quad J_3=(deg-n)/2,$$ where $\omega$ is the Kahler form of the torus, $deg$ is the bidegree and $n$ the complex dimension? I would expect a spin $(g/2)$ representation given by $$(1,\omega,\ldots,\omega^g),$$ which I cannot relate to $I_g.$

My attempt at an ANSWER is that $$(1,\omega,\ldots,\omega^g)$$ is only the highest spin representation $(g/2)$ inside $H^\bullet(T^{2g}),$ because one can imagine to start, e.g., from $\mathrm{d}z_i$ or $\mathrm{d}z_i\wedge\mathrm{d}z_j$ instead of 1, and then construct a shorter representation, to get that $I_g$ is indeed a sum of irreps. I'd like to know if this reasoning is correct.