Definition of the Berkovich spectrum

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I am trying to read these notes:

http://www-personal.umich.edu/~takumim/Berkovich.pdf

Regarding the Berkovich spectrum. In definition [2.24] it says that the spectrum is the set of bounded (non-trivial) multiplicative semi-norms. But in the explicit definition, elements in the Berkovich spectrum are said to be bounded by the norm of the ring $A$, and it says that we can assume this because we can replace the norm with an equivalent norm.

Initially I did not see the importance of this fact and assumed one can simply discuss bounded semi-norms, and not just norms bounded by the ring norm. However this part of the definition becomes essential when discussing the Gelfand transform.

I do not fully understand the remark above, and how this equivalent norm should be taken. Right now this condition seems to me to make the Berkovich spectrum much less rich than simply bounded multiplicative semi-norms.

I would appreciate it if someone could point to what I'm missing.

spectra geometric-functional-analysis

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asked Jul 23 at 15:44

Keen-ameteur

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up vote
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I am trying to read these notes:

http://www-personal.umich.edu/~takumim/Berkovich.pdf

Regarding the Berkovich spectrum. In definition [2.24] it says that the spectrum is the set of bounded (non-trivial) multiplicative semi-norms. But in the explicit definition, elements in the Berkovich spectrum are said to be bounded by the norm of the ring $A$, and it says that we can assume this because we can replace the norm with an equivalent norm.

Initially I did not see the importance of this fact and assumed one can simply discuss bounded semi-norms, and not just norms bounded by the ring norm. However this part of the definition becomes essential when discussing the Gelfand transform.

I do not fully understand the remark above, and how this equivalent norm should be taken. Right now this condition seems to me to make the Berkovich spectrum much less rich than simply bounded multiplicative semi-norms.

I would appreciate it if someone could point to what I'm missing.

spectra geometric-functional-analysis

share|cite|improve this question

asked Jul 23 at 15:44

Keen-ameteur

639213

add a comment | 

up vote
4
down vote

favorite

up vote
4
down vote

favorite

I am trying to read these notes:

http://www-personal.umich.edu/~takumim/Berkovich.pdf

Regarding the Berkovich spectrum. In definition [2.24] it says that the spectrum is the set of bounded (non-trivial) multiplicative semi-norms. But in the explicit definition, elements in the Berkovich spectrum are said to be bounded by the norm of the ring $A$, and it says that we can assume this because we can replace the norm with an equivalent norm.

Initially I did not see the importance of this fact and assumed one can simply discuss bounded semi-norms, and not just norms bounded by the ring norm. However this part of the definition becomes essential when discussing the Gelfand transform.

I do not fully understand the remark above, and how this equivalent norm should be taken. Right now this condition seems to me to make the Berkovich spectrum much less rich than simply bounded multiplicative semi-norms.

I would appreciate it if someone could point to what I'm missing.

spectra geometric-functional-analysis

share|cite|improve this question

asked Jul 23 at 15:44

Keen-ameteur

639213

I am trying to read these notes:

http://www-personal.umich.edu/~takumim/Berkovich.pdf

Regarding the Berkovich spectrum. In definition [2.24] it says that the spectrum is the set of bounded (non-trivial) multiplicative semi-norms. But in the explicit definition, elements in the Berkovich spectrum are said to be bounded by the norm of the ring $A$, and it says that we can assume this because we can replace the norm with an equivalent norm.

Initially I did not see the importance of this fact and assumed one can simply discuss bounded semi-norms, and not just norms bounded by the ring norm. However this part of the definition becomes essential when discussing the Gelfand transform.

I do not fully understand the remark above, and how this equivalent norm should be taken. Right now this condition seems to me to make the Berkovich spectrum much less rich than simply bounded multiplicative semi-norms.

I would appreciate it if someone could point to what I'm missing.

spectra geometric-functional-analysis

share|cite|improve this question

asked Jul 23 at 15:44

Keen-ameteur

639213

share|cite|improve this question

share|cite|improve this question

asked Jul 23 at 15:44

Keen-ameteur

639213

asked Jul 23 at 15:44

Keen-ameteur

639213

asked Jul 23 at 15:44

Keen-ameteur

639213

639213

add a comment | 

add a comment | 

1 Answer
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I don't exactly recall what was meant by that remark, but here is an observation from [Berkovich, p. 12].

Let $(mathscrA,lVert cdot rVert)$ be the Banach ring in question, and let $lvert cdot rvert$ be a multiplicative seminorm on $mathscrA$ such that there exists a constant $C > 0$ for which $lvert f rvert le C cdot lVert f rVert$ for all $f in mathscrA$.

Claim. $lvert f rvert le lVert f rVert$ for all $f in mathscrA$.

Proof. For every integer $n ge 1$, we have
$$lvert f rvert^n = lvert f^n rvert le C cdot lVert f^n rVert le C cdot lVert f rVert^n$$
since $lvert cdot rvert$ is multiplicative and $lVert cdot rVert$ is submultiplicative. Taking $n$th roots, we then have
$$lvert f rvert le sqrt[n]C lVert f rVert$$
for every integer $n ge 1$, hence $lvert f rvert le lVert f rVert$. $blacksquare$

In other words, there is no replacement by an equivalent norm necessary.

EDIT. This observation has been incorporated into the newest version of the notes.

share|cite|improve this answer

edited Aug 6 at 20:36

answered Jul 24 at 18:17

Takumi Murayama

5,73111545

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
3
down vote



accepted

I don't exactly recall what was meant by that remark, but here is an observation from [Berkovich, p. 12].

