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Let $X$ be a smooth projective variety over $\mathbb{C}$. The following information is all equivalent (any of these numbers can be computed by a linear equation from any of the others):

  1. the arithmetic genus of $X$
  2. the constant coefficient of the Hilbert polynomial of $X$
  3. $\chi(X, \mathscr{O}_X)$
  4. the "Todd genus" $\int_X \operatorname{td}(T_X)$, where $T_X$ is the tangent bundle of $X$ and $\operatorname{td}$ denotes the Todd class.

Is there a geometric characterization for any of these numbers?

If I understand correctly, characteristic classes (and in particular, Todd classes) can be defined entirely from the topology of $X$, or at least its structure as a smooth manifold. [Edit: This is not true--see the answer of "anonymous." If I understand correctly, the Todd class of a complex vector bundle is a smooth invariant. However, different complex structures on the same real manifold $X$ can give rise to non-isomorphic complex vector bundle structures on $T_X$; in fact, a complex vector bundle structure on $T_X$ is, by definition, an almost complex structure on a real manifold $X$.] Thus, in some sense, item 4 provides a "geometric characterization" for the arithmetic genus of $X$ (and the other items on the list). However, I personally find this description so far abstracted from actual geometric properties of $X$ as to be hardly geometric at all. (If anyone disagrees with me and can articulate a geometric intuition for the Todd genus, that would be a reasonable answer.)

By comparison, I do consider the following characterizations of various properties "geometric":

  • The self-intersection number of the diagonal embedding of $X$ into $X \times X$. (the Euler characteristic)
  • The number of points in which a general linear space of complementary dimension meets $X \subset \mathbb P^n$. (the degree of $X \hookrightarrow \mathbb P^n$)
  • The genus of the curve $X \cap L$, where $L$ is a general linear space of dimension one greater than $\operatorname{codim} X$. (I don't know of a standard name for this, but in a particular sense, it is one of the coefficients of the Hilbert polynomial of $X$.)
  • The maximum number of copies of $S^1$ that can be removed from $X$ without disconnecting it. (the genus of $X$ if $X$ is a smooth curve, i.e., Riemann surface)

Note that either the first or the last point gives a geometric characterization for (information equivalent to) the genus of a curve. Without one of these, I would not consider the third bullet a "geometric characterization" of anything. In a way, this provides part of my motivation for asking this question. Let $L_k$ be a general linear space of codimension $k$ in $\mathbb P^n$. Unless I am mistaken, knowing the Hilbert polynomial for $X$ is equivalent to knowing the arithmetic genus of $X \cap L_k$, for every $k \leq n$ such that this intersection is nonempty, via the formula $$ \chi(\mathscr O_X(n)) = \sum_{k \ge 0} \chi(\mathscr O_{X \cap L_k}) \binom{n+k-1}{k}\;\text.$$

Thus, a geometric characterization for arithmetic genus would automatically give a geometric characterization for the Hilbert polynomial. (Again, in some sense, this is already provided by the Hirzebruch-Riemann-Roch Theorem; but I find this formula so abstracted as to be hardly geometric at all.)

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Just a comment which probably won't help you much, but if you have a curve and you are able to "compute" a finite morphism to $\mathbf{P}^1$, then you can determine the genus. "Computing" a finite morphism means determining its degree and ramification type. This is possible in some cases (for example modular curves) and gives you a (geometric?) characterization of the genus. –  Ariyan Javanpeykar Apr 1 '12 at 10:52
    
I consider that the (arithmetic) genus of $X$ already has at least two geometric characterizations when $X$ is a curve. I'm primarily interested in characterizations that work when $X$ is not a curve. –  Charles Staats Apr 1 '12 at 12:25
    
Another comment: I'm asking for a geometric characterization, not simply a geometric property or a method of computation in some cases (although these are, of course, good to know). For instance, I consider the last bullet point a very nice geometric characterization for the genus of a real closed 2-manifold, as opposed to e.g. a characterization in terms of Betti numbers, but I imagine that this is rarely, if ever, a good definition for computing the genus. –  Charles Staats Apr 1 '12 at 14:29

2 Answers 2

up vote 27 down vote accepted

First let me note that there is an unfortunate clash in terminology: the arithmetic genus of a smooth complex projective variety $X$ of dimension $n$ can mean either

a) The number $\chi (X, \mathcal O_X)$: the Hirzebruch arithmetic number in which you are interested .
b) The number $p_a(X)=(-1)^n(\chi (X, \mathcal O_X)-1)$, the Severi arithmetic genus, which has historical precedence but of course was defined non-cohomologically.
For example, for projective space we have $\chi (\mathbb P^n, \mathcal O_{\mathbb P^n})=1$ but $p_a (\mathbb P^n)=0$.

Hirzebruch introduced his definition mainly because it has the powerful multiplicativity property $$\chi (X\times Y,\mathcal O_{X\times Y})= \chi (X,\mathcal O_{X})\cdot \chi ( Y,\mathcal O_{ Y})$$ which certainly is a step toward the geometric interpretation you are seeking.

