What's the difference between $S^1$ and $x^(2pi i)/log2:xin(Bbb Z_2^times)t$?
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What's the difference between $S^1$ and $x^(2pi i)/log2:xin(Bbb Z_2^times)t$ ?
What elements are in one and not the other?
If $f(x)=x^(2pi i)/log2$ then I have the following results:
This sends e.g. $2Bbb N-1$ to a dense set in the unit circle.
$f(2^nx)=f(x)$, for all $xinBbb R$, for all $ninBbb Z$.
This latter bullet means every branch of the logarithm sits in precisely one 2-adic valuation implying that the set $[f(x)]$ is indexed by a single value satisfying $lvert f(x)rvert_2=1$. I was first expecting the preimage of $S^1$ to be the 2-adic units given by $[f(x)]inBbb Z_2^times$, because this map takes out the powers of $2$.
But I can create counterexamples which would not conventionally have a 2-adic valuation, e.g. $pi^frac2pi ilog2$ for which $f^-1(x)=2^npi:ninBbb Z$.
It is not by any means obvious we can choose $pi$ to be the indexing element of that set, satisfying $lvertpirvert_2=1$ but perhaps with a bit of work a valuation can be achieved.
What form do any counterexamples take? Is it simply all the numbers not having a 2-adic valuation? And can the 2-adic valuation be extended by this means to a larger set?
functions complex-numbers circle p-adic-number-theory
asked Aug 23 at 22:10
Robert Frost
3,9251036
 |Â
show 6 more comments
up vote
-6
down vote
favorite
What's the difference between $S^1$ and $x^(2pi i)/log2:xin(Bbb Z_2^times)t$ ?
What elements are in one and not the other?
If $f(x)=x^(2pi i)/log2$ then I have the following results:
This sends e.g. $2Bbb N-1$ to a dense set in the unit circle.
$f(2^nx)=f(x)$, for all $xinBbb R$, for all $ninBbb Z$.
This latter bullet means every branch of the logarithm sits in precisely one 2-adic valuation implying that the set $[f(x)]$ is indexed by a single value satisfying $lvert f(x)rvert_2=1$. I was first expecting the preimage of $S^1$ to be the 2-adic units given by $[f(x)]inBbb Z_2^times$, because this map takes out the powers of $2$.
But I can create counterexamples which would not conventionally have a 2-adic valuation, e.g. $pi^frac2pi ilog2$ for which $f^-1(x)=2^npi:ninBbb Z$.
It is not by any means obvious we can choose $pi$ to be the indexing element of that set, satisfying $lvertpirvert_2=1$ but perhaps with a bit of work a valuation can be achieved.
What form do any counterexamples take? Is it simply all the numbers not having a 2-adic valuation? And can the 2-adic valuation be extended by this means to a larger set?
functions complex-numbers circle p-adic-number-theory
asked Aug 23 at 22:10
Robert Frost
3,9251036
1
Why two downvotes?
– Chase Ryan Taylor
Aug 23 at 22:27
ok I've found e.g. $pi^frac2pi ilog2$ whose preimage is $2^npi:ninBbb Z$ which would not be in $Bbb Z_2^times$. But I'm still struggling to see why such numbers are not represented
– Robert Frost
Aug 23 at 22:40
2
pi and i aren't in $mathbbQ_2$ much less $mathbbZ_2^times$.
– Steven
Aug 24 at 8:08
@Steven $Bbb Q_2$ I meant to cover by $f(2^nx)=f(x)$ so I'm talking modulo powers of $2$. It's the $pi$ category of numbers I don't understand.
– Robert Frost
Aug 24 at 8:11
1
@Steven I've put a bit of work into this and basically, yes, I agree the inverse map of what I've put can't discriminate the values of 2-adics such as your example. Your example $left(sum_n=0^infty 2^n! right)^2 pi i / log 2$ limits to $theta=pi$ in this map so it's indistinguishable from the number $1/2$. Thank-you for your help.
– Robert Frost
Aug 28 at 16:34
 |Â
show 6 more comments
up vote
-6
down vote
favorite
up vote
-6
down vote
favorite
What's the difference between $S^1$ and $x^(2pi i)/log2:xin(Bbb Z_2^times)t$ ?
