In general, let $L$ be a line bundle on a complex torus $X=V/ \Lambda$ of dimension $g$ and let $H$ be the hermitian form corresponding to the first Chern class $c_1(L)$. The imaginary part $E:= \textrm{Im}(H)$ is an alternating form which is integer-valued on the lattice $\Lambda$.
By elementary linear algebra there is a basis of $\Lambda$ with respect to which $E$ is given by the matrix $$\left(\begin{matrix}0 & D \cr - D & 0 \end{matrix}\right),$$ where $D=\textrm{diag}(d_1, \ldots, d_g)$ and the $d_i$ are strictly positive integers satisfying $d_i|d_i+1$ d_i|d_{i+1}$for all$i=1, \ldots ,g-1$. If$L$is positive-definite then one shows that $$h^0(X, L)=\textrm{Pf}(E)=\det(D).$$ The proof consists in explicitly writing a basis for$H^0(X, L)$by using canonical theta functions, as in Sebastian's answer. If$X=J(C)=H^0(\omega_C)^*/H_1(C, \mathbb{Z})$is the Jacobian of a smooth curve, then the theta divisor$\Theta$is a principal polarization, i.e.$D$is the identity matrix. This can be seen by taking a standard homology basis for$H_1(C, \mathbb{Z})$. It follows$h^0(X, \Theta)=1$. See [Birkenhake-Lange, Complex Abelian Varieties, Chapters 3 and 11] for further details. 3 added 2 characters in body In general, let$L$be a line bundle on a complex torus$X=V/ \Lambda$of dimension$g$and let$H$be the hermitian form corresponding to its the first Chern class .$c_1(L)$. The imaginary part$E:= \textrm{Im}(H)$is an alternating form which is integer-valued on the lattice$\Lambda$. By elementary linear algebra there is a basis of$\Lambda$with respect to which$E$is given by the matrix $$\left(\begin{matrix}0 & \Delta D \cr - \Delta D & 0 \end{matrix}\right),$$ where$D=\textrm{diag}(d_1, \ldots, d_g)$and the$d_i$are strictly positive integers satisfying$d_i|d_i+1$for all$i=1, \ldots ,g-1$. If$L$is positive-definite then one shows that $$h^0(X, L)=\textrm{Pf}(E)=\det(D).$$ The proof consists in explicitly writing a basis for$H^0(X, L)$by using canonical theta functions, as in Sebastian's answer. If$X=J(C)=H^0(\omega_C)^*/H_1(C, \mathbb{Z})$is the Jacobian of a smooth curve, then the theta divisor$\Theta$is a principal polarization, i.e.$D$is the identity matrix. This can be seen by taking a standard homology basis for$H_1(C, \mathbb{Z})$. It follows$h^0(X, \Theta)=1$. See [Birkenhake-Lange, Complex Abelian Varieties, Chapters 3 and 11] for further details. 2 added 179 characters in body In general, let$L$be a line bundle on a complex torus$X=V/ \Lambda$of dimension$g$and let$H$the hermitian form corresponding to its first Chern class. The imaginary part$E:= \textrm{Im}(H)$is an alternating form which is integer-valued on the lattice$\Lambda$. By elementary linear algebra there is a basis of$\Lambda$with respect to which$E$is given by the matrix $$\left(\begin{matrix}0 & \Delta \cr - \Delta & 0 \end{matrix}\right),$$ where$D=\textrm{diag}(d_1, \ldots, d_g)$and the$d_i$are strictly positive integers satisfying$d_i|d_i+1$for all$i=1, \ldots ,g-1$. If$L$is positive-definite then one shows that $$h^0(X, L)=\textrm{Pf}(E)=\det(D).$$ The proof consists in explicitly writing a basis for$H^0(X, L)$by using canonical theta functions, as in Sebastian Sebastian's answer. Saying that If$X=J(C)=H^0(\omega_C)^*/H_1(C, \mathbb{Z})$is the Jacobian of a smooth curve, then the theta divisor$\Theta$is a principal polarizationis equivalent to say that , i.e.$D$is the identity matrix, so . This can be seen by taking a standard homology basis for$H_1(C, \mathbb{Z})$. It follows$h^0(X, \Theta)=1\$.