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Note: this is a revision of an earlier post. It was kindly pointed out that my initial proposed metric was in fact not a metric, so this is a revised version.

I was thinking how to say that two sets of prime numbers are close to each other, and came up with this.

NOTATION

• $\ \Delta(A, B)\ :=\ (A\setminus B)\cup(B\setminus A)\ \ $ is the symmetric difference of sets $A$ and $B$;
• $\ \mathbb P := \{2, 3, \ldots\}\ $ is the set of all rational primes;
• Fix a real number $s > 1$. For any rational prime $p$, define $h_s(p) = \log((1-p^{-s})^{-1}) = -\log(1-p^{-s})$;
• For any subset $P \subset \mathbb P$, define $h_s(P) = \sum_{p \in P} h_s(p)$;
• Finally, for any two subsets $P$ and $Q$ of $\mathbb P$, define $d_s(P, Q) = h_s(\Delta(P, Q))$.

For example, $\ d_s(S, T)\ =\ \log(\zeta(s))$ for every partition $S, T$ of $\mathbb P$.

I claim that $d_s$ is a metric on the set of subsets of rational primes. Clearly, $d_s(P, P) = 0$. Symmetry follows by virtue of the symmetry in the definition of $d_s$. So it remains to prove the triangle inequality, that $d_s(P, R) \leq d_s(P, Q) + d_s(Q, R)$, for any three subsets $P$, $Q$, and $R$ of $\mathbb P$.

To see this, observe that if $p \in \Delta(P, R)$, then either $p \in \Delta(P, Q)$ or $p \in \Delta(Q, R)$. Consequently, $\Delta(P, R) \subset \Delta(P, Q) \cup \Delta(Q, R)$. Therefore, $d_s(P, R) \leq d_s(P, Q) + d_s(Q, R)$ as required.

Now, we can now make a statement along these lines meaningfully: $\{2, p\} \rightarrow \{2\}$ as $p \rightarrow \infty$.

Is it of any interest? Has this already been defined and is it called something?

My hope is that this may find use as follows. Suppose that there is a set of objects such that for each object in the set there is a set of associated prime numbers, and suppose that each object is uniquely determined by its set of associated primes. Then the metric defined above can be used to construct a metric on this set.

For example, consider the set G of Galois extensions of Q. Each extension K/Q in G has an associated set of its unramified primes, say U(K). Define an equivalence relation ~ on G by saying that two extensions are equivalent if they have the same set of unramified primes. Consider the set of equivalence classe, H = G/~, and for an extension K/Q denote its equivalence class by K/~. Then there is a metric e on H defined by e(K/~, L/~) = d(U(K), U(L)).

Any pointers would be welcomed! Thanks, Martin

Note: this is a revision of an earlier post. It was kindly pointed out that my initial proposed metric was in fact not a metric, so this is a revised version.

I was thinking how to say that two sets of prime numbers are close to each other, and came up with this.

NOTATION

• $\ \Delta(A, B)\ :=\ (A\setminus B)\cup(B\setminus A)\ \ $ is the symmetric difference of sets $A$ and $B$;
• $\ \mathbb P := \{2, 3, \ldots\}\ $ is the set of all rational primes;
• Fix a real number $s > 1$. For any rational prime $p$, define $h_s(p) = \log((1-p^{-s})^{-1}) = -\log(1-p^{-s})$;
• For any subset $P \subset \mathbb P$, define $h_s(P) = \sum_{p \in P} h_s(p)$;
• Finally, for any two subsets $P$ and $Q$ of $\mathbb P$, define $d_s(P, Q) = h_s(\Delta(P, Q))$.

For example, $\ d_s(S, T)\ =\ \log(\zeta(s))$ for every partition $S, T$ of $\mathbb P$.

I claim that $d_s$ is a metric on the set of subsets of rational primes. Clearly, $d_s(P, P) = 0$. Symmetry follows by virtue of the symmetry in the definition of $d_s$. So it remains to prove the triangle inequality, that $d_s(P, R) \leq d_s(P, Q) + d_s(Q, R)$, for any three subsets $P$, $Q$, and $R$ of $\mathbb P$.