Another step in the right direction is that for a finite covering $X\to Y$ of degree $d$ we have the pleasant relation $\chi (X,\mathcal O_{X})=d\cdot \chi ( Y,\mathcal O_{ Y})$.
But the most important geometric property is that $\chi (X,\mathcal O_{X})$ is a birational invariant, because each number $dim_\mathbb C H^i(X,\mathcal O_{X})$ is already a birational invariant.

Arithmetic genus is reasonably easy to compute: for a hypersurface $H\subset \mathbb P^n$ of degree $d$ you have $p_a(H)=\binom {d-1}{n}$, which for $n=2$ gives the well-known elementary formula $p_a(C)=\frac {(d-1)(d-2)}{2}$ for the plane curve $C$.
[This formula (and others) can be found in Hartshorne, Chapter I, Exercise 7.2, page 54]
For a surface you have Max Noether's formula $\chi (S, \mathcal O_S)=\frac {c_1^2(S)+c_2(S)}{12}$, where $c_2(S)$ (=second Chern class of $S$) is also the purely topological Euler-Poincaré characteristic of $S$, equal to the alternating sum of the Betti numbers of the underlying toplogical space.$S_{top}$.

Finally, Fulton has given an axiomatic characterization of the arithmetic genus in algebraic geometry over an arbitrary algebraically closed field here.
In a sense it may be considered an explanation of the geometric significance of the arithmetic genus: if you want it to satisfy certain geometric properties, the definition is forced upon you.

Edit (added by Charles with Georges's permission): Fulton's axiomatic characterization may be described as follows: There is a unique assignment of a number $\mathcal{A}(X)$ to every [smooth irreducible projective variety over a fixed algebraically closed field] (hereafter simply "variety"), such that the following three axioms are satisfied:

  1. $\mathcal{A}$ respects isomorphism classes.
  2. If $X$ is a point, then $\mathcal{A}(X) = 1$.
  3. Let $X$, $Y$, and $Z$ be (smooth) varieties of the same dimension. Suppose that $X$, $Y$, and $Z$ can be embedded as codimension-one subvarieties of a common (smooth) variety $W$, such that

    • $X$ and $Y+Z$ are linearly equivalent as divisors in $W$, and
    • $Y$ and $Z$ intersect transversely in a disjoint union of (smooth) varieties $V_1, \dotsc, V_{\ell}$.

    Then $$\mathcal{A}(X) = \mathcal{A}(Y) + \mathcal{A}(Z) - \sum_i \mathcal{A}(V_i).$$

This assignment takes $X$ to its "Hirzebruch arithmetic number" $\mathcal{A}(X) = \chi(X, \mathcal{O}_X)$.

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Assuming you know the dimension of $X$ (which I suppose I am assuming implicitly), a) and b) contain exactly the same information, so I am equally interested in both of them. –  Charles Staats Apr 1 '12 at 12:21
    
Dear @Charles: oh yes, sure. I only mentioned both definitions as a warning to be careful with the literature. Severi's definition is still used due to the weight of tradition, or because some formulas are prettier with that definition, or because it is nice that arithmetic genus coincides with geometric genus for smooth curves, or for other reasons. –  Georges Elencwajg Apr 1 '12 at 12:49
    
I find Fulton's characterization the most convincing as an answer to my question, simply because it is a characterization and not merely a property. The axioms Fulton gives are (at least in my mind) clearly geometric in nature. I would prefer a more direct definition, rather than simply an axiomatic characterization, but I consider this answer worthy of acceptance if nothing more direct turns up. In the mean time, I would appreciate it if you would either include Fulton's axioms, or give me license to do so. (I would not want to make a change of this magnitude without your permission.) –  Charles Staats Apr 1 '12 at 14:23
1  
Dear Charles: yes, I confirm that Fulton means Hirzebruch genus when he says arithmetic genus. So, sweet dreams and thanks for the good work! –  Georges Elencwajg Apr 2 '12 at 7:31
4  
Wonderful answer! Notice Fulton's axioms give an inductive procedure for actually calculating the arithmetic genus of a plane curve by degenerating to one of lower degree plus a line. I.e. (d-1)(d-2)/2 is the only formula obeying the axioms. If there are nodes, one can blow up and work on the resulting smooth surface W. If a general curve, project into the plane. The same approach computes Noether's formula for a smooth surface in 3-space (with a little topology and adjunction), but projecting introduces more singularities. –  roy smith Apr 2 '12 at 21:11

It is not true that quantities like the Todd genus (or the arithmetic genus) depend only on the underlying smooth manifold. There are examples of diffeomorphic Kaehler manifolds with different Todd genera. Check the following papers for further information:

D. Kotschick, Topologically invariant Chern numbers of projective varieties, Adv. Math. 2012, doi 10.1016/j.aim.2011.10.020

D. Kotschick and S. Schreieder, The Hodge ring of Kaehler manifolds, Compos. Math. 2013, doi 10.1112/S0010437X12000759

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A worthwhile point. I assume what is happening here is that Chern classes are an invariant of the complex vector bundle $T_X$--but different complex structures on the same smooth manifold $X$ can give rise to non-isomorphic descriptions $T_X$ as a complex vector bundle over $X$. –  Charles Staats Mar 30 '13 at 17:52

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