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How can a spectrum object be constructed?

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2

1

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There are so many spectrum objects in mathematics.

• spectrum of a linear operator


• spectrum of a ring


• spectrum of a graph


• spectrum in algebraic topology


• spectrum of a sentence (mathematical logic)

There has to be some universal property uniting all of these spectrum objects. How do you use a universal construction to derive each of the five?

category-theory

share|cite|improve this question

edited Feb 27 at 10:14

Tomislav Ostojich

asked Feb 27 at 9:40

Tomislav OstojichTomislav Ostojich

759718

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  The spectrum of a graph is the spectrum of a linear operator, and the fact that there is also a spectrum of ring is just a coincidence as far as I know. I doubt there's any concept underlying both cases.
  $endgroup$
  – Arnaud D.
  Feb 27 at 10:03
  

  
  
  
  


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  $begingroup$
  @Arnaud D. There's also a spectrum in algebraic topology, so I don't think it's a coincidence.
  $endgroup$
  – Tomislav Ostojich
  Feb 27 at 10:04
  

  
  
  
  


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  3
  

  
  
  
  
  
  

  
  $begingroup$
  Sounds like even more of a coincidence to me...
  $endgroup$
  – Arnaud D.
  Feb 27 at 10:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ArnaudD Spectrum of a sentence (mathematical logic)
  $endgroup$
  – Tomislav Ostojich
  Feb 27 at 10:13
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  Spectra in algebraic topology and spectra of sentences are totally unrelated to the other notions of spectra you mentioned.
  $endgroup$
  – Eric Wofsey
  Mar 15 at 6:47
  

  
  
  
  

 | 
show 4 more comments

2

1

$begingroup$

There are so many spectrum objects in mathematics.

• spectrum of a linear operator


• spectrum of a ring


• spectrum of a graph


• spectrum in algebraic topology


• spectrum of a sentence (mathematical logic)

There has to be some universal property uniting all of these spectrum objects. How do you use a universal construction to derive each of the five?

category-theory

share|cite|improve this question

edited Feb 27 at 10:14

Tomislav Ostojich

asked Feb 27 at 9:40

Tomislav OstojichTomislav Ostojich

759718

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The spectrum of a graph is the spectrum of a linear operator, and the fact that there is also a spectrum of ring is just a coincidence as far as I know. I doubt there's any concept underlying both cases.
  $endgroup$
  – Arnaud D.
  Feb 27 at 10:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Arnaud D. There's also a spectrum in algebraic topology, so I don't think it's a coincidence.
  $endgroup$
  – Tomislav Ostojich
  Feb 27 at 10:04
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  Sounds like even more of a coincidence to me...
  $endgroup$
  – Arnaud D.
  Feb 27 at 10:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ArnaudD Spectrum of a sentence (mathematical logic)
  $endgroup$
  – Tomislav Ostojich
  Feb 27 at 10:13
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  Spectra in algebraic topology and spectra of sentences are totally unrelated to the other notions of spectra you mentioned.
  $endgroup$
  – Eric Wofsey
  Mar 15 at 6:47
  

  
  
  
  

 | 
show 4 more comments

$begingroup$

There are so many spectrum objects in mathematics.

• spectrum of a linear operator


• spectrum of a ring


• spectrum of a graph


• spectrum in algebraic topology


• spectrum of a sentence (mathematical logic)

There has to be some universal property uniting all of these spectrum objects. How do you use a universal construction to derive each of the five?

category-theory

share|cite|improve this question

edited Feb 27 at 10:14

Tomislav Ostojich

asked Feb 27 at 9:40

Tomislav OstojichTomislav Ostojich

759718

$endgroup$

share|cite|improve this question

edited Feb 27 at 10:14

Tomislav Ostojich

asked Feb 27 at 9:40

Tomislav OstojichTomislav Ostojich

759718

share|cite|improve this question

edited Feb 27 at 10:14

Tomislav Ostojich

asked Feb 27 at 9:40

Tomislav OstojichTomislav Ostojich

759718

asked Feb 27 at 9:40

Tomislav OstojichTomislav Ostojich

759718

asked Feb 27 at 9:40

Tomislav OstojichTomislav Ostojich

759718

1 Answer
1

active

oldest

votes

4

$begingroup$

This does not answer the question (I'd be surprised if there was a categorical way to define a "spectrum" which somehow encompasses the weird definitions in logic, topology, etc.) but it's a bit too long to be a comment.

