[proofplan] We identify the tangent space to deformations with $H^1(C,T_C)$ and compute the cotangent map of the period map as the multiplication map $\operatorname{Sym}^2H^0(C,K_C)\to H^0(C,K_C^{\otimes 2})$. Since duality turns injectivity of the differential into surjectivity of this multiplication map, the theorem becomes a statement about quadratic products of holomorphic differentials. The non-hyperelliptic case is exactly Max Noether's theorem for canonical curves. The hyperelliptic case is computed from the double cover $C\to \mathbb{P}^1$: products of canonical forms land only in the invariant summand of $H^0(C,K_C^{\otimes 2})$, which is all of it in genus $2$ but has positive codimension for genus at least $3$. [/proofplan]

[step:Identify the cotangent map with multiplication of canonical forms]Let $K_C$ denote the holomorphic cotangent bundle of $C$, and let $T_C$ denote the holomorphic tangent bundle. The Kodaira-Spencer identification gives \begin{align*} T_{[C]}\mathcal{M}\cong H^1(C,T_C). \end{align*} For curves, the infinitesimal period map is the map whose action on a deformation class $\xi\in H^1(C,T_C)$ sends a holomorphic one-form $\omega\in H^0(C,K_C)$ to the class obtained by contracting $\xi$ with $\omega$ in $H^1(C,\mathcal O_C)$. Equivalently, its dual map is \begin{align*} \mu:\operatorname{Sym}^2H^0(C,K_C)\longrightarrow H^0(C,K_C^{\otimes 2}), \end{align*} where $\mu$ is defined on pure symmetric tensors by \begin{align*} \mu(\omega\eta)=\omega\otimes\eta \end{align*} for $\omega,\eta\in H^0(C,K_C)$, using the natural multiplication of sections of the line bundle $K_C$. We justify the duality statement explicitly. Serre duality for the compact Riemann surface $C$ gives a perfect pairing \begin{align*} H^1(C,T_C)\times H^0(C,K_C^{\otimes 2})\longrightarrow \mathbb C, \end{align*} because $T_C^*\otimes K_C\cong K_C^{\otimes 2}$. It also gives a perfect pairing \begin{align*} H^1(C,\mathcal O_C)\times H^0(C,K_C)\longrightarrow \mathbb C. \end{align*} Under these pairings, the [bilinear form](/page/Bilinear%20Form) associated to the differential of the period map is \begin{align*} (\xi,\omega,\eta)\longmapsto \langle \xi,\omega\eta\rangle, \end{align*} where $\xi\in H^1(C,T_C)$ and $\omega,\eta\in H^0(C,K_C)$. Therefore the transpose of $d\mathcal P_{[C]}$ is precisely $\mu$. Consequently, \begin{align*} d\mathcal P_{[C]}\text{ is injective}\quad \iff \quad \mu\text{ is surjective}. \end{align*}[/step]

[guided]The period map records the Hodge structure on $H^1(C,\mathbb C)$, so its first derivative is controlled by how first-order deformations move holomorphic one-forms. The deformation tangent space is \begin{align*} T_{[C]}\mathcal{M}\cong H^1(C,T_C), \end{align*} by the [Kodaira-Spencer correspondence](/theorems/9117). For a class $\xi\in H^1(C,T_C)$ and a holomorphic one-form $\omega\in H^0(C,K_C)$, the first variation of $\omega$ is obtained by contracting the tangent-vector part of $\xi$ with $\omega$. This gives a class in $H^1(C,\mathcal O_C)$. Thus the infinitesimal period map can be viewed as \begin{align*} d\mathcal P_{[C]}:H^1(C,T_C)\longrightarrow \operatorname{Hom}(H^0(C,K_C),H^1(C,\mathcal O_C)). \end{align*} To compute injectivity, it is better to dualize. Serre duality gives a perfect pairing \begin{align*} H^1(C,T_C)\times H^0(C,T_C^*\otimes K_C)\longrightarrow \mathbb C. \end{align*} Since $T_C^*\cong K_C$, this target is \begin{align*} H^0(C,T_C^*\otimes K_C)=H^0(C,K_C^{\otimes 2}). \end{align*} Serre duality also gives a perfect pairing \begin{align*} H^1(C,\mathcal O_C)\times H^0(C,K_C)\longrightarrow \mathbb C. \end{align*} Under these two pairings, evaluating the dual of the infinitesimal period map on two holomorphic one-forms $\omega,\eta\in H^0(C,K_C)$ gives the quadratic differential $\omega\eta\in H^0(C,K_C^{\otimes 2})$. Thus the cotangent map is the multiplication map \begin{align*} \mu:\operatorname{Sym}^2H^0(C,K_C)\longrightarrow H^0(C,K_C^{\otimes 2}). \end{align*} The passage from injectivity to surjectivity is ordinary finite-dimensional linear algebra: a [linear map](/page/Linear%20Map) between finite-dimensional complex vector spaces is injective if and only if its dual map is surjective. Hence proving the theorem is equivalent to determining exactly when $\mu$ is surjective.[/guided]

