an inverse of the Artin-Hasse exponential?
In the p-adic world the Artin-Hasse exponential is the sollowing power series:
$$
E_p(x)= exp left( sum_{n=0}^{infty}frac{x^{p^n}}{p^n} right)
$$
where $E_p(x)in 1+xmathbb{Z}_{(p)}[[x]]$ with radius of convergence $r=1$.
My question is : as the classical exponential it is possibile define a sort of 'logarithm' which invert this series?
Thanks for the suggestions!
number-theory power-series exponential-function p-adic-number-theory
edited Dec 19 '18 at 2:21
Will Fisher
4,0231032
asked Dec 4 '18 at 9:16
andres
1757
add a comment |
In the p-adic world the Artin-Hasse exponential is the sollowing power series:
$$
E_p(x)= exp left( sum_{n=0}^{infty}frac{x^{p^n}}{p^n} right)
$$
where $E_p(x)in 1+xmathbb{Z}_{(p)}[[x]]$ with radius of convergence $r=1$.
My question is : as the classical exponential it is possibile define a sort of 'logarithm' which invert this series?
Thanks for the suggestions!
number-theory power-series exponential-function p-adic-number-theory
edited Dec 19 '18 at 2:21
Will Fisher
4,0231032
asked Dec 4 '18 at 9:16
andres
1757
add a comment |
In the p-adic world the Artin-Hasse exponential is the sollowing power series:
$$
E_p(x)= exp left( sum_{n=0}^{infty}frac{x^{p^n}}{p^n} right)
$$
where $E_p(x)in 1+xmathbb{Z}_{(p)}[[x]]$ with radius of convergence $r=1$.
My question is : as the classical exponential it is possibile define a sort of 'logarithm' which invert this series?
Thanks for the suggestions!
number-theory power-series exponential-function p-adic-number-theory
edited Dec 19 '18 at 2:21
Will Fisher
4,0231032
asked Dec 4 '18 at 9:16
andres
1757
In the p-adic world the Artin-Hasse exponential is the sollowing power series:
$$
E_p(x)= exp left( sum_{n=0}^{infty}frac{x^{p^n}}{p^n} right)
$$
where $E_p(x)in 1+xmathbb{Z}_{(p)}[[x]]$ with radius of convergence $r=1$.
My question is : as the classical exponential it is possibile define a sort of 'logarithm' which invert this series?
Thanks for the suggestions!
number-theory power-series exponential-function p-adic-number-theory
number-theory power-series exponential-function p-adic-number-theory
edited Dec 19 '18 at 2:21
Will Fisher
4,0231032
asked Dec 4 '18 at 9:16
andres
1757
edited Dec 19 '18 at 2:21
Will Fisher
4,0231032
asked Dec 4 '18 at 9:16
andres
1757
edited Dec 19 '18 at 2:21
Will Fisher
4,0231032
edited Dec 19 '18 at 2:21
Will Fisher
4,0231032
edited Dec 19 '18 at 2:21
Will Fisher
4,0231032
4,0231032
asked Dec 4 '18 at 9:16
andres
1757
asked Dec 4 '18 at 9:16
andres
1757
asked Dec 4 '18 at 9:16
andres
1757
1757
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
Let’s consider two formal groups over $Bbb Z_{(p)}$ [this is the rationals with no $p$ in the denominator], I’ll call them $mathcal M$ and $F$. They are both of height one, say when reduced modulo $p$. The easy one is $mathcal M$, the formal group of multiplication, ${mathcal M}(x,y)=x+y+xy=(1+x)(1+y)-1$. It has a logarithm ($Bbb Q$-formal-group isomorphism with the additive formal group ${mathcal A}(x,y)=x+y$), namely $x-x^2/2+x^3/3-x^4/4+cdots$, which you know all about; in particular, you know that its inverse is $exp(x)-1$.
The second formal group $F$ is, as I said above, also of height one, and is best described by means of its logarithm:
$$
log_F(x)=x+frac{x^p}p+frac{x^{p^2}}{p^2}+cdots=sum_{n=0}^inftyfrac{x^{p^n}}{p^n},.
$$
Now, these two formal groups are alike in one other important respect: both have the property that their $[p]$-endomorphism is $[p](x)equiv x^ppmod p$. When this happens, the formal groups are $Bbb Z_p$-isomorphic, and indeed $Bbb Z_{(p)}$-isomorphic, since both are defined over that ring.
