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Chapter 14 Frobenius actions and the Lefschetz–Monsky trace formula

Original lecture date: November 20, 2019.

In this lecture, we will give an example of a Frobenius action on the cohomology of the affine piece of a hyperelliptic curve, and give a proof of the Lefschetz trace formula in the \(p\)-adic setting due to Monsky.

Readings 14.0.1.

The Frobenius action on a hyperelliptic curve is presented as in [71]. The Lefschetz trace formula is presented as in Monsky [101]. See also [75].

Section 14.1

Example 14.1.1.

Let \(k=\FF_q\) be a finite field with characteristic \(p\) not equal to 2. Let \(P(x) \in k[x]\) be a monic polynomial of degree \(2g+1\) with no repeated roots. Let \(\overline{A}\) be the ring

\begin{equation*} \overline{A}:= k[x,y,z]/(y^2-P(x), yz-1) \end{equation*}

and let \(X\) the affine curve \(X := \Spec \overline{A}\text{;}\) \(X\) is then a hyperelliptic curve of genus \(g\) with one rational Weierstrass point at infinity, with all of its Weierstrass points removed. Choose \(\tilde{P}(x)\in W(k)[x]\) a monic polynomial lifting \(P(x)\text{;}\) note that \(\tilde{P}(x)\) has no repeated roots, since it is a lift of something with no repeated roots. Define the lift

\begin{equation*} A := W(k)[x,y,z]/(y^2-\tilde{P}(x), yz-1). \end{equation*}

Remark 14.1.2.

The space \(H_{\dR}^1(\Spec A[\frac{1}{p}])\) has a basis given by

\begin{align*} x^i\frac{dx}{y} & \qquad (i=0,\dots,2g-1), \\ x^i\frac{dx}{y^2} & \qquad (i=0,\dots,2g) \end{align*}

(see Set 5 exercises). Since \(X\) is hyperelliptic, it has an involution

\begin{align*} x &\mapsto x\\ y &\mapsto -y\\ z &\mapsto -z. \end{align*}

Lifting this involution, we find that the \(x^i\frac{dx}{y}\) form the minus eigenspace and that the \(x^i\frac{dx}{y^2}\) form the plus eigenspace.

Definition 14.1.3.

With \(A,k\) as before, let \(A^{\dagger}\) be the set of all sums of the form

\begin{equation*} \sum_{n=-\infty}^{\infty}\frac{Q_n(x)}{y^n}, \end{equation*}

with \(Q_n(x)\in W(k)[x]\) such that \(\deg \,Q_n(x) \le 2g\text{.}\) Moreover, we require as a convergence condition that there exist positive integers \(a,b\) such that the \(p\)-adic valuation of \(Q_n\) is at least \(a|n|-b\text{.}\)

Note first that via the relation \(yz=1\) in \(A\) that every element in \(A\) is a polynomial in \(x,y,y^{-1}\text{.}\) We may then successively take remainders modulo \(\tilde{P}(x)\) and use the relation \(y^2 = \tilde{P}(x)\) to reduce the degrees of everything to at most \(2g\text{.}\)

To compute the action of the \(q\)-power Frobenius on \(H_{\MW}^1(X),\) it suffices to find a map \(Q:A^{\dagger}\rightarrow A^{\dagger}\) lifting the action of Frobenius on \(\overline{A}\text{.}\) To do this, let \(C\) be the projective curve obtained from \(X\) by adding back in the Weierstrass points. We then get the following diagram:

where the vertical arrow is the two-to-one cover of \(\PP^1\) by the hyperelliptic curve \(C\text{.}\) Since we took out the Weierstrass points in defining \(X\text{,}\) the arrow \(X\rightarrow \PP^1\) is étale, so we may lift the standard Frobenius map \(x\mapsto x^p\) on \(\PP^1\) to \(X\text{.}\) Using that \((A^{\dagger}, (p))\) is a henselian pair, we may then lift the induced Frobenius action on \(X\) to \(A^{\dagger}\text{.}\) We may compute out this action explicitly, as

\begin{align*} x &\mapsto x^q\\ y &\mapsto \sqrt{\tilde{P}(x^q)}\\ &= \sqrt{\tilde{P}(x)^q + p(*)}\\ &= \sqrt{y^{2g} + p(*)}\\ &= y^q(1 + p(*)z^{2q})^{1/2} \\ z &\mapsto y^{-1}, \end{align*}

where the expression in the last line for \(y\) may be expanded out using the generalized binomial theorem and the instances of \((*)\) denote polynomial quantities which may be computed explicitly.

Remark 14.1.5.

This example of the use of Monsky–Washnitzer cohomology for algorithmic computation is taken from [71]. Subsequently, Tuitman [121], [122] adapted this method to work for arbitrary curves. (As an aside, note that this also gives a method to compute Coleman's \(p\)-adic path integrals [7]; such computations play a pivotal role in the Chabauty–Coleman–Kim approach to finding rational points on curves, as in [6].)

In higher dimension, Abbott–Kedlaya–Roe [1] involves similar computations for smooth hypersurfaces in \(\PP^n\text{,}\) using the explicit description of de Rham cohomology by Griffiths [52], [53]. This has been further adapted to smooth nondegenerate hypersurfaces in toric varieties by Costa–Harvey–Kedlaya (unpublished, but see [25] for a preview).

We now turn our attention to proving the Lefschetz trace formula.

We start with two key reductions.