To see this, observe that if $p \in \Delta(P, R)$, then either $p \in \Delta(P, Q)$ or $p \in \Delta(Q, R)$. Consequently, $\Delta(P, R) \subset \Delta(P, Q) \cup \Delta(Q, R)$. Therefore, $d_s(P, R) \leq d_s(P, Q) + d_s(Q, R)$ as required.

Now, we can now make a statement along these lines meaningfully: $\{2, p\} \rightarrow \{2\}$ as $p \rightarrow \infty$.

Is it of any interest? Has this already been defined and is it called something?

My hope is that this may find use as follows. Suppose that there is a set of objects such that for each object in the set there is a set of associated prime numbers, and suppose that each object is uniquely determined by its set of associated primes. Then the metric defined above can be used to construct a metric on this set.

For example, consider the set G of Galois extensions of Q. Each extension K/Q in G has an associated set of the rational primes that split completely in K, say Spl(K), and this set uniquely characterizes K. Then there is a metric e on G defined by e(K, L) = d(Spl(K), spl(L)).

Any pointers would be welcomed! Thanks, Martin

Note: this is a revision of an earlier post. It was kindly pointed out that my initial proposed metric was in fact not a metric, so this is a revised version.

I was thinking how to say that two sets of prime numbers are close to each other, and came up with this.

NOTATION

• $\ \Delta(A, B)\ :=\ (A\setminus B)\cup(B\setminus A)\ \ $ is the symmetric difference of sets $A$ and $B$;
• $\ \mathbb P := \{1\ 2\ \ldots\}\ $ is the set of all rational primes;
• Fix a real number $s > 1$. For any rational prime $p$, define $h_s(p) = \log((1-p^{-s})^{-1}) = -\log(1-p^{-s})$;
• For any subset $P \subset \mathbb P$, define $h_s(P) = \sum_{p \in P} h_s(p)$;
• Finally, for any two subsets $P$ and $Q$ of $\mathbb P$, define $d_s(P, Q) = h_s(\Delta(P, Q))$.

For example, $\ d_s(S, T)\ =\ \log(\zeta(s))$ for every partition $S, T$ of $\mathbb P$.

I claim that $d_s$ is a metric on the set of subsets of rational primes. Clearly, $d_s(P, P) = 0$. Symmetry follows by virtue of the symmetry in the definition of $d_s$. So it remains to prove the triangle inequality, that $d_s(P, R) \leq d_s(P, Q) + d_s(Q, R)$, for any three subsets $P$, $Q$, and $R$ of $\mathbb P$.

To see this, observe that if $p \in \Delta(P, R)$, then either $p \in \Delta(P, Q)$ or $p \in \Delta(Q, R)$. Consequently, $\Delta(P, R) \subset \Delta(P, Q) \cup \Delta(Q, R)$. Therefore, $d_s(P, R) \leq d_s(P, Q) + d_s(Q, R)$ as required.

Now, we can now make a statement along these lines meaningfully: $\{2, p\} \rightarrow \{2\}$ as $p \rightarrow \infty$.

Is it of any interest? Has this already been defined and is it called something?

My hope is that this may find use as follows. Suppose that there is a set of objects such that for each object in the set there is a set of associated prime numbers, and suppose that each object is uniquely determined by its set of associated primes. Then the metric defined above can be used to construct a metric on this set.

For example, consider the set G of Galois extensions of Q. Each extension K/Q in G has an associated set of its unramified primes, say U(K). Define an equivalence relation ~ on G by saying that two extensions are equivalent if they have the same set of unramified primes. Consider the set of equivalence classe, H = G/~, and for an extension K/Q denote its equivalence class by K/~. Then there is a metric e on H defined by e(K/~, L/~) = d(U(K), U(L)).