Let $(mathscrA,lVert cdot rVert)$ be the Banach ring in question, and let $lvert cdot rvert$ be a multiplicative seminorm on $mathscrA$ such that there exists a constant $C > 0$ for which $lvert f rvert le C cdot lVert f rVert$ for all $f in mathscrA$.

Claim. $lvert f rvert le lVert f rVert$ for all $f in mathscrA$.

Proof. For every integer $n ge 1$, we have
$$lvert f rvert^n = lvert f^n rvert le C cdot lVert f^n rVert le C cdot lVert f rVert^n$$
since $lvert cdot rvert$ is multiplicative and $lVert cdot rVert$ is submultiplicative. Taking $n$th roots, we then have
$$lvert f rvert le sqrt[n]C lVert f rVert$$
for every integer $n ge 1$, hence $lvert f rvert le lVert f rVert$. $blacksquare$

In other words, there is no replacement by an equivalent norm necessary.

EDIT. This observation has been incorporated into the newest version of the notes.

share|cite|improve this answer

edited Aug 6 at 20:36

answered Jul 24 at 18:17

Takumi Murayama

5,73111545

add a comment | 

up vote
3
down vote



accepted

I don't exactly recall what was meant by that remark, but here is an observation from [Berkovich, p. 12].

Let $(mathscrA,lVert cdot rVert)$ be the Banach ring in question, and let $lvert cdot rvert$ be a multiplicative seminorm on $mathscrA$ such that there exists a constant $C > 0$ for which $lvert f rvert le C cdot lVert f rVert$ for all $f in mathscrA$.

Claim. $lvert f rvert le lVert f rVert$ for all $f in mathscrA$.

Proof. For every integer $n ge 1$, we have
$$lvert f rvert^n = lvert f^n rvert le C cdot lVert f^n rVert le C cdot lVert f rVert^n$$
since $lvert cdot rvert$ is multiplicative and $lVert cdot rVert$ is submultiplicative. Taking $n$th roots, we then have
$$lvert f rvert le sqrt[n]C lVert f rVert$$
for every integer $n ge 1$, hence $lvert f rvert le lVert f rVert$. $blacksquare$

In other words, there is no replacement by an equivalent norm necessary.

EDIT. This observation has been incorporated into the newest version of the notes.

share|cite|improve this answer

edited Aug 6 at 20:36

answered Jul 24 at 18:17

Takumi Murayama

5,73111545

add a comment | 

up vote
3
down vote



accepted

up vote
3
down vote



accepted

I don't exactly recall what was meant by that remark, but here is an observation from [Berkovich, p. 12].

Let $(mathscrA,lVert cdot rVert)$ be the Banach ring in question, and let $lvert cdot rvert$ be a multiplicative seminorm on $mathscrA$ such that there exists a constant $C > 0$ for which $lvert f rvert le C cdot lVert f rVert$ for all $f in mathscrA$.

Claim. $lvert f rvert le lVert f rVert$ for all $f in mathscrA$.

Proof. For every integer $n ge 1$, we have
$$lvert f rvert^n = lvert f^n rvert le C cdot lVert f^n rVert le C cdot lVert f rVert^n$$
since $lvert cdot rvert$ is multiplicative and $lVert cdot rVert$ is submultiplicative. Taking $n$th roots, we then have
$$lvert f rvert le sqrt[n]C lVert f rVert$$
for every integer $n ge 1$, hence $lvert f rvert le lVert f rVert$. $blacksquare$

In other words, there is no replacement by an equivalent norm necessary.

EDIT. This observation has been incorporated into the newest version of the notes.

share|cite|improve this answer

edited Aug 6 at 20:36

answered Jul 24 at 18:17

Takumi Murayama

5,73111545

I don't exactly recall what was meant by that remark, but here is an observation from [Berkovich, p. 12].

Let $(mathscrA,lVert cdot rVert)$ be the Banach ring in question, and let $lvert cdot rvert$ be a multiplicative seminorm on $mathscrA$ such that there exists a constant $C > 0$ for which $lvert f rvert le C cdot lVert f rVert$ for all $f in mathscrA$.

Claim. $lvert f rvert le lVert f rVert$ for all $f in mathscrA$.

Proof. For every integer $n ge 1$, we have
$$lvert f rvert^n = lvert f^n rvert le C cdot lVert f^n rVert le C cdot lVert f rVert^n$$
since $lvert cdot rvert$ is multiplicative and $lVert cdot rVert$ is submultiplicative. Taking $n$th roots, we then have
$$lvert f rvert le sqrt[n]C lVert f rVert$$
for every integer $n ge 1$, hence $lvert f rvert le lVert f rVert$. $blacksquare$

In other words, there is no replacement by an equivalent norm necessary.

EDIT. This observation has been incorporated into the newest version of the notes.

share|cite|improve this answer

edited Aug 6 at 20:36

answered Jul 24 at 18:17

Takumi Murayama

5,73111545

share|cite|improve this answer

share|cite|improve this answer

edited Aug 6 at 20:36

edited Aug 6 at 20:36

edited Aug 6 at 20:36

answered Jul 24 at 18:17

Takumi Murayama

5,73111545

answered Jul 24 at 18:17

Takumi Murayama

5,73111545

answered Jul 24 at 18:17

Takumi Murayama

5,73111545

5,73111545

add a comment | 

add a comment | 

 

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