Let $X$ be a compact Kähler manifold and $alb \colon X \to \mathrm{Alb}(X)$ be the Albanese morphism. I am interested in a number of questions about relations between the geometry of $X$ and the geometry of $alb$. There is a number of more or less obious considerations (such as: $alb$ induces isomorphisms on $H^1$ and $Pic^0$; its fibers contain each point together with its rational connected component, etc.) However, I can't find even particular answers to plenty of natural questions. For example here are the simplest:

  • When is the Albanese map surjective?

  • And when is it injective?

  • When is the image smooth? I've heard that it can be singular, though I am not able to construct a counter-example nor to find it in the literature.

I suspect that no simple answers can be giving to these questions, but I'd be glad to hear any necessar and/or sufficient conditions on $X$ for the questions above.

As I have mentioned, these are only the simplest questions and I am interested in any non-trivial results on Albanese mappings.

  • 1
    $\begingroup$ There is a nice discussion of properties of the Albanese mapping in Ueno's Classification theory of algebraic varieties and compact complex manifolds. $\endgroup$
    – Ben McKay
    Commented Sep 6, 2017 at 19:47
  • $\begingroup$ Perhaps you know this work, and the conjecture that -K nef implies surjectivity. ams.org/journals/proc/2003-131-02/S0002-9939-02-06702-3/… $\endgroup$
    – roy smith
    Commented Sep 6, 2017 at 23:47
  • $\begingroup$ apparently knowing some plurigenera = 1 suffices for subjectivity. I am a novice, but this is a nice review on the paper of Ein and Lazarsfeld on Singularities of Theta divisors. books.google.com/… $\endgroup$
    – roy smith
    Commented Sep 6, 2017 at 23:58

2 Answers 2


The first natural observation is the following. Set as usual $q(X):=h^{1, \,0}(X)$, so that $q(X)= \dim \mathrm{Alb}(X)$. Then the Albanese map of $X$ cannot be surjective if $\dim X < q(X)$, and it cannot be injective if $\dim X > q(X)$.

Moreover, the Albanese map is never injective if $X$ contains some rational curve, because any map from $\mathbb{P}^1$ to an abelian variety is necessarily constant.

Let me now give an example related to your third question, showing that the answer can be quite subtle in general.

Let $C$ be a smooth curve of genus $3$ and let $X:= \mathrm{Sym}^2(C)$ be its second symmetric product. Then $X$ is a smooth, minimal surface of general type with $p_g=q=3$, $K^2=6$.

Therefore $\mathrm{Alb}(X)$ is an abelian threefold, and we can show that the Albanese map $$a_X \colon X \longrightarrow \mathrm{Alb}(X)$$ is a birational morphism onto its image $\Sigma \subset \mathrm{Alb}(X)$, which is a principal polarization.


  • if $C$ is non-hyperelliptic then $a_X$ is an immersion and so $\Sigma$ is smooth;
  • if $C$ is hyperelliptic then $a_X$ contracts the unique $(-2)$-curve in $X$ corresponding to the $g_2^1$ on $C$; in this case, $\Sigma$ has an ordinary double point as its unique singularity.
  • $\begingroup$ In regard to this lovely answer, one may look also at the same map in the genus 2 case, where the g(1,2) is represented by the unique (-1) curve on the symmetric square of C, which then collapses to a smooth point on the Jacobian by the surjective Albanese map. I.e. contracting rational curves negates injectivity, but may or may not produce singularities. $\endgroup$
    – roy smith
    Commented Sep 6, 2017 at 23:40
  • 1
    $\begingroup$ If you want a reference to the literature, the paper of Smith and Varley in the International Journal of Math in 2002, on a Torelli type result for Prym varieties gives the generalization of this example, that the Abel-Prym map for any doubly covered non h.e. curve of genus ≥ 4, is an albanese map with image the Prym theta divisor. Hence in general the image is singular but the source is not. worldscientific.com/doi/abs/10.1142/S0129167X02001137. Of course the direct generalization woulld be to the Abel parametrization of a Jacobian theta divisor. $\endgroup$
    – roy smith
    Commented Sep 7, 2017 at 0:25

Let me give an answer to your first question.

Let $\pi: X \to Y$ be a dominant morphism of smooth projective varieties over $\mathbf C$ with connected fibers; $E$ a prime divisor on $Y$. The multiplicity of $\pi$ along $E$ is defined by $$ m(E) \overset{\text{def}}= \inf\{m_j\}, \quad \pi^\ast(E) = \sum_j m_j D_j, $$ and $$ \Delta_\pi \overset{\text{def}}= \sum_i \left(1 - \frac1{m(E_i)}\right) E_i $$ is called the multiplicity divisor associated to $\pi$.

Let $\pi: X \dashrightarrow Y$ be a dominant rational map of smooth projective varieties over $\mathbf C$ with connected fibers. The Kodaira dimension of $\pi$, denoted by $\kappa(\pi)$, is defined to be $\inf\{\kappa(Y^\prime, K_{Y^\prime} + \Delta_{\pi^\prime})\}$, where $\pi^\prime: X^\prime \to Y^\prime$ is taken over all dominant morphisms such that there exist birational maps $u: X \dashrightarrow X^\prime$ and $v:Y \dashrightarrow Y^\prime$ satisfying $\pi^\prime \circ u = v \circ \pi$.

Let $\pi: X \dashrightarrow Y$ be a dominant rational map of smooth projective varieties over $\mathbf C$ with connected fibers. $\pi$ is said to be of general type if $\kappa(\pi) = \dim(Y)$.

Let $X$ be a smooth projective variety over $\mathbf C$. $X$ is said to be special if there is no dominant rational map of general type with connected fibers from $X$ to any smooth projective variety $Y$ with $\dim(Y) > 0$.

Theorem. If $X$ is special, then the Albanese morphism $\alpha: X \to A$ is dominant with connected fibers and $\Delta_\alpha = 0$.

Proof. [CAM] Proposition 5.3.

Theorem. If $X$ is rationally connected, then $X$ is special (but of course the Albanese morphism is trivial in this case).

Theorem. If $\kappa(X) = 0$, then $X$ is special.

Theorem. If $-K_X$ is nef, then $X$ is special.

Theorem. If $X$ is special, then any finite étale covering of $X$ is also special.

Theorem. For any $n > 0$ and $\kappa \in \{-\infty, 0, \dots, n - 1\}$, there exists a special variety with dimension $n$ and Kodaira dimension $\kappa$.

Conjecture. $X$ is special if and only if the Kobayashi pseudo-metric on $X$ is trivial.

Conjecture. If $X$ is defined over a number field $K$, then $X_{\mathbf C}$ is special if and only if $X(L)$ is Zariski dense for some finite extension $L \mid K$.

[CAM] Frédéric Campana. Orbifolds, Special Varieties and Classification Theory.


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