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Proposition. For compact complex manifolds of dimension $1$ (a.k.a. complex Riemann surfaces) the Hodge numbers are topological invariants.

Proof. The Hodge numbers are determined by the genus, which happens to be a topological invariant.

In a more general setting it is true that

Theorem. Let $f:X\to B$ be a family of complex manifolds and assume that $X_0$ is Kähler for some $0\in B$. Then for $b$ in a neighbourhood of $0$ in $B$, the Hodge numbers of $X_b$ are the same as the Hodge numbers of $X_0$.

Proof. See C.Voisin, Hodge Theory and Complex Algebraic Geometry, I, p.235.

Remark. It actually follows that all nearby fibers are Kähler, but it is not needed in the proof.

From this point I can only speculate: I don't think the Hodge numbers are topological invariants. If they were, then the assumption that one fiber is Kähler is superfluous. So, this suggests that it seems likely that there may be topologically equivalent complex manifolds with different Hodge numbers. I would even expect Kähler ones and perhaps even diffeomorphic ones, but that might be going out on a limb.

Catanese and Manetti gave various examples of orientedly diffeomorphic but not deformation equivalent smooth projective algebraic surfaces of general type. I wonder if those would perhaps have different Hodge numbers. The above theorem definitely does not apply. That in itself, of course, does not prove anything...

EDIT: (Added later) As pointed out by Greg Kuperberg here, this same question had been debated before here.

Proposition. For compact complex manifolds of dimension $1$ (a.k.a. complex Riemann surfaces) the Hodge numbers are topological invariants.

Proof. The Hodge numbers are determined by the genus, which happens to be a topological invariant.

In a more general setting it is true that

Theorem. Let $f:X\to B$ be a family of complex manifolds and assume that $X_0$ is Kähler for some $0\in B$. Then for $b$ in a neighbourhood of $0$ in $B$, the Hodge numbers of $X_b$ are the same as the Hodge numbers of $X_0$.

Proof. See C.Voisin, Hodge Theory and Complex Algebraic Geometry, I, p.235.

Remark. It actually follows that all nearby fibers are Kähler, but it is not needed in the proof.

From this point I can only speculate: I don't think the Hodge numbers are topological invariants. If they were, then the assumption that one fiber is Kähler is superfluous. So, this suggests that it seems likely that there may be topologically equivalent complex manifolds with different Hodge numbers. I would even expect Kähler ones and perhaps even diffeomorphic ones, but that might be going out on a limb.

Catanese and Manetti gave various examples of orientedly diffeomorphic but not deformation equivalent smooth projective algebraic surfaces of general type. I wonder if those would perhaps have different Hodge numbers. The above theorem definitely does not apply. That in itself, of course, does not prove anything...

EDIT: (Added later) As pointed out by Greg Kuperberg here, this same question had been debated before in another forum.

Proposition. For compact complex manifolds of dimension $1$ (a.k.a. complex Riemann surfaces) the Hodge numbers are topological invariants.

Proof. The Hodge numbers are determined by the genus, which happens to be a topological invariant.

In a more general setting it is true that

Theorem. Let $f:X\to B$ be a family of complex manifolds and assume that $X_0$ is Kähler for some $0\in B$. Then for $b$ in a neighbourhood of $0$ in $B$, the Hodge numbers of $X_b$ are the same as the Hodge numbers of $X_0$.

Proof. See C.Voisin, Hodge Theory and Complex Algebraic Geometry, I, p.235.

Remark. It actually follows that all nearby fibers are Kähler, but it is not needed in the proof.

From this point I can only speculate: I don't think the Hodge numbers are topological invariants. If they were, then the assumption that one fiber is Kähler is superfluous. So, this suggests that it seems likely that there may be topologically equivalent complex manifolds with different Hodge numbers. I would even expect Kähler ones and perhaps even diffeomorphic ones, but that might be going out on a limb.

Catanese and Manetti gave various examples of orientedly diffeomorphic but not deformation equivalent smooth projective algebraic surfaces of general type. I wonder if those would perhaps have different Hodge numbers. The above theorem definitely does not apply. That in itself, of course, does not prove anything...

EDIT: (Added later) As pointed out by Greg Kuperberg here, this same question had been debated before here.

