The transfer map from $K_2L$ to $K_2F$ for a galois extension $L/F$
3
$begingroup$
In J. Tate's "Relations between $K_2$ and Galois Cohomology" Lemma 3.4 J. Tate talks about the transfer map in k-theory, and references the Milnor's "Introduction to Algebraic K-Theory", but in the proof of the mentioned lemma Tate uses a property of $K_2$-transfer
$$f(text{tr}(y))=sum_{sin G}sy$$ In the milnor's book there's not a proof of this fact, moreover, there's not a mention for the action of $G=text{Gal}(L/F)$. Then, how acts G over $K_2$?, for the Matsumoto's Theorem, $K_2L$ has a presentation by the symbols ${alpha,beta}$ with $alpha,betain L^{bullet}$. How acts $G$ over the symbols?.
Any hint for the proof of the formula above?
Thanks.
algebraic-number-theory algebraic-k-theory
asked Dec 13 '18 at 2:03
Elvis Torres PérezElvis Torres Pérez
532
$endgroup$
|
show 2 more comments
3
$begingroup$
In J. Tate's "Relations between $K_2$ and Galois Cohomology" Lemma 3.4 J. Tate talks about the transfer map in k-theory, and references the Milnor's "Introduction to Algebraic K-Theory", but in the proof of the mentioned lemma Tate uses a property of $K_2$-transfer
$$f(text{tr}(y))=sum_{sin G}sy$$ In the milnor's book there's not a proof of this fact, moreover, there's not a mention for the action of $G=text{Gal}(L/F)$. Then, how acts G over $K_2$?, for the Matsumoto's Theorem, $K_2L$ has a presentation by the symbols ${alpha,beta}$ with $alpha,betain L^{bullet}$. How acts $G$ over the symbols?.
Any hint for the proof of the formula above?
Thanks.
algebraic-number-theory algebraic-k-theory
asked Dec 13 '18 at 2:03
Elvis Torres PérezElvis Torres Pérez
532
$endgroup$
$begingroup$
For the definition and properties of the transfer map for $K_2$ (called also trace), you should consult "A reciprocity law for K2-traces" , John Tate; Shmuel Rosset Commentarii mathematici Helvetici (1983) Volume: 58, page 38-47
$endgroup$
– nguyen quang do
Dec 13 '18 at 17:15
$begingroup$
Thanks for the response, i has read the article, the formula for $f(text{tr}(y))$ above is used but is not proved, and the action is used over the cyclic algebras $sigma(a,b)=(sigma a,sigma b)$, is possible that the action is the same for the symbol ${a,b}$.
$endgroup$
– Elvis Torres Pérez
Dec 13 '18 at 19:13
$begingroup$
So what prevents you to use the definition to derive the desired formula ? After all, the Milnor symbols are nothing but tensor products modulo relations, so the Galois action is deduced from the natural (diagonal) action on $otimes$
$endgroup$
– nguyen quang do
Dec 13 '18 at 21:11
$begingroup$
Certainly, in the definition of transfer first define $f^{#}:E(n,L) to E(F)$ where $E(n,L)$ is the group generated by the elementary $ntimes n$ matrices, and $E(F)$ is the direct limit of $E(n,F)$. Finally uses the schur multiplier $H_2(cdot)$ (second group of homology) on $f^{#}$ and takes direct limit... maybe i can use the homological transfer property, ill try it... thanks.
$endgroup$
– Elvis Torres Pérez
Dec 13 '18 at 23:32
$begingroup$
You could use the (co)homological transfer, but in view of Tate's goal (show the Licht.-Quillen conjecture for $K_2$ of global fields), this would be tautological ! Of course, the L-Q. conjecture is now proved in general thanks to Voevodsky et.al.
$endgroup$
– nguyen quang do
Dec 14 '18 at 7:44
|
show 2 more comments
3
3
3
$begingroup$
In J. Tate's "Relations between $K_2$ and Galois Cohomology" Lemma 3.4 J. Tate talks about the transfer map in k-theory, and references the Milnor's "Introduction to Algebraic K-Theory", but in the proof of the mentioned lemma Tate uses a property of $K_2$-transfer
$$f(text{tr}(y))=sum_{sin G}sy$$ In the milnor's book there's not a proof of this fact, moreover, there's not a mention for the action of $G=text{Gal}(L/F)$. Then, how acts G over $K_2$?, for the Matsumoto's Theorem, $K_2L$ has a presentation by the symbols ${alpha,beta}$ with $alpha,betain L^{bullet}$. How acts $G$ over the symbols?.
Any hint for the proof of the formula above?
Thanks.
algebraic-number-theory algebraic-k-theory
asked Dec 13 '18 at 2:03
Elvis Torres PérezElvis Torres Pérez
532
$endgroup$
In J. Tate's "Relations between $K_2$ and Galois Cohomology" Lemma 3.4 J. Tate talks about the transfer map in k-theory, and references the Milnor's "Introduction to Algebraic K-Theory", but in the proof of the mentioned lemma Tate uses a property of $K_2$-transfer
$$f(text{tr}(y))=sum_{sin G}sy$$ In the milnor's book there's not a proof of this fact, moreover, there's not a mention for the action of $G=text{Gal}(L/F)$. Then, how acts G over $K_2$?, for the Matsumoto's Theorem, $K_2L$ has a presentation by the symbols ${alpha,beta}$ with $alpha,betain L^{bullet}$. How acts $G$ over the symbols?.
