= Solution
Let $X$ be a connected <compact Riemann surface> of <genus> $g$. Choose an oriented <symplectic basis> $a_1,\ldots,a_g,b_1,\ldots,b_g$ of $H_1(X,\mathbb Z)$, with $a_i\cdot b_j=\delta_{ij}$. For a closed complex <differential form> $\alpha$, write $A_i(\alpha)=\int_{a_i}\alpha$ and $B_i(\alpha)=\int_{b_i}\alpha$. The <Riemann bilinear relations> begin with the identity
$$
\int_X\alpha\wedge\beta=\sum_{i=1}^g\bigl(A_i(\alpha)B_i(\beta)-B_i(\alpha)A_i(\beta)\bigr).
$$
To prove it, cut $X$ along the chosen cycles to obtain a fundamental polygon $P$. On this simply connected polygon, a closed <differential form> has a primitive $F$, with $dF=\alpha$. <Stokes theorem> gives $\int_P\alpha\wedge\beta=\int_{\partial P}F\beta$. The two copies of each edge have opposite orientations, and their values of $F$ differ by the period acquired in passing between them. Pairing the copies of the $a_i$ and $b_i$ edges gives, respectively, $-B_i(\alpha)A_i(\beta)$ and $A_i(\alpha)B_i(\beta)$. Summing proves the identity; the convention $a_i\cdot b_i=1$ fixes its sign.
For <holomorphic one-forms> $\omega,\eta$, their <wedge product> is zero because locally both are multiples of $dz$. Thus the first <holomorphic> <Riemann bilinear relations> is
$$
\sum_i\bigl(A_i(\omega)B_i(\eta)-B_i(\omega)A_i(\eta)\bigr)=0.
$$
For a nonzero <holomorphic one-form> $\omega=f(z)\,dz$, the second relation follows by applying the bilinear identity to $\omega,\overline\omega$:
$$
i\sum_i\bigl(A_i(\omega)\overline{B_i(\omega)}-B_i(\omega)\overline{A_i(\omega)}\bigr)
=i\int_X\omega\wedge\overline\omega
=2\int_X|f|^2\,dx\,dy>0.
$$
In particular, a <holomorphic one-form> with all $a$-periods zero must vanish. Since $\dim H^0(X,K_X)=g$, the map taking a <holomorphic one-form> to its $a$-periods is an <isomorphism>. We can therefore choose normalized <holomorphic one-forms> $\omega_1,\ldots,\omega_g$ with $\int_{a_i}\omega_j=\delta_{ij}$. Put $B_{ij}=\int_{b_i}\omega_j$. The first relation applied to $\omega_j,\omega_k$ gives $B_{jk}=B_{kj}$, and the second applied to $\sum_jc_j\omega_j$ gives $2\overline c^{,t}(\operatorname{Im}B)c>0$ for $c\ne0$. Consequently
$$
\boxed{B=B^t,\qquad Y=\operatorname{Im}B>0.}
$$
These conclusions make the construction of the <Jacobian variety> work. Integration identifies the image of $H_1(X,\mathbb Z)$ in $H^0(X,K_X)^*$ with the <period lattice> $\Lambda=\mathbb Z^g+B\mathbb Z^g$. The real-linear map $(m,n)\mapsto m+Bn$ is invertible: its imaginary part is $Yn$, and $Y$ is invertible. Hence $\Lambda$ is a discrete lattice of full real <rank> $2g$, and
$$
\boxed{\operatorname{Jac}(X)=H^0(X,K_X)^*/H_1(X,\mathbb Z)\simeq\mathbb C^g/(\mathbb Z^g+B\mathbb Z^g)}
$$
is a <compact> <complex torus>, rather than a non-Hausdorff quotient. Changing the <symplectic basis> changes the displayed coordinates but not this intrinsic quotient.
The <Riemann bilinear relations> also supply its <polarization of a complex torus>. Use the <Hermitian form> $H(z,w)=z^tY^{-1}\overline w$, linear in the first argument, and set $E=-\operatorname{Im}H$. Symmetry of $B$ gives
$$
E(m+Bn,m'+Bn')=m^tn'-n^tm',\qquad E(v,iv)=H(v,v)>0.
$$
Thus $E$ is an integral positive <Riemann form on a complex torus>. Its matrix on the displayed lattice basis is the unimodular <symplectic matrix>. The corresponding positive <holomorphic line bundle>, equivalently the one constructed from theta automorphy in Solution 4, gives a principal <polarization of a complex torus>. By the <Kodaira embedding theorem> this makes the <complex torus> projective, hence an <abelian variety>. The normalization $E=-\operatorname{Im}H$ here is important because $H$ was chosen linear in its first argument. If $g=0$, the spaces of <holomorphic one-forms> and periods are zero and the <Jacobian variety> is a point.
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