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Example of complex projective variety without a real descent
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I encountered the problem of trying to find the examples of complex varieties which are not the complexification of a real variety, when I considered the Exercise II.4.7.(a) in Hartshorne's book. I wonder whether there are indeed some interesting complex varieties which don't have the so called semi-linear involution in Hartshorne's book.
For rational affine curve, I have found the following example $$A=mathbbC[x,frac1x-1,frac1x-i,frac1x-i-1]$$
by a little hard calculation, considering the possible semi-linear involutions of the field $mathbbCleft(xright)$ induced by the complex conjugation of $mathbbC$ .
But I really wonder
whether there are some interesting examples of complex projective varieties without real descent.
I have poor knowledge on the classical topics in algebraic geometry, therefore I find it difficult to work out the problem by myself. Could anyone shed some light on my confusion? Thanks a lot!
algebraic-geometry
asked Jul 25 at 10:20
Tibeku
663
add a comment |Â
up vote
2
down vote
favorite
I encountered the problem of trying to find the examples of complex varieties which are not the complexification of a real variety, when I considered the Exercise II.4.7.(a) in Hartshorne's book. I wonder whether there are indeed some interesting complex varieties which don't have the so called semi-linear involution in Hartshorne's book.
For rational affine curve, I have found the following example $$A=mathbbC[x,frac1x-1,frac1x-i,frac1x-i-1]$$
by a little hard calculation, considering the possible semi-linear involutions of the field $mathbbCleft(xright)$ induced by the complex conjugation of $mathbbC$ .
But I really wonder
whether there are some interesting examples of complex projective varieties without real descent.
I have poor knowledge on the classical topics in algebraic geometry, therefore I find it difficult to work out the problem by myself. Could anyone shed some light on my confusion? Thanks a lot!
algebraic-geometry
asked Jul 25 at 10:20
Tibeku
663
2
If $X$ is a complex algebraic variety with a model over $mathbbR$, then $X$ is isomorphic over $mathbbC$ to its conjugate $X^sigma$. (Here, $sigma:Spec mathbbCto Spec mathbbC$ is complex conjugation, and $X^sigma = Xtimes_mathbbC,sigmamathbbC$.) So, "interesting" examples of complex algebraic varieties which can't be defined over $mathbbR$ can be constructed by simply writing down objects which aren't isomorphic to their conjugate. For instance, let $E$ be an elliptic curve with $j$-invariant $j(E)$...cont'd
– Ariyan Javanpeykar
Jul 25 at 22:23
2
...Suppose that $overlinej(E)neq j(E)$. Now, since $j(E^sigma) = overlinej(E)$, it follows that $E$ and $E^sigma$ are not isomorphic. For an explicit example: take $E$ to be an elliptic curve with $j$-invariant equal to $sqrt-1$. This elliptic curve can't be defined over $mathbbR$. You can find higher genus curves by using instead of elliptic curves higher genus hyperelliptic curves.
– Ariyan Javanpeykar
Jul 25 at 22:23
This is indeed a nice illustration for my confusion. I think I will understand the problem better after learning some about elliptic curves. Thanks lot!
– Tibeku
Jul 26 at 2:07
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I encountered the problem of trying to find the examples of complex varieties which are not the complexification of a real variety, when I considered the Exercise II.4.7.(a) in Hartshorne's book. I wonder whether there are indeed some interesting complex varieties which don't have the so called semi-linear involution in Hartshorne's book.
For rational affine curve, I have found the following example $$A=mathbbC[x,frac1x-1,frac1x-i,frac1x-i-1]$$
by a little hard calculation, considering the possible semi-linear involutions of the field $mathbbCleft(xright)$ induced by the complex conjugation of $mathbbC$ .
But I really wonder
whether there are some interesting examples of complex projective varieties without real descent.
