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Question regarding moduli space of a Calabi-Yau manifold

+ 2 like - 0 dislike
183 views

On page 132 of "Introduction to Supergravity" by Horiatiu Nastase, the author says:

On $M = CY_3$ (Calabi-Yau space) there are $b_3$ topologically nontrivial 3-surfaces, for which we can define a basis $(A_I, B^J)$, where $I, J = 1, \ldots, b_3/2$, such that $A_I \cap A_J = B^I \cap B^J = 0$ and $A_I \cap B^J = -B^J \cap A_I = \delta_I^J$.

What does the intersection symbol denote here? I understand that $A_I$ and $B^J$ are $(b_3/2, 0)$ and $(0, b_3/2)$ forms.

What does the minus sign in $A_I \cap B^J = -B^J \cap A_I = \delta_I^J$ mean?

This post imported from StackExchange Physics at 2015-06-07 10:46 (UTC), posted by SE-user leastaction
asked Jun 7, 2015 in Theoretical Physics by leastaction (425 points) [ no revision ]

1 Answer

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Here $A_I$ and $B^J$ are not differential forms. There are elements of the third homology group $H_3(X, \mathbb{Z})$ of $X$. In general, if $X$ is a manifold, one can define its homology groups $H_i(X,\mathbb{Z})$, for $i$ integer running from $0$ to the dimension $d$ of $X$, as follows. Consider all the possible submanifolds of $X$ of dimension $i$ and then all the formal linear combinations with integer coefficients of such submanifolds. As it is possible to add and substract formal linear combinations, one obtains a commutative group which is in general "huge" because there are many submanifolds in $X$: given a submanifold and moving  it a bit you obtain another submanifold. The idea of homology, which goes back to Poincaré, is to reduce this hugeness by declaring as trivial a submanifold of dimension $i$ which is the boundary of a domain of dimension $(i+1)$ in $X$. The homology group $H_i(X, \mathbb{Z})$ is defined as the quotient of the huge group of linear combinations of submanifolds of dimension $i$ by the subgroup generated by the submanifolds which are trivial in the above sense, i.e. boundaries. (Beware: this description is technically wrong and has for only goal to give a rough idea of what homology is). One can show that the homology groups $H_i(X,\mathbb{Z})$ are not huge but of reasonable size: there are commutative groups of finite type. Under some technical assumption (no torsion), this means that they are of the form $H_i(X,\mathbb{Z})=\mathbb{Z}^{b_i}$ for some integers $b_i$, i.e. it is possible to find a basis of $H_i(X,\mathbb{Z})$ made of $b_i$ elements. The numbers $b_0$,$b_1$,...,$b_d$ are called the Betti numbers of $X$.

Some examples: for $X$ a 2-sphere, of dimension $d=2$, we have $b_0=1$,$b_1=0$, $b_2=1$. For $X$ a 2-torus, of dimension $d=2$, we have $b_0=1$, $b_1=2$, $b_2=1$. For $X$ a surface of genus $g$, of dimension $d=2$, we have $b_0=1$,$b_1=2g$, $b_2=1$. For $X$ a quintic hypersurface in $\mathbb{CP}^4$, an example of Calabi-Yau 3-fold, of dimension $d=6$, we have $b_0=1$,$b_1=0$,$b_2=1$,$b_3=204$, $b_4=1$,$b_5=0$,$b_6=1$.

