an inverse of the Artin-Hasse exponential?
$begingroup$
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
4,0331032
asked Dec 4 '18 at 9:16
andresandres
1758
$endgroup$
add a comment |
$begingroup$
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
4,0331032
asked Dec 4 '18 at 9:16
andresandres
1758
$endgroup$
add a comment |
$begingroup$
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
4,0331032
asked Dec 4 '18 at 9:16
andresandres
1758
$endgroup$
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
4,0331032
asked Dec 4 '18 at 9:16
andresandres
1758
edited Dec 19 '18 at 2:21
Will Fisher
4,0331032
asked Dec 4 '18 at 9:16
andresandres
1758
edited Dec 19 '18 at 2:21
Will Fisher
4,0331032
edited Dec 19 '18 at 2:21
Will Fisher
4,0331032
edited Dec 19 '18 at 2:21
Will Fisher
4,0331032
4,0331032
asked Dec 4 '18 at 9:16
andresandres
1758
asked Dec 4 '18 at 9:16
andresandres
1758
asked Dec 4 '18 at 9:16
andresandres
1758
1758
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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
LubinLubin
44.1k44585
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3025329%2fan-inverse-of-the-artin-hasse-exponential%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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
LubinLubin
44.1k44585
$endgroup$
add a comment |
$begingroup$
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
LubinLubin
44.1k44585
$endgroup$
add a comment |
$begingroup$
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
LubinLubin
44.1k44585
$endgroup$
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
LubinLubin
44.1k44585
edited Dec 16 '18 at 1:07
answered Dec 16 '18 at 0:59
LubinLubin
44.1k44585
answered Dec 16 '18 at 0:59
LubinLubin
44.1k44585
answered Dec 16 '18 at 0:59
LubinLubin
44.1k44585
44.1k44585
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3025329%2fan-inverse-of-the-artin-hasse-exponential%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
