## October 16, 2011

### Spectra of Operators and Rings

#### Posted by Tom Leinster

There are many uses of the word ‘spectrum’ in mathematics, and most of them are related. (The main exception that I’m aware of is spectra in the sense of homotopy theory.) In particular, the spectrum of a linear operator on a finite-dimensional vector space — that is, its set of eigenvalues — can be seen as the spectrum of an associated commutative ring.

I’ll explain how that story goes. But let me state up front the question that I want to ask. Although it was motivated by my desire to link up the various notions of spectra, it doesn’t actually mention spectra at all.

Let $V$ be a finite-dimensional vector space over a field $k$. Let $T$ be an operator on $V$, in other words, a linear map $V \to V$. Write $\chi(x) \in k[x]$ for its characteristic polynomial. What is the significance of the ring

$k[x]/(\chi(x))?$

Let me explain the kind of answer I’m hoping for. Suppose we replace the characteristic polynomial $\chi$ by the minimal polynomial $m$. Then the ring

$k[x]/(m(x))$

can be seen as the ring of polynomials in $T$. Precisely: write $End(V)$ for the $k$-algebra of operators (endomorphisms) on $V$, with composition as its product. The homomorphism $k[x] \to End(V)$ sending $x$ to $T$ has $(m(x))$ as its kernel, and the image is the subalgebra of $End(V)$ generated by $T$. Hence $k[x]/(m(x))$ is isomorphic to this subalgebra, which consists of the polynomials in $T$.

But is there a nice way to think about $k[x]/(\chi(x))$?

I said that the spectrum of a linear operator $T$ was its set of eigenvalues, but it should really be thought of as a set with multiplicity. The multiplicities here are the algebraic multiplicities. I’ll write $Spec(T)$ for the spectrum of $T$.

Usually people define the algebraic multiplicity of an eigenvalue $\lambda$ as the power of $(x - \lambda)$ appearing in the characteristic polynomial. In the first post I ever wrote for the Café, Linear Algebra Done Right, I explained a more conceptual way to think about the algebraic multiplicity. It’s the dimension of the eventual kernel of $T - \lambda$, defined as

$evKer(T - \lambda) = \bigcup_{i \in \mathbb{N}} Ker((T - \lambda)^i).$

(I’m assuming that we’re over an algebraically closed field.) For that reason, I wanted to use the term ‘dynamic multiplicity’ instead. But I’ll just say ‘multiplicity’.

In any case, we can write the characteristic polynomial of $T$ as

$\chi(x) = (x - \lambda_1)^{\alpha_1} \cdots (x - \lambda_k)^{\alpha_k}$

where $\lambda_1, \ldots, \lambda_k$ are the distinct eigenvalues of $T$ and $\alpha_1, \ldots, \alpha_k$ are their multiplicities. We can form the ring

$R(T) = k[x]/(\chi(x)),$

which is commutative. So, starting from a linear operator $T$, we’ve defined a commutative ring $R(T)$.

Every commutative ring $R$ has a spectrum, $Spec(R)$. At its most basic, the spectrum of a ring is the set of prime ideals; but it also carries a topology and a sheaf of rings. In the jargon, $Spec(R)$ is a ‘ringed space’.

So: given a linear operator $T$, we get two things called ‘spectra’: the set-with-multiplicities $Spec(T)$ of eigenvalues, and the ringed space $Spec(R(T))$. How are they related?

The answer was worked out long ago, and I guess all algebraic geometers know it. Nevertheless, I think there are plenty of people who know both uses of the word ‘spectrum’, but aren’t aware that there’s a close connection between them. So here goes.

For our ring $R(T)$, the (mere) set $Spec(R(T))$ of prime ideals is

$Spec(R(T)) = \{ (x - \lambda_1), \ldots, (x - \lambda_k) \}.$

The points of $Spec(R(T))$ are, therefore, in natural, one-to-one correspondence with the points of $Spec(T)$. That’s an encouraging start.

As a topological space, $Spec(R(T))$ is discrete: every subset is open. A sheaf of widgets on a discrete space $S$ is just an $S$-indexed family of widgets (namely, the stalks of the sheaf). Here, ‘widget’ is ‘ring’. So, to describe the sheaf $O$ with which $Spec(R(T))$ comes equipped, we just have to describe one ring $O_p$ for each point $p$ of $Spec(R(T))$. And in fact, the ring associated to the point $(x - \lambda_i)$ of $Spec(R(T))$ is just

$k[x]/((x - \lambda_i)^{\alpha_i}).$

The point now is that you can recover the multiplicity $\alpha_i$ from this ring. At least, you can do so if you’re allowed to know its $k$-algebra structure: for $\alpha_i$ is just its dimension as a vector space over $k$.

