The Hilbert symbols of quaternion algebras over a totally real field
Clash Royale CLAN TAG#URR8PPP
$begingroup$
Let $k$ be a totally real number field. Every quaternion algebra $B$ over $k$ can be written as
$$B = left(fraca,bkright), $$
for some constants $a,b in k^times$. My question is, can I always choose these constants so that $sigma(a) > 0$ for all real embeddings $sigma colon k hookrightarrow mathbbR$ at which $B$ splits?
A priori, one only has $sigma(a) > 0$ or $sigma(b) > 0$ for each of the embeddings $sigma colon k hookrightarrow mathbbR$ which split $B$. But in many examples one can use the symmetries $bigl(fraca,,bkbigr) = bigl(fracb,,akbigr) = bigl(fraca,,-abkbigr)$ to achieve the above condition.
algebraic-number-theory quaternions quaternion-algebras
edited Feb 19 at 10:04
YCor
28.2k483136
asked Feb 19 at 9:35
AbenthyAbenthy
28016
$endgroup$
add a comment |
$begingroup$
Let $k$ be a totally real number field. Every quaternion algebra $B$ over $k$ can be written as
$$B = left(fraca,bkright), $$
for some constants $a,b in k^times$. My question is, can I always choose these constants so that $sigma(a) > 0$ for all real embeddings $sigma colon k hookrightarrow mathbbR$ at which $B$ splits?
A priori, one only has $sigma(a) > 0$ or $sigma(b) > 0$ for each of the embeddings $sigma colon k hookrightarrow mathbbR$ which split $B$. But in many examples one can use the symmetries $bigl(fraca,,bkbigr) = bigl(fracb,,akbigr) = bigl(fraca,,-abkbigr)$ to achieve the above condition.
algebraic-number-theory quaternions quaternion-algebras
edited Feb 19 at 10:04
YCor
28.2k483136
asked Feb 19 at 9:35
AbenthyAbenthy
28016
$endgroup$
add a comment |
$begingroup$
Let $k$ be a totally real number field. Every quaternion algebra $B$ over $k$ can be written as
$$B = left(fraca,bkright), $$
for some constants $a,b in k^times$. My question is, can I always choose these constants so that $sigma(a) > 0$ for all real embeddings $sigma colon k hookrightarrow mathbbR$ at which $B$ splits?
A priori, one only has $sigma(a) > 0$ or $sigma(b) > 0$ for each of the embeddings $sigma colon k hookrightarrow mathbbR$ which split $B$. But in many examples one can use the symmetries $bigl(fraca,,bkbigr) = bigl(fracb,,akbigr) = bigl(fraca,,-abkbigr)$ to achieve the above condition.
algebraic-number-theory quaternions quaternion-algebras
edited Feb 19 at 10:04
YCor
28.2k483136
asked Feb 19 at 9:35
AbenthyAbenthy
28016
$endgroup$
Let $k$ be a totally real number field. Every quaternion algebra $B$ over $k$ can be written as
$$B = left(fraca,bkright), $$
for some constants $a,b in k^times$. My question is, can I always choose these constants so that $sigma(a) > 0$ for all real embeddings $sigma colon k hookrightarrow mathbbR$ at which $B$ splits?
A priori, one only has $sigma(a) > 0$ or $sigma(b) > 0$ for each of the embeddings $sigma colon k hookrightarrow mathbbR$ which split $B$. But in many examples one can use the symmetries $bigl(fraca,,bkbigr) = bigl(fracb,,akbigr) = bigl(fraca,,-abkbigr)$ to achieve the above condition.
algebraic-number-theory quaternions quaternion-algebras
algebraic-number-theory quaternions quaternion-algebras
edited Feb 19 at 10:04
YCor
28.2k483136
asked Feb 19 at 9:35
AbenthyAbenthy
28016
edited Feb 19 at 10:04
YCor
28.2k483136
asked Feb 19 at 9:35
AbenthyAbenthy
28016
edited Feb 19 at 10:04
YCor
28.2k483136
edited Feb 19 at 10:04
YCor
28.2k483136
edited Feb 19 at 10:04
YCor
28.2k483136
28.2k483136
asked Feb 19 at 9:35
AbenthyAbenthy
28016
asked Feb 19 at 9:35
AbenthyAbenthy
28016
asked Feb 19 at 9:35
AbenthyAbenthy
28016
28016
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Yes.
First choose $a$. You can take any $a$ such that $K = k(sqrta)$ is a splitting field of $B$, so by Grunwald--Wang (or elementary congruences and sign conditions) you can enforce $sigma(a)>0$ at all places where $B$ splits.
