Why the $2 pi i$ in the Tate twist?
I am reading here about Hodge structures and in (1.4) the Tate Hodge-structure $mathbb{Z}(n)$ and the Tate twist $V(n)$ on a Hodge structure $V$ are defined. I understand that one might want to consider only morphisms between Hodge structures preserving the weight and therefore the twist allows one to define morphisms between structures of different weights by changing the weight formally. But why do we need the $(2 pi i)^n$? Can't we just take $mathbb{Z}(n):=mathbb{Z}$ for all $n$ as $mathbb{Z}-$modules and simply define the Hodge structures as $H^{-n,-n}(mathbb{Z}(n)):=mathbb{C}$ (and $0$ otherwise)? Do we use the $(2 pi i)^n$ simply as an indicator to remind ourselves of what $n$ is, or does it play any deeper role in the theory?
algebraic-geometry hodge-theory
asked Nov 29 '18 at 10:44
57Jimmy57Jimmy
3,340422
add a comment |
I am reading here about Hodge structures and in (1.4) the Tate Hodge-structure $mathbb{Z}(n)$ and the Tate twist $V(n)$ on a Hodge structure $V$ are defined. I understand that one might want to consider only morphisms between Hodge structures preserving the weight and therefore the twist allows one to define morphisms between structures of different weights by changing the weight formally. But why do we need the $(2 pi i)^n$? Can't we just take $mathbb{Z}(n):=mathbb{Z}$ for all $n$ as $mathbb{Z}-$modules and simply define the Hodge structures as $H^{-n,-n}(mathbb{Z}(n)):=mathbb{C}$ (and $0$ otherwise)? Do we use the $(2 pi i)^n$ simply as an indicator to remind ourselves of what $n$ is, or does it play any deeper role in the theory?
algebraic-geometry hodge-theory
asked Nov 29 '18 at 10:44
57Jimmy57Jimmy
3,340422
4
The Tate twist sounds like a dance from the 1950s. And with $2pi i$ in there, it definitely includes an imaginary circular motion.
– Asaf Karagila♦
Nov 29 '18 at 10:46
1
I think it's for people who study periods, so the integration map get rid of the annoying constant (but I might be wrong)
– Nicolas Hemelsoet
Nov 29 '18 at 11:45
1
Tate twists in singular cohomology are a device for dealing with factors of $2 pi i$ which come up whenever one compares singular and de Rham cohomology of complex projective varieties. In the case of singular cohomology, Tate twists are largely a matter of normalising things conveniently. Without them, one could just write out factors of $2 pi i$ everywhere. On the other hand, there is also a notion of Tate twists for $ell$-adic cohomology, which cannot be omitted so easily,
– Kevin
Nov 29 '18 at 12:23
add a comment |
I am reading here about Hodge structures and in (1.4) the Tate Hodge-structure $mathbb{Z}(n)$ and the Tate twist $V(n)$ on a Hodge structure $V$ are defined. I understand that one might want to consider only morphisms between Hodge structures preserving the weight and therefore the twist allows one to define morphisms between structures of different weights by changing the weight formally. But why do we need the $(2 pi i)^n$? Can't we just take $mathbb{Z}(n):=mathbb{Z}$ for all $n$ as $mathbb{Z}-$modules and simply define the Hodge structures as $H^{-n,-n}(mathbb{Z}(n)):=mathbb{C}$ (and $0$ otherwise)? Do we use the $(2 pi i)^n$ simply as an indicator to remind ourselves of what $n$ is, or does it play any deeper role in the theory?
algebraic-geometry hodge-theory
asked Nov 29 '18 at 10:44
57Jimmy57Jimmy
3,340422
I am reading here about Hodge structures and in (1.4) the Tate Hodge-structure $mathbb{Z}(n)$ and the Tate twist $V(n)$ on a Hodge structure $V$ are defined. I understand that one might want to consider only morphisms between Hodge structures preserving the weight and therefore the twist allows one to define morphisms between structures of different weights by changing the weight formally. But why do we need the $(2 pi i)^n$? Can't we just take $mathbb{Z}(n):=mathbb{Z}$ for all $n$ as $mathbb{Z}-$modules and simply define the Hodge structures as $H^{-n,-n}(mathbb{Z}(n)):=mathbb{C}$ (and $0$ otherwise)? Do we use the $(2 pi i)^n$ simply as an indicator to remind ourselves of what $n$ is, or does it play any deeper role in the theory?
algebraic-geometry hodge-theory
algebraic-geometry hodge-theory
asked Nov 29 '18 at 10:44
57Jimmy57Jimmy
3,340422
asked Nov 29 '18 at 10:44
57Jimmy57Jimmy
3,340422
asked Nov 29 '18 at 10:44
57Jimmy57Jimmy
3,340422
asked Nov 29 '18 at 10:44
57Jimmy57Jimmy
3,340422
asked Nov 29 '18 at 10:44
57Jimmy57Jimmy
3,340422
3,340422
4
The Tate twist sounds like a dance from the 1950s. And with $2pi i$ in there, it definitely includes an imaginary circular motion.
