# Why are period integrals naïve periods?

Apologies for the long question.

I recall the definition of a (naïve) period according to Kontsevitch and Zagier [KS]:

A (naïve) period is a complex number whose real and imaginary parts are absolutely convergent integrals of rational functions with rational coefficients on domains of $\mathbb R^d$ bounded by polynomial inequalities with rational coefficients.

For example,

$$\pi = \iint_{x^2+y^2 \leq 1} dx dy$$

is a period, whereas $e$ is conjecturally not a period. As [KS] point out, the following definition is equivalent:

A period is a complex number which is an absolutely convergent integral of an algebraic function defined over $\mathbb Q$, on a domains of $\mathbb R^d$ bounded by polynomial inequalities with real algebraic coefficients.

To illustrate, remark that for $\lambda>1 \in \mathbb Q$,

$$2\int_0^1 \frac{dx}{\sqrt{x(x-1)(x-\lambda)}}$$

is a period according to the second definition, but not obviously to the first one. But it can be rewritten as

$$\iint_{0 \leq x \leq 1, y^2 \leq x(x-1)(x-\lambda)} dx dy,$$

and so it is a period also according to the first definition.

Kontsevitch and Zagier point out that the periods of algebraic varieties in the classical sense are naïve periods. Let $X/\mathbb Q$ be a smooth projective variety over $\mathbb Q$ of dimension $d$, and $\omega \in H^0(X, \Omega^d_{X/\mathbb Q})$ a global algebraic differential form of top degree on $X$. Being of top degree, the form $\omega$ is automatically closed, and gives rise to a de Rham cohomology class $[\omega] \in H^d_{dR}(X/\mathbb C)$ in the middle cohomology of the variety $X(\mathbb C)$, which has real dimension $2d$. Now let $D$ be a divisor on $X$ with normal crossings, and let $[\sigma] \in H_d(X(\mathbb C), D(\mathbb C), \mathbb Q)$ be a homology class relative to $D$, represented by a $d$-chain whose boundary lies on $D$. Then the integral

$$\int_\sigma \omega|_{\sigma}$$

depends only on the relative homology class of $\sigma$ and on the cohomology class of $\omega$.

Claim: the integral $\int_\sigma \omega|_{\sigma}$ is a period.

Kontsevitch and Zagier seem to treat this as an obvious fact. It seems to me like a fairly difficult theorem. Am I overlooking a simple proof?

To illustrate, let $X$ is the elliptic curve $y^2z=x(x-z)(x-\lambda z)$, $\omega = dx/y$ is the invariant differential on $X$, and $D$ is the empty divisor, then the periods of $\omega$ in this sense are precisely the linear combinations of the classical period integrals

$$2\int_0^1 \frac{dx}{\sqrt{x(x-1)(x-\lambda)}}, 2\int_\lambda^\infty \frac{dx}{\sqrt{x(x-1)(x-\lambda)}},$$

which shows that the statement is true as both of these integrals are naïve periods. In order to get these integral formulas, we treat $X$ as a degree $2$ ramified covering of $\mathbb P^1$ via $(x,y) \mapsto x$; then $y$ becomes a multi-valued function on $\mathbb A^1$ with branch points at $\lambda = 0,1,\lambda, \infty$. The first homology of $X(\mathbb C)$ is generated by two cycles whose image in $\mathbb P^1(\mathbb C)$ circle the branch cuts $[0,1]$ and $[\lambda, \infty] \subseteq \mathbb P^1(\mathbb R)$ clockwise. Thus it suffices to see that the integrals of $\omega$ along these cycles are periods. Consider a $1$-cycle $\sigma$ which circles the branch cut $[0,1]$ clockwise on the branch of $y$ which is positive for real $x$ large enough. This cycle is homologous to the cycle which goes from $0$ to $1$ along the positive branch of $y$, then goes back to $1$ along the negative branch. Thus

$$\int_\sigma \omega = \int_0^1 \frac{dx}{\sqrt{x(x-1)(x-\lambda)}} - \int_0^1 \frac{dx}{-\sqrt{x(x-1)(x-\lambda)}} = 2 \int_0^1 \frac{dx}{\sqrt{x(x-1)(x-\lambda)}}.$$

Very good. But what about a variety of dimension $d>1$? I am happy with supposing that the divisor $D$ is empty, so that $[\sigma]$ is the homology class of a $d$-cycle. Here is how I thought one might prove that the integral $\int_\sigma \omega$ is a period. Let $U$ be an affine open subvariety of $X$ such that $U(\mathbb C)$ contains the image of the $d$-cycle $\sigma$. By the Noether normalization lemma, there is a finite map $U \to \mathbb A^d$, and an open $V\subseteq \mathbb A^d$ such that the restriction $U' \to V$ is finite étale of some degree $n$. Let us suppose still that $U'(\mathbb C)$ contains the image of $\sigma$. Locally in the complex topology of $V(\mathbb C)$, the morphism $U' \to V$ has $n$ sections along which we can pull back the differential $\omega$, to get a multivalued differential $\omega$ on $V$. Let $\sigma'$ denote the $d$-cycle on $V(\mathbb C)$ obtained by composing $\sigma$ with $U'(\mathbb C) \to V(\mathbb C)$; the integral

$$\int_{\sigma'}\omega$$

has $n$ possible values, depending on the branch of $\omega$ which is chosen (and then analytically continued along $\sigma'$).

It should therefore be proven that the integrals $\int_{\sigma'} \omega$ are periods. The problem is that the cycle $\sigma$ is only differentiable. For instance, in the example above, if we re-parametrize the integral

$$\int_0^1 \frac{dx}{\sqrt{x(x-1)(x-\lambda)}}$$

by a diffeomorphism $t \mapsto x(t)$ of the interval, preserving the boundary, then we get

$$\int_0^1 \frac{x'(t) dt }{\sqrt{x(t)(x(t)-1)(x(t)-\lambda)}}$$

which is not recognizable as a period anymore. Hence, it appears crucial that the homology class of a path $\sigma$ circling the branch cut $[0,1]$ can be represented also by a path which is piecewise-linear, namely the path which goes from $0$ to $1$ in constant time on the top sheet and goes back to $0$ in constant time on the bottom sheet.

It appears to me that the statement that $\int_{\sigma'} \omega$ is a period depends on the fact that any $d$-cycle $\sigma$ on $V(\mathbb C) \subseteq \mathbb A^d(\mathbb C)$ is homologous to one which is piecewise-linear, or even piecewise-polynomial, with algebraic coefficients. It seems to me that this is true, although it is probably a fairly difficult theorem of algebraic topology.

Is my way of approaching this problem correct, or am I overlooking something much simpler?

Thank you for reading, and for your ideas.

## 1 Answer

I confess I never looked into the details, but the equivalence between naïve periods and periods seen as integrals in algebraic varieties is supposed to be worked out by Benjamin Friedrich in his thesis (http://arxiv.org/abs/math/0506113).

There is also a very nice book being written by Huber and Müller-Stach where you can find this and much more stuff about periods (http://home.mathematik.uni-freiburg.de/arithgeom/preprints/buch/buch-v1.pdf).

• While this link may answer the question, it is better to include the essential parts of the answer here and provide the link for reference. Link-only answers can become invalid if the linked page changes. - From Review – Ben Sheller Feb 21 '16 at 1:49
• You are right. I can't give a description since I never worked out the details myself, as I told above. I tried to post this "answer" simply as a comment but I don't have enough reputation. – Docteur Cottard Feb 21 '16 at 10:46