Any pointers would be welcomed! Thanks, Martin

Note: this is a revision of an earlier post. It was kindly pointed out that my initial proposed metric was in fact not a metric, so this is a revised version.

I was thinking how to say that two sets of prime numbers are close to each other, and came up with this.

NOTATION

• $\ \Delta(A, B)\ :=\ (A\setminus B)\cup(B\setminus A)\ \ $ is the symmetric difference of sets $A$ and $B$;
• $\ \mathbb P := \{2, 3, \ldots\}\ $ is the set of all rational primes;
• Fix a real number $s > 1$. For any rational prime $p$, define $h_s(p) = \log((1-p^{-s})^{-1}) = -\log(1-p^{-s})$;
• For any subset $P \subset \mathbb P$, define $h_s(P) = \sum_{p \in P} h_s(p)$;
• Finally, for any two subsets $P$ and $Q$ of $\mathbb P$, define $d_s(P, Q) = h_s(\Delta(P, Q))$.

For example, $\ d_s(S, T)\ =\ \log(\zeta(s))$ for every partition $S, T$ of $\mathbb P$.

I claim that $d_s$ is a metric on the set of subsets of rational primes. Clearly, $d_s(P, P) = 0$. Symmetry follows by virtue of the symmetry in the definition of $d_s$. So it remains to prove the triangle inequality, that $d_s(P, R) \leq d_s(P, Q) + d_s(Q, R)$, for any three subsets $P$, $Q$, and $R$ of $\mathbb P$.

To see this, observe that if $p \in \Delta(P, R)$, then either $p \in \Delta(P, Q)$ or $p \in \Delta(Q, R)$. Consequently, $\Delta(P, R) \subset \Delta(P, Q) \cup \Delta(Q, R)$. Therefore, $d_s(P, R) \leq d_s(P, Q) + d_s(Q, R)$ as required.

Now, we can now make a statement along these lines meaningfully: $\{2, p\} \rightarrow \{2\}$ as $p \rightarrow \infty$.

Is it of any interest? Has this already been defined and is it called something?

My hope is that this may find use as follows. Suppose that there is a set of objects such that for each object in the set there is a set of associated prime numbers, and suppose that each object is uniquely determined by its set of associated primes. Then the metric defined above can be used to construct a metric on this set.

For example, consider the set G of Galois extensions of Q. Each extension K/Q in G has an associated set of its unramified primes, say U(K). Define an equivalence relation ~ on G by saying that two extensions are equivalent if they have the same set of unramified primes. Consider the set of equivalence classe, H = G/~, and for an extension K/Q denote its equivalence class by K/~. Then there is a metric e on H defined by e(K/~, L/~) = d(U(K), U(L)).

Any pointers would be welcomed! Thanks, Martin

Note: this is a revision of an earlier post. It was kindly pointed out that my initial proposed metric was in fact not a metric, so this is a revised version.

I was thinking how to say that two sets of prime numbers are close to each other, and came up with this.

NOTATION

• $\ \Delta(A, B)\ :=\ (A\setminus B)\cup(B\setminus A)\ \ $ is the symmetric difference of sets $A$ and $B$;
• $\ \mathbb P := \{1\ 2\ \ldots\}\ $ is the set of all rational primes;
• Fix a real number $s > 1$. For any rational prime $p$, define $h_s(p) = \log((1-p^{-s})^{-1}) = -\log(1-p^{-s})$;
• For any subset $P \subset \mathbb P$, define $h_s(P) = \sum_{p \in P} h_s(p)$;
• Finally, for any two subsets $P$ and $Q$ of $\mathbb P$, define $d_s(P, Q) = h_s(\Delta(P, Q))$.

For example, $\ d_s(S, T)\ =\ \log(\zeta(s))$ for every partition $S, T$ of $\mathbb P$.

I claim that $d_s$ is a metric on the set of subsets of rational primes. Clearly, $d_s(P, P) = 0$. Symmetry follows by virtue of the symmetry in the definition of $d_s$. So it remains to prove the triangle inequality, that $d_s(P, R) \leq d_s(P, Q) + d_s(Q, R)$, for any three subsets $P$, $Q$, and $R$ of $\mathbb P$.

