Topology of algebraic variety Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Variety and algebraic curvesHow to tell if algebraic set is a variety?Rational parametrization of algebraic varietyQing Liu's definition of an algebraic variety, a non-separated lineDegree of an algebraic variety given by a polynomial parametrizationHow does a complex algebraic variety know about its analytic topology?$GL(mathbb C)$ as an Algebraic Groupalgebraic variety of dimension 0When is a complex manifold of the form $mathbbC^g/Lambda$ an algebraic variety?Why is the algebraic torus an affine variety?
How can I make names more distinctive without making them longer?
What is the role of the transistor and diode in a soft start circuit?
How do I keep my slimes from escaping their pens?
When do you get frequent flier miles - when you buy, or when you fly?
When a candle burns, why does the top of wick glow if bottom of flame is hottest?
Dating a Former Employee
Why did the rest of the Eastern Bloc not invade Yugoslavia?
Why are Kinder Surprise Eggs illegal in the USA?
What's the meaning of 間時肆拾貳 at a car parking sign
Resolving to minmaj7
Why are there no cargo aircraft with "flying wing" design?
Why didn't this character "real die" when they blew their stack out in Altered Carbon?
What does "fit" mean in this sentence?
What's the purpose of writing one's academic biography in the third person?
How to tell that you are a giant?
What would be the ideal power source for a cybernetic eye?
How do pianists reach extremely loud dynamics?
Seeking colloquialism for “just because”
Why aren't air breathing engines used as small first stages
Check which numbers satisfy the condition [A*B*C = A! + B! + C!]
Error "illegal generic type for instanceof" when using local classes
How widely used is the term Treppenwitz? Is it something that most Germans know?
Can I cast Passwall to drop an enemy into a 20-foot pit?
Using audio cues to encourage good posture
Topology of algebraic variety
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Variety and algebraic curvesHow to tell if algebraic set is a variety?Rational parametrization of algebraic varietyQing Liu's definition of an algebraic variety, a non-separated lineDegree of an algebraic variety given by a polynomial parametrizationHow does a complex algebraic variety know about its analytic topology?$GL(mathbb C)$ as an Algebraic Groupalgebraic variety of dimension 0When is a complex manifold of the form $mathbbC^g/Lambda$ an algebraic variety?Why is the algebraic torus an affine variety?
$begingroup$
I have an algebraic variety given by a polynomial $1+x(1+y)^2$ over $mathbb C$. Is there any reasonable way to see how topologically the algebraic variety defined by this polynomial looks like?
algebraic-geometry
$endgroup$
migrated from mathoverflow.net Mar 26 at 16:53
This question came from our site for professional mathematicians.
add a comment |
$begingroup$
I have an algebraic variety given by a polynomial $1+x(1+y)^2$ over $mathbb C$. Is there any reasonable way to see how topologically the algebraic variety defined by this polynomial looks like?
algebraic-geometry
$endgroup$
migrated from mathoverflow.net Mar 26 at 16:53
This question came from our site for professional mathematicians.
1
$begingroup$
In your case it's easy to see that your curve is isomorphic to $Bbb C^*$
$endgroup$
– Nicolas Hemelsoet
Mar 27 at 10:27
1
$begingroup$
Your curve is $ (frac-1z^2, z-1), z in BbbC^*$ which is isomorphic to $BbbC^*$. To see it, look first at $ (frac1z, z), z in BbbC^*$ then apply $(u,v) to (-u^2,v-1)$
$endgroup$
– reuns
Mar 27 at 11:41
add a comment |
$begingroup$
I have an algebraic variety given by a polynomial $1+x(1+y)^2$ over $mathbb C$. Is there any reasonable way to see how topologically the algebraic variety defined by this polynomial looks like?
algebraic-geometry
$endgroup$
I have an algebraic variety given by a polynomial $1+x(1+y)^2$ over $mathbb C$. Is there any reasonable way to see how topologically the algebraic variety defined by this polynomial looks like?
algebraic-geometry
algebraic-geometry
asked Mar 26 at 15:22
Adam Klah
migrated from mathoverflow.net Mar 26 at 16:53
This question came from our site for professional mathematicians.
migrated from mathoverflow.net Mar 26 at 16:53
This question came from our site for professional mathematicians.
