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For an abelian variety $A_{/\mathbb{Q}}$, its volume $vol(A(\mathbb{R}))$ appears in the conjectured Birch Swinnerton-Dyer formula for the L-series at 1.

I am having trouble in understanding the size of this volume, as a function of the (minimal) equations defining the variety. Hence the question:

Say we have a hyperelliptic curve given by $$y^2 = x^{2g+1}+a_{2g}x^{2g}+\ldots +a_0$$

for simplicity, assume that $a_i\approx M > 0$ for all $i$. What is a good crude estimate for the volume of the Jacobian?

Note: for an elliptic curve this is easy, since there is only one differential and hence only one simple integral.

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If you really want an answer and none is forthcoming you might consider asking on MO. –  Qiaochu Yuan Dec 26 '10 at 20:54
    
I'm afraid it'll be closed for not being research level :/ –  simplequestions Dec 26 '10 at 23:16
    
How do you define the volume of an abelian variety? –  Kevin H. Lin Dec 27 '10 at 2:02
    
$A_{/\mathbb{Q}} \rightarrow A_{/\mathbb{C}}$, the volume is then the determinant of the pairing matrix between a basis for the real differentials and a basis for the real homology (loops in complex homology fixed by complex conjugation). –  simplequestions Dec 30 '10 at 12:18

1 Answer 1

up vote 2 down vote accepted

I assumed above that all coefficients are positive, but this isn't easiest to compute with, so apply the isomorphism $x \mapsto -x$, and now the coefficients are alternating. The volume is unchanged.

A basis for the differentials is $\{\omega_i : 1\le i \le g\}$, where $$\omega_i = x^{i-1}\frac{dx}{2y}$$

For $M$ large, there is a positive root $\alpha_1$ of size $M$, and the other $2g$ roots are small. We now need to figure out a basis for the real homology (loops fixed by complex conjugation). There are $g-1$ independent loops $\gamma_{j+1}$, the $j$-th loop being from $\alpha_{2j}$ to $\alpha_{2j+1}$ and back around the "hole" (for example, if all roots are real, these are simply the paths where the polynomial is positive). One more independent loop is from $\alpha_1$ to $\infty$ and back (being real, this is the easiest to understand).

The volume of the jacobian of the curve is $|det((\int_{\gamma_j} \omega_i)_{ij})|$. So let's approximate these integrals.

For any $j>1$, since the roots involved are $O(1)$, the loop $\gamma_j$ is close to the origin. In this vicinity the root $\alpha_1$ contributes to $\int_{\gamma_j} \omega_i$ a factor of $M^{-1/2}$, and if we factor it out (recall that we only want a crude approximation), we are left with an integral close to the origin over a function that is relatively small. Hence $\int_{\gamma_j} \omega_i \approx M^{-1/2}$. Note that this is the same for any $i$ since close to the origin $x^{i-1}$ doesn't add much.

For $j=1$, the integral can be approximated as $$\int_{\gamma_1} \omega_i = 2\int_M^\infty \frac{x^{i-1} dx}{2y} \approx \int_M^{2M} \frac{x^{i-1} dx}{\sqrt{x^{2g+1}+\cdot a_0}}$$

The small roots give each the denominator a factor of $M^{1/2}$, so we can approximate further to arrive at: $$M^{-g} \int_M^{2M} \frac{x^{i-1} dx}{\sqrt{x-M}} \approx M^{1/2+i-1-g}$$

We have approximated all entries of the matrix. The top row's largest coefficient is of size $M^{-1/2}$, and hence the largest product of a generalised diagonal is $(M^{-1/2})^g$, since the entries in the other rows are also of size $M^{-1/2}$.

The crude approximation for the volume of the jacobian of a hyperelliptic curve given by a polynomial as above with positive coefficients of size $M$, a large number, is: $$Vol(J(\mathbb{R})) \approx M^{-g/2}$$

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