  • First, both sides of the desired equality are additive for a scissors decomposition (i.e. a union of an open set and its closed complement): this is obvious for the left side, and for the right side it follows from the excision exact sequence.

  • Second, if \(X = \Spec(\FF_q)\text{,}\) then \(H_{\MW}^0(X) = K\text{,}\) \(H_{\MW}^i(X) = 0\) for \(i>0\text{,}\) and the action of \(F\) on \(H_{\MW}^0(X)\) is the identity map. Hence both sides of the desired equality equal 1.

From these observations, it follows that we may reduce the general case of the theorem to the case where \(\#X(\FF_q) = 0\text{.}\) This equality has a key algebraic consequence: it implies that the ideal in \(\overline{A}\) generated by all elements of the form \(f^q - f\) for \(f\in \overline{A}\) is trivial (see supplementary exercises). That is, we can find an equality of the form

\begin{equation*} 1 = \sum \overline{a}_i (\overline{b}_i^q - \overline{b}_i) \end{equation*}

for some \(\overline{a}_i, \overline{b}_i\in\overline{A}\text{;}\) we will use this equality in a crucial way to see that the right-hand side of the trace formula is also zero.

By Theorem 13.1.2, we may choose a lift \(\phi\colon A^{\dagger}\rightarrow A^{\dagger}\) of Frobenius to \(A^{\dagger}\text{.}\) We may then find elements \(a_i, b_i \in A^{\dagger}\) such that

\begin{equation*} 1 \equiv \sum a_i (\varphi(b_i) - b_i) \mod p. \end{equation*}

Since \(p\) belongs to the Jacobson radical of \(A^\dagger\text{,}\) we may multiply all of the \(a_i\) by a suitable unit to ensure that in fact

\begin{equation*} 1 = \sum a_i (\varphi(b_i) - b_i). \end{equation*}

Note that \(\varphi\colon A^{\dagger} \rightarrow A^{\dagger}\) is finite flat of degree \(q^n\) (this is true mod \(p\text{,}\) and one can argue then that it is true in general). Define \(\psi\colon A^{\dagger}\rightarrow A^{\dagger}\) to be the reduced trace of \(\varphi\text{,}\) i.e.

\begin{equation*} \psi = \frac{1}{q^n}\cdot \Trace(\varphi), \qquad \psi \circ \varphi = \text{id}. \end{equation*}

Since we know by Berthelot that \(\dim H_{\MW}^i(X) \lt \infty\text{,}\) \(\psi\) is not just a left inverse but a genuine inverse to \(\varphi\) by linear algebra. Thus, the actions of \(q^n F^{-1}\) and \(q^n\psi\) coincide, so we want to show that the alternating sum of traces of \(\psi\) is 0; in fact, we will show that each trace of \(\psi\) vanishes in this situation.

Note that \(\psi(\varphi(a)b) = a\psi(b)\text{.}\) Therefore, if we define \(L_a\colon \Omega^+ \rightarrow \Omega^+\) on the de Rham complex to be multiplication by \(a\text{,}\) we then have

\begin{equation*} \psi \circ L_{\varphi(a)} = L_a \circ \psi, \end{equation*}

implying that

\begin{equation*} \sum_i L_{a_i} \circ \psi \circ L_{\varphi(b_i)} = \sum_i L_{a_i} \circ L_{b_i} \circ \psi. \end{equation*}

Take the traces of both sides to obtain

\begin{align*} \Trace\left(\sum_i L_{a_i} \circ \psi \circ L_{\varphi(b_i)} \right) &= \Trace\left(\sum_i L_{a_i} \circ L_{b_i} \circ \psi \right)\\ &= \Trace\left(\sum_i L_{a_i b_i}\circ \psi \right) \end{align*}

since composition of the multiplication operator becomes multiplication by the product. On the other hand, we may apply the fact that the trace of a product is invariant under cyclic permutations to get that

\begin{equation*} \Trace\left(\sum_i L_{a_i}\circ \psi \circ L_{\varphi(b_i)}\right) = \Trace\left(\sum_i L_{\varphi(b_i)} \circ L_{a_i} \circ \psi\right). \end{equation*}

We thus get the equality

\begin{equation*} \Trace\left(\sum_i L_{\varphi(b_i)} \circ L_{a_i} \circ \psi \right) = \Trace\left(\sum_i L_{a_i b_i}\circ \psi \right), \end{equation*}

so that

\begin{align*} 0 &= \Trace\left(\sum_i (L_{\varphi(b_i)} \circ L_{a_i} \circ \psi - L_{a_i b_i}\circ \psi) \right)\\ &= \Trace\left(\sum_i L_{a_i\varphi(b_i) - b_i} \circ \psi \right)\\ &= \Trace\left(L_{\sum_i a_i(\varphi(b_i)-b_i)}\circ \psi \right)\\ &= \Trace(\psi) \end{align*}

since \(\sum_i a_i(\varphi(b_i) - b_i) = 1\) by prior arrangement. Thus, \(\Trace(\psi) = 0\) as desired.

Remark 14.1.7.

Recall that Monsky did not have the finite-dimensionality of \(\dim H_{\MW}^i(X)\) at his disposal when he originally devised this argument. This required him to be more careful in two aspects. First, he had to introduce a suitable topology in order to argue that he could take traces (recall that there was a corresponding step in Dwork's proof of the rationality of zeta functions). Second, he could not assume that \(F\) is invertible, and so he had to set up the formula in a more cautious way.