Now pick any $b$ such that $(fraca,bk)cong B$. Multiplying $b$ by any norm from $K$ does not change the isomorphism class of resulting algebra. So all you need is to find an element $cin K^times$ whose norm down to $k$ has prescribed signs at the real places that split $B$. At those places, $K/k$ is split, so it reduces the problem to prescribing the signs of $sigma(c)$ for every real embedding $sigma$ of $K$ above the real places of $k$ that split $B$. This amounts to finding an element of $K$ that lies in some open cone in $Kotimes_mathbbQmathbbR$, which is always possible.
edited Feb 19 at 12:14
KConrad
30.4k5132202
answered Feb 19 at 11:09
AurelAurel
2,76121224
$endgroup$
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: "504"
;
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%2fmathoverflow.net%2fquestions%2f323551%2fthe-hilbert-symbols-of-quaternion-algebras-over-a-totally-real-field%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
$begingroup$
Yes.
First choose $a$. You can take any $a$ such that $K = k(sqrta)$ is a splitting field of $B$, so by Grunwald--Wang (or elementary congruences and sign conditions) you can enforce $sigma(a)>0$ at all places where $B$ splits.
Now pick any $b$ such that $(fraca,bk)cong B$. Multiplying $b$ by any norm from $K$ does not change the isomorphism class of resulting algebra. So all you need is to find an element $cin K^times$ whose norm down to $k$ has prescribed signs at the real places that split $B$. At those places, $K/k$ is split, so it reduces the problem to prescribing the signs of $sigma(c)$ for every real embedding $sigma$ of $K$ above the real places of $k$ that split $B$. This amounts to finding an element of $K$ that lies in some open cone in $Kotimes_mathbbQmathbbR$, which is always possible.
edited Feb 19 at 12:14
KConrad
30.4k5132202
answered Feb 19 at 11:09
AurelAurel
2,76121224
$endgroup$
add a comment |
$begingroup$
Yes.
First choose $a$. You can take any $a$ such that $K = k(sqrta)$ is a splitting field of $B$, so by Grunwald--Wang (or elementary congruences and sign conditions) you can enforce $sigma(a)>0$ at all places where $B$ splits.
Now pick any $b$ such that $(fraca,bk)cong B$. Multiplying $b$ by any norm from $K$ does not change the isomorphism class of resulting algebra. So all you need is to find an element $cin K^times$ whose norm down to $k$ has prescribed signs at the real places that split $B$. At those places, $K/k$ is split, so it reduces the problem to prescribing the signs of $sigma(c)$ for every real embedding $sigma$ of $K$ above the real places of $k$ that split $B$. This amounts to finding an element of $K$ that lies in some open cone in $Kotimes_mathbbQmathbbR$, which is always possible.
edited Feb 19 at 12:14
KConrad
30.4k5132202
answered Feb 19 at 11:09
AurelAurel
2,76121224
$endgroup$
add a comment |
$begingroup$
Yes.
First choose $a$. You can take any $a$ such that $K = k(sqrta)$ is a splitting field of $B$, so by Grunwald--Wang (or elementary congruences and sign conditions) you can enforce $sigma(a)>0$ at all places where $B$ splits.
Now pick any $b$ such that $(fraca,bk)cong B$. Multiplying $b$ by any norm from $K$ does not change the isomorphism class of resulting algebra. So all you need is to find an element $cin K^times$ whose norm down to $k$ has prescribed signs at the real places that split $B$. At those places, $K/k$ is split, so it reduces the problem to prescribing the signs of $sigma(c)$ for every real embedding $sigma$ of $K$ above the real places of $k$ that split $B$. This amounts to finding an element of $K$ that lies in some open cone in $Kotimes_mathbbQmathbbR$, which is always possible.
edited Feb 19 at 12:14
KConrad
30.4k5132202
answered Feb 19 at 11:09
AurelAurel
2,76121224
$endgroup$
Yes.
First choose $a$. You can take any $a$ such that $K = k(sqrta)$ is a splitting field of $B$, so by Grunwald--Wang (or elementary congruences and sign conditions) you can enforce $sigma(a)>0$ at all places where $B$ splits.
Now pick any $b$ such that $(fraca,bk)cong B$. Multiplying $b$ by any norm from $K$ does not change the isomorphism class of resulting algebra. So all you need is to find an element $cin K^times$ whose norm down to $k$ has prescribed signs at the real places that split $B$. At those places, $K/k$ is split, so it reduces the problem to prescribing the signs of $sigma(c)$ for every real embedding $sigma$ of $K$ above the real places of $k$ that split $B$. This amounts to finding an element of $K$ that lies in some open cone in $Kotimes_mathbbQmathbbR$, which is always possible.
edited Feb 19 at 12:14
KConrad
30.4k5132202
answered Feb 19 at 11:09
AurelAurel
2,76121224
edited Feb 19 at 12:14
KConrad
30.4k5132202
edited Feb 19 at 12:14
KConrad
30.4k5132202
edited Feb 19 at 12:14
KConrad
30.4k5132202
30.4k5132202
answered Feb 19 at 11:09
AurelAurel
2,76121224
answered Feb 19 at 11:09
AurelAurel
2,76121224
answered Feb 19 at 11:09
AurelAurel
2,76121224
2,76121224
add a comment |
add a comment |
Thanks for contributing an answer to MathOverflow!
- 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%2fmathoverflow.net%2fquestions%2f323551%2fthe-hilbert-symbols-of-quaternion-algebras-over-a-totally-real-field%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