So you see that the Artin-Hasse is just the unique $Bbb Z_{(p)}$-isomorphism $psi:Ftomathcal M$ such that $psi'(0)=1$. Note that since it’s a $Bbb Z_p$-power series, it’s automatically convergent for all $z$ with $v_p(z)>0$. To describe it a bit more explicitly, I suppose you could appeal to the “exponential series” of $F$, which is just $log_F^{-1}inBbb Q[[x]]$, same domain of convergence as the ordinary exponential, and write your desired inverse as $log_F^{-1}circlog_{mathcal M}>$ …
answered Dec 16 '18 at 0:59
Lubin
43.8k44585
add a comment |
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Let’s consider two formal groups over $Bbb Z_{(p)}$ [this is the rationals with no $p$ in the denominator], I’ll call them $mathcal M$ and $F$. They are both of height one, say when reduced modulo $p$. The easy one is $mathcal M$, the formal group of multiplication, ${mathcal M}(x,y)=x+y+xy=(1+x)(1+y)-1$. It has a logarithm ($Bbb Q$-formal-group isomorphism with the additive formal group ${mathcal A}(x,y)=x+y$), namely $x-x^2/2+x^3/3-x^4/4+cdots$, which you know all about; in particular, you know that its inverse is $exp(x)-1$.
The second formal group $F$ is, as I said above, also of height one, and is best described by means of its logarithm:
$$
log_F(x)=x+frac{x^p}p+frac{x^{p^2}}{p^2}+cdots=sum_{n=0}^inftyfrac{x^{p^n}}{p^n},.
$$
Now, these two formal groups are alike in one other important respect: both have the property that their $[p]$-endomorphism is $[p](x)equiv x^ppmod p$. When this happens, the formal groups are $Bbb Z_p$-isomorphic, and indeed $Bbb Z_{(p)}$-isomorphic, since both are defined over that ring.
So you see that the Artin-Hasse is just the unique $Bbb Z_{(p)}$-isomorphism $psi:Ftomathcal M$ such that $psi'(0)=1$. Note that since it’s a $Bbb Z_p$-power series, it’s automatically convergent for all $z$ with $v_p(z)>0$. To describe it a bit more explicitly, I suppose you could appeal to the “exponential series” of $F$, which is just $log_F^{-1}inBbb Q[[x]]$, same domain of convergence as the ordinary exponential, and write your desired inverse as $log_F^{-1}circlog_{mathcal M}>$ …
answered Dec 16 '18 at 0:59
Lubin
43.8k44585
add a comment |
Let’s consider two formal groups over $Bbb Z_{(p)}$ [this is the rationals with no $p$ in the denominator], I’ll call them $mathcal M$ and $F$. They are both of height one, say when reduced modulo $p$. The easy one is $mathcal M$, the formal group of multiplication, ${mathcal M}(x,y)=x+y+xy=(1+x)(1+y)-1$. It has a logarithm ($Bbb Q$-formal-group isomorphism with the additive formal group ${mathcal A}(x,y)=x+y$), namely $x-x^2/2+x^3/3-x^4/4+cdots$, which you know all about; in particular, you know that its inverse is $exp(x)-1$.
The second formal group $F$ is, as I said above, also of height one, and is best described by means of its logarithm:
$$
log_F(x)=x+frac{x^p}p+frac{x^{p^2}}{p^2}+cdots=sum_{n=0}^inftyfrac{x^{p^n}}{p^n},.
$$
Now, these two formal groups are alike in one other important respect: both have the property that their $[p]$-endomorphism is $[p](x)equiv x^ppmod p$. When this happens, the formal groups are $Bbb Z_p$-isomorphic, and indeed $Bbb Z_{(p)}$-isomorphic, since both are defined over that ring.
