Polishcuk ex. 7 pg. 149


E is an elliptic curve and S is the Fourier Mukai transform with the poincare as its kernel:

\mathscr{S} \colon D^b(E) \to D^b(\widehat{E})

Let L = \mathcal{O}_E(e) be the line bundle associated to the divisor e, which is the identity of E.  Note L^3 = \mathcal{O}_E(3e) which is very ample (which can be proved by the Riemann Roch theorem and criterion for a line bundle to give a closed immersion see for example, pg 163 of Miranda).  Define S_L, T_L \colon D^b(E) \to D^b(E) by

S_L(F) = \phi_L^*\mathscr{S}(F)

T_L(F) = F \otimes L

Where the pullback and tensor are both derived.  Let


S = \left(\begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array}\right)

T = \left(\begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array}\right)

A coherent locally free sheaf F has a rank \mbox{rk} F \in \mathbb{N}.  Also if \mbox{rk} F = r, then \bigwedge^r F is a line bundle, which corresponds to a divisor which has a degree, so set \deg F = \deg \bigwedge^r F \in \mathbb{Z}.  Also any sheaf on A has a characteristic \chi(F) = \sum_i (-1)^i \dim_k H^i(E,F).  These notions are extended to any F^\bullet \in D^b(E) by 

\mbox{rk} F^\bullet = \sum_i (-1)^i \mbox{rk} F^i

\deg F^\bullet = \sum_i (-1)^i \deg F^i

\chi( F^\bullet) = \sum_i (-1)^i \chi( F^i)

The Exercise

There is a map

D^b(E) \xrightarrow{(\deg, \mbox{rk})} \mathbb{Z}^2.  

Abbreviate the name of this map to (d,r).  The point of this exercise is to show the functors S_L[1] = [1] \circ S_L and T_L induce the action of S,T on \mathbb{Z}^2.  Most of the argument consists of 

Lemma: For F \in D^b(A)

\deg F = - \mbox{rk} F

\mbox{rk} \mathscr{S}(F) = \deg F

proof: First replace F by a finite resolution of locally free sheaves of finite rank (this is doable by, for example, prop. 3.26 pg 77 of Huybrechts).  Thus the lemma can be reduced to the case where F is a locally free sheaf on A.

The next big result to use is that \deg F = \chi(F) this is proved vectbund.  Also there is (11.3.7) from Polischuk: \mbox{rk} \mathscr{S}(F) = \chi (F) and \chi(\mathscr{S}(F)) = (-1)^g \mbox{rk} F, where g is the dimension of the abel. var.

In this case, g = 1, so I compute

\deg \mathscr{S}(F) = \chi (\mathscr{S}(F)) = - \mbox{rk} F

\mbox{rk} \mathscr{S}(F) = \chi(F) = \deg F 



The action of S_L[1]

Now the action of S_L[1] is easy to compute.  Indeed, if (d,r)F = (a,b)  the lemma says (d,r)\mathscr{S}(F) = (-b, a).  Pulling back a locally free sheaf does not change its rank and by the Grothendiek-Riemann-Roch theorem (ex. 5 Polishcuk pg. 149) \chi(\phi_L^*\mathscr{S}(F)) = \deg(\phi_L)\chi(\mathscr{S}(F)) = 1\cdot \chi(\mathscr{S}(F)), which shows (d,r) S_L(F) = (-b, a).  Now recalling the alternating sum definition of rank and degree it follows that shifting the complex by 1 introduces a (-1) factor hence

(d,r)S_L[1](F) = (b, -a)

as required.

The action of T_L

Its clear that T_L(F) has the same rank as F, so it must be shown that \deg T_L(F) = \deg F + \mbox{rk} F.  For an invertible sheaf corresponding to an effective divisor consisting of only smooth points, there is an exact sequence 

0 \to \mathcal{O}_E \to L(D) \to \mbox{Sky}_D \to 0

Where \mbox{Sky}_D is a skyscraper sheaf that has support on D = \sum n_p p.  In particular, the stalk (\mbox{Sky}_D)_p = k(p)^{n_p}.  The maps are as follows.  The invertible sheaf L(D) corresponds to a closed subscheme whose ideal sheaf is locally generated at p by a function that vanishes to order n_p \ge 0, say the function is f_p, in a local coordinate (since E is a curve just need one coordinate) x, the function is f_p = x^{n_p}.  Then L(D) is locally generated near p by 1/x^{n_p}.  So, locally,  the first have of the above sequence is

0 \to \mathcal{O}_{E,p} \to 1/x^{n_p} \mathcal{O}_{E,p}

Or equivalently, since x is not a zero divisor 

0 \mathcal{O}_{E,p} \to \mathcal{O}_{E,p}

1 \mapsto x^{n_p}

The assumption that the points of D are smooth points says \mathcal{O}_{E,p} is a regular local ring of dimension 1; in particular, the maximal ideal is principle.  So \mathcal{O}_{E,p} \cong k[x]_{(x)}.  From this it follows that the cokernel k[x]_{(x)}/(x^{n_p}) = k[x]/x^{n_p} is a k(p) vector space of dimension n_p with basis 1, x, \dotsc, x^{n_p - 1} which give the desired surjection to (\mbox{Sky}_D)_p, so the map of sheaves of above is exact as it is exact on stalks.  

In the case at hand, D = e and the sequence is 0 \to \mathcal{O}_E \to L(e) \to \mbox{Sky}_e \to 0.  Then tensoring with a locally free sheaf F, I get

0 \to F \to F\otimes L(e) \to F \otimes \mbox{Sky}_e \to 0

From which I conclude \chi(F \otimes L(e)) = \chi( F) + \chi(F \otimes \mbox{Sky}_e).  Now for any skyscraper sheaf H^i(E, \mbox{Sky}) = 0 for i \ne 0 (Miranda pg. 303).  Thus \chi(F \otimes L(e)) = \dim \Gamma(E, F \otimes L(e)) = \mbox{rk} F since \deg e = 1.  Thus

\deg (F \otimes L(e)) = \chi(F \otimes L(e)) = \chi(F) + \mbox{rk} F = \deg F + \mbox{rk} F

Thus the action of T_L is correct for locally free sheaves.  For a general complex of locally free sheaves F^\bullet I have 

\deg (F^\bullet \otimes L(e) ) = \sum_i (-1)^i \chi(F^i \otimes L(e))

= \sum_i (-1)^i \chi(F^i) + \sum_i (-1)^i \mbox{rk} F^i

= \deg(F^\bullet) + \mbox{rk} F^\bullet

As required. 


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