Revisions to An Expectation of Cohen-Lenstra Measure

replaced http://front.math.ucdavis.edu/ with https://arxiv.org/abs/

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edited Jun 22, 2022 at 7:16

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The Cohen-Lenstra measure on the set of abelian p-groups assigns $\mathbb{P}(G) = \prod_{i \geq 1} \left( 1 - \frac{1}{p^i}\right) \cdot |\mathrm{Aut}(G)|^{-1} $. Apparently, this is equivalent to taking cokernels of random maps $f: (\mathbb{Z}_p)^N \to (\mathbb{Z}_p)^N$ and letting $N \to \infty$. These are the p-adics and there is a Haar measure on this linear space of maps. Alternatively, choose random maps between the finite groups: $f: (\mathbb{Z}/p^k\mathbb{Z})^N \to (\mathbb{Z}/p^k \mathbb{Z})^N$ and let $k, N \to \infty$.

Q: If G is a random abelian p-group according to the Cohen-Lenstra measure and A is a deterministic, why is the expected number of surjections $\phi: G \to A$ equal to 1? In fact, if G were deterministic I don't think this number could ever be 1 unless |G| = 1.

For references see Section 8 of Homological stability for Hurwitz spaces and the Cohen-Lenstra conjecture over function fields Homological stability for Hurwitz spaces and the Cohen-Lenstra conjecture over function fields by Ellenberg, Venkatesh and Westerland or Terry Tao's blog entry At the AustMS conference.

The Cohen-Lenstra measure on the set of abelian p-groups assigns $\mathbb{P}(G) = \prod_{i \geq 1} \left( 1 - \frac{1}{p^i}\right) \cdot |\mathrm{Aut}(G)|^{-1} $. Apparently, this is equivalent to taking cokernels of random maps $f: (\mathbb{Z}_p)^N \to (\mathbb{Z}_p)^N$ and letting $N \to \infty$. These are the p-adics and there is a Haar measure on this linear space of maps. Alternatively, choose random maps between the finite groups: $f: (\mathbb{Z}/p^k\mathbb{Z})^N \to (\mathbb{Z}/p^k \mathbb{Z})^N$ and let $k, N \to \infty$.

Q: If G is a random abelian p-group according to the Cohen-Lenstra measure and A is a deterministic, why is the expected number of surjections $\phi: G \to A$ equal to 1? In fact, if G were deterministic I don't think this number could ever be 1 unless |G| = 1.

For references see Section 8 of Homological stability for Hurwitz spaces and the Cohen-Lenstra conjecture over function fields by Ellenberg, Venkatesh and Westerland or Terry Tao's blog entry At the AustMS conference.

The Cohen-Lenstra measure on the set of abelian p-groups assigns $\mathbb{P}(G) = \prod_{i \geq 1} \left( 1 - \frac{1}{p^i}\right) \cdot |\mathrm{Aut}(G)|^{-1} $. Apparently, this is equivalent to taking cokernels of random maps $f: (\mathbb{Z}_p)^N \to (\mathbb{Z}_p)^N$ and letting $N \to \infty$. These are the p-adics and there is a Haar measure on this linear space of maps. Alternatively, choose random maps between the finite groups: $f: (\mathbb{Z}/p^k\mathbb{Z})^N \to (\mathbb{Z}/p^k \mathbb{Z})^N$ and let $k, N \to \infty$.

Q: If G is a random abelian p-group according to the Cohen-Lenstra measure and A is a deterministic, why is the expected number of surjections $\phi: G \to A$ equal to 1? In fact, if G were deterministic I don't think this number could ever be 1 unless |G| = 1.

For references see Section 8 of Homological stability for Hurwitz spaces and the Cohen-Lenstra conjecture over function fields by Ellenberg, Venkatesh and Westerland or Terry Tao's blog entry At the AustMS conference.

corrected spelling of Venkatesh's name

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edited Dec 28, 2009 at 21:45

JSE

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The Cohen-Lenstra measure on the set of abelian p-groups assigns $\mathbb{P}(G) = \prod_{i \geq 1} \left( 1 - \frac{1}{p^i}\right) \cdot |\mathrm{Aut}(G)|^{-1} $. Apparently, this is equivalent to taking cokernels of random maps $f: (\mathbb{Z}_p)^N \to (\mathbb{Z}_p)^N$ and letting $N \to \infty$. These are the p-adics and there is a Haar measure on this linear space of maps. Alternatively, choose random maps between the finite groups: $f: (\mathbb{Z}/p^k\mathbb{Z})^N \to (\mathbb{Z}/p^k \mathbb{Z})^N$ and let $k, N \to \infty$.

Q: If G is a random abelian p-group according to the Cohen-Lenstra measure and A is a deterministic, why is the expected number of surjections $\phi: G \to A$ equal to 1? In fact, if G were deterministic I don't think this number could ever be 1 unless |G| = 1.

For references see Section 8 of Homological stability for Hurwitz spaces and the Cohen-Lenstra conjecture over function fields by Ellenberg, VenateshVenkatesh and Westerland or Terry Tao's blog entry At the AustMS conference.

The Cohen-Lenstra measure on the set of abelian p-groups assigns $\mathbb{P}(G) = \prod_{i \geq 1} \left( 1 - \frac{1}{p^i}\right) \cdot |\mathrm{Aut}(G)|^{-1} $. Apparently, this is equivalent to taking cokernels of random maps $f: (\mathbb{Z}_p)^N \to (\mathbb{Z}_p)^N$ and letting $N \to \infty$. These are the p-adics and there is a Haar measure on this linear space of maps. Alternatively, choose random maps between the finite groups: $f: (\mathbb{Z}/p^k\mathbb{Z})^N \to (\mathbb{Z}/p^k \mathbb{Z})^N$ and let $k, N \to \infty$.

