In relation to [this question](https://mathoverflow.net/questions/318900/3-fold-of-general-type-homeomorphic-to-rational-3-fold), I would like to ask for examples of (complex) threefolds of general type with no (nontrivial) holomorphic form.



Threefolds of general type with no holomorphic forms?
One simply connected example arises as the total space of a family of Godeaux surfaces over $\mathbb{P}^1\_k$ with sufficiently positive discriminant.


Let $k$ be a field of characteristic different from $5$. Let $\mathbb{P}^1\_k$ be $\text{Proj}\ k[R,S]$. Let $\mathbb{P}^3\_k$ be $\text{Proj}\ k[T\_0,T\_1,T\_2,T\_3]$. Let $e$ be a nonnegative integer. Let $b\_0,b\_1,b\_2,b\_3 \in k[R,S]\_e$ be a $4$-tuple of homogeneous polynomials of degree $e$ on $\mathbb{P}^1\_k$.


**Hypothesis 1.** The integer $e$ is $\geq 3$. Every zero in $\mathbb{P}^1\_k$ of each polynomial $b\_i$ is a simple zero, and there is no simultaneous zero of two or more of the polynomials $b\_i,b\_j$.


**Notation 2.** Denote by $f$ be the section of $\text{pr}\_{\mathbb{P}^1}^\*\mathcal{O}(e)\otimes \text{pr}\_{\mathbb{P}^3}^\*\mathcal{O}(5)$, $$f = b\_0 T\_0^5 + b\_1 T\_1^5 + b\_2 T\_2^5 + b\_3 T\_3^5.$$ Denote by $Y$ be the zero scheme of $f$ as a hypersurface in $\mathbb{P}^1\_k\times\_{\text{Spec}\ k} \mathbb{P}^3\_k$. For each $i=0,1,2,3$, denote by $p\_i$ the $k$-point of $\mathbb{P}^3\_k$ where $T\_j$ vanishes for every $j\neq i$. Also denote by $Y\_i\subset Y$ the product $\text{Zero}(b\_i)\times\{p\_i\}$. Finally, denote the restrictions of $\text{pr}\_{\mathbb{P}^1}$ and $\text{pr}\_{\mathbb{P}^3}$ to $Y$ as follows, $$\pi:Y\to \mathbb{P}^1\_k, \ \ \rho:Y\to \mathbb{P}^3\_k.$$


Let $\mu\_5$ denote the group $k$-scheme of $5^{\text{th}}$ roots of unity in $\mathbb{G}\_{m,k}$. Since $\text{char}(k)$ is not $5$, this is $k$-smooth. Let $\mu\_5$ act on $\mathbb{P}^1\_k\times\_{\text{Spec}\ k}\mathbb{P}^3\_k$ as follows, $$\zeta \bullet([S\_0,S\_1],[T\_0,T\_1,T\_2,T\_3]) = ([S\_0,S\_1],[T\_0,\zeta T\_1, \zeta^2 T\_2, \zeta^3 T\_3]).$$ The polynomial $f$ is invariant for this action. Thus, there is an induced action on $Y$.


**Notation 3.** Denote the quotient of this $\mu\_5$-action on $Y$ by $\nu:Y\to X'$. Denote by $\phi:X\to X'$ any strong desingularization of $X'$ that is projective.


**Proposition 4.** The projective $k$-scheme $X$ is simply connected and of general type. Moreover, for $r=1,2,3$, the only global section of $\Omega^r\_{X/k}$ is the zero section.


This will be proved in stages.


**Lemma 5.** The hypersurface $Y$ is smooth. The singular locus of the projection $\pi$ equals $Y\_0\cup Y\_1\cup Y\_2 \cup Y\_3$. The projection $\rho$ is smooth over $p\_i$ for every $i=0,1,2,3$. The dualizing sheaf of $Y$ equals the restriction of $\text{pr}\_{\mathbb{P}^1}^\*\mathcal{O}(e-2)\otimes \text{pr}\_{\mathbb{P}^3}^\*\mathcal{O}(1)$. 


**Proof.** By the Jacobian criterion, the projection $\pi$ is smooth away from $\cup\_i Y\_i$. The fiber of $\rho$ over $p\_i$ is the zero scheme of $b\_i$, and this is reduced (hence smooth) by hypothesis. Thus, the projection morphism $\rho$ is smooth at every point of $\cup\_i Y\_i$. For every point of $Y$, either $\pi$ or $\rho$ is smooth, and hence $Y$ is everywhere smooth. The remaining computations are straightforward. **QED**


**Lemma 6.** The action of $\mu\_5$ on $Y\setminus \cup\_i Y\_i$ is free. For every point of $\cup\_i Y\_i$, the *age* (in the sense of Reid -- Shepherd-Barron -- Tai) is $> 1$.