It's a bit misleading to think of the first three constructions as the same universal construction going on in three different categories, because they are in fact the same construction in one category: the category of rings. (I wouldn't be surprised if the spectrum in the sense of topology is also the special case of the spectrum of a ring, but I don't know anything about that.)

Let $k$ be an algebraically closed field (characteristic $0$ if you like), and $T$ a linear operator acting on a finite-dimensional $k$-vector space. Then $operatornameSpec T$ is, by definition, the space of "generalized eigenvalues" (i.e. diagonal entries in the Jordan canonical form of $T$) with repetition. But if $chi_T(x) in k[x]$ is the characteristic polynomial of $T$ then the roots of $chi_T(x)$ are precisely the generalized eigenvalues with repetition, so the affine scheme $operatornameSpec k[x]/(chi_T(x)) = operatornameSpec T$ in a natural way: namely, its points are generalized eigenvalues of $T$, and we can recover their multiplicities as the dimension of the localization as a $k$-algebra. Therefore the spectrum of a ring is a generalization of a spectrum of a linear operator.

Since the "spectrum" of a graph is the spectrum of its adjacency matrix $M$, this is obviously a special case of the spectrum of a linear operator. So it is the spectrum of the ring $mathbb C[x]/(chi_M(x))$.

share|cite|improve this answer

answered Mar 15 at 6:13

Aidan BackusAidan Backus

13916

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What is the universal property for JCF? "because they are in fact the same construction in one category: the category of rings." Why are vector spaces in the category of rings? They are in the category of R-modules?
  $endgroup$
  – Tomislav Ostojich
  Mar 15 at 16:47
  

  
  
  
  

add a comment | 

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4

$begingroup$

This does not answer the question (I'd be surprised if there was a categorical way to define a "spectrum" which somehow encompasses the weird definitions in logic, topology, etc.) but it's a bit too long to be a comment.

It's a bit misleading to think of the first three constructions as the same universal construction going on in three different categories, because they are in fact the same construction in one category: the category of rings. (I wouldn't be surprised if the spectrum in the sense of topology is also the special case of the spectrum of a ring, but I don't know anything about that.)

Let $k$ be an algebraically closed field (characteristic $0$ if you like), and $T$ a linear operator acting on a finite-dimensional $k$-vector space. Then $operatornameSpec T$ is, by definition, the space of "generalized eigenvalues" (i.e. diagonal entries in the Jordan canonical form of $T$) with repetition. But if $chi_T(x) in k[x]$ is the characteristic polynomial of $T$ then the roots of $chi_T(x)$ are precisely the generalized eigenvalues with repetition, so the affine scheme $operatornameSpec k[x]/(chi_T(x)) = operatornameSpec T$ in a natural way: namely, its points are generalized eigenvalues of $T$, and we can recover their multiplicities as the dimension of the localization as a $k$-algebra. Therefore the spectrum of a ring is a generalization of a spectrum of a linear operator.

Since the "spectrum" of a graph is the spectrum of its adjacency matrix $M$, this is obviously a special case of the spectrum of a linear operator. So it is the spectrum of the ring $mathbb C[x]/(chi_M(x))$.

share|cite|improve this answer

answered Mar 15 at 6:13

Aidan BackusAidan Backus

13916

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What is the universal property for JCF? "because they are in fact the same construction in one category: the category of rings." Why are vector spaces in the category of rings? They are in the category of R-modules?
  $endgroup$
  – Tomislav Ostojich
  Mar 15 at 16:47
  

  
  
  
  

add a comment | 

4

$begingroup$

This does not answer the question (I'd be surprised if there was a categorical way to define a "spectrum" which somehow encompasses the weird definitions in logic, topology, etc.) but it's a bit too long to be a comment.