I have poor knowledge on the classical topics in algebraic geometry, therefore I find it difficult to work out the problem by myself. Could anyone shed some light on my confusion? Thanks a lot!
algebraic-geometry
asked Jul 25 at 10:20
Tibeku
663
I encountered the problem of trying to find the examples of complex varieties which are not the complexification of a real variety, when I considered the Exercise II.4.7.(a) in Hartshorne's book. I wonder whether there are indeed some interesting complex varieties which don't have the so called semi-linear involution in Hartshorne's book.
For rational affine curve, I have found the following example $$A=mathbbC[x,frac1x-1,frac1x-i,frac1x-i-1]$$
by a little hard calculation, considering the possible semi-linear involutions of the field $mathbbCleft(xright)$ induced by the complex conjugation of $mathbbC$ .
But I really wonder
whether there are some interesting examples of complex projective varieties without real descent.
I have poor knowledge on the classical topics in algebraic geometry, therefore I find it difficult to work out the problem by myself. Could anyone shed some light on my confusion? Thanks a lot!
algebraic-geometry
asked Jul 25 at 10:20
Tibeku
663
asked Jul 25 at 10:20
Tibeku
663
asked Jul 25 at 10:20
Tibeku
663
asked Jul 25 at 10:20
Tibeku
663
663
2
If $X$ is a complex algebraic variety with a model over $mathbbR$, then $X$ is isomorphic over $mathbbC$ to its conjugate $X^sigma$. (Here, $sigma:Spec mathbbCto Spec mathbbC$ is complex conjugation, and $X^sigma = Xtimes_mathbbC,sigmamathbbC$.) So, "interesting" examples of complex algebraic varieties which can't be defined over $mathbbR$ can be constructed by simply writing down objects which aren't isomorphic to their conjugate. For instance, let $E$ be an elliptic curve with $j$-invariant $j(E)$...cont'd
– Ariyan Javanpeykar
Jul 25 at 22:23
2
...Suppose that $overlinej(E)neq j(E)$. Now, since $j(E^sigma) = overlinej(E)$, it follows that $E$ and $E^sigma$ are not isomorphic. For an explicit example: take $E$ to be an elliptic curve with $j$-invariant equal to $sqrt-1$. This elliptic curve can't be defined over $mathbbR$. You can find higher genus curves by using instead of elliptic curves higher genus hyperelliptic curves.
– Ariyan Javanpeykar
Jul 25 at 22:23
This is indeed a nice illustration for my confusion. I think I will understand the problem better after learning some about elliptic curves. Thanks lot!
– Tibeku
Jul 26 at 2:07
add a comment |Â
2
If $X$ is a complex algebraic variety with a model over $mathbbR$, then $X$ is isomorphic over $mathbbC$ to its conjugate $X^sigma$. (Here, $sigma:Spec mathbbCto Spec mathbbC$ is complex conjugation, and $X^sigma = Xtimes_mathbbC,sigmamathbbC$.) So, "interesting" examples of complex algebraic varieties which can't be defined over $mathbbR$ can be constructed by simply writing down objects which aren't isomorphic to their conjugate. For instance, let $E$ be an elliptic curve with $j$-invariant $j(E)$...cont'd
– Ariyan Javanpeykar
Jul 25 at 22:23
2
...Suppose that $overlinej(E)neq j(E)$. Now, since $j(E^sigma) = overlinej(E)$, it follows that $E$ and $E^sigma$ are not isomorphic. For an explicit example: take $E$ to be an elliptic curve with $j$-invariant equal to $sqrt-1$. This elliptic curve can't be defined over $mathbbR$. You can find higher genus curves by using instead of elliptic curves higher genus hyperelliptic curves.
– Ariyan Javanpeykar
Jul 25 at 22:23
This is indeed a nice illustration for my confusion. I think I will understand the problem better after learning some about elliptic curves. Thanks lot!