**Proof.** The fixed points of the action on $\mathbb{P}^3\_k$ are the points $p\_i$. Since the projection $\rho$ is equivariant, every fixed point in $Y$ maps to $p\_i$ for some $i$. Thus, the fixed locus is contained in $\cup\_i Y\_i$.


Since the projection $\rho$ is étale at every point of $\cup\_i Y\_i$, the age of that fixed point in $Y$ equals the age of the image fixed point, say $p\_i$, in $\mathbb{P}^3\_k$. By the choice of action of $\mu\_5$ on $\mathbb{P}^3\_k$, every eigenvalue that occurs in $T\_{\mathbb{P}^3,p\_i}$ has multiplicity $1$. Moreover, since the fixed points are isolated, none of these eigenvalues equals $1$. Thus, for each choice of primitive generator $\zeta$ of $\mu\_5$, the eigenvalues $\zeta^a,\zeta^b,\zeta\_c$ that occur have $0<a<b<c <5$, up to rearrangement. Thus, the sum $a+b+c$ is at least $1+2+3=6$. Therefore the age $\alpha(\mathbb{P}^3,p\_i) = (a+b+c)/5$ is strictly larger than $1$. **QED** 


**Lemma 7.** The $k$-scheme $X'$ is projective, normal, $\mathbb{Q}$-Gorenstein, has ample $\mathbb{Q}$-canonical divisor class, and has only terminal finite quotient singularities. Finally, $X'$ is simply connected. Thus, every projective desingularization $X$ of $X'$ is a simply connected, projective $3$-fold of general type.


**Proof.** The only nontrivial claim is that $X'$ has only terminal singularities. This follows by the Reid--Shepherd-Barron--Tai criterion and the computation of the age above. 
Since $Y$ is simply connected, also $X'$ is simply connected.
**QED**


Since the morphism $\pi$ is $\mu\_5$-invariant, there is a unique $k$-morphism, $$\pi':X'\to \mathbb{P}^1\_k,$$ such that $\pi'\circ \nu$ equals $\pi$. By Lemma 3, the singular locus of $\pi'$ equals $X'\_0\cup X'\_1\cup X'\_2 \cup X'\_3$, for $X'\_i:= \nu(Y\_i)$. Thus, the smooth locus of $\pi$ equals the open complement $U$ of $\cup\_i X'\_i$ in $X'$. On $U$, the transitivity exact sequence of $\pi$ gives a short exact sequence, $$0 \to \pi^\*\Omega\_{\mathbb{P}^1/k}|\_U \to \Omega\_{U/k} \to \Omega\_{\pi}|\_U \to 0.$$ This induces a short exact sequence, $$0 \to \pi^\*\Omega\_{\mathbb{P}^1/k}\otimes\Omega\_{\pi}|\_U \to \Omega^2\_{U/k} \to \Omega^2\_{\pi}|\_U \to 0.$$ This also induces an isomorphism, $$\Omega^3\_{U/k} \cong \pi^\*\Omega\_{\mathbb{P}^1/k}\otimes \Omega^2\_{\pi}|\_U.$$


**Lemma 8.** For $r=1,2$, the only global section of $\Omega^r\_\pi$ on $U$ is the zero section. 


**Proof.** These coherent sheaves are locally free sheaves on the smooth $k$-scheme $U$. Every nonzero section on $U$ restricts to a nonzero section of the restriction on the generic fiber $F$ of $\pi$. The restriction of $\Omega^r\_\pi$ to the generic fiber $F$ equals $\Omega^r\_{F/k(\mathbb{P}^1)}$. The generic fiber $F$ of $\pi$ is a Godeaux surface over $k(\mathbb{P}^1)$. Thus, $\Omega^r\_{F/k(\mathbb{P}^1)}$ has only the zero global section for $r=1$ and $r=2$. **QED**


**Proof of Proposition 4.** It only remains to prove that every global section of $\Omega^r\_{X/k}$ is the zero section for $r=1,2,3$. Since $\phi$ is an isomorphism over $U$, it suffices to prove that every global section of $\Omega^r\_{U/k}$ is the zero section.


Since the global sections of $\Omega^r\_\pi$ on $U$ are zero for $r>0$, the only possibility for a nonzero global section of $\Omega^r\_{U/k}$ is a global section of $\pi^\*\Omega\_{\mathbb{P}^1\_k/k}$ with $r=1$. Of course $\pi^\*\Omega\_{\mathbb{P}^1\_k/k}$ equals $\pi^\*\mathcal{O}(-2)$. Since $X'$ is normal and since $\cup\_i X'\_i$ consists of finitely many points, every global section of $\pi^\*\mathcal{O}(-2)$ on $U$ is the restriction of a unique global section on all of $X'$ of $\pi^\*\mathcal{O}(-2)$. By the projection formula, every such global section is the zero global section. **QED** 


