# 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?