Lemma 12.31.2.711AF6E8C0D8 The extension $U$ does not split. For every $s\geq 0$,
\[ H^0(E,\operatorname {Sym}^ sU)=\mathbb C h^ s. \]
For $k\geq 1$ and $s\geq 0$,
\[ \begin{aligned} H^0(E,A^{-k}\otimes \operatorname {Sym}^ sU)& =0,\\ H^1(E,A^ k\otimes \operatorname {Sym}^ sU)& =0,\\ h^0(E,A^ k\otimes \operatorname {Sym}^ sU)& =kd(s+1), \end{aligned} \]
where $h^0=\dim H^0$. For $s\geq 1$, taking the coefficient of $u^ s$ defines a surjection
\[ H^0(E,A\otimes \operatorname {Sym}^ sU)\longrightarrow H^0(E,A), \]
whose kernel is $hH^0(E,A\otimes \operatorname {Sym}^{s-1}U)$.
Proof.
A splitting of $U$ would give an invariant section $u+a(z)h$ lifting $1$, with $a$ entire and
\[ a(z+1)=a(z),\qquad a(z+\tau )=a(z)-2\pi \mathrm{i}d. \]
The derivative $a'$ would then be lattice periodic, hence constant on the compact torus. The first identity forces this constant to vanish, contradicting the second. Thus the extension does not split.
More generally, write an invariant polynomial of degree $s$ as
\[ F(z,h,u)=\sum _{j=0}^ s a_ j(z)h^{s-j}u^ j. \]
If its largest nonzero $u$-degree is $j>0$, then $a_ j$ is lattice periodic, hence a nonzero constant. Comparing the next coefficients gives
\[ a_{j-1}(z+1)=a_{j-1}(z),\qquad a_{j-1}(z+\tau )=a_{j-1}(z)-j(2\pi \mathrm{i}d)a_ j, \]
and the same derivative argument gives a contradiction. Hence $F=a_0h^ s$ with $a_0$ constant; this also covers $s=0$.
The powers of the invariant line $\mathcal O_ Eh$ give a filtration of $\operatorname {Sym}^ sU$ whose $s+1$ successive quotients are trivial line bundles. A line bundle of negative degree on a compact curve has no nonzero section, since the zero divisor of a nonzero section has nonnegative degree. Hence the filtration twisted by $A^{-k}$ has no global sections. Serre duality and $\Omega _ E^1\simeq \mathcal O_ E$ give $H^1(E,A^ k)=0$, and the cohomology sequences of the filtration twisted by $A^ k$ then give $H^1(E,A^ k\otimes \operatorname {Sym}^ sU)=0$. Riemann–Roch on the curve of genus one, for a bundle of rank $s+1$ and degree $kd(s+1)$, gives the dimension formula.
Finally, the leading coefficient gives the exact sequence of bundles
\[ 0\longrightarrow A\otimes \operatorname {Sym}^{s-1}U \xrightarrow {\ h\ }A\otimes \operatorname {Sym}^ sU \longrightarrow A\longrightarrow 0. \]
Indeed, translating $u$ by a multiple of $h$ does not change the leading coefficient, and a polynomial with zero leading coefficient is divisible by $h$. The vanishing of $H^1(E,A\otimes \operatorname {Sym}^{s-1}U)$ just proved gives the asserted surjectivity and kernel on global sections.
$\square$
Pome
Comments
Corrections, questions, and mathematical discussion are welcome. Sign in with GitHub to comment. Comments are public.