Thus, even if you don’t know $T$, knowing $Spec(T)$ tells you what $Spec(R(T))$ is, and vice versa. In that sense, $Spec(T)$ and $Spec(R(T))$ contain equivalent information about the operator $T$.

This is satisfying, but one question remains. How do we understand the ring $R(T)$? As I said at the beginning, it would be easily understandable if we’d used $k[x]/(m(x))$ instead of $k[x]/(\chi(x))$; but then the spectrum would tell us the powers of $(x - \lambda_i)$ in the minimal polynomial, rather than the characteristic polynomial, and that’s not such important information.

Maybe an example will make the problem clearer. In fact, a really trivial example will do: $T = 0$. Say $dim(V) = N$. Then

$m(x) = x, \quad \chi(x) = x^N.$

Hence

$k[x]/(m(x)) = k, \quad R(T) = k[x]/(\chi(x)) = k[x]/(x^N).$

Conceptually, what is the process that turns the zero operator on $k^N$ into the ring $k[x]/(x^N)$?

Posted at October 16, 2011 1:58 PM UTC

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### Re: Spectra of Operators and Rings

Deformation? Given a linear operator you can consider the moduli space of all linear operators with the same characteristic polynomial. Generically they should have minimal polynomial the characteristic polynomial, so an arbitrarily small deformation of a given operator (preserving its characteristic polynomial) will generate an algebra isomorphic to the quotient by the characteristic polynomial. I don’t know how satisfying this is, though.

Posted by: Qiaochu Yuan on October 16, 2011 6:47 PM | Permalink | Reply to this

### Re: Spectra of Operators and Rings

Thanks, Qiaochu, but I’m not sure I understand you.

First, I guess your opening sentence “Deformation?” is the answer to the closing sentence of my post. Is my guess right? If so, I’m not really sure how to assemble the grammar of your answer. “The deformation of the zero operator on $k^N$ is the ring $k[x]/(x^N)$” doesn’t really sound right.

Second, you wrote:

an arbitrarily small deformation of a given operator […] will generate an algebra …

What do you mean by “generate an algebra” here? That is, what process are you using to turn an operator into an algebra? My guess is that you’re assigning to an operator $S$ the ring $k[x]/(m_S(x))$, where $m_S$ is the minimum polynomial of $S$. Is that right?

Thanks!

Posted by: Tom Leinster on October 16, 2011 8:30 PM | Permalink | Reply to this

### Re: Spectra of Operators and Rings

Sorry, I guess I worded my reply confusingly. I think Andrew Stacey’s version of this argument probably has the best ratio of explanatory power to background necessary, so here’s my spin on it:

Given any linear operator T (I’ve forgotten how to typeset math here!) one can consider operators close to T, say in some matrix norm. Generically these operators have distinct eigenvalues, so their characteristic and minimal polynomials coincide. In particular we can find a sequence of such operators converging to T. Of course their characteristic polynomials converge to the characteristic polynomial of T, but it’s not necessary for their minimal polynomials to converge to the minimal polynomial of T; this is the sense in which the minimal polynomial behaves poorly as a function on the space of matrices.

So the characteristic polynomial is what the minimal polynomial “should be” if T had more room to move around, in the same way (as in Jason Starr’s response) that the intersection multiplicity of, say, two algebraic curves is what the number of intersections “should be” if the curves had more room to move around.

Posted by: Qiaochu Yuan on October 20, 2011 7:03 PM | Permalink | Reply to this

### Re: Spectra of Operators and Rings

Great; thanks.

I’ve forgotten how to typeset math here!

Just enclose it between dollar signs as usual. (There are a few quirks.)

The only thing to know is that when you’re writing your comment, you have to choose an appropriate “text filter” from the dropdown menu. I guess anything containing “itex” will work. I use “itex to MathML with parbreaks”.

Re characteristic and minimal polynomials, I get the point now. Thanks. But remember the original question: I feel like I have a good intuitive understanding of what algebraic multiplicity is, and therefore, for the most part, what the characteristic polynomial is. What I wanted to do was to transform that into a good understanding of what $k[x]/(\chi(x))$ is.

Posted by: Tom Leinster on October 20, 2011 9:15 PM | Permalink | Reply to this

### Re: Spectra of Operators and Rings

I agree with Qiaochu Yuan. The assignment to every linear operator $T$ of the ring $k[x]/\chi_T(x)$ is the only assignment of quotient rings which “generically” is $k[x]/m_T(x)$ and which is “flat” in $T$. This is precisely the same issue as comes up in assigning multiplicities to non-transverse intersections in algebraic geometry. It might seem non-intuitive to “count” certain solutions with multiplicities, but it is the only way to count which is “continuous” in the parameters.

