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Dan Petersen
Dan Petersen
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Why doesdo H_4 and M_4 have the same virtual Euler characteristic?

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Dan Petersen
Dan Petersen
  • 40.2k
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  • 114
  • 201

Here's a funny coincidence:

The virtual (or "orbifold") Euler characteristic of $\mathcal M_g$ is known by the work of Harer and Zagier: one has $\chi(\mathcal M_g) = \zeta(1-2g)/(2-2g)$.

Now consider the virtual Euler characteristic of the moduli space $\mathcal H_g$ of hyperelliptic curves. Quotienting by the hyperelliptic involution gives a map $\mathcal H_g \to \mathcal M_{0,2g+2}/\Sigma_{2g+2}$ (sending each curve to its set of branch points), and conversely the hyperelliptic curve can be recovered from its branch locus. Hence (since the hyperelliptic curve will have twice as big automorphism group) $$\chi(\mathcal H_g) = \frac 1 2 \chi(\mathcal M_{0,2g+2}/\Sigma_{2g+2}) = \frac{1}{2(2g+2)!}\chi(\mathcal M_{0,2g+2}) = -\frac{(2g-1)!}{2(2g+2)!}.$$ As a sanity check, these two numbers agree for $g=2$. However, what is more surprising is that the two numbers agree also for $g=4$: we have $\chi(\mathcal M_4) = \chi(\mathcal H_4) = -1/1440$. Is there any "reason" for this to occur, or is it just a coincidence?

Edit: This is what the numbers look like for some other values of $g$.

sage: [bernoulli(2*g)/(4*g*(g-1)) for g in range(2,10)]

[-1/240, 1/1008, -1/1440, 1/1056, -691/327600, 1/144, -3617/114240, 43867/229824]

sage: [-factorial(2*g-1)/(2*factorial(2*g+2)) for g in range(2,10)]

[-1/240, -1/672, -1/1440, -1/2640, -1/4368, -1/6720, -1/9792, -1/13680]

Here's a funny coincidence:

The virtual (or "orbifold") Euler characteristic of $\mathcal M_g$ is known by the work of Harer and Zagier: one has $\chi(\mathcal M_g) = \zeta(1-2g)/(2-2g)$.

Now consider the virtual Euler characteristic of the moduli space $\mathcal H_g$ of hyperelliptic curves. Quotienting by the hyperelliptic involution gives a map $\mathcal H_g \to \mathcal M_{0,2g+2}/\Sigma_{2g+2}$ (sending each curve to its set of branch points), and conversely the hyperelliptic curve can be recovered from its branch locus. Hence (since the hyperelliptic curve will have twice as big automorphism group) $$\chi(\mathcal H_g) = \frac 1 2 \chi(\mathcal M_{0,2g+2}/\Sigma_{2g+2}) = \frac{1}{2(2g+2)!}\chi(\mathcal M_{0,2g+2}) = -\frac{(2g-1)!}{2(2g+2)!}.$$ As a sanity check, these two numbers agree for $g=2$. However, what is more surprising is that the two numbers agree also for $g=4$: we have $\chi(\mathcal M_4) = \chi(\mathcal H_4) = -1/1440$. Is there any "reason" for this to occur, or is it just a coincidence?

Here's a funny coincidence:

The virtual (or "orbifold") Euler characteristic of $\mathcal M_g$ is known by the work of Harer and Zagier: one has $\chi(\mathcal M_g) = \zeta(1-2g)/(2-2g)$.

Now consider the virtual Euler characteristic of the moduli space $\mathcal H_g$ of hyperelliptic curves. Quotienting by the hyperelliptic involution gives a map $\mathcal H_g \to \mathcal M_{0,2g+2}/\Sigma_{2g+2}$ (sending each curve to its set of branch points), and conversely the hyperelliptic curve can be recovered from its branch locus. Hence (since the hyperelliptic curve will have twice as big automorphism group) $$\chi(\mathcal H_g) = \frac 1 2 \chi(\mathcal M_{0,2g+2}/\Sigma_{2g+2}) = \frac{1}{2(2g+2)!}\chi(\mathcal M_{0,2g+2}) = -\frac{(2g-1)!}{2(2g+2)!}.$$ As a sanity check, these two numbers agree for $g=2$. However, what is more surprising is that the two numbers agree also for $g=4$: we have $\chi(\mathcal M_4) = \chi(\mathcal H_4) = -1/1440$. Is there any "reason" for this to occur, or is it just a coincidence?

Edit: This is what the numbers look like for some other values of $g$.

sage: [bernoulli(2*g)/(4*g*(g-1)) for g in range(2,10)]

[-1/240, 1/1008, -1/1440, 1/1056, -691/327600, 1/144, -3617/114240, 43867/229824]

sage: [-factorial(2*g-1)/(2*factorial(2*g+2)) for g in range(2,10)]

[-1/240, -1/672, -1/1440, -1/2640, -1/4368, -1/6720, -1/9792, -1/13680]

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Dan Petersen
Dan Petersen
  • 40.2k
  • 2
  • 114
  • 201

Why does H_4 and M_4 have the same virtual Euler characteristic?

Here's a funny coincidence:

The virtual (or "orbifold") Euler characteristic of $\mathcal M_g$ is known by the work of Harer and Zagier: one has $\chi(\mathcal M_g) = \zeta(1-2g)/(2-2g)$.

Now consider the virtual Euler characteristic of the moduli space $\mathcal H_g$ of hyperelliptic curves. Quotienting by the hyperelliptic involution gives a map $\mathcal H_g \to \mathcal M_{0,2g+2}/\Sigma_{2g+2}$ (sending each curve to its set of branch points), and conversely the hyperelliptic curve can be recovered from its branch locus. Hence (since the hyperelliptic curve will have twice as big automorphism group) $$\chi(\mathcal H_g) = \frac 1 2 \chi(\mathcal M_{0,2g+2}/\Sigma_{2g+2}) = \frac{1}{2(2g+2)!}\chi(\mathcal M_{0,2g+2}) = -\frac{(2g-1)!}{2(2g+2)!}.$$ As a sanity check, these two numbers agree for $g=2$. However, what is more surprising is that the two numbers agree also for $g=4$: we have $\chi(\mathcal M_4) = \chi(\mathcal H_4) = -1/1440$. Is there any "reason" for this to occur, or is it just a coincidence?