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Realizing the Berkovich affine line as a union of Berkovich spectrums
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I am trying to understand what is the relation of the affine Berkovich space to the Berkovich space on an appropriate polynomial ring. A more exact version of the question is as follows:
Let $(K,Vert cdot Vert)$ be a complete valued field. The affine Berkovich space is defined as:
$ A_K^n, textan:=
beginBmatrix
& vert f+gvert_x leq vert fvert_x +vert gvert_x \
vert cdot vert _x:K[x_1,...,x_n]rightarrow [0,infty) quad Bigg vert & vert fcdot gvert_x= vert fvert_x cdot vert gvert_x \
& vert avert_x= Vert aVert ; textfor all ; ain K
endBmatrix $
i.e, non-trivial multiplicative semi-norms on $K[x_1,...,x_n]$ extending $Vert cdot Vert$. Given a Banach ring $(A,Vert cdot Vert)$, it's Berkovich spectrum is:
$ mathcalM(A):= beginBmatrix
textmultiplicative bounded seminorms \
vert cdot vert : A rightarrow [0,infty)
endBmatrix $
They both have the weakest topology induced by evaluations.
I have been trying to understand what is the relation between $A_K^n, textan$ and a Berkovich spectrum of $K[x_1,...,x_n]$. They would coincide were it not for the extra assumption of boundedness. I saw somewhere that the Berkovich affine space is not a Berkovich spectrum, but instead an increasing union of Berkovich spectrums of power rings.
I have not found thus far how these spectrums are defined, and would appreciate anyone explaining, or directing to a reference where it is indeed written.
general-topology algebraic-geometry spectra geometric-functional-analysis
asked Jul 17 at 14:42
Keen-ameteur
644213
add a comment |Â
up vote
3
down vote
favorite
I am trying to understand what is the relation of the affine Berkovich space to the Berkovich space on an appropriate polynomial ring. A more exact version of the question is as follows:
Let $(K,Vert cdot Vert)$ be a complete valued field. The affine Berkovich space is defined as:
$ A_K^n, textan:=
beginBmatrix
& vert f+gvert_x leq vert fvert_x +vert gvert_x \
vert cdot vert _x:K[x_1,...,x_n]rightarrow [0,infty) quad Bigg vert & vert fcdot gvert_x= vert fvert_x cdot vert gvert_x \
& vert avert_x= Vert aVert ; textfor all ; ain K
endBmatrix $
i.e, non-trivial multiplicative semi-norms on $K[x_1,...,x_n]$ extending $Vert cdot Vert$. Given a Banach ring $(A,Vert cdot Vert)$, it's Berkovich spectrum is:
$ mathcalM(A):= beginBmatrix
textmultiplicative bounded seminorms \
vert cdot vert : A rightarrow [0,infty)
endBmatrix $
They both have the weakest topology induced by evaluations.
I have been trying to understand what is the relation between $A_K^n, textan$ and a Berkovich spectrum of $K[x_1,...,x_n]$. They would coincide were it not for the extra assumption of boundedness. I saw somewhere that the Berkovich affine space is not a Berkovich spectrum, but instead an increasing union of Berkovich spectrums of power rings.
I have not found thus far how these spectrums are defined, and would appreciate anyone explaining, or directing to a reference where it is indeed written.
general-topology algebraic-geometry spectra geometric-functional-analysis
asked Jul 17 at 14:42
Keen-ameteur
644213
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I am trying to understand what is the relation of the affine Berkovich space to the Berkovich space on an appropriate polynomial ring. A more exact version of the question is as follows:
Let $(K,Vert cdot Vert)$ be a complete valued field. The affine Berkovich space is defined as:
$ A_K^n, textan:=
beginBmatrix
& vert f+gvert_x leq vert fvert_x +vert gvert_x \
vert cdot vert _x:K[x_1,...,x_n]rightarrow [0,infty) quad Bigg vert & vert fcdot gvert_x= vert fvert_x cdot vert gvert_x \
& vert avert_x= Vert aVert ; textfor all ; ain K
endBmatrix $
i.e, non-trivial multiplicative semi-norms on $K[x_1,...,x_n]$ extending $Vert cdot Vert$. Given a Banach ring $(A,Vert cdot Vert)$, it's Berkovich spectrum is:
$ mathcalM(A):= beginBmatrix
textmultiplicative bounded seminorms \
vert cdot vert : A rightarrow [0,infty)
endBmatrix $
They both have the weakest topology induced by evaluations.
