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Albanese Map of a Riemannian Manifold

Given a compact Riemannian manifold $M$ such that the first Betti number $b$ is nonzero, there exists a map $\mathrm{alb}:M\to {\mathbb{R}}^b/{\mathbb{Z}}^b$, called the Albanese map. The aim of post is to prove the smoothness of the Albanese map.

Author

Affiliation

Fei Ye

QCC-CUNY

Let $C$ be a Riemann surface of genus $g$ and $p_0$ be point in $C$. There is a map $C\to \mathrm{Jac}(C)=H^1(C,\mathbb{C}{)}^{\ast }/H_1(C,\mathbb{Z})$ defined by $p\mapsto ({\int }_{p_0}^p{\omega }_1,\dots ,{\int }_{p_0}^p{\omega }_g)$. This map is called a Abel-Jacobi map. The generalization of this construction in a higher dimension is also known as the Albanese map. Albanese maps have many good applications. Interested reader may check the discussions What is the Albanese map good for on math stackexchange. For a heuristic discussion of Abel-Jacobi map, we refer the reader to the section on Abel’s theorem in (Griffiths and Harris 1994).

In this post, we will generalized the construction of Albanese map from complex manifolds to real manifolds. Let $M$ be a compact smooth manifold. Fix a Riemannian metric on $M$.

1 First Cohomology group

We first recall de Rham Theorem and Hodge Theorem on Riemannian manifolds.

Theorem 1 (de Rham Theorem) The following map is an isomorphism $$\iota :H_{\mathrm{dR}}^p(M,\mathbb{R})\to H^p(M,\mathbb{R})\simeq {\mathrm{Hom}}_{\mathbb{R}}(H_p(M,\mathbb{R}),\mathbb{R}),$$ where $\iota$ is defined by $\iota ([\omega ])(\gamma )={\int }_{\gamma }\omega$.

Consider harmonic forms which are differential forms annihilated by the Laplacian operator $\mathrm{\Delta }=\mathrm{d}\delta +\delta \mathrm{d}$.

Theorem 2 (Hodge Theorem) In each de Rham cohomology class $[\omega ]$ of a differential $p$-form $\omega$, there is an unique harmonic form ${\omega }_{\mathcal{H}}$ such that $[\omega ]=[{\omega }_{\mathcal{H}}]$.

Denote by ${\mathcal{H}}^1(M,\mathbb{R})$ the space of harmonic 1-forms on $M$. Then $${\mathcal{H}}^1(M,\mathbb{R})\simeq H_{\mathrm{dR}}^1(M,\mathbb{R}).$$

2 Albanese Map

Let ${\omega }_1$, $\dots$, ${\omega }_b$ be a basis of ${\mathcal{H}}^1(M,\mathbb{R})$, where $b$ is the first betti number of $M$. Choose a basis ${\gamma }_1$, $\dots$, ${\gamma }_b$ of $H_1(M,\mathbb{Z})$, we can define a $\mathbb{Z}$-linear map $H_1(M,\mathbb{Z})\to {\mathcal{H}}^1(M,\mathbb{R}{)}^{\ast }={\mathbb{R}}^b$ by $$({\gamma }_1,\dots ,{\gamma }_b)\mapsto ({\int }_{{\gamma }_1},\dots ,{\int }_{{\gamma }_b}).$$

The linear maps ${\int }_{{\gamma }_i}$ are linearly independent over $\mathbb{R}$. Indeed, if $\sum _{i=1}^b{a}_i{\int }_{{\gamma }_i}=0$, then $\sum _{i=1}^b{a}_i{\int }_{{\gamma }_i}\omega =0$ for any harmonic 1-form. Hence, $\sum _{i=1}^b{a}_i{\gamma }_i=0$. Since ${\gamma }_i$ is a basis, then $a_i=0$ for all $i$. Therefore, the image $\mathrm{\Lambda }\subset {\mathcal{H}}^1(M,\mathbb{R}{)}^{\ast }$ of $H_1(M,\mathbb{Z})$ is a lattice isomorphic to the additive subgroup ${\mathbb{Z}}^b\subset {\mathbb{R}}^b$.

Fix a point $p_0$ in $M$. There is a map $${\mathrm{alb}}_M:M\to \mathrm{Alb}(M):={\mathcal{H}}^1(M,\mathbb{R}{)}^{\ast }/\mathrm{\Lambda },$$ called the Albanese map, defined by $${\mathrm{alb}}_M(p)=({\int }_{{\gamma }_p}{\omega }_1,\dots ,{\int }_{{\gamma }_p}{\omega }_b),$$ where ${\gamma }_p$ is a path from $p_0$ to $p$. The map is well-defined, i.e., independent of the choice of the path. Because, if ${\gamma }_x^{\prime }$ is another path, then ${\gamma }_x-{\gamma }_x^{\prime }$ is a loop in $H_1(M,\mathbb{Z})$. The difference is then an element in the lattice $\mathrm{\Lambda }$ which is isomorphic to ${\mathbb{Z}}^b$. The isomorphism between $\mathrm{\Lambda }$ and ${\mathbb{Z}}^b$ follows from the linear independency of the images of a basis of $H_1(M,\mathbb{Z})$.