– Tibeku
Jul 26 at 2:07
2
2
If $X$ is a complex algebraic variety with a model over $mathbbR$, then $X$ is isomorphic over $mathbbC$ to its conjugate $X^sigma$. (Here, $sigma:Spec mathbbCto Spec mathbbC$ is complex conjugation, and $X^sigma = Xtimes_mathbbC,sigmamathbbC$.) So, "interesting" examples of complex algebraic varieties which can't be defined over $mathbbR$ can be constructed by simply writing down objects which aren't isomorphic to their conjugate. For instance, let $E$ be an elliptic curve with $j$-invariant $j(E)$...cont'd
– Ariyan Javanpeykar
Jul 25 at 22:23
If $X$ is a complex algebraic variety with a model over $mathbbR$, then $X$ is isomorphic over $mathbbC$ to its conjugate $X^sigma$. (Here, $sigma:Spec mathbbCto Spec mathbbC$ is complex conjugation, and $X^sigma = Xtimes_mathbbC,sigmamathbbC$.) So, "interesting" examples of complex algebraic varieties which can't be defined over $mathbbR$ can be constructed by simply writing down objects which aren't isomorphic to their conjugate. For instance, let $E$ be an elliptic curve with $j$-invariant $j(E)$...cont'd
– Ariyan Javanpeykar
Jul 25 at 22:23
2
2
...Suppose that $overlinej(E)neq j(E)$. Now, since $j(E^sigma) = overlinej(E)$, it follows that $E$ and $E^sigma$ are not isomorphic. For an explicit example: take $E$ to be an elliptic curve with $j$-invariant equal to $sqrt-1$. This elliptic curve can't be defined over $mathbbR$. You can find higher genus curves by using instead of elliptic curves higher genus hyperelliptic curves.
– Ariyan Javanpeykar
Jul 25 at 22:23
...Suppose that $overlinej(E)neq j(E)$. Now, since $j(E^sigma) = overlinej(E)$, it follows that $E$ and $E^sigma$ are not isomorphic. For an explicit example: take $E$ to be an elliptic curve with $j$-invariant equal to $sqrt-1$. This elliptic curve can't be defined over $mathbbR$. You can find higher genus curves by using instead of elliptic curves higher genus hyperelliptic curves.
– Ariyan Javanpeykar
Jul 25 at 22:23
This is indeed a nice illustration for my confusion. I think I will understand the problem better after learning some about elliptic curves. Thanks lot!
– Tibeku
Jul 26 at 2:07
This is indeed a nice illustration for my confusion. I think I will understand the problem better after learning some about elliptic curves. Thanks lot!
– Tibeku
Jul 26 at 2:07
add a comment |Â
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2
If $X$ is a complex algebraic variety with a model over $mathbbR$, then $X$ is isomorphic over $mathbbC$ to its conjugate $X^sigma$. (Here, $sigma:Spec mathbbCto Spec mathbbC$ is complex conjugation, and $X^sigma = Xtimes_mathbbC,sigmamathbbC$.) So, "interesting" examples of complex algebraic varieties which can't be defined over $mathbbR$ can be constructed by simply writing down objects which aren't isomorphic to their conjugate. For instance, let $E$ be an elliptic curve with $j$-invariant $j(E)$...cont'd
– Ariyan Javanpeykar
Jul 25 at 22:23
2
...Suppose that $overlinej(E)neq j(E)$. Now, since $j(E^sigma) = overlinej(E)$, it follows that $E$ and $E^sigma$ are not isomorphic. For an explicit example: take $E$ to be an elliptic curve with $j$-invariant equal to $sqrt-1$. This elliptic curve can't be defined over $mathbbR$. You can find higher genus curves by using instead of elliptic curves higher genus hyperelliptic curves.
– Ariyan Javanpeykar
Jul 25 at 22:23
This is indeed a nice illustration for my confusion. I think I will understand the problem better after learning some about elliptic curves. Thanks lot!
– Tibeku
Jul 26 at 2:07