It's a bit misleading to think of the first three constructions as the same universal construction going on in three different categories, because they are in fact the same construction in one category: the category of rings. (I wouldn't be surprised if the spectrum in the sense of topology is also the special case of the spectrum of a ring, but I don't know anything about that.)

Let $k$ be an algebraically closed field (characteristic $0$ if you like), and $T$ a linear operator acting on a finite-dimensional $k$-vector space. Then $operatornameSpec T$ is, by definition, the space of "generalized eigenvalues" (i.e. diagonal entries in the Jordan canonical form of $T$) with repetition. But if $chi_T(x) in k[x]$ is the characteristic polynomial of $T$ then the roots of $chi_T(x)$ are precisely the generalized eigenvalues with repetition, so the affine scheme $operatornameSpec k[x]/(chi_T(x)) = operatornameSpec T$ in a natural way: namely, its points are generalized eigenvalues of $T$, and we can recover their multiplicities as the dimension of the localization as a $k$-algebra. Therefore the spectrum of a ring is a generalization of a spectrum of a linear operator.

Since the "spectrum" of a graph is the spectrum of its adjacency matrix $M$, this is obviously a special case of the spectrum of a linear operator. So it is the spectrum of the ring $mathbb C[x]/(chi_M(x))$.

share|cite|improve this answer

answered Mar 15 at 6:13

Aidan BackusAidan Backus

13916

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What is the universal property for JCF? "because they are in fact the same construction in one category: the category of rings." Why are vector spaces in the category of rings? They are in the category of R-modules?
  $endgroup$
  – Tomislav Ostojich
  Mar 15 at 16:47
  

  
  
  
  

add a comment | 

$begingroup$

This does not answer the question (I'd be surprised if there was a categorical way to define a "spectrum" which somehow encompasses the weird definitions in logic, topology, etc.) but it's a bit too long to be a comment.

It's a bit misleading to think of the first three constructions as the same universal construction going on in three different categories, because they are in fact the same construction in one category: the category of rings. (I wouldn't be surprised if the spectrum in the sense of topology is also the special case of the spectrum of a ring, but I don't know anything about that.)

Let $k$ be an algebraically closed field (characteristic $0$ if you like), and $T$ a linear operator acting on a finite-dimensional $k$-vector space. Then $operatornameSpec T$ is, by definition, the space of "generalized eigenvalues" (i.e. diagonal entries in the Jordan canonical form of $T$) with repetition. But if $chi_T(x) in k[x]$ is the characteristic polynomial of $T$ then the roots of $chi_T(x)$ are precisely the generalized eigenvalues with repetition, so the affine scheme $operatornameSpec k[x]/(chi_T(x)) = operatornameSpec T$ in a natural way: namely, its points are generalized eigenvalues of $T$, and we can recover their multiplicities as the dimension of the localization as a $k$-algebra. Therefore the spectrum of a ring is a generalization of a spectrum of a linear operator.

Since the "spectrum" of a graph is the spectrum of its adjacency matrix $M$, this is obviously a special case of the spectrum of a linear operator. So it is the spectrum of the ring $mathbb C[x]/(chi_M(x))$.

share|cite|improve this answer

answered Mar 15 at 6:13

Aidan BackusAidan Backus

13916

$endgroup$

share|cite|improve this answer

answered Mar 15 at 6:13

Aidan BackusAidan Backus

13916

answered Mar 15 at 6:13

Aidan BackusAidan Backus

13916

answered Mar 15 at 6:13

Aidan BackusAidan Backus

13916