Start with a closed 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 each $T^{2g}$ is a representation of $SU(2),$ precisely it is $I\_g=I\_1^{\otimes g},$ where $I\_1=(1/2)\oplus2(0),$ and I'm using $SU(2)$ representations.


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.e. 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.


Note that this has a double on MSE: as soon as I get an answer on one, I'll delete the other.



SU(2) Lefschetz decomposition for cohomology of Riemann surface Jacobian
Your description of the states for g=1 is correct and $dz |0>$ , $\overline{dz} |0>$
are spin 0 representations whereas $|0>$ and $dz \wedge \overline{dz} |0>$ gives a 
spin 1/2 representation : the creation operator is the cup-product by the class
$dz \wedge \overline{dz}$.


The "attempt of answer" is correct : I\_g decomposes as a sum of irreducible representations
and the obvious one of spin (g/2) is the one of highest weight.
The precise formula for the decomposition of I\_g is formula 4.8 in this paper


<http://arxiv.org/pdf/hep-th/9910181v2.pdf>


(which I think is the best reference for this SU(2) Gopakumar-Vafa story)


