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March 16, 2004

Counting Points

Urs Schreiber asked me to explain what “count[ing] the points on the Calabi-Yau, defined over the finite field F p kF_{p^k}” means. I started to respond with a comment, but realized this might work better as a full-fledged post.

Consider the equation

(1)x 1 5+x 2 5+x 3 5+x 4 5+x 5 55ψx 1x 2x 3x 4x 5=0 x_1^5+ x_2^5+ x_3^5+ x_4^5+ x_5^5- 5\psi x_1 x_2 x_3 x_4 x_5 = 0

Algebraic geometry is the study of the geometry of the space of solutions to algebraic equations such as this one. We might be interested in the affine variety, where (x 1,x 2,x 3,x 4,x 5) 5(x_1,x_2,x_3,x_4,x_5)\in \mathbb{C}^5. Alternatively, we might take the (x 1,x 2,x 3,x 4,x 5)(0,0,0,0,0)(x_1,x_2,x_3,x_4,x_5)\neq(0,0,0,0,0), and identify points (x 1,x 2,x 3,x 4,x 5)(λx 1,λx 2,λx 3,λx 4,λx 5)(x_1,x_2,x_3,x_4,x_5)\sim (\lambda x_1,\lambda x_2,\lambda x_3,\lambda x_4,\lambda x_5), λ *\forall \lambda\in \mathbb{C}^*, a nonzero complex number. This yields the projective variety, the quintic hypersurface in P 4\mathbb{C}P^4. This latter is a Calabi-Yau manifold, which makes it rather interesting for physicists.

Algebraic geometry is a hard subject. It’s hard because algebraic geometers don’t want to restrict themselves to the space (affine or projective) of solutions over the complexes. They’d like to study the space of solutions over arbitrary fields. So they need to set up the tools of geometry to work even when the equations are defined, say over a finite field.

One such field is F pF_p. Here, pp is a prime, and we do arithmetic over the integers modulo pp. In F 7F_7, 5=25=-2 and 5=1/35 =1/3 (since 5+2=05+2 = 0 mod 77 and 5×3=1 5 \times 3 =1 mod 77. Since pp is a prime, every nonzero integer in /p\mathbb{Z}/p\mathbb{Z} is invertible modulo pp.

A field is said to have characteristic-kk if adding the multiplicative identity element, 11, to itself kk times gives the additive identity element, 00. If you never get the additive identity element, the field is said to have characteristic-0. \mathbb{Q}, \mathbb{R} and \mathbb{C} are fields of characteristic-0. F pF_p is a field of characteristic-pp, with pp elements.

How about some more fields of characteristic pp? Pick a polynomial P(x)P(x) of degree nn, with coefficient in F pF_p, which is is irreducible (i.e., which cannot be factored into lower-degree polynomials with coefficients in F pF_p). Define

(2)F p n=F p[x]/(P(x)) F_{p^n} = F_p[x]/(P(x))

the ring of polynomials (with coefficients in F pF_p), modulo the ideal generated by our chosen P(x)P(x). Each equivalence class of polynomials has a representative of degree less than nn. Moreover, since P(x)P(x) was irreducible, each polynomial has a multiplicative inverse. So F p nF_{p^n} is a field of characteristic-pp, with p np^n elements.

Exercise: Construct F 3 2F_{3^2}, using the polynomial x 2+1x^2+1, which is irreducible in F 3F_3. Write out the 9 linear polynomials representing F 3[x]/(x 2+1)F_3[x]/(x^2+1) and construct their multiplication table.

Now think about redoing everything you know about geometry (cohomology, vector bundles, sheaves, …) in characteristic-pp. Number Theorists are typically interested in things like counting the number of solutions to equations like the quintic above, and avail themselves of the powerful tools of algebraic geometry to do it.

Whew!

OK, back to the quintic, defined with coefficients in F pF_p (or F p nF_{p^n}). Since the field is finite, so must the number of solutions of the quintic equation, in F pF_p. We can consider ν(ψ)\nu(\psi), the number of solutions to the affine equation, or N(ψ)N(\psi), the number of solutions to the projective equation. They are simply related:

(3)N(ψ)=ν(ψ)1p1 N(\psi) = \frac{\nu(\psi) -1}{p-1}

(we remove the origin and mod out by rescalings by nonzero elements of F pF_p), but Candelas and company prefer to write formulæ for ν(ψ)\nu(\psi), rather than for N(ψ)N(\psi).

Now, here’s where the magic comes in. The periods of the holomorphic 3-form on the quintic Calabi-Yau, integrated over some basis of 3-cycles satisfy a Picard-Fuchs equation,

(4)[(λddλ) 45λ k=1 4(5λddλ+k)]ϖ=0 \left[\left(\lambda \frac{d}{d\lambda}\right)^4 -5 \lambda\prod_{k=1}^4\left (5 \lambda \frac{d}{d\lambda} +k\right)\right] \varpi =0

in the variable λ=1/(5ψ) 5\lambda=1/(5\psi)^5. The independent solutions can be written as

(5)ϖ 0 = f 0(λ) ϖ 1 = f 0(λ)logλ+f 1(λ) ϖ 2 = f 0(λ)log 2λ+2f 1(λ)logλ+f 2(λ) ϖ 3 = f 0(λ)log 3λ+3f 1(λ)logλ+2f 2(λ)logλ+f 3(λ) \array{\arrayopts{\colalign{right center left}} \varpi_0 &=& f_0(\lambda)\\ \varpi_1 &=& f_0(\lambda)\log \lambda + f_1(\lambda)\\ \varpi_2 &=& f_0(\lambda)\log^2 \lambda + 2f_1(\lambda)\log\lambda +f_2(\lambda)\\ \varpi_3 &=& f_0(\lambda)\log^3 \lambda + 3f_1(\lambda)\log\lambda +2 f_2(\lambda)\log\lambda +f_3(\lambda) }

where the f j(λ)f_j(\lambda) are certain power series in λ\lambda.

(6)f 0(λ)= m=0 (5m)!(m!) 5λ m f_0(\lambda)= \sum_{m=0}^\infty\frac{(5m)!}{(m!)^5} \lambda^m

Let f j (n)f^{(n)}_j be the power series truncated to the first n+1n+1 terms.

Candelas et al can write down an exact expression for ν(ψ)\nu(\psi) in terms of the f j (n)f^{(n)}_j. The full formula is a little complicated, but the first approximation to it is easy to state:

(7)ν(ψ)=f 0 ([p/5])(λ)modp \nu(\psi) = f^{([p/5])}_0(\lambda) \quad \text{mod}\, p

where [p/5][p/5] is the integer part of p/5p/5.

Of course, there’s nothing special about the quintic. They can do similar things for more complicated Calabi-Yau’s, and they can also get results over the fields F p nF_{p^n}.

All of this is bound up in some mysterious way with Mirror Symmetry.

I don’t know what it all means, and neither do they, but their papers (I,II) make very intriguing reading.

Posted by distler at March 16, 2004 11:19 PM

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Re: Counting Points

Wow. My maths is very under excercised, to the point where even first-order differentiation isn’t easy, but I tried to follow, and found that interesting.

Thanks for the good read

Posted by: Tom on March 23, 2004 10:42 PM | Permalink | Reply to this

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