I have been trying to understand what is the relation between $A_K^n, textan$ and a Berkovich spectrum of $K[x_1,...,x_n]$. They would coincide were it not for the extra assumption of boundedness. I saw somewhere that the Berkovich affine space is not a Berkovich spectrum, but instead an increasing union of Berkovich spectrums of power rings.
I have not found thus far how these spectrums are defined, and would appreciate anyone explaining, or directing to a reference where it is indeed written.
general-topology algebraic-geometry spectra geometric-functional-analysis
asked Jul 17 at 14:42
Keen-ameteur
644213
I am trying to understand what is the relation of the affine Berkovich space to the Berkovich space on an appropriate polynomial ring. A more exact version of the question is as follows:
Let $(K,Vert cdot Vert)$ be a complete valued field. The affine Berkovich space is defined as:
$ A_K^n, textan:=
beginBmatrix
& vert f+gvert_x leq vert fvert_x +vert gvert_x \
vert cdot vert _x:K[x_1,...,x_n]rightarrow [0,infty) quad Bigg vert & vert fcdot gvert_x= vert fvert_x cdot vert gvert_x \
& vert avert_x= Vert aVert ; textfor all ; ain K
endBmatrix $
i.e, non-trivial multiplicative semi-norms on $K[x_1,...,x_n]$ extending $Vert cdot Vert$. Given a Banach ring $(A,Vert cdot Vert)$, it's Berkovich spectrum is:
$ mathcalM(A):= beginBmatrix
textmultiplicative bounded seminorms \
vert cdot vert : A rightarrow [0,infty)
endBmatrix $
They both have the weakest topology induced by evaluations.
I have been trying to understand what is the relation between $A_K^n, textan$ and a Berkovich spectrum of $K[x_1,...,x_n]$. They would coincide were it not for the extra assumption of boundedness. I saw somewhere that the Berkovich affine space is not a Berkovich spectrum, but instead an increasing union of Berkovich spectrums of power rings.
I have not found thus far how these spectrums are defined, and would appreciate anyone explaining, or directing to a reference where it is indeed written.
general-topology algebraic-geometry spectra geometric-functional-analysis
asked Jul 17 at 14:42
Keen-ameteur
644213
asked Jul 17 at 14:42
Keen-ameteur
644213
asked Jul 17 at 14:42
Keen-ameteur
644213
asked Jul 17 at 14:42
Keen-ameteur
644213
644213
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
3
down vote
accepted
Let me see if I can help a bit; please let me know if you need more clarification. First of all, I don't know of many basic references on Berkovich spaces. One you might consider is the first two chapters in Baker–Rumely (which works over an algebraically closed field). For another reference, you might find this set of notes (especially §5) and the citations therein helpful.
To make this discussion a bit simpler, we will work over a complete non-Archimedean valued field $(K,lvertcdotrvert)$, and only consider the case when $n = 1$.
Let's first recall that Berkovich's notion of a spectrum $mathcalM(A)$ needs $A$ to be a Banach ring. To make sense of the Berkovich spectrum of $K[x]$, then, one would want to put a norm on this ring that makes it complete. The usual choice (the "Gauss norm")
$$biggllVert sum_i=0^infty a_ix^i biggrrVert = max_i lvert a_i rvert$$
would make it so that the polynomial ring is not complete unless the field $K$ is trivially valued. In general, we therefore consider the Banach ring (the "Tate algebra")
$$Kx := biggl f = sum_i=0^infty a_ix^i biggmvert a_i in K, lim_i to infty lvert a_i rvert = 0 biggr$$
which can be thought of as the "ring of convergent power series on the closed disc of radius $1$." Note that in the trivially valued case, this ring is just the usual polynomial ring.
