Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

What is the generalization of eigenvalues/vectors to modules?

To be specific, given a "vector" v in a module over some ring, and a linear "operator" O from the module to itself (please feel free to correct my terminology :-) ), I would like to learn what we know about problems of the form

O v = k v

where k is a member of the same ring.

I have been looking through a lot of books and online resources about modules, but I am having trouble finding the answer to this question, and I am guessing that it is probably because I don't know what the name of the thing is that I should be looking for.

Edit: Fixed a typo -- thanks Boris! (I said that O was a map from the ring to itself when I meant it was a map from the module to itself.)

Update: To be clear, I would also be happy with an answer of the form: there is not a good generalization of eigenevalues for modules with no additional structure at all, but there is if you can assume the additional structure X, where X is, say, a dot product, a norm, an involution operator, etc.

share|improve this question
1  
I don't really understand your question (in fact, I do not see what the question is!). What exactly do you want to generalize? How will you tell if a proposed generalization is good? What do you want to do with it? –  Mariano Suárez-Alvarez Nov 24 '11 at 6:17
    
In any case the question should restrict to the case of commutative rings. I suppose you want your operator to be at least $R$-linear, which means that is commutes with all scalar multiplications by elements of $R$. However if $k$ is not in the center of $R$, then multiplication by $k$ does not have this property, so there is no hope of finding any solutions for such $k$. –  Marc van Leeuwen Nov 24 '11 at 13:36
add comment

3 Answers

In the case of commutative rings, you can view spectra as points in the quotient of $End_R(M)$ by the conjugation action of $Aut_R(M)$. You can use the tensor product to turn this into a quotient of a scheme by a group. If $M$ is locally free of rank $n$, the coefficients of the characteristic polynomial (say, viewed as traces of $\wedge^i O$ on $\wedge^i M$) give you a map to affine $n$-space over the spectrum of $R$. You can think of this as a space of elementary symmetric polynomials in eigenvalues. If you take your operator $O \in End_R(M)$, and send it to a point in this space, I suppose an eigenvalue is what you get by lifting to an element in the $S_n$-orbit in the affine space of roots, and projecting to a coordinate. These don't exist globally.

This sort of construction arises when studying the Hitchin map.

(Minor comment: Darij's claim that the eigenvalue map is discontinuous uses an implicit assumption that the set with three elements should be endowed with the discrete topology.)

share|improve this answer
    
Hmm, I'll need to think about that. Thank you! –  Gregory Crosswhite Nov 24 '11 at 6:35
add comment

There is another way to look at this. Let $K$ be a commutative ring and $M$ a $K$-module. Then giving a $K$-linear endomorphism of $M$ is equivalent to an action of the polynomial ring $K[x]$ on $M$. Then the question becomes: what is the structure of $M$ as a $K[T]$-module? The classification of finitely generated $K[x]$-modules in the case $K$ is a field is a well-known result that used to be taught to undergraduates.

share|improve this answer
2  
I would go so far as to say that the correct generalization of an eigenvalue is an irreducible $K[T]$-module, whatever those turn out to be for a particular $K$. For instance, this is one way to interpret why we can talk about the complex eigenvalues of a real matrix. –  Will Sawin Nov 24 '11 at 12:09
    
@Will Yes, exactly. You also need to think about whether you want the irreducible representations that occur as composition factors or whether you want the decomposition into indecomposable modules. –  Bruce Westbury Nov 24 '11 at 13:35
add comment

This just means that the submodule generated by $v$, i.e. $\lbrace rv\mid r\in R\rbrace$, is invariant with respect to the operator $O$ (which acts rather in the module than the ring $R$), no more.

share|improve this answer
2  
I think you won't have a good theory mainly because the theory of eigenvalues and eigenvectors is uncontinuous/unfunctorial. Let me try to clarify this on the example of the Jordan normal form of a $2\times 2$ matrix. It can have three different forms: either a diagonal matrix with distinct elements on the diagonal, in which case the diagonalization is essentially unique (i. e., every eigenvalue has only one eigenvector up to scalar multiplication), or a diagonal matrix with equal elements on the diagonal, in which case the diagonalization gives a lot of freedom (because the matrix we ... –  darij grinberg Nov 24 '11 at 3:46
1  
... are diagonalizing is a scalar multiple of the unity matrix), or a Jordan block, in which case we have more freedom than in the first but less freedom than in the second case. So the Jordan form gives us a map from the matrix ring $\mathrm M_2\left(k\right)$ into the three-element set $\left\{1,2,3\right\}$, which simply says what case we are in. Now, this map is highly un-continuous, and there is a yoga that uncontinuous maps can only rarely be reasonably defined over rings. For example, what is the "rank" of a matrix over a ring? You can define it as the maximum $r$ for which the ... –  darij grinberg Nov 24 '11 at 3:48
1  
... matrix has a nonvanishing $r\times r$ minor, but much of the rank theory that was easy over rings becomes harder here. Basically, there is nothing interesting to say about matrices of rank $m$ over a ring; still there are theorems about matrices of rank $\leq m$ over a ring, since rank is, while not continuous, at least semicontinuous. As for combinatorial properties of the Jordan normal form, I guess they are not even semicontinuous, so we shouldn't expect much to hold here. –  darij grinberg Nov 24 '11 at 3:50
1  
What IS possible for a general ring is to adjoin the "formal" eigenvalues of a matrix to the ring. By "formal" eigenvalues, I just mean the roots of the characteristic polynomial. This is possible because the characteristic polynomial is monic. The ring extension that will contain these roots will be a free module over the ring of rank $n!$, where $n$ is the degree of the characteristic polynomial (i. e., the size of the matrix). I can go into details if you are interested in that. But the eigenvalues-eigenvectors relation breaks down, since in the case of fields it relies on the ... –  darij grinberg Nov 24 '11 at 3:53
2  
@Darij: that seems long enough that you could write it up as an answer, for better visibility –  Yemon Choi Nov 24 '11 at 4:22
show 11 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.