[step:Apply Max Noether to the non-hyperelliptic case] Assume $g\ge 3$ and $C$ is non-hyperelliptic. Max Noether's theorem for canonical curves states that, for a non-hyperelliptic compact Riemann surface of genus at least $3$, the canonical ring \begin{align*} \bigoplus_{m=0}^{\infty}H^0(C,K_C^{\otimes m}) \end{align*} is generated in degree $1$. In degree $2$, this says precisely that \begin{align*} \mu:\operatorname{Sym}^2H^0(C,K_C)\longrightarrow H^0(C,K_C^{\otimes 2}) \end{align*} is surjective. By the cotangent identification above, $d\mathcal P_{[C]}$ is injective. [/step]

[step:Compute canonical sections on a hyperelliptic curve]Assume now that $C$ is hyperelliptic of genus $g\ge 2$. Let \begin{align*} \pi:C\longrightarrow \mathbb P^1 \end{align*} be the hyperelliptic double cover, and let $\iota:C\to C$ be the hyperelliptic involution. Let $L$ be the line bundle $\mathcal O_{\mathbb P^1}(g+1)$ associated with the double cover, so that \begin{align*} \pi_*\mathcal O_C\cong \mathcal O_{\mathbb P^1}\oplus L^{-1} \end{align*} and the second summand is the $(-1)$-eigensheaf for $\iota$. The canonical bundle formula for the double cover gives \begin{align*} K_C\cong \pi^*(K_{\mathbb P^1}\otimes L)\cong \pi^*\mathcal O_{\mathbb P^1}(g-1). \end{align*} Therefore the projection formula gives \begin{align*} H^0(C,K_C)\cong H^0(\mathbb P^1,\mathcal O_{\mathbb P^1}(g-1))\oplus H^0(\mathbb P^1,\mathcal O_{\mathbb P^1}(-2)). \end{align*} Since $H^0(\mathbb P^1,\mathcal O_{\mathbb P^1}(-2))=0$, every holomorphic one-form on $C$ lies in the same eigenspace for the hyperelliptic involution. Thus every product of two holomorphic one-forms lies in the invariant summand of $H^0(C,K_C^{\otimes 2})$. For the square of the canonical bundle, the same computation gives \begin{align*} K_C^{\otimes 2}\cong \pi^*\mathcal O_{\mathbb P^1}(2g-2). \end{align*} Hence \begin{align*} H^0(C,K_C^{\otimes 2})\cong H^0(\mathbb P^1,\mathcal O_{\mathbb P^1}(2g-2))\oplus H^0(\mathbb P^1,\mathcal O_{\mathbb P^1}(g-3)). \end{align*} The first summand is invariant under $\iota$, and the second summand is anti-invariant.[/step]

[guided]The hyperelliptic curve comes with a degree-two map \begin{align*} \pi:C\longrightarrow \mathbb P^1 \end{align*} and an involution $\iota:C\to C$ exchanging the two sheets. The structure sheaf of $C$ decomposes after pushing forward to $\mathbb P^1$: \begin{align*} \pi_*\mathcal O_C\cong \mathcal O_{\mathbb P^1}\oplus L^{-1}, \end{align*} where $L\cong\mathcal O_{\mathbb P^1}(g+1)$. The summand $\mathcal O_{\mathbb P^1}$ consists of invariant functions, while $L^{-1}$ consists of anti-invariant functions. The canonical bundle formula for this double cover is \begin{align*} K_C\cong \pi^*(K_{\mathbb P^1}\otimes L). \end{align*} Since $K_{\mathbb P^1}\cong\mathcal O_{\mathbb P^1}(-2)$ and $L\cong\mathcal O_{\mathbb P^1}(g+1)$, we get \begin{align*} K_C\cong \pi^*\mathcal O_{\mathbb P^1}(g-1). \end{align*} Now apply the projection formula to the line bundle $\mathcal O_{\mathbb P^1}(g-1)$. It gives \begin{align*} H^0(C,K_C)\cong H^0(\mathbb P^1,\mathcal O_{\mathbb P^1}(g-1))\oplus H^0(\mathbb P^1,\mathcal O_{\mathbb P^1}(-2)). \end{align*} The second summand vanishes because $\mathcal O_{\mathbb P^1}(-2)$ has negative degree. Therefore all holomorphic one-forms on $C$ come from the first summand. In particular, when two such forms are multiplied, their product is invariant under the hyperelliptic involution. For quadratic differentials, the same method gives \begin{align*} K_C^{\otimes 2}\cong \pi^*\mathcal O_{\mathbb P^1}(2g-2). \end{align*} Using the projection formula again, \begin{align*} H^0(C,K_C^{\otimes 2})\cong H^0(\mathbb P^1,\mathcal O_{\mathbb P^1}(2g-2))\oplus H^0(\mathbb P^1,\mathcal O_{\mathbb P^1}(g-3)). \end{align*} The first summand is invariant and the second is anti-invariant. This decomposition is the decisive point: multiplication of holomorphic one-forms can only reach the invariant part, so injectivity can fail exactly when the anti-invariant part is nonzero.[/guided]