Now borrowing the usual intuition from algebraic geometry, the ring $A$ that we take the spectrum of should consist of all regular functions on the space $mathcalM(A)$. In this case, $mathcalM(Kx)$ should therefore be thought of as a closed disc of radius $1$; after all, if you take an element of $Kx$ and plug an element of $K$ with norm $>1$, then the power series can diverge!
To get closed discs with larger radii, then, Berkovich changes the norm and considers the ring
$$Kr^-1x := biggl f = sum_i=0^infty a_ix^i biggmvert a_i in K, lim_i to infty lvert a_i rvert r^i = 0 biggr$$
for each $r > 0$, which now can be thought of as the "ring of convergent power series on the closed disc of radius $r$." The closed disc of radius $r$ is
$$E(r) := mathcalM(Kr^-1x),$$
and we can define the affine line as
$$mathbfA^1_mathrmBerk := bigcup_r > 0 E(r).$$
answered Jul 17 at 19:03
Takumi Murayama
5,74111545
1
Also, when OP says the ``Berkovich spectrum of $K[x_1,ldots,x_n]$'', this is not well-defined-you need to equip the polynomial ring with this norm. One choice of norm is a Gauss norm $| cdot |_mathbfr$ associated to the tuple $mathbfr = (r_1,ldots,r_n)$. In this case, the Berkovich spectrum of $(K[x_1,ldots,x_n], | cdot |_mathbfr)$ is the spectrum of its completion (i.e, the Tate algebra $K r^-1 x $ as above), giving the closed disc, as in Takumi's answer. Taking the union over all possible choices of norms on $K[x_1,ldots,x_n]$ gives $mathbfA^n_mathrmBerk$.
– msteve
Jul 18 at 19:06
Thanks for the answer, and for the Baker Rumley refrence.
– Keen-ameteur
Jul 22 at 12:27
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Let me see if I can help a bit; please let me know if you need more clarification. First of all, I don't know of many basic references on Berkovich spaces. One you might consider is the first two chapters in Baker–Rumely (which works over an algebraically closed field). For another reference, you might find this set of notes (especially §5) and the citations therein helpful.
To make this discussion a bit simpler, we will work over a complete non-Archimedean valued field $(K,lvertcdotrvert)$, and only consider the case when $n = 1$.
Let's first recall that Berkovich's notion of a spectrum $mathcalM(A)$ needs $A$ to be a Banach ring. To make sense of the Berkovich spectrum of $K[x]$, then, one would want to put a norm on this ring that makes it complete. The usual choice (the "Gauss norm")
$$biggllVert sum_i=0^infty a_ix^i biggrrVert = max_i lvert a_i rvert$$
would make it so that the polynomial ring is not complete unless the field $K$ is trivially valued. In general, we therefore consider the Banach ring (the "Tate algebra")
$$Kx := biggl f = sum_i=0^infty a_ix^i biggmvert a_i in K, lim_i to infty lvert a_i rvert = 0 biggr$$
which can be thought of as the "ring of convergent power series on the closed disc of radius $1$." Note that in the trivially valued case, this ring is just the usual polynomial ring.
Now borrowing the usual intuition from algebraic geometry, the ring $A$ that we take the spectrum of should consist of all regular functions on the space $mathcalM(A)$. In this case, $mathcalM(Kx)$ should therefore be thought of as a closed disc of radius $1$; after all, if you take an element of $Kx$ and plug an element of $K$ with norm $>1$, then the power series can diverge!