Proposition. For compact complex manifolds of dimension $1$ (a.k.a. complex Riemann surfaces) the Hodge numbers are topological invariants.

Proof. The Hodge numbers are determined by the genus, which happens to be a topological invariant.

In a more general setting it is true that

Theorem. Let $f:X\to B$ be a family of complex manifolds and assume that $X_0$ is Kähler for some $0\in B$. Then for $b$ in a neighbourhood of $0$ in $B$, the Hodge numbers of $X_b$ are the same as the Hodge numbers of $X_0$.

Proof. See C.Voisin, Hodge Theory and Complex Algebraic Geometry, I, p.235.

Remark. It actually follows that all nearby fibers are Kähler, but it is not needed in the proof.

From this point I can only speculate: I don't think the Hodge numbers are topological invariants. If they were, then the assumption that one fiber is Kähler is superfluous. So, this suggests that it seems likely that there may be topologically equivalent complex manifolds with different Hodge numbers. I would even expect Kähler ones and perhaps even diffeomorphic ones, but that might be going out on a limb.

Catanese and Manetti gave various examples of orientedly diffeomorphic but not deformation equivalent smooth projective algebraic surfaces of general type. I wonder if those would perhaps have different Hodge numbers. The above theorem definitely does not apply. That in itself, of course, does not prove anything...

EDIT: (Added later) As pointed out by Greg Kuperberg here, this same question had been debated before here.

I am not sure that I would go as far as saying that Hodge numbers are topological invariants or that they are independent of the complex structure, but there are certain statements that are definitely true.

Proposition. For compact complex manifolds of dimension $1$ (a.k.a. complex Riemann surfaces) the Hodge numbers are topological invariants.

Proof. The Hodge numbers are determined by the genus, which happens to be a topological invariant.

In a more general setting it is true that

Theorem. Let $f:X\to B$ be a family of complex manifolds and assume that $X_0$ is Kähler for some $0\in B$. Then for $b$ in a neighbourhood of $0$ in $B$, the Hodge numbers of $X_b$ are the same as the Hodge numbers of $X_0$.

Proof. See C.Voisin, Hodge Theory and Complex Algebraic Geometry, I, p.235.

Remark. It actually follows that all nearby fibers are Kähler, but it is not needed in the proof.

From this point I can only speculate: If the Hodge numbers were topological invariants, then the assumption that one fiber is Kähler is superfluous. So, this suggests that it seems likely that there may be topologically equivalent complex manifolds with different Hodge numbers. I would even expect Kähler ones and perhaps even diffeomorphic ones, but that might be going out on a limb.

Catanese and Manetti gave various examples of orientedly diffeomorphic but not deformation equivalent smooth projective algebraic surfaces of general type. I wonder if those would perhaps have different Hodge numbers. The above theorem definitely does not apply. That in itself, of course, does not prove anything...

EDIT: (Added later) As pointed out by Greg Kuperberg here, this same question had been debated before here.

Proposition. For compact complex manifolds of dimension $1$ (a.k.a. complex Riemann surfaces) the Hodge numbers are topological invariants.

Proof. The Hodge numbers are determined by the genus, which happens to be a topological invariant.

In a more general setting it is true that

Theorem. Let $f:X\to B$ be a family of complex manifolds and assume that $X_0$ is Kähler for some $0\in B$. Then for $b$ in a neighbourhood of $0$ in $B$, the Hodge numbers of $X_b$ are the same as the Hodge numbers of $X_0$.

Proof. See C.Voisin, Hodge Theory and Complex Algebraic Geometry, I, p.235.

Remark. It actually follows that all nearby fibers are Kähler, but it is not needed in the proof.

From this point I can only speculate: I don't think the Hodge numbers are topological invariants. If they were, then the assumption that one fiber is Kähler is superfluous. So, this suggests that it seems likely that there may be topologically equivalent complex manifolds with different Hodge numbers. I would even expect Kähler ones and perhaps even diffeomorphic ones, but that might be going out on a limb.

Catanese and Manetti gave various examples of orientedly diffeomorphic but not deformation equivalent smooth projective algebraic surfaces of general type. I wonder if those would perhaps have different Hodge numbers. The above theorem definitely does not apply. That in itself, of course, does not prove anything...

EDIT: (Added later) As pointed out by Greg Kuperberg here, this same question had been debated before here.