Q: If G is a random abelian p-group according to the Cohen-Lenstra measure and A is a deterministic, why is the expected number of surjections $\phi: G \to A$ equal to 1? In fact, if G were deterministic I don't think this number could ever be 1 unless |G| = 1.

For references see Section 8 of Homological stability for Hurwitz spaces and the Cohen-Lenstra conjecture over function fields by Ellenberg, Venatesh and Westerland or Terry Tao's blog entry At the AustMS conference.

The Cohen-Lenstra measure on the set of abelian p-groups assigns $\mathbb{P}(G) = \prod_{i \geq 1} \left( 1 - \frac{1}{p^i}\right) \cdot |\mathrm{Aut}(G)|^{-1} $. Apparently, this is equivalent to taking cokernels of random maps $f: (\mathbb{Z}_p)^N \to (\mathbb{Z}_p)^N$ and letting $N \to \infty$. These are the p-adics and there is a Haar measure on this linear space of maps. Alternatively, choose random maps between the finite groups: $f: (\mathbb{Z}/p^k\mathbb{Z})^N \to (\mathbb{Z}/p^k \mathbb{Z})^N$ and let $k, N \to \infty$.

Q: If G is a random abelian p-group according to the Cohen-Lenstra measure and A is a deterministic, why is the expected number of surjections $\phi: G \to A$ equal to 1? In fact, if G were deterministic I don't think this number could ever be 1 unless |G| = 1.

For references see Section 8 of Homological stability for Hurwitz spaces and the Cohen-Lenstra conjecture over function fields by Ellenberg, Venkatesh and Westerland or Terry Tao's blog entry At the AustMS conference.

Fixed typo: should be |Aut(G)|^{-1}, not |Aut(G)|

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edited Dec 28, 2009 at 19:14

David E Speyer

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The Cohen-Lenstra measure on the set of abelian p-groups assigns $\mathbb{P}(G) = \prod_{i \geq 1} \left( 1 - \frac{1}{p^i}\right) \cdot |\mathrm{Aut}(G)| $$\mathbb{P}(G) = \prod_{i \geq 1} \left( 1 - \frac{1}{p^i}\right) \cdot |\mathrm{Aut}(G)|^{-1} $. Apparently, this is equivalent to taking cokernels of random maps $f: (\mathbb{Z}_p)^N \to (\mathbb{Z}_p)^N$ and letting $N \to \infty$. These are the p-adics and there is a Haar measure on this linear space of maps. Alternatively, choose random maps between the finite groups: $f: (\mathbb{Z}/p^k\mathbb{Z})^N \to (\mathbb{Z}/p^k \mathbb{Z})^N$ and let $k, N \to \infty$.

Q: If G is a random abelian p-group according to the Cohen-Lenstra measure and A is a deterministic, why is the expected number of surjections $\phi: G \to A$ equal to 1? In fact, if G were deterministic I don't think this number could ever be 1 unless |G| = 1.

For references see Section 8 of Homological stability for Hurwitz spaces and the Cohen-Lenstra conjecture over function fields by Ellenberg, Venatesh and Westerland or Terry Tao's blog entry At the AustMS conference.

The Cohen-Lenstra measure on the set of abelian p-groups assigns $\mathbb{P}(G) = \prod_{i \geq 1} \left( 1 - \frac{1}{p^i}\right) \cdot |\mathrm{Aut}(G)| $. Apparently, this is equivalent to taking cokernels of random maps $f: (\mathbb{Z}_p)^N \to (\mathbb{Z}_p)^N$ and letting $N \to \infty$. These are the p-adics and there is a Haar measure on this linear space of maps. Alternatively, choose random maps between the finite groups: $f: (\mathbb{Z}/p^k\mathbb{Z})^N \to (\mathbb{Z}/p^k \mathbb{Z})^N$ and let $k, N \to \infty$.

Q: If G is a random abelian p-group according to the Cohen-Lenstra measure and A is a deterministic, why is the expected number of surjections $\phi: G \to A$ equal to 1? In fact, if G were deterministic I don't think this number could ever be 1 unless |G| = 1.

For references see Section 8 of Homological stability for Hurwitz spaces and the Cohen-Lenstra conjecture over function fields by Ellenberg, Venatesh and Westerland or Terry Tao's blog entry At the AustMS conference.

The Cohen-Lenstra measure on the set of abelian p-groups assigns $\mathbb{P}(G) = \prod_{i \geq 1} \left( 1 - \frac{1}{p^i}\right) \cdot |\mathrm{Aut}(G)|^{-1} $. Apparently, this is equivalent to taking cokernels of random maps $f: (\mathbb{Z}_p)^N \to (\mathbb{Z}_p)^N$ and letting $N \to \infty$. These are the p-adics and there is a Haar measure on this linear space of maps. Alternatively, choose random maps between the finite groups: $f: (\mathbb{Z}/p^k\mathbb{Z})^N \to (\mathbb{Z}/p^k \mathbb{Z})^N$ and let $k, N \to \infty$.

Q: If G is a random abelian p-group according to the Cohen-Lenstra measure and A is a deterministic, why is the expected number of surjections $\phi: G \to A$ equal to 1? In fact, if G were deterministic I don't think this number could ever be 1 unless |G| = 1.

For references see Section 8 of Homological stability for Hurwitz spaces and the Cohen-Lenstra conjecture over function fields by Ellenberg, Venatesh and Westerland or Terry Tao's blog entry At the AustMS conference.