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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.

Moreover:

• 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.

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.

Moreover:

• 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.

The first natural observations is the following: the Albanese map of $X$ cannot be surjective if $\dim \,X < q(X) = \dim \mathrm{Alb}(X)$, 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.

Moreover:

• 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.

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.

Moreover:

• 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.

The answer to these questions is not simple in general. Let me give an example related to your third question that shows some kind of situation we can encounter.

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

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

Moreover:

• if $C$ is non-hyperelliptic then $a_S$ is an immersion and so $\Sigma$ is smooth;
• if $C$ is hyperelliptic then $a_S$ contracts the unique $(-2)$-curve in $S$ corresponding to the $g_2^1$ on $C$; in this case, $\Sigma$ has an ordinary double point as its unique singularity.

The first natural observations is the following: the Albanese map of $X$ cannot be surjective if $\dim \,X < q(X) = \dim \mathrm{Alb}(X)$, 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.

Moreover:

• 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.