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
    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
    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
    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
    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
    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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.