To see this, observe that if $p \in \Delta(P, R)$, then either $p \in \Delta(P, Q)$ or $p \in \Delta(Q, R)$. Consequently, $\Delta(P, R) \subset \Delta(P, Q) \cup \Delta(Q, R)$. Therefore, $d_s(P, R) \leq d_s(P, Q) + d_s(Q, R)$ as required.

Is it of any interest? Has this already been defined and is it called something?

My hope is that this may find use as follows. Suppose that there is a set of objects such that for each object in the set there is a set of associated prime numbers, and suppose that each object is uniquely determined by its set of associated primes. Then the metric defined above can be used to construct a metric on this set.

For example, consider the set G of Galois extensions of Q. Each extension K/Q in G has an associated set of its unramified primes, say U(K). Define an equivalence relation ~ on G by saying that two extensions are equivalent if they have the same set of unramified primes. Consider the set of equivalence classe, H = G/~, and for an extension K/Q denote its equivalence class by K/~. Then there is a metric e on H defined by e(K/~, L/~) = d(U(K), U(L)).

Any pointers would be welcomed! Thanks, Martin

Note: this is a revision of an earlier post. It was kindly pointed out that my initial proposed metric was in fact not a metric, so this is a revised version.

I was thinking how to say that two sets of prime numbers are close to each other, and came up with this.

NOTATION

• $\ \Delta(A, B)\ :=\ (A\setminus B)\cup(B\setminus A)\ \ $ is the symmetric difference of sets $A$ and $B$;
• $\ \mathbb P := \{1\ 2\ \ldots\}\ $ is the set of all rational primes;
• Fix a real number $s > 1$. For any rational prime $p$, define $h_s(p) = \log((1-p^{-s})^{-1}) = -\log(1-p^{-s})$;
• For any subset $P \subset \mathbb P$, define $h_s(P) = \sum_{p \in P} h_s(p)$;
• Finally, for any two subsets $P$ and $Q$ of $\mathbb P$, define $d_s(P, Q) = h_s(\Delta(P, Q))$.

For example, $\ d_s(S, T)\ =\ \log(\zeta(s))$ for every partition $S, T$ of $\mathbb P$.

I claim that $d_s$ is a metric on the set of subsets of rational primes. Clearly, $d_s(P, P) = 0$. Symmetry follows by virtue of the symmetry in the definition of $d_s$. So it remains to prove the triangle inequality, that $d_s(P, R) \leq d_s(P, Q) + d_s(Q, R)$, for any three subsets $P$, $Q$, and $R$ of $\mathbb P$.

To see this, observe that if $p \in \Delta(P, R)$, then either $p \in \Delta(P, Q)$ or $p \in \Delta(Q, R)$. Consequently, $\Delta(P, R) \subset \Delta(P, Q) \cup \Delta(Q, R)$. Therefore, $d_s(P, R) \leq d_s(P, Q) + d_s(Q, R)$ as required.

Now, we can now make a statement along these lines meaningfully: $\{2, p\} \rightarrow \{2\}$ as $p \rightarrow \infty$.

Is it of any interest? Has this already been defined and is it called something?

My hope is that this may find use as follows. Suppose that there is a set of objects such that for each object in the set there is a set of associated prime numbers, and suppose that each object is uniquely determined by its set of associated primes. Then the metric defined above can be used to construct a metric on this set.

For example, consider the set G of Galois extensions of Q. Each extension K/Q in G has an associated set of its unramified primes, say U(K). Define an equivalence relation ~ on G by saying that two extensions are equivalent if they have the same set of unramified primes. Consider the set of equivalence classe, H = G/~, and for an extension K/Q denote its equivalence class by K/~. Then there is a metric e on H defined by e(K/~, L/~) = d(U(K), U(L)).

Any pointers would be welcomed! Thanks, Martin