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Let $C$ be a smooth projective curve over $\mathbb{C}$. Let $A$ be a degree $d$ line bundle on $C$, and $M$ be a degree 0 line bundle on $C$ such that $M^2=\mathcal{O}_C$, that is, it is a 2-torsion line bundle. Therefore, we have that $deg(A)=deg(A\otimes M)$. What is $h^0(C,A\otimes M)$ in general? Can we say that it is zero?

More precisely,

We have the subvariety $W^r_d(C)=\{A\in Pic^d(C)|h^0(C,A)\geq r+1\}$ of $Pic^d(C)$. A 2-torsion line $M$ acts on $Pic^d(C)$. Assume $0< d < g$ Then what will the action of $M$ on $Pic^d(C)$ do to $W^r_d(C)\setminus W^{r+1}_d(C)$ ? In fact consider $W^0_d(C)=\{A\in Pic^d(C):h^0(C,A)\geq 1\}$. This has expected dimension $\rho(g,0,d)<g$ when $d<g$. This will get translated to some other variety under translation by $M$ right? So that $h^0(A\otimes M)=0$. How do we prove this?

I have attempted the following: Consider $t_M:Pic^d(C)\longrightarrow Pic^d(C)$ given by $B\mapsto B\otimes M$. This is an isomorphism. Now $A\in Pic^d(C)$ is such that $h^0(A)\geq 1$. That is $A\in W^0_d(C)$.We have that $W^0_d(C)\subset Pic^d(C)$. Now, every component of $W^0_d(C)$ has expected dimension (when $d<g$): $\rho(g,0,d)=g-1(g-d+0)=d< g$.\ Now $Pic^d(C)$ is a $g$-dimensional variety, so under translation by $M\in Pic^0(C)$, a general element of the subvariety $W^0_d(C)$ of $Pic^d(C)$ moves to another outside $W^0_d(C)$. Hence for a general $A\in W^0_d(C)$, $h^0(C,A\otimes M)=0$.

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    $\begingroup$ Certainly not. Take for $C$ a hyperelliptic curve: then $W^1_2$ has only one point, which is not preserved by tensor product with any line bundle of order 2. In fact, I believe the situation you describe never happens when $\dim W^r_d<g$. $\endgroup$
    – abx
    Commented Oct 2, 2015 at 6:05
  • $\begingroup$ Thanks @abx, if we consider $W^{g-2}_{2g-2}\setminus W^{g-1}_{2g-2}$ this will be preserved by tensor product by a order 2 line bundle. That's why I had this doubt. How do we prove your claim? Can you give me any leads? $\endgroup$
    – gradstudent
    Commented Oct 2, 2015 at 15:41
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    $\begingroup$ Have you heard of Prym differentials? They are sections of the tensor product of the canonical bundle with a 2 torsion line bundle. This 2 torsion bundle determines a double cover D of the curve C, and the Prym differentials on C correspond to differentials on D that are skew symmetric for the sheet exchange involution of D. Maybe that pattern generalizes. (Oh and there are always only g-1 indept Prym diferentials on a curve of genus g.) $\endgroup$
    – roy smith
    Commented Oct 2, 2015 at 15:44
  • $\begingroup$ One statement in the question is wrong $W^0_d$ has expected (and actual) dimension d. $\endgroup$
    – meh
    Commented Nov 4, 2015 at 14:08
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    $\begingroup$ Unless I misunderstand your notation $W_0^d = {A \mbox{ s.t. } h^0(A) \geq 1 } = {C_d)} $ and the expected dimension is d . In addition, plugging r=0 into the formula $g -(g-d+r)(r+1) gives the Brill-Noether number of effective line bundles is d, which is to be 'expected'. $\endgroup$
    – meh
    Commented Nov 4, 2015 at 17:17

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For a definite class of counterexamples, look at theta characteristics. These are defined as the line bundles whose square is the canonical class. So, by definition, any two theta characteristics differ by a two-torsion line bundle. A theta characteristic $L$ is called even or odd according to whether $H^0(C,L)$ is even or odd. The parity of a theta characteristic is a non-trivial quadratic form modulo 2 on the space of theta characteristics and, in particular, is non-constant. (See Mumford Ann Sci ENS 1971). This shows that $W^0_{g-1}$ is not invariant by translation by two torsion.

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