To get closed discs with larger radii, then, Berkovich changes the norm and considers the ring
$$Kr^-1x := biggl f = sum_i=0^infty a_ix^i biggmvert a_i in K, lim_i to infty lvert a_i rvert r^i = 0 biggr$$
for each $r > 0$, which now can be thought of as the "ring of convergent power series on the closed disc of radius $r$." The closed disc of radius $r$ is
$$E(r) := mathcalM(Kr^-1x),$$
and we can define the affine line as
$$mathbfA^1_mathrmBerk := bigcup_r > 0 E(r).$$
answered Jul 17 at 19:03
Takumi Murayama
5,74111545
1
Also, when OP says the ``Berkovich spectrum of $K[x_1,ldots,x_n]$'', this is not well-defined-you need to equip the polynomial ring with this norm. One choice of norm is a Gauss norm $| cdot |_mathbfr$ associated to the tuple $mathbfr = (r_1,ldots,r_n)$. In this case, the Berkovich spectrum of $(K[x_1,ldots,x_n], | cdot |_mathbfr)$ is the spectrum of its completion (i.e, the Tate algebra $K r^-1 x $ as above), giving the closed disc, as in Takumi's answer. Taking the union over all possible choices of norms on $K[x_1,ldots,x_n]$ gives $mathbfA^n_mathrmBerk$.
– msteve
Jul 18 at 19:06
Thanks for the answer, and for the Baker Rumley refrence.
– Keen-ameteur
Jul 22 at 12:27
add a comment |Â
up vote
3
down vote
accepted
Let me see if I can help a bit; please let me know if you need more clarification. First of all, I don't know of many basic references on Berkovich spaces. One you might consider is the first two chapters in Baker–Rumely (which works over an algebraically closed field). For another reference, you might find this set of notes (especially §5) and the citations therein helpful.
To make this discussion a bit simpler, we will work over a complete non-Archimedean valued field $(K,lvertcdotrvert)$, and only consider the case when $n = 1$.
Let's first recall that Berkovich's notion of a spectrum $mathcalM(A)$ needs $A$ to be a Banach ring. To make sense of the Berkovich spectrum of $K[x]$, then, one would want to put a norm on this ring that makes it complete. The usual choice (the "Gauss norm")
$$biggllVert sum_i=0^infty a_ix^i biggrrVert = max_i lvert a_i rvert$$
would make it so that the polynomial ring is not complete unless the field $K$ is trivially valued. In general, we therefore consider the Banach ring (the "Tate algebra")
$$Kx := biggl f = sum_i=0^infty a_ix^i biggmvert a_i in K, lim_i to infty lvert a_i rvert = 0 biggr$$
which can be thought of as the "ring of convergent power series on the closed disc of radius $1$." Note that in the trivially valued case, this ring is just the usual polynomial ring.
Now borrowing the usual intuition from algebraic geometry, the ring $A$ that we take the spectrum of should consist of all regular functions on the space $mathcalM(A)$. In this case, $mathcalM(Kx)$ should therefore be thought of as a closed disc of radius $1$; after all, if you take an element of $Kx$ and plug an element of $K$ with norm $>1$, then the power series can diverge!
To get closed discs with larger radii, then, Berkovich changes the norm and considers the ring
$$Kr^-1x := biggl f = sum_i=0^infty a_ix^i biggmvert a_i in K, lim_i to infty lvert a_i rvert r^i = 0 biggr$$
for each $r > 0$, which now can be thought of as the "ring of convergent power series on the closed disc of radius $r$." The closed disc of radius $r$ is
$$E(r) := mathcalM(Kr^-1x),$$
and we can define the affine line as
$$mathbfA^1_mathrmBerk := bigcup_r > 0 E(r).$$
answered Jul 17 at 19:03
Takumi Murayama
5,74111545
1
Also, when OP says the ``Berkovich spectrum of $K[x_1,ldots,x_n]$'', this is not well-defined-you need to equip the polynomial ring with this norm. One choice of norm is a Gauss norm $| cdot |_mathbfr$ associated to the tuple $mathbfr = (r_1,ldots,r_n)$. In this case, the Berkovich spectrum of $(K[x_1,ldots,x_n], | cdot |_mathbfr)$ is the spectrum of its completion (i.e, the Tate algebra $K r^-1 x $ as above), giving the closed disc, as in Takumi's answer. Taking the union over all possible choices of norms on $K[x_1,ldots,x_n]$ gives $mathbfA^n_mathrmBerk$.
– msteve
Jul 18 at 19:06
Thanks for the answer, and for the Baker Rumley refrence.
