PomeExample 12.29.2.423CBD89023E The factors
\[ b_{m\tau +n}(z)=\exp \bigl (-\pi \mathrm{i}(m^2\tau +2mz)\bigr ) \]
define a line bundle of degree one on $E$.
Example 12.29.2423CBD89023E
Proof. For $\gamma =r\tau +s$, the cocycle identity follows from
\[ \begin{aligned} b_{m\tau +n}(z+\gamma )b_{r\tau +s}(z) & =b_{(m+r)\tau +n+s}(z)\exp (-2\pi \mathrm{i}ms)\\ & =b_{(m+r)\tau +n+s}(z). \end{aligned} \]
Put $v=\operatorname {Im}\tau $. The squared norm
\[ |t|^2\exp \bigl (-2\pi (\operatorname {Im}z)^2/v\bigr ) \]
is invariant under the action, because $\operatorname {Im}(z+m\tau +n)=\operatorname {Im}z+mv$ and
\[ |b_{m\tau +n}(z)|^2=\exp \bigl (2\pi (m^2v+2m\operatorname {Im}z)\bigr ). \]
Its first Chern form is
\[ \frac{\mathrm{i}}{2\pi }\partial \bar\partial \left(\frac{2\pi (\operatorname {Im}z)^2}{v}\right) =\frac{\mathrm{i}}{2v}\, \mathrm{d}z\wedge \mathrm{d}\bar z. \]
This positive form integrates to one over a fundamental parallelogram, which has area $v$. Thus the degree is one. $\square$
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