So you see that the Artin-Hasse is just the unique $Bbb Z_{(p)}$-isomorphism $psi:Ftomathcal M$ such that $psi'(0)=1$. Note that since it’s a $Bbb Z_p$-power series, it’s automatically convergent for all $z$ with $v_p(z)>0$. To describe it a bit more explicitly, I suppose you could appeal to the “exponential series” of $F$, which is just $log_F^{-1}inBbb Q[[x]]$, same domain of convergence as the ordinary exponential, and write your desired inverse as $log_F^{-1}circlog_{mathcal M}>$ …
answered Dec 16 '18 at 0:59
Lubin
43.8k44585
add a comment |
Let’s consider two formal groups over $Bbb Z_{(p)}$ [this is the rationals with no $p$ in the denominator], I’ll call them $mathcal M$ and $F$. They are both of height one, say when reduced modulo $p$. The easy one is $mathcal M$, the formal group of multiplication, ${mathcal M}(x,y)=x+y+xy=(1+x)(1+y)-1$. It has a logarithm ($Bbb Q$-formal-group isomorphism with the additive formal group ${mathcal A}(x,y)=x+y$), namely $x-x^2/2+x^3/3-x^4/4+cdots$, which you know all about; in particular, you know that its inverse is $exp(x)-1$.
The second formal group $F$ is, as I said above, also of height one, and is best described by means of its logarithm:
$$
log_F(x)=x+frac{x^p}p+frac{x^{p^2}}{p^2}+cdots=sum_{n=0}^inftyfrac{x^{p^n}}{p^n},.
$$
Now, these two formal groups are alike in one other important respect: both have the property that their $[p]$-endomorphism is $[p](x)equiv x^ppmod p$. When this happens, the formal groups are $Bbb Z_p$-isomorphic, and indeed $Bbb Z_{(p)}$-isomorphic, since both are defined over that ring.
So you see that the Artin-Hasse is just the unique $Bbb Z_{(p)}$-isomorphism $psi:Ftomathcal M$ such that $psi'(0)=1$. Note that since it’s a $Bbb Z_p$-power series, it’s automatically convergent for all $z$ with $v_p(z)>0$. To describe it a bit more explicitly, I suppose you could appeal to the “exponential series” of $F$, which is just $log_F^{-1}inBbb Q[[x]]$, same domain of convergence as the ordinary exponential, and write your desired inverse as $log_F^{-1}circlog_{mathcal M}>$ …
answered Dec 16 '18 at 0:59
Lubin
43.8k44585
Let’s consider two formal groups over $Bbb Z_{(p)}$ [this is the rationals with no $p$ in the denominator], I’ll call them $mathcal M$ and $F$. They are both of height one, say when reduced modulo $p$. The easy one is $mathcal M$, the formal group of multiplication, ${mathcal M}(x,y)=x+y+xy=(1+x)(1+y)-1$. It has a logarithm ($Bbb Q$-formal-group isomorphism with the additive formal group ${mathcal A}(x,y)=x+y$), namely $x-x^2/2+x^3/3-x^4/4+cdots$, which you know all about; in particular, you know that its inverse is $exp(x)-1$.
The second formal group $F$ is, as I said above, also of height one, and is best described by means of its logarithm:
$$
log_F(x)=x+frac{x^p}p+frac{x^{p^2}}{p^2}+cdots=sum_{n=0}^inftyfrac{x^{p^n}}{p^n},.
$$
Now, these two formal groups are alike in one other important respect: both have the property that their $[p]$-endomorphism is $[p](x)equiv x^ppmod p$. When this happens, the formal groups are $Bbb Z_p$-isomorphic, and indeed $Bbb Z_{(p)}$-isomorphic, since both are defined over that ring.
So you see that the Artin-Hasse is just the unique $Bbb Z_{(p)}$-isomorphism $psi:Ftomathcal M$ such that $psi'(0)=1$. Note that since it’s a $Bbb Z_p$-power series, it’s automatically convergent for all $z$ with $v_p(z)>0$. To describe it a bit more explicitly, I suppose you could appeal to the “exponential series” of $F$, which is just $log_F^{-1}inBbb Q[[x]]$, same domain of convergence as the ordinary exponential, and write your desired inverse as $log_F^{-1}circlog_{mathcal M}>$ …
answered Dec 16 '18 at 0:59
Lubin
43.8k44585
edited Dec 16 '18 at 1:07
answered Dec 16 '18 at 0:59
Lubin
43.8k44585
answered Dec 16 '18 at 0:59
Lubin
43.8k44585
answered Dec 16 '18 at 0:59
Lubin
43.8k44585
43.8k44585
add a comment |
add a comment |
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