What elements are in one and not the other?
If $f(x)=x^(2pi i)/log2$ then I have the following results:
This sends e.g. $2Bbb N-1$ to a dense set in the unit circle.
$f(2^nx)=f(x)$, for all $xinBbb R$, for all $ninBbb Z$.
This latter bullet means every branch of the logarithm sits in precisely one 2-adic valuation implying that the set $[f(x)]$ is indexed by a single value satisfying $lvert f(x)rvert_2=1$. I was first expecting the preimage of $S^1$ to be the 2-adic units given by $[f(x)]inBbb Z_2^times$, because this map takes out the powers of $2$.
But I can create counterexamples which would not conventionally have a 2-adic valuation, e.g. $pi^frac2pi ilog2$ for which $f^-1(x)=2^npi:ninBbb Z$.
It is not by any means obvious we can choose $pi$ to be the indexing element of that set, satisfying $lvertpirvert_2=1$ but perhaps with a bit of work a valuation can be achieved.
What form do any counterexamples take? Is it simply all the numbers not having a 2-adic valuation? And can the 2-adic valuation be extended by this means to a larger set?
functions complex-numbers circle p-adic-number-theory
asked Aug 23 at 22:10
Robert Frost
3,9251036
What's the difference between $S^1$ and $x^(2pi i)/log2:xin(Bbb Z_2^times)t$ ?
What elements are in one and not the other?
If $f(x)=x^(2pi i)/log2$ then I have the following results:
This sends e.g. $2Bbb N-1$ to a dense set in the unit circle.
$f(2^nx)=f(x)$, for all $xinBbb R$, for all $ninBbb Z$.
This latter bullet means every branch of the logarithm sits in precisely one 2-adic valuation implying that the set $[f(x)]$ is indexed by a single value satisfying $lvert f(x)rvert_2=1$. I was first expecting the preimage of $S^1$ to be the 2-adic units given by $[f(x)]inBbb Z_2^times$, because this map takes out the powers of $2$.
But I can create counterexamples which would not conventionally have a 2-adic valuation, e.g. $pi^frac2pi ilog2$ for which $f^-1(x)=2^npi:ninBbb Z$.
It is not by any means obvious we can choose $pi$ to be the indexing element of that set, satisfying $lvertpirvert_2=1$ but perhaps with a bit of work a valuation can be achieved.
What form do any counterexamples take? Is it simply all the numbers not having a 2-adic valuation? And can the 2-adic valuation be extended by this means to a larger set?
functions complex-numbers circle p-adic-number-theory
asked Aug 23 at 22:10
Robert Frost
3,9251036
edited Aug 24 at 8:58
asked Aug 23 at 22:10
Robert Frost
3,9251036
asked Aug 23 at 22:10
Robert Frost
3,9251036
asked Aug 23 at 22:10
Robert Frost
3,9251036
3,9251036
1
Why two downvotes?
– Chase Ryan Taylor
Aug 23 at 22:27
ok I've found e.g. $pi^frac2pi ilog2$ whose preimage is $2^npi:ninBbb Z$ which would not be in $Bbb Z_2^times$. But I'm still struggling to see why such numbers are not represented
– Robert Frost
Aug 23 at 22:40
2
pi and i aren't in $mathbbQ_2$ much less $mathbbZ_2^times$.
– Steven
Aug 24 at 8:08
@Steven $Bbb Q_2$ I meant to cover by $f(2^nx)=f(x)$ so I'm talking modulo powers of $2$. It's the $pi$ category of numbers I don't understand.
– Robert Frost
Aug 24 at 8:11
1
@Steven I've put a bit of work into this and basically, yes, I agree the inverse map of what I've put can't discriminate the values of 2-adics such as your example. Your example $left(sum_n=0^infty 2^n! right)^2 pi i / log 2$ limits to $theta=pi$ in this map so it's indistinguishable from the number $1/2$. Thank-you for your help.
– Robert Frost
Aug 28 at 16:34
 |Â
show 6 more comments
1
Why two downvotes?
– Chase Ryan Taylor
Aug 23 at 22:27
ok I've found e.g. $pi^frac2pi ilog2$ whose preimage is $2^npi:ninBbb Z$ which would not be in $Bbb Z_2^times$. But I'm still struggling to see why such numbers are not represented
– Robert Frost
Aug 23 at 22:40
2
pi and i aren't in $mathbbQ_2$ much less $mathbbZ_2^times$.
– Steven
Aug 24 at 8:08
@Steven $Bbb Q_2$ I meant to cover by $f(2^nx)=f(x)$ so I'm talking modulo powers of $2$. It's the $pi$ category of numbers I don't understand.
– Robert Frost
Aug 24 at 8:11
1
@Steven I've put a bit of work into this and basically, yes, I agree the inverse map of what I've put can't discriminate the values of 2-adics such as your example. Your example $left(sum_n=0^infty 2^n! right)^2 pi i / log 2$ limits to $theta=pi$ in this map so it's indistinguishable from the number $1/2$. Thank-you for your help.
– Robert Frost
Aug 28 at 16:34
1
1
Why two downvotes?
– Chase Ryan Taylor
Aug 23 at 22:27
Why two downvotes?
– Chase Ryan Taylor
Aug 23 at 22:27
ok I've found e.g. $pi^frac2pi ilog2$ whose preimage is $2^npi:ninBbb Z$ which would not be in $Bbb Z_2^times$. But I'm still struggling to see why such numbers are not represented
– Robert Frost
Aug 23 at 22:40
ok I've found e.g. $pi^frac2pi ilog2$ whose preimage is $2^npi:ninBbb Z$ which would not be in $Bbb Z_2^times$. But I'm still struggling to see why such numbers are not represented
– Robert Frost
Aug 23 at 22:40
2
2
pi and i aren't in $mathbbQ_2$ much less $mathbbZ_2^times$.
– Steven
Aug 24 at 8:08
pi and i aren't in $mathbbQ_2$ much less $mathbbZ_2^times$.
– Steven
Aug 24 at 8:08
@Steven $Bbb Q_2$ I meant to cover by $f(2^nx)=f(x)$ so I'm talking modulo powers of $2$. It's the $pi$ category of numbers I don't understand.
– Robert Frost
Aug 24 at 8:11
@Steven $Bbb Q_2$ I meant to cover by $f(2^nx)=f(x)$ so I'm talking modulo powers of $2$. It's the $pi$ category of numbers I don't understand.
– Robert Frost
Aug 24 at 8:11
1
1
@Steven I've put a bit of work into this and basically, yes, I agree the inverse map of what I've put can't discriminate the values of 2-adics such as your example. Your example $left(sum_n=0^infty 2^n! right)^2 pi i / log 2$ limits to $theta=pi$ in this map so it's indistinguishable from the number $1/2$. Thank-you for your help.
– Robert Frost
Aug 28 at 16:34
@Steven I've put a bit of work into this and basically, yes, I agree the inverse map of what I've put can't discriminate the values of 2-adics such as your example. Your example $left(sum_n=0^infty 2^n! right)^2 pi i / log 2$ limits to $theta=pi$ in this map so it's indistinguishable from the number $1/2$. Thank-you for your help.
– Robert Frost
Aug 28 at 16:34
 |Â
show 6 more comments
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1
Why two downvotes?
– Chase Ryan Taylor
Aug 23 at 22:27
ok I've found e.g. $pi^frac2pi ilog2$ whose preimage is $2^npi:ninBbb Z$ which would not be in $Bbb Z_2^times$. But I'm still struggling to see why such numbers are not represented
– Robert Frost
Aug 23 at 22:40
2
pi and i aren't in $mathbbQ_2$ much less $mathbbZ_2^times$.
– Steven
Aug 24 at 8:08
@Steven $Bbb Q_2$ I meant to cover by $f(2^nx)=f(x)$ so I'm talking modulo powers of $2$. It's the $pi$ category of numbers I don't understand.
– Robert Frost
Aug 24 at 8:11
1
@Steven I've put a bit of work into this and basically, yes, I agree the inverse map of what I've put can't discriminate the values of 2-adics such as your example. Your example $left(sum_n=0^infty 2^n! right)^2 pi i / log 2$ limits to $theta=pi$ in this map so it's indistinguishable from the number $1/2$. Thank-you for your help.
– Robert Frost
Aug 28 at 16:34