Any hint for the proof of the formula above?
Thanks.
algebraic-number-theory algebraic-k-theory
algebraic-number-theory algebraic-k-theory
asked Dec 13 '18 at 2:03
Elvis Torres PérezElvis Torres Pérez
532
asked Dec 13 '18 at 2:03
Elvis Torres PérezElvis Torres Pérez
532
asked Dec 13 '18 at 2:03
Elvis Torres PérezElvis Torres Pérez
532
asked Dec 13 '18 at 2:03
Elvis Torres PérezElvis Torres Pérez
532
asked Dec 13 '18 at 2:03
Elvis Torres PérezElvis Torres Pérez
532
532
$begingroup$
For the definition and properties of the transfer map for $K_2$ (called also trace), you should consult "A reciprocity law for K2-traces" , John Tate; Shmuel Rosset Commentarii mathematici Helvetici (1983) Volume: 58, page 38-47
$endgroup$
– nguyen quang do
Dec 13 '18 at 17:15
$begingroup$
Thanks for the response, i has read the article, the formula for $f(text{tr}(y))$ above is used but is not proved, and the action is used over the cyclic algebras $sigma(a,b)=(sigma a,sigma b)$, is possible that the action is the same for the symbol ${a,b}$.
$endgroup$
– Elvis Torres Pérez
Dec 13 '18 at 19:13
$begingroup$
So what prevents you to use the definition to derive the desired formula ? After all, the Milnor symbols are nothing but tensor products modulo relations, so the Galois action is deduced from the natural (diagonal) action on $otimes$
$endgroup$
– nguyen quang do
Dec 13 '18 at 21:11
$begingroup$
Certainly, in the definition of transfer first define $f^{#}:E(n,L) to E(F)$ where $E(n,L)$ is the group generated by the elementary $ntimes n$ matrices, and $E(F)$ is the direct limit of $E(n,F)$. Finally uses the schur multiplier $H_2(cdot)$ (second group of homology) on $f^{#}$ and takes direct limit... maybe i can use the homological transfer property, ill try it... thanks.
$endgroup$
– Elvis Torres Pérez
Dec 13 '18 at 23:32
$begingroup$
You could use the (co)homological transfer, but in view of Tate's goal (show the Licht.-Quillen conjecture for $K_2$ of global fields), this would be tautological ! Of course, the L-Q. conjecture is now proved in general thanks to Voevodsky et.al.
$endgroup$
– nguyen quang do
Dec 14 '18 at 7:44
|
show 2 more comments
$begingroup$
For the definition and properties of the transfer map for $K_2$ (called also trace), you should consult "A reciprocity law for K2-traces" , John Tate; Shmuel Rosset Commentarii mathematici Helvetici (1983) Volume: 58, page 38-47
$endgroup$
– nguyen quang do
Dec 13 '18 at 17:15
$begingroup$
Thanks for the response, i has read the article, the formula for $f(text{tr}(y))$ above is used but is not proved, and the action is used over the cyclic algebras $sigma(a,b)=(sigma a,sigma b)$, is possible that the action is the same for the symbol ${a,b}$.
$endgroup$
– Elvis Torres Pérez
Dec 13 '18 at 19:13
$begingroup$
So what prevents you to use the definition to derive the desired formula ? After all, the Milnor symbols are nothing but tensor products modulo relations, so the Galois action is deduced from the natural (diagonal) action on $otimes$
$endgroup$
– nguyen quang do
Dec 13 '18 at 21:11
$begingroup$
Certainly, in the definition of transfer first define $f^{#}:E(n,L) to E(F)$ where $E(n,L)$ is the group generated by the elementary $ntimes n$ matrices, and $E(F)$ is the direct limit of $E(n,F)$. Finally uses the schur multiplier $H_2(cdot)$ (second group of homology) on $f^{#}$ and takes direct limit... maybe i can use the homological transfer property, ill try it... thanks.
$endgroup$
– Elvis Torres Pérez
Dec 13 '18 at 23:32
$begingroup$
You could use the (co)homological transfer, but in view of Tate's goal (show the Licht.-Quillen conjecture for $K_2$ of global fields), this would be tautological ! Of course, the L-Q. conjecture is now proved in general thanks to Voevodsky et.al.