Equipped with the quotient Euclidean metric, the quotient $\mathrm{Alb}(M)={\mathbb{R}}^b/{\mathbb{Z}}^b$ is a flat torus, called the Jacobi torus or Albanese torus.

Theorem 3 Let $M$ be a compact Riemannian manifold. The Albanese map ${\mathrm{alb}}_M:M\to \mathrm{Alb}(M)$ is smooth map.

Proof. Let $\rho :\tilde{M}\to M$ be the universal covering. Since $M$ is a smooth manifold, $\tilde{M}$ admit a unique smooth structure such that $\pi$ is a smooth covering map (see for example (Lee 2009) Theorem 1.86).

Fix a point ${\tilde{p}}_0\in {\pi }^{-1}(p_0)$. We define a map $$\alpha :\tilde{M}\to {\mathcal{H}}^1(M,\mathbb{R}{)}^{\ast }$$ by $\alpha (\tilde{p})(\omega )={\int }_{{\gamma }_{\tilde{p}}}{\rho }^{\ast }\omega$, where ${\gamma }_{\tilde{p}}$ is a path from ${\tilde{p}}_0$ to $\tilde{p}$. The map $\alpha$ is well-defined because $\tilde{M}$ is contractible.

Then the following diagram commutes $$\begin{array}{ccc}\tilde{M}& \stackrel{\alpha }{\to }& {\mathcal{H}}^1(M,\mathbb{R}{)}^{\ast }\\ \rho \downarrow & & \downarrow q& \\ M& \stackrel{\mathrm{alb}}{\to }& \mathrm{Alb}M\end{array}$$ Because the quotient map $q$ and the covering map $\pi$ are both smooth, and moreover, $\pi$ is locally diffeomorphic, to show that $\mathrm{alb}$ is smooth, it suffices to show that $\alpha$ is smooth. Note that the linear map $F:{\mathcal{H}}^1(M,\mathbb{R}{)}^{\ast }\to {\mathbb{R}}^b$ defined $F(l)=(l({\omega }_1)),\dots ,l({\omega }_b))$ is an isomorphism as it has the full rank. To show that $\alpha$ is smooth, it suffices to show that $F\circ \alpha$ is smooth.

For any harmonic 1-form $\omega \in {\mathcal{H}}^1(M,\mathbb{R})$, we define $${\alpha }_{\omega }:\tilde{M}\to \mathbb{R}$$ by ${\alpha }_{\omega }(\tilde{p})=\alpha (\tilde{p})(\omega )$.

To show $F\circ \alpha$ is smooth, it suffices to show that ${\alpha }_{\omega }$ is smooth for any $\omega$.

Because $\tilde{M}$ is contractible, the 1-form ${\rho }^{\ast }\omega$ is exact. Then there is a smooth function $\beta$ such that ${\rho }^{\ast }\omega =\mathrm{d}\beta$. Therefore, by the Stock’s theorem $${\alpha }_{\omega }(\tilde{p})=\beta (\tilde{p})-\beta ({\tilde{p}}_0).$$ Therefore, ${\alpha }_{\omega }$ is smooth for any $\omega \in {\mathcal{H}}^1(M,\mathbb{R})$.

It follows that $\alpha$ is a smooth map.

The Albanese map indeed is a harmonic map. Moreover, it satisfies the following universal property.

Theorem 4 (Universal Property of Albanese Map) If $f:M\to T$ is a smooth map from $M$ to a flat torus $T$ such that $f(p_0)=0$, then there exists a unique smooth map of flat tori $g:\mathrm{Alb}(M)\to T$ such that $g\circ {\mathrm{alb}}_M=f$, i.e., the following diagram commutes. $$\begin{array}{ccc}M& \stackrel{\mathrm{alb}}{\to }& \mathrm{Alb}(M)\\ f\downarrow & \swarrow g& \\ T& & \end{array}$$

For proofs of the harmonicity and universal property of the Albanese map, we refer the reader to (Nagano and Smyth 1975).

The Albanese map is a special case of Abel-Jacobi map which sends codimenion $k$ cycles to intermediate Jacobi varieties. Interested reader may find details from Chapter 12 in (Voisin 2002) or Chapter V in [Beauville2003].

References

Griffiths, Phillip, and Joe Harris. 1994. Principles of Algebraic Geometry. Wiley classics libr. ed. Wiley Classics Library. New York, NY: Wiley.

Lee, Jeffrey M. 2009. Manifolds and Differential Geometry. Graduate Studies in Mathematics, v. 107. Providence, R.I: American Mathematical Society.

Nagano, Tadashi, and Brian Smyth. 1975. “Minimal Varieties and Harmonic Maps in Tori.” Commentarii Mathematici Helvetici 50 (1): 249–65. https://doi.org/bpthcs.

Voisin, Claire. 2002. Hodge Theory and Complex Algebraic Geometry I. Translated by Leila Schneps. Vol. 1. Cambridge Studies in Advanced Mathematics. Cambridge: Cambridge University Press. https://doi.org/10.1017/CBO9780511615344.