[step:Show the hyperelliptic multiplication map has cokernel of dimension $g-2$] Under the identification \begin{align*} H^0(C,K_C)\cong H^0(\mathbb P^1,\mathcal O_{\mathbb P^1}(g-1)), \end{align*} the multiplication map $\mu$ becomes the ordinary polynomial multiplication map \begin{align*} \operatorname{Sym}^2H^0(\mathbb P^1,\mathcal O_{\mathbb P^1}(g-1))\longrightarrow H^0(\mathbb P^1,\mathcal O_{\mathbb P^1}(2g-2)) \end{align*} on the invariant summand of $H^0(C,K_C^{\otimes 2})$. This map is surjective because every homogeneous polynomial of degree $2g-2$ in two variables is a linear combination of products of homogeneous polynomials of degree $g-1$. Thus \begin{align*} \operatorname{im}\mu=H^0(\mathbb P^1,\mathcal O_{\mathbb P^1}(2g-2)) \end{align*} inside the invariant summand. The cokernel is therefore \begin{align*} H^0(\mathbb P^1,\mathcal O_{\mathbb P^1}(g-3)). \end{align*} Since \begin{align*} \dim_{\mathbb C}H^0(\mathbb P^1,\mathcal O_{\mathbb P^1}(d))=d+1 \end{align*} for $d\ge 0$, this cokernel has dimension \begin{align*} \dim_{\mathbb C}H^0(\mathbb P^1,\mathcal O_{\mathbb P^1}(g-3))=g-2. \end{align*} For $g\ge 3$, this dimension is positive, so $\mu$ is not surjective. By duality, $d\mathcal P_{[C]}$ is not injective. [/step]

[step:Check genus $2$ separately] Assume $g=2$. The hyperelliptic computation above gives \begin{align*} H^0(C,K_C)\cong H^0(\mathbb P^1,\mathcal O_{\mathbb P^1}(1)) \end{align*} and \begin{align*} H^0(C,K_C^{\otimes 2})\cong H^0(\mathbb P^1,\mathcal O_{\mathbb P^1}(2))\oplus H^0(\mathbb P^1,\mathcal O_{\mathbb P^1}(-1)). \end{align*} The second summand vanishes, so \begin{align*} H^0(C,K_C^{\otimes 2})\cong H^0(\mathbb P^1,\mathcal O_{\mathbb P^1}(2)). \end{align*} The multiplication map is \begin{align*} \operatorname{Sym}^2H^0(\mathbb P^1,\mathcal O_{\mathbb P^1}(1))\longrightarrow H^0(\mathbb P^1,\mathcal O_{\mathbb P^1}(2)). \end{align*} If $u,v$ are homogeneous coordinates on $\mathbb P^1$, then $u^2$, $uv$, and $v^2$ form a basis of $H^0(\mathbb P^1,\mathcal O_{\mathbb P^1}(2))$, and each is a product of two sections of $\mathcal O_{\mathbb P^1}(1)$. Hence $\mu$ is surjective. Therefore $d\mathcal P_{[C]}$ is injective in genus $2$. [/step]

[step:Combine the three cases] If $g\ge 3$ and $C$ is non-hyperelliptic, Max Noether's theorem makes $\mu$ surjective, so the differential of the period map is injective. If $g=2$, the explicit hyperelliptic computation also makes $\mu$ surjective, so the differential is injective. If $g\ge 3$ and $C$ is hyperelliptic, the cokernel of $\mu$ has dimension $g-2>0$, so $\mu$ is not surjective and the differential is not injective. These are exactly the alternatives stated in the theorem. [/step]