$endgroup$
– nguyen quang do
Dec 14 '18 at 7:44
$begingroup$
For the definition and properties of the transfer map for $K_2$ (called also trace), you should consult "A reciprocity law for K2-traces" , John Tate; Shmuel Rosset Commentarii mathematici Helvetici (1983) Volume: 58, page 38-47
$endgroup$
– nguyen quang do
Dec 13 '18 at 17:15
$begingroup$
For the definition and properties of the transfer map for $K_2$ (called also trace), you should consult "A reciprocity law for K2-traces" , John Tate; Shmuel Rosset Commentarii mathematici Helvetici (1983) Volume: 58, page 38-47
$endgroup$
– nguyen quang do
Dec 13 '18 at 17:15
$begingroup$
Thanks for the response, i has read the article, the formula for $f(text{tr}(y))$ above is used but is not proved, and the action is used over the cyclic algebras $sigma(a,b)=(sigma a,sigma b)$, is possible that the action is the same for the symbol ${a,b}$.
$endgroup$
– Elvis Torres Pérez
Dec 13 '18 at 19:13
$begingroup$
Thanks for the response, i has read the article, the formula for $f(text{tr}(y))$ above is used but is not proved, and the action is used over the cyclic algebras $sigma(a,b)=(sigma a,sigma b)$, is possible that the action is the same for the symbol ${a,b}$.
$endgroup$
– Elvis Torres Pérez
Dec 13 '18 at 19:13
$begingroup$
So what prevents you to use the definition to derive the desired formula ? After all, the Milnor symbols are nothing but tensor products modulo relations, so the Galois action is deduced from the natural (diagonal) action on $otimes$
$endgroup$
– nguyen quang do
Dec 13 '18 at 21:11
$begingroup$
So what prevents you to use the definition to derive the desired formula ? After all, the Milnor symbols are nothing but tensor products modulo relations, so the Galois action is deduced from the natural (diagonal) action on $otimes$
$endgroup$
– nguyen quang do
Dec 13 '18 at 21:11
$begingroup$
Certainly, in the definition of transfer first define $f^{#}:E(n,L) to E(F)$ where $E(n,L)$ is the group generated by the elementary $ntimes n$ matrices, and $E(F)$ is the direct limit of $E(n,F)$. Finally uses the schur multiplier $H_2(cdot)$ (second group of homology) on $f^{#}$ and takes direct limit... maybe i can use the homological transfer property, ill try it... thanks.
$endgroup$
– Elvis Torres Pérez
Dec 13 '18 at 23:32
$begingroup$
Certainly, in the definition of transfer first define $f^{#}:E(n,L) to E(F)$ where $E(n,L)$ is the group generated by the elementary $ntimes n$ matrices, and $E(F)$ is the direct limit of $E(n,F)$. Finally uses the schur multiplier $H_2(cdot)$ (second group of homology) on $f^{#}$ and takes direct limit... maybe i can use the homological transfer property, ill try it... thanks.
$endgroup$
– Elvis Torres Pérez
Dec 13 '18 at 23:32
$begingroup$
You could use the (co)homological transfer, but in view of Tate's goal (show the Licht.-Quillen conjecture for $K_2$ of global fields), this would be tautological ! Of course, the L-Q. conjecture is now proved in general thanks to Voevodsky et.al.
$endgroup$
– nguyen quang do
Dec 14 '18 at 7:44
$begingroup$
You could use the (co)homological transfer, but in view of Tate's goal (show the Licht.-Quillen conjecture for $K_2$ of global fields), this would be tautological ! Of course, the L-Q. conjecture is now proved in general thanks to Voevodsky et.al.
$endgroup$
– nguyen quang do
Dec 14 '18 at 7:44
|
show 2 more comments
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$begingroup$
For the definition and properties of the transfer map for $K_2$ (called also trace), you should consult "A reciprocity law for K2-traces" , John Tate; Shmuel Rosset Commentarii mathematici Helvetici (1983) Volume: 58, page 38-47
$endgroup$
– nguyen quang do
Dec 13 '18 at 17:15
$begingroup$
Thanks for the response, i has read the article, the formula for $f(text{tr}(y))$ above is used but is not proved, and the action is used over the cyclic algebras $sigma(a,b)=(sigma a,sigma b)$, is possible that the action is the same for the symbol ${a,b}$.
$endgroup$
– Elvis Torres Pérez
Dec 13 '18 at 19:13
$begingroup$
So what prevents you to use the definition to derive the desired formula ? After all, the Milnor symbols are nothing but tensor products modulo relations, so the Galois action is deduced from the natural (diagonal) action on $otimes$
$endgroup$
– nguyen quang do
Dec 13 '18 at 21:11
$begingroup$
Certainly, in the definition of transfer first define $f^{#}:E(n,L) to E(F)$ where $E(n,L)$ is the group generated by the elementary $ntimes n$ matrices, and $E(F)$ is the direct limit of $E(n,F)$. Finally uses the schur multiplier $H_2(cdot)$ (second group of homology) on $f^{#}$ and takes direct limit... maybe i can use the homological transfer property, ill try it... thanks.
$endgroup$
– Elvis Torres Pérez
Dec 13 '18 at 23:32
$begingroup$
You could use the (co)homological transfer, but in view of Tate's goal (show the Licht.-Quillen conjecture for $K_2$ of global fields), this would be tautological ! Of course, the L-Q. conjecture is now proved in general thanks to Voevodsky et.al.
$endgroup$
– nguyen quang do
Dec 14 '18 at 7:44