3
$\begingroup$

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.

Improve this question
$\endgroup$
4
  • 2
    $\begingroup$ 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? $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Nov 24, 2011 at 6:17
  • $\begingroup$ 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 $k$. $\endgroup$
    – Marc van Leeuwen
    Commented Nov 24, 2011 at 13:36
  • $\begingroup$ @MarcvanLeeuwen, why is there no hope of finding such solutions? That multiplication by $k$ is not linear surely doesn't mean that we can't ask for solutions of $Ov = k v$ for linear $O$ (any more than non-linearity of a quadratic form $q$ means that we can't ask for solutions to $Lv = qv$). $\endgroup$
    – LSpice
    Commented May 22, 2020 at 17:26
  • 1
    $\begingroup$ @LSpice: I'm not saying you cannot pose or solve the equation ($v=0$ always works), but it is not very interesting. You would like to find a nonzero sub-module $M$ on which $O$ acts like multiplication by a scalar $k$. This means that for all $r\in R$ and $m\in M$ one requires $(kr)m=k(rm)=O(rm)=r(Om)=r(km)=(rk)m$ so the sumodule must be annihilated by $kr-rk$ for all $r$; this is very restrictive. $\endgroup$
    – Marc van Leeuwen
    Commented May 28, 2020 at 13:45

3 Answers 3

Reset to default
8
$\begingroup$

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.

Improve this answer
$\endgroup$
2
  • 5
    $\begingroup$ 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. $\endgroup$
    – Will Sawin
    Commented Nov 24, 2011 at 12:09
  • 2
    $\begingroup$ @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. $\endgroup$
    – Bruce Westbury
    Commented Nov 24, 2011 at 13:35
6
$\begingroup$

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.)

Improve this answer
$\endgroup$
2
  • $\begingroup$ Hmm, I'll need to think about that. Thank you! $\endgroup$
    – Gregory Crosswhite
    Commented Nov 24, 2011 at 6:35
  • $\begingroup$ @darijgrinberg's claim that the eigenvalue (I think rather Jordan-form) map is discontinuous: mathoverflow.net/questions/81768/… . $\endgroup$
    – LSpice
    Commented May 22, 2020 at 17:33
2
$\begingroup$

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.

Improve this answer
$\endgroup$
16
  • 3
    $\begingroup$ 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 ... $\endgroup$
    – darij grinberg
    Commented Nov 24, 2011 at 3:46
  • 2
    $\begingroup$ ... 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 ... $\endgroup$
    – darij grinberg
    Commented Nov 24, 2011 at 3:48
  • 2
    $\begingroup$ ... 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. $\endgroup$
    – darij grinberg
    Commented Nov 24, 2011 at 3:50
  • 1
    $\begingroup$ 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 ... $\endgroup$
    – darij grinberg
    Commented Nov 24, 2011 at 3:53
  • 3
    $\begingroup$ @Darij: that seems long enough that you could write it up as an answer, for better visibility $\endgroup$
    – Yemon Choi
    Commented Nov 24, 2011 at 4:22

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .