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 Post subject: smooth vs analytic
PostPosted: Sat, 17 Sep 2011 04:12:57 UTC 
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Suppose f\colon\mathbb{R}\to\mathbb{R} is smooth with f(\mathbb{Q})\subseteq\mathbb{Q}. Is f\in\mathbb{Q}(x) necessary?

Suppose f\colon\mathbb{C}\to\mathbb{C} is analytic with f(\mathbb{Q}[i])\subseteq\mathbb{Q}[i]. Is f\in\mathbb{C}(x) necessary?

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\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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 Post subject: Re: smooth vs analytic
PostPosted: Sat, 17 Sep 2011 10:00:38 UTC 
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outermeasure wrote:
Suppose f\colon\mathbb{R}\to\mathbb{R} is smooth with f(\mathbb{Q})\subseteq\mathbb{Q}. Is f\in\mathbb{Q}(x) necessary?

Suppose f\colon\mathbb{C}\to\mathbb{C} is analytic with f(\mathbb{Q}[i])\subseteq\mathbb{Q}[i]. Is f\in\mathbb{C}(x) necessary?


The first one:

Spoiler:
No. Hit any element of \mathbb{Q}(x) (assumed to have no real zeros for the denominator) with a partition of unity on a bounded interval, so that it vanishes outside.

Second one:

Spoiler:
Lemma: If f\in\mathbb{C}(x) is necessary, then f\in\mathbb{Q}(i)[x] is necessary.

Proof: It is clear that if f(x)\in \mathbb{C}(x) is necessary then really f(x)\in\mathbb{C}[x] is necessary, since any finite pole of a rational function would break our assumption that f is analytic on all of \mathbb{C}. It is easy to see from here that if this is the case, then really f\in\mathbb{Q}(i)[x]--just test the polynomial on \deg (f)+1 rational points and solve the resulting system of linear equations over the vector space \mathbb{Q}(i).\;\square

I'm going to think more on this tomorrow, it's 4:18 am here and I just pulled these two out, but I think the second one is going to require me to remember a couple more facts.


Great problem outermeasure! I like how the cases are so starkly different.

EDIT:
Spoiler:
realized I meant \mathbb{Q}(i)[x], not \mathbb{Q}[x]

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 Post subject: Re: smooth vs analytic
PostPosted: Sat, 17 Sep 2011 10:18:51 UTC 
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Shadow wrote:
outermeasure wrote:
Suppose f\colon\mathbb{R}\to\mathbb{R} is smooth with f(\mathbb{Q})\subseteq\mathbb{Q}. Is f\in\mathbb{Q}(x) necessary?

Suppose f\colon\mathbb{C}\to\mathbb{C} is analytic with f(\mathbb{Q}[i])\subseteq\mathbb{Q}[i]. Is f\in\mathbb{C}(x) necessary?


The first one:

Spoiler:
No. Hit any element of \mathbb{Q}(x) (assumed to have no real zeros for the denominator) with a bump function to make it zero outside some compact interval


Why would a bump function \eta sends \mathbb{Q} to \mathbb{Q}?

Spoiler:
In addition to using bump functions, you want to enumerate \mathbb{Q}(x) and use diagonal argument.

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\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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 Post subject: Re: smooth vs analytic
PostPosted: Sat, 17 Sep 2011 10:21:44 UTC 
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outermeasure wrote:
Shadow wrote:
outermeasure wrote:
Suppose f\colon\mathbb{R}\to\mathbb{R} is smooth with f(\mathbb{Q})\subseteq\mathbb{Q}. Is f\in\mathbb{Q}(x) necessary?

Suppose f\colon\mathbb{C}\to\mathbb{C} is analytic with f(\mathbb{Q}[i])\subseteq\mathbb{Q}[i]. Is f\in\mathbb{C}(x) necessary?


The first one:

Spoiler:
No. Hit any element of \mathbb{Q}(x) (assumed to have no real zeros for the denominator) with a bump function to make it zero outside some compact interval


Why would a bump function \eta sends \mathbb{Q} to \mathbb{Q}?

Spoiler:
In addition to using bump functions, you want to enumerate \mathbb{Q}(x) and use diagonal argument.



Oh wow, I didn't even see you had responded, I actually modified my earlier argument, I might still need to do what you said, but I don't think so at a glance.

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 Post subject: Re: smooth vs analytic
PostPosted: Sun, 18 Sep 2011 06:48:38 UTC 
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Shadow wrote:
Second one:

Spoiler:
Lemma: If f\in\mathbb{C}(x) is necessary, then f\in\mathbb{Q}(i)[x] is necessary.

Proof: It is clear that if f(x)\in \mathbb{C}(x) is necessary then really f(x)\in\mathbb{C}[x] is necessary, since any finite pole of a rational function would break our assumption that f is analytic on all of \mathbb{C}. It is easy to see from here that if this is the case, then really f\in\mathbb{Q}(i)[x]--just test the polynomial on \deg (f)+1 rational points and solve the resulting system of linear equations over the vector space \mathbb{Q}(i).\;\square

I'm going to think more on this tomorrow, it's 4:18 am here and I just pulled these two out, but I think the second one is going to require me to remember a couple more facts.


Great problem outermeasure! I like how the cases are so starkly different.


Yes, that's why I wrote \mathbb{C}(x) instead of \mathbb{Q}[i](x) or \mathbb{Q}[i][x].

Actually the cases are not that much different! :twisted:

_________________
\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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 Post subject: Re: smooth vs analytic
PostPosted: Sun, 18 Sep 2011 07:01:05 UTC 
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outermeasure wrote:
Shadow wrote:
Second one:

Spoiler:
Lemma: If f\in\mathbb{C}(x) is necessary, then f\in\mathbb{Q}(i)[x] is necessary.

Proof: It is clear that if f(x)\in \mathbb{C}(x) is necessary then really f(x)\in\mathbb{C}[x] is necessary, since any finite pole of a rational function would break our assumption that f is analytic on all of \mathbb{C}. It is easy to see from here that if this is the case, then really f\in\mathbb{Q}(i)[x]--just test the polynomial on \deg (f)+1 rational points and solve the resulting system of linear equations over the vector space \mathbb{Q}(i).\;\square

I'm going to think more on this tomorrow, it's 4:18 am here and I just pulled these two out, but I think the second one is going to require me to remember a couple more facts.


Great problem outermeasure! I like how the cases are so starkly different.


Yes, that's why I wrote \mathbb{C}(x) instead of \mathbb{Q}[i](x) or \mathbb{Q}[i][x].

Actually the cases are not that much different! :twisted:


Really now? I know at least I cannot pull the same trick in the analytic context because of Liouville, but perhaps you mean there's an approach which handles both simultaneously which is not mine?

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 Post subject: Re: smooth vs analytic
PostPosted: Sun, 18 Sep 2011 07:30:23 UTC 
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Shadow wrote:
Really now? I know at least I cannot pull the same trick in the analytic context because of Liouville, but perhaps you mean there's an approach which handles both simultaneously which is not mine?


Yes, there is an approach without using bump functions.

Spoiler:
Enumerating \mathbb{Q} or \mathbb{Q}[i] is still crucial.


BTW, yes, there exists a bump function \eta such that \eta(\mathbb{Q})\subset\mathbb{Q} (but not every bump function has this property).

_________________
\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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 Post subject: Re: smooth vs analytic
PostPosted: Mon, 19 Sep 2011 05:32:05 UTC 
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Constructing a bump function with \eta(\mathbb{Q})\subset\mathbb{Q} (gist):

Spoiler:
Enumerate the rationals q_n. Take a bump function \eta_0. If \eta_0(q_1)\in\mathbb{Q}, define \eta_1=\eta_0, else: choose a small interval I_1=(q_1-\delta_1,q_1+\delta_1) with \overline{I_1}\cap\eta_0^{-1}(\{1\})=\varnothing, I_1\subset\mathop{\mathrm{supp}}\eta_0, and a small \epsilon_1>0 such that \eta_0(q_1)+\epsilon_1\in\mathbb{Q}, \eta_0(I_1)\subset(0,1-\epsilon_1). Let \tilde{\eta}_1 be a bump function centred at q_1, vanishing outside I_1, with derivatives suitably bounded, and let \eta_1=\eta_0+\tilde{\eta}_1. Inductively construct \eta_n, making sure we avoid touching those q_n previously done and that \epsilon_n is small enough to guarantee \tilde{\eta}_n has sufficiently small C^n-norm (\sum\lVert\tilde{\eta}_n\rVert_{C^n(\mathbb{R})}<\infty) and \eta_n(q_n)\in\mathbb{Q}. Finally, let \eta=\lim\eta_n.


Here is the common approach:

Spoiler:
Let x_j be an enumeration of \mathbb{Q}(i) (or \mathbb{Q} --- I'll just drop the mention unless it is significantly different). Recall g_n(x)=\prod_{j=1}^n(x-x_j) vanishes at x=x_1,x_2,\dots,x_n, and this serves as our building block.

Start with f_1(x)=\epsilon_1 g_1(x), f_{n+1}(x)=f_n(x)+\epsilon_{n+1} g_{n+1}(x). There are now two variants to how we choose the sequence \epsilon_n --- either we use a cardinality argument, or we construct a power series directly.

Cardinality argument: We choose \epsilon_n\in\mathbb{Q}(i) sufficiently small so that f_n converges uniformly on compacts, e.g., let a_n be a sequence of positive reals such that \sum a_n<+\infty and \lvert\epsilon_n\rvert<\dfrac{a_n}{\sup\{\lvert g_n(x)\rvert\colon \lvert x\rvert\leq n\}}. Then the limit \lim f_n=:f is analytic and f(x_j)\in\mathbb{Q}(i) for all j (since only finitely many summands are nonzero at x_j). Now the choice of sequence (\epsilon_n) bijects with the continuum, but there are only countably many polynomials in \mathbb{Q}(i)[x] (or \mathbb{Q}(x) for rational case).

Of course, you can dress that up in a diagonal argument instead --- choose your \epsilon_n sufficiently small and yet f_n(x_n) disagrees with p_n(x_n), where (p_n(x)) is an enumeration of \mathbb{Q}(i)[x] (or \mathbb{Q}(x)).

Constructing power series directly: Note that \deg g_n=n, so if we let \epsilon_n=\epsilon'_n x^{n^2} (say) we have the coefficients of various g_n's will not interfere with each other, so it remains to choose the coefficients \epsilon'_n. Now choose \epsilon'_n\in\mathbb{Q}(i)^\times with size sufficiently rapidly decreasing so the f_n converges on compacts (and for the rational case, it is easy to make it transcendence --- indeed in this case this is guaranteed by the arbitrarily long string of zeros).


There is nothing special about \mathbb{Q} or \mathbb{Q}(i) in this problem, except they are countable subrings with an element inside the punctured unit disc, so we can replace them with, e.g. \mathbb{Z}[\frac{1}{2}] or the algebraic closure of \mathbb{Q}(e).

Edit: fix subscripts and typo.

_________________
\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


Last edited by outermeasure on Mon, 19 Sep 2011 08:01:22 UTC, edited 1 time in total.
fix subscripts and typo


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 Post subject: Re: smooth vs analytic
PostPosted: Mon, 19 Sep 2011 05:37:53 UTC 
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outermeasure wrote:
Constructing a bump function with \eta(\mathbb{Q})\subset\mathbb{Q} (gist):

Spoiler:
Enumerate the rationals q_n. Take a bump function \eta_1. If \eta_1(q_1)\in\mathbb{Q}, define \eta_2=\eta_1, else: choose a small interval I_1=(q_1-\delta_1,q_1+\delta_1) with \overline{I_1}\cap\eta^{-1}(\{1\})=\varnothing, I_1\subset\mathop{\mathrm{supp}}\eta_1, and a small \epsilon_1>0 such that \eta_1(q_1)+\epsilon_1\in\mathbb{Q}, \eta(I_1)\subset(0,1-\epsilon_1). Let \tilde{\eta}_1 be a bump function centred at q_1, vanishing outside I_1, with derivatives suitably bounded, and let \eta_2=\eta_1+\tilde{\eta}_1. Inductively construct \eta_n, making sure we avoid touching those q_n previously done and that \epsilon_n is small enough to guarantee \tilde{\eta}_n has sufficiently small C^n-norm (\sum\lVert\tilde{\eta}_n\rVert_{C^n(\mathbb{R})}<\infty) and \eta_n(q_n)\in\mathbb{Q}. Finally, let \eta=\lim\eta_n.


Here is the common approach:

Spoiler:
Let x_j be an enumeration of \mathbb{Q}(i) (or \mathbb{Q} --- I'll just drop the mention unless it is significantly different). Recall g_n(x)=\prod_{j=1}^n(x-x_j) vanishes at x=x_1,x_2,\dots,x_n, and this serves as our building block.

Start with f_1(x)=\epsilon_1 g_1(x), f_{n+1}(x)=f_n(x)+\epsilon_{n+1} g_{n+1}(x). There are now two variants to how we choose the sequence \epsilon_n --- either we use a cardinality argument, or we construct a power series directly.

Cardinality argument: We choose \epsilon_n\in\mathbb{Q}(i) sufficiently small so that f_n converges uniformly on compacts, e.g., let a_n be a sequence of positive reals such that \sum a_n<+\infty and \lvert\epsilon_n\rvert<\dfrac{a_n}{\sup\{\lvert g_n(x)\rvert\colon \lvert x\rvert\leq n\}}. Then the limit \lim f_n=:f is analytic and f(x_j)\in\mathbb{Q}(i) for all j (since only finitely many summands are nonzero at x_j). Now the choice of sequence (\epsilon_n) bijects with the continuum, but there are only countably many polynomials in \mathbb{Q}(i)[x] (or \mathbb{Q}(x) for rational case).

Of course, you can dress that up in a diagonal argument instead --- choose your \epsilon_n sufficiently small and yet f_n(x_n) disagrees with p_n(x_n), where (p_n(x)) is an enumeration of \mathbb{Q}(i)[x] (or \mathbb{Q}(x)).

Constructing power series directly: Note that \deg g_n=n, so if we let \epsilon_n=\epsilon'_n x^{n^2} (say) we have the coefficients of various g_n's will not interfere with each other, so it remains to choose the coefficients \epsilon'_n. Now choose \epsilon'_n\in\mathbb{Q}(i)^\times with size sufficiently rapidly decreasing so the f_n converges on compacts (and for the rational case, it is easy to make it transcendence --- indeed in this case this is guaranteed by the arbitrarily long sting of zeros).


There is nothing special about \mathbb{Q} or \mathbb{Q}(i) in this problem, except they are countable subrings with an element inside the punctured unit disc, so we can replace them with, e.g. \mathbb{Z}[\frac{1}{2}] or the algebraic closure of \mathbb{Q}(e).


Ah I see, I was trying to use the interpolation to produce a sequence, but I couldn't see the bit with the \epsilon_n for convergence. Also, I like that bump construction. Nice problem!

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