Given a submanifold $A$ of dimension $i$ and a submanifold $B$ of dimension $d-i$, we expect them to intersect in some union of points. One I say "we expect", it is a generic linear algebra expectation: in $\mathbb{R}^d$, a $i$-plane and an $d-i$-plane in general position intersect only in one point. Example: two lines in a plane, one line and a plane in 3-space... This linear algebra picture is a local description of what happens near an intersection point of our two submanifolds. So if the two manifolds $A$ and $B$ are "in general positions", their intersection is some union of points and one would like to define their intersection number $A \cap B$ as the number of these intersection points. There are several problems with this tentative definition. First to eliminate the problems of things "going to infinity", let me assume that $X$ is compact. Second, $A$ and $B$ are not necessarely in general position, their intersection could be non transverse and this can be the case even at the linear algebra level (example: a line inside a plane in 3-space). So the intersection number of two submanifolds is not well-defined in general. But the key point is that it is possible to make it well-defined at the level of homology. If $A$ and $B$ do not intersect transversally, one can show that it is possible to deform $A$ in $A'$ in the same homology class as $A$, and $B$ in $B'$, in the same homology class as $B$, such that $A'$ and $B'$ intersect transversally. So we are tempted to define the intersection number of the homology classes of $A$ and $B$ as the intersection number of $A'$ and $B'$. But there is still a problem: maybe this intersection number depends on our precise choice of deformation of $A$ and $B$ in $A'$ and $B'$. How to be sure that that the intersection number is independent of the choice of representatives in the homology classes? It is for this reason that intersection numbers have to be defined not just as a naive counting of intersection points but as a counting with signs and to make sense of that one has to assume that $X$ is oriented and that we are considering oriented submanifolds. If $A$ and $B$ intersect transversally at a point $p$, then let $e_1$, ..., $e_i$ an oriented basis of the tangent space of $A$ at $p$ and $e_{i+1}$, ...,$e_n$ is an oriented basis of the tangent space of $B$ at $p$ then we say that the intersection number $A \cap B$ at $p$ is $1$ if  $e_1$, ..., $e_i$, $e_{i+1}$, ...,$e_n$ is an oriented basis of the tangent space of $X$ at $p$ and is $-1$ if this basis defines the reverse orientation of $X$. The signs give a way to eliminate the contribution of the "spurious" intersections, i.e. the ones which disappear under deformation. The upshot is a well-defined map $\cap \colon H_i(X,\mathbb{Z}) \times H_{d-i}(X,\mathbb{Z}) \rightarrow \mathbb{Z}$, $(A,B) \mapsto A \cap B$. Remark that the definition of $A \cap B$ depends on a choice of ordering of $A$ and $B$ so in general $A \cap B \neq B \cap A$. More precisely to go from one to the other, one has to go from the basis  $e_1$, ..., $e_i$, $e_{i+1}$, ...,$e_n$ to the basis $e_{i+1}$, ...,$e_n$,$e_1$, ..., $e_i$ and this changes the orientation by a sign $(-1)^{i(d-i)}$ and so $A \cap B = (-1)^{i(d-i)} B \cap A$.

If $X$ is a Calabi-Yau 3-fold, in particular of complex dimension $3$ and so of real dimension $d=6$, and if we take $i=3$ and so $d-i=3$, we obtain an intersection pairing: $\cap \colon H_3(X,\mathbb{Z}) \times H_3(X, \mathbb{Z}) \rightarrow \mathbb{Z}$ which is skew symmetric: $A \cap B = -B \cap A$. According to Poincaré duality this pairing is non-degenerate, i.e. is symplectic, and it is an elementary fact in linear algebra to check that there exists a basis $A_I, B^J$ with the properties mentionned in the question (a symplectic basis).

Exactly the same structure appears on $H_1(X,\mathbb{Z})$ of a real surface and it is maybe a case to understand first before going to higher dimension. There the $A_I, B^J$ are a choice of $g$ parallels and $g$ meridians on a genus $g$ surface.

Poincaré duality gives a duality between homology and cohomology. If $X$ is an oriented compact manifold of dimension $d$ then an homology class $A$ of degree $i$ has a Poincaré dual cohomology class $\alpha$ of degree $d-i$. In the de Rham description of cohomology, $\alpha$ can be represented by a differential form of degree $d-i$. Under this duality, the intersection product of homology classes is mapped to the cup product of cohomology classes, i.e. $A \cap B = \int_X \alpha \wedge \beta$. From this point of view, the relation $A \cap B = (-1)^{i(d-i)} B \cap A$ is a simple consequence of the analogue result for the wedge product of differential forms.

answered Jun 7, 2015 by 40227 (4,660 points) [ revision history ]
edited Jun 7, 2015 by 40227

Thank you for the detailed response!

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