– Asaf Karagila♦
Nov 29 '18 at 10:46
1
I think it's for people who study periods, so the integration map get rid of the annoying constant (but I might be wrong)
– Nicolas Hemelsoet
Nov 29 '18 at 11:45
1
Tate twists in singular cohomology are a device for dealing with factors of $2 pi i$ which come up whenever one compares singular and de Rham cohomology of complex projective varieties. In the case of singular cohomology, Tate twists are largely a matter of normalising things conveniently. Without them, one could just write out factors of $2 pi i$ everywhere. On the other hand, there is also a notion of Tate twists for $ell$-adic cohomology, which cannot be omitted so easily,
– Kevin
Nov 29 '18 at 12:23
add a comment |
4
The Tate twist sounds like a dance from the 1950s. And with $2pi i$ in there, it definitely includes an imaginary circular motion.
– Asaf Karagila♦
Nov 29 '18 at 10:46
1
I think it's for people who study periods, so the integration map get rid of the annoying constant (but I might be wrong)
– Nicolas Hemelsoet
Nov 29 '18 at 11:45
1
Tate twists in singular cohomology are a device for dealing with factors of $2 pi i$ which come up whenever one compares singular and de Rham cohomology of complex projective varieties. In the case of singular cohomology, Tate twists are largely a matter of normalising things conveniently. Without them, one could just write out factors of $2 pi i$ everywhere. On the other hand, there is also a notion of Tate twists for $ell$-adic cohomology, which cannot be omitted so easily,
– Kevin
Nov 29 '18 at 12:23
4
4
The Tate twist sounds like a dance from the 1950s. And with $2pi i$ in there, it definitely includes an imaginary circular motion.
– Asaf Karagila♦
Nov 29 '18 at 10:46
The Tate twist sounds like a dance from the 1950s. And with $2pi i$ in there, it definitely includes an imaginary circular motion.
– Asaf Karagila♦
Nov 29 '18 at 10:46
1
1
I think it's for people who study periods, so the integration map get rid of the annoying constant (but I might be wrong)
– Nicolas Hemelsoet
Nov 29 '18 at 11:45
I think it's for people who study periods, so the integration map get rid of the annoying constant (but I might be wrong)
– Nicolas Hemelsoet
Nov 29 '18 at 11:45
1
1
Tate twists in singular cohomology are a device for dealing with factors of $2 pi i$ which come up whenever one compares singular and de Rham cohomology of complex projective varieties. In the case of singular cohomology, Tate twists are largely a matter of normalising things conveniently. Without them, one could just write out factors of $2 pi i$ everywhere. On the other hand, there is also a notion of Tate twists for $ell$-adic cohomology, which cannot be omitted so easily,
– Kevin
Nov 29 '18 at 12:23
Tate twists in singular cohomology are a device for dealing with factors of $2 pi i$ which come up whenever one compares singular and de Rham cohomology of complex projective varieties. In the case of singular cohomology, Tate twists are largely a matter of normalising things conveniently. Without them, one could just write out factors of $2 pi i$ everywhere. On the other hand, there is also a notion of Tate twists for $ell$-adic cohomology, which cannot be omitted so easily,
– Kevin
Nov 29 '18 at 12:23
add a comment |
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4
The Tate twist sounds like a dance from the 1950s. And with $2pi i$ in there, it definitely includes an imaginary circular motion.
– Asaf Karagila♦
Nov 29 '18 at 10:46
1
I think it's for people who study periods, so the integration map get rid of the annoying constant (but I might be wrong)
– Nicolas Hemelsoet
Nov 29 '18 at 11:45
1
Tate twists in singular cohomology are a device for dealing with factors of $2 pi i$ which come up whenever one compares singular and de Rham cohomology of complex projective varieties. In the case of singular cohomology, Tate twists are largely a matter of normalising things conveniently. Without them, one could just write out factors of $2 pi i$ everywhere. On the other hand, there is also a notion of Tate twists for $ell$-adic cohomology, which cannot be omitted so easily,
– Kevin
Nov 29 '18 at 12:23