– Keen-ameteur
Jul 22 at 12:27
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Let me see if I can help a bit; please let me know if you need more clarification. First of all, I don't know of many basic references on Berkovich spaces. One you might consider is the first two chapters in Baker–Rumely (which works over an algebraically closed field). For another reference, you might find this set of notes (especially §5) and the citations therein helpful.
To make this discussion a bit simpler, we will work over a complete non-Archimedean valued field $(K,lvertcdotrvert)$, and only consider the case when $n = 1$.
Let's first recall that Berkovich's notion of a spectrum $mathcalM(A)$ needs $A$ to be a Banach ring. To make sense of the Berkovich spectrum of $K[x]$, then, one would want to put a norm on this ring that makes it complete. The usual choice (the "Gauss norm")
$$biggllVert sum_i=0^infty a_ix^i biggrrVert = max_i lvert a_i rvert$$
would make it so that the polynomial ring is not complete unless the field $K$ is trivially valued. In general, we therefore consider the Banach ring (the "Tate algebra")
$$Kx := biggl f = sum_i=0^infty a_ix^i biggmvert a_i in K, lim_i to infty lvert a_i rvert = 0 biggr$$
which can be thought of as the "ring of convergent power series on the closed disc of radius $1$." Note that in the trivially valued case, this ring is just the usual polynomial ring.
Now borrowing the usual intuition from algebraic geometry, the ring $A$ that we take the spectrum of should consist of all regular functions on the space $mathcalM(A)$. In this case, $mathcalM(Kx)$ should therefore be thought of as a closed disc of radius $1$; after all, if you take an element of $Kx$ and plug an element of $K$ with norm $>1$, then the power series can diverge!
To get closed discs with larger radii, then, Berkovich changes the norm and considers the ring
$$Kr^-1x := biggl f = sum_i=0^infty a_ix^i biggmvert a_i in K, lim_i to infty lvert a_i rvert r^i = 0 biggr$$
for each $r > 0$, which now can be thought of as the "ring of convergent power series on the closed disc of radius $r$." The closed disc of radius $r$ is
$$E(r) := mathcalM(Kr^-1x),$$
and we can define the affine line as
$$mathbfA^1_mathrmBerk := bigcup_r > 0 E(r).$$
answered Jul 17 at 19:03
Takumi Murayama
5,74111545
Let me see if I can help a bit; please let me know if you need more clarification. First of all, I don't know of many basic references on Berkovich spaces. One you might consider is the first two chapters in Baker–Rumely (which works over an algebraically closed field). For another reference, you might find this set of notes (especially §5) and the citations therein helpful.
To make this discussion a bit simpler, we will work over a complete non-Archimedean valued field $(K,lvertcdotrvert)$, and only consider the case when $n = 1$.
Let's first recall that Berkovich's notion of a spectrum $mathcalM(A)$ needs $A$ to be a Banach ring. To make sense of the Berkovich spectrum of $K[x]$, then, one would want to put a norm on this ring that makes it complete. The usual choice (the "Gauss norm")
$$biggllVert sum_i=0^infty a_ix^i biggrrVert = max_i lvert a_i rvert$$
would make it so that the polynomial ring is not complete unless the field $K$ is trivially valued. In general, we therefore consider the Banach ring (the "Tate algebra")
$$Kx := biggl f = sum_i=0^infty a_ix^i biggmvert a_i in K, lim_i to infty lvert a_i rvert = 0 biggr$$
which can be thought of as the "ring of convergent power series on the closed disc of radius $1$." Note that in the trivially valued case, this ring is just the usual polynomial ring.
Now borrowing the usual intuition from algebraic geometry, the ring $A$ that we take the spectrum of should consist of all regular functions on the space $mathcalM(A)$. In this case, $mathcalM(Kx)$ should therefore be thought of as a closed disc of radius $1$; after all, if you take an element of $Kx$ and plug an element of $K$ with norm $>1$, then the power series can diverge!