1
$begingroup$
In your case it's easy to see that your curve is isomorphic to $Bbb C^*$
$endgroup$
– Nicolas Hemelsoet
Mar 27 at 10:27
1
$begingroup$
Your curve is $ (frac-1z^2, z-1), z in BbbC^*$ which is isomorphic to $BbbC^*$. To see it, look first at $ (frac1z, z), z in BbbC^*$ then apply $(u,v) to (-u^2,v-1)$
$endgroup$
– reuns
Mar 27 at 11:41
add a comment |
1
$begingroup$
In your case it's easy to see that your curve is isomorphic to $Bbb C^*$
$endgroup$
– Nicolas Hemelsoet
Mar 27 at 10:27
1
$begingroup$
Your curve is $ (frac-1z^2, z-1), z in BbbC^*$ which is isomorphic to $BbbC^*$. To see it, look first at $ (frac1z, z), z in BbbC^*$ then apply $(u,v) to (-u^2,v-1)$
$endgroup$
– reuns
Mar 27 at 11:41
1
1
$begingroup$
In your case it's easy to see that your curve is isomorphic to $Bbb C^*$
$endgroup$
– Nicolas Hemelsoet
Mar 27 at 10:27
$begingroup$
In your case it's easy to see that your curve is isomorphic to $Bbb C^*$
$endgroup$
– Nicolas Hemelsoet
Mar 27 at 10:27
1
1
$begingroup$
Your curve is $ (frac-1z^2, z-1), z in BbbC^*$ which is isomorphic to $BbbC^*$. To see it, look first at $ (frac1z, z), z in BbbC^*$ then apply $(u,v) to (-u^2,v-1)$
$endgroup$
– reuns
Mar 27 at 11:41
$begingroup$
Your curve is $ (frac-1z^2, z-1), z in BbbC^*$ which is isomorphic to $BbbC^*$. To see it, look first at $ (frac1z, z), z in BbbC^*$ then apply $(u,v) to (-u^2,v-1)$
$endgroup$
– reuns
Mar 27 at 11:41
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The open sets of that variety are the variety itself take away finite number of (algebraic) curves and points. Is this what you mean by "topologically see"?
answered Mar 27 at 9:44
quantumquantum
538210
$endgroup$
add a comment |
Your Answer
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%2f3163464%2ftopology-of-algebraic-variety%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
$begingroup$
The open sets of that variety are the variety itself take away finite number of (algebraic) curves and points. Is this what you mean by "topologically see"?
answered Mar 27 at 9:44
quantumquantum
538210
$endgroup$
add a comment |
$begingroup$
The open sets of that variety are the variety itself take away finite number of (algebraic) curves and points. Is this what you mean by "topologically see"?
answered Mar 27 at 9:44
quantumquantum
538210
$endgroup$
add a comment |
$begingroup$
The open sets of that variety are the variety itself take away finite number of (algebraic) curves and points. Is this what you mean by "topologically see"?
answered Mar 27 at 9:44
quantumquantum
538210
$endgroup$
The open sets of that variety are the variety itself take away finite number of (algebraic) curves and points. Is this what you mean by "topologically see"?
answered Mar 27 at 9:44
quantumquantum
538210
answered Mar 27 at 9:44
quantumquantum
538210
answered Mar 27 at 9:44
quantumquantum
538210
answered Mar 27 at 9:44
quantumquantum
538210
538210
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%2f3163464%2ftopology-of-algebraic-variety%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
1
$begingroup$
In your case it's easy to see that your curve is isomorphic to $Bbb C^*$
$endgroup$
– Nicolas Hemelsoet
Mar 27 at 10:27
1
$begingroup$
Your curve is $ (frac-1z^2, z-1), z in BbbC^*$ which is isomorphic to $BbbC^*$. To see it, look first at $ (frac1z, z), z in BbbC^*$ then apply $(u,v) to (-u^2,v-1)$
$endgroup$
– reuns
Mar 27 at 11:41