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In an article about toric manifolds, I have seen the following notions, which I don't understand. We view a symplectic toric manifold $(M,\omega)$ as a Kähler manifold with Kähler form $\omega$, and first Chern class $c$.


It is said that the cohomology class $[\omega]$ and the first Chern class $c$ are effective, which means that they are Poincaré dual to some holomorphic hypersurfaces.

Question 1: why is that true ?


We call a homology class in $H_2(M,\mathbb{Z})$ effective if it has non-negative intersection indices with fundamental cycles of all compact holomorphic hypersurfaces in $M$.

Question 2: What kind of intersection index are we talking about here?

Would this definition be equivalent to saying that a homology class $A \in H_2(M,\mathbb{Z})$ is effective if for any compact holomorphic hypersurface $B \in H_{n-2}(M, \mathbb{Z})$, the natural pairing $\langle B, A \rangle$ is non-negative ?


From now on, we suppose that the cohomology class $[\omega]$ is integral, that is, it is the image of an element in $H_2(M,\mathbb{Z})$.


Question 3: Is there any relation between the quotient $\underset{A \text{ effective}}{\max} \langle c, A \rangle / \langle [\omega], A \rangle$ and the cup-length of the toric manifold $M$ ?

Question 4: In the article, the cohomology class $[\omega]$ is assumed to be primitive. What does this mean ?


Toric manifolds can actually be constructed via a process called symplectic reduction: given the linear action of the torus $(S^1)^n$ on $\mathbb{C}^n$, and a subtorus $\mathbb{T}^k \subset (S^1)^n$, one can choose a regular value $p \in \text{Lie}(\mathbb{T}^k)^* \simeq \mathbb{R}^k$ of the moment map $P$ associated with the $\mathbb{T}^k$ action on $\mathbb{C}^n$, and define a toric manifold a the quotient $M_p := P^{-1}(p) / \mathbb{T}^k$. Given this construction, there is an isomorphism $$\mathbb{Z}^k \simeq H_2(M, \mathbb{Z}),$$ which identifies the set of effective homology classes with the intersection of the lattice $\mathbb{Z}^k$ with the first orthant in $\mathbb{R}^n \simeq \text{Lie}((S^1)^n)$.

Question 5: what is this isomorphism, and why does this identification hold ?

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  • $\begingroup$ The first Chern class equals the cycle class of the sum of all irreducible components of the open torus orbit. So, in projective space $\mathbb{P}^n$, that is the sum of the (cycle classes) of the $n+1$ coordinate hyperplanes. $\endgroup$ – Jason Starr Oct 19 '18 at 11:15
  • $\begingroup$ Thanks @Jason Starr. I don't understand what your claim is. Does any sum of irreductible components of the torus orbit correspond to holomorphic hypersurface ? How I can we see that ? $\endgroup$ – BrianT Oct 19 '18 at 11:21
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    $\begingroup$ "How I can we see that?" For a holomorphic action of a torus $T\cong (\mathbb{C}^\times)^r$ on a complex manifold $X$, the "derivative" of the action is an induced holomorphic map of holomorphic vector bundles $u:\mathfrak{t}\otimes_{\mathbb{C}}\mathcal{O}_X \to T_X$, where $\mathfrak{t}$ is the Lie algebra of $T$ and where $T_X$ is the (holomorphic) tangent bundle of $X$. Assuming that there exists an open, free $T$-orbit, i.e., $X$ is a toric manifold, then the determinant of $u$ is (locally) a holomorphic function whose zero locus is the sum of these irreducible components. $\endgroup$ – Jason Starr Oct 19 '18 at 11:31
  • $\begingroup$ Thank you very much. Do you have a reference for the answers to my questions ? Do you know why we use the term "effective" ? $\endgroup$ – BrianT Oct 19 '18 at 11:32
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    $\begingroup$ Grauert's book "Coherent Analytic Sheaves" is an excellent reference about coherent analytic sheaves such as $\mathcal{O}_X$, $T_X$, and $\mathfrak{t}\otimes_{\mathbb{C}}\mathcal{O}_X$. A cohomology class on a compact Kaehler manifolds is "effective" if it equals the (Poincare dual) of a sum of cycle classes of closed analytic subvarieties of the Kaehler manifold. For instance, the zero locus of the "holomorphic function" $\text{det}(u)$ is a closed analytic subvariety whose cycle class represents the first Chern class. $\endgroup$ – Jason Starr Oct 19 '18 at 11:45

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