To get closed discs with larger radii, then, Berkovich changes the norm and considers the ring
$$Kr^-1x := biggl f = sum_i=0^infty a_ix^i biggmvert a_i in K, lim_i to infty lvert a_i rvert r^i = 0 biggr$$
for each $r > 0$, which now can be thought of as the "ring of convergent power series on the closed disc of radius $r$." The closed disc of radius $r$ is
$$E(r) := mathcalM(Kr^-1x),$$
and we can define the affine line as
$$mathbfA^1_mathrmBerk := bigcup_r > 0 E(r).$$
answered Jul 17 at 19:03
Takumi Murayama
5,74111545
answered Jul 17 at 19:03
Takumi Murayama
5,74111545
answered Jul 17 at 19:03
Takumi Murayama
5,74111545
answered Jul 17 at 19:03
Takumi Murayama
5,74111545
5,74111545
1
Also, when OP says the ``Berkovich spectrum of $K[x_1,ldots,x_n]$'', this is not well-defined-you need to equip the polynomial ring with this norm. One choice of norm is a Gauss norm $| cdot |_mathbfr$ associated to the tuple $mathbfr = (r_1,ldots,r_n)$. In this case, the Berkovich spectrum of $(K[x_1,ldots,x_n], | cdot |_mathbfr)$ is the spectrum of its completion (i.e, the Tate algebra $K r^-1 x $ as above), giving the closed disc, as in Takumi's answer. Taking the union over all possible choices of norms on $K[x_1,ldots,x_n]$ gives $mathbfA^n_mathrmBerk$.
– msteve
Jul 18 at 19:06
Thanks for the answer, and for the Baker Rumley refrence.
– Keen-ameteur
Jul 22 at 12:27
add a comment |Â
1
Also, when OP says the ``Berkovich spectrum of $K[x_1,ldots,x_n]$'', this is not well-defined-you need to equip the polynomial ring with this norm. One choice of norm is a Gauss norm $| cdot |_mathbfr$ associated to the tuple $mathbfr = (r_1,ldots,r_n)$. In this case, the Berkovich spectrum of $(K[x_1,ldots,x_n], | cdot |_mathbfr)$ is the spectrum of its completion (i.e, the Tate algebra $K r^-1 x $ as above), giving the closed disc, as in Takumi's answer. Taking the union over all possible choices of norms on $K[x_1,ldots,x_n]$ gives $mathbfA^n_mathrmBerk$.
– msteve
Jul 18 at 19:06
Thanks for the answer, and for the Baker Rumley refrence.
– Keen-ameteur
Jul 22 at 12:27
1
1
Also, when OP says the ``Berkovich spectrum of $K[x_1,ldots,x_n]$'', this is not well-defined-you need to equip the polynomial ring with this norm. One choice of norm is a Gauss norm $| cdot |_mathbfr$ associated to the tuple $mathbfr = (r_1,ldots,r_n)$. In this case, the Berkovich spectrum of $(K[x_1,ldots,x_n], | cdot |_mathbfr)$ is the spectrum of its completion (i.e, the Tate algebra $K r^-1 x $ as above), giving the closed disc, as in Takumi's answer. Taking the union over all possible choices of norms on $K[x_1,ldots,x_n]$ gives $mathbfA^n_mathrmBerk$.
– msteve
Jul 18 at 19:06
Also, when OP says the ``Berkovich spectrum of $K[x_1,ldots,x_n]$'', this is not well-defined-you need to equip the polynomial ring with this norm. One choice of norm is a Gauss norm $| cdot |_mathbfr$ associated to the tuple $mathbfr = (r_1,ldots,r_n)$. In this case, the Berkovich spectrum of $(K[x_1,ldots,x_n], | cdot |_mathbfr)$ is the spectrum of its completion (i.e, the Tate algebra $K r^-1 x $ as above), giving the closed disc, as in Takumi's answer. Taking the union over all possible choices of norms on $K[x_1,ldots,x_n]$ gives $mathbfA^n_mathrmBerk$.
– msteve
Jul 18 at 19:06
Thanks for the answer, and for the Baker Rumley refrence.
– Keen-ameteur
Jul 22 at 12:27
Thanks for the answer, and for the Baker Rumley refrence.
– Keen-ameteur
Jul 22 at 12:27
add a comment |Â
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