Posted by: Jason Starr` on October 16, 2011 8:08 PM | Permalink | Reply to this

### Re: Spectra of Operators and Rings

Thanks, Jason. By the way, to type Latex here, you need to select an appropriate option from the drop-down menu marked “Text filter” in the comment form. I use “itex to MathML with parbreaks”.

The message I’m getting is that you understand $k[x]/(\chi_T(x))$ by thinking not only about $T$ itself, but also about “nearby” operators. I don’t really know what “nearby” means, but I guess that’s because I don’t understand what “flat” means in this context. Let me see. The process

$T \mapsto k[x]/(\chi_T(x))$

takes an element of the endomorphism ring $End(V)$ and turns it into a ring. So, it’s an $End(V)$-indexed family of rings. (Or you could replace “ring” by “$k$-algebra” throughout the previous two sentences.)

If I understand you correctly, whenever $R$ is a ring and $(A_r)_{r \in R}$ is an $R$-indexed family of rings, there must be a definition of what it means for such a family to be “flat”. Am I warm?

Posted by: Tom Leinster on October 16, 2011 8:37 PM | Permalink | Reply to this

### Re: Spectra of Operators and Rings

I searched for “flat family” in Wikipedia, but it just asked me if I meant “fat family”. Nope.

Posted by: Tom Leinster on October 16, 2011 10:31 PM | Permalink | Reply to this

### Re: Spectra of Operators and Rings

I would recommend that you look in the index of any commutative algebra book under the word “flat”.

Posted by: Jason Starr on October 21, 2011 2:53 AM | Permalink | Reply to this

### Re: Spectra of Operators and Rings

Jason, your first comment was helpful, but this one not so much so.

I guess you think I’m being lazy, asking people to explain stuff that I could look up in any commutative algebra book. But you’re overestimating the clarity of your original comment.

I know the definitions of flatness for modules over rings, for homomorphisms of rings, and for algebras. Sure, a random commutative algebra book will include those definitions. But as far as I can tell, none of those three definitions covers the notion of flatness invoked in your comment:

The assignment to every linear operator $T$ of the ring $k[x]/\chi_T(x)$ is the only assignment of quotient rings which “generically” is $k[x]/m_T(x)$ and which is “flat” in $T$.

You’re saying, among other things, that the assignment $T \mapsto k[x]/(\chi_T(x))$ is “flat”. Now $T$ is an element of the endomorphism ring of $V$, and $k[x]/(\chi_T(x))$ is a ring, so I guessed that there was some definition of what it means for a ring-indexed family of rings to be “flat”. By a “ring-indexed family of rings” I mean a ring $R$ together with a ring $A_r$ for each $r \in R$. I asked you in this comment whether that was the correct interpretation of your comment, but I still don’t know the answer.

So, let me ask again: are you alluding to some notion of flatness of a ring-indexed family of rings? If so, what is that definition?

Perhaps there’s a canonical way of turning a ring-indexed family of rings into a module, and then it would be clear what it meant for it to be flat. Perhaps my guessed interpretation of what you wrote was wrong. In any case, if your second comment is correct, the notion of flatness used in your first comment must somehow be a case of the notion of flatness for modules.

But “look in the index of any commutative algebra book under the word ‘flat’” is simply unhelpful.

Posted by: Tom Leinster on October 21, 2011 4:22 PM | Permalink | Reply to this

### Re: Spectra of Operators and Rings

You are correct: I think you could look this up in any commutative algebra textbook.

Posted by: Jason Starr on October 22, 2011 4:51 AM | Permalink | Reply to this

### Re: Spectra of Operators and Rings

Eisenbud? Atiyah-Macdonald? Schenck’s little blue book?

Posted by: Yemon Choi on October 22, 2011 5:35 PM | Permalink | Reply to this

### Re: Spectra of Operators and Rings

I tried Matsumura, to no avail. I’m absolutely willing to look stuff up, but at present I don’t know what to look up in order to understand Jason’s original comment. If someone knows and can tell me, I’d be interested, and it would be good to get something positive out of this conversation.

All usages of “flat” appearing in your average commutative algebra book (e.g. flat homomorphism of rings, flat algebra) are derived from the notion of flatness for a module. So, Jason seems to be saying (i) that from the family

$\Bigl( k[x]/\chi_T(x) \Bigr)_{T \in End(V)},$

one can derive a module, and (ii) that module is flat. It’s (i) that’s a mystery to me, and which no one seems able or willing to explain.

On the other hand, David R’s brief comment suggests that one needs to invoke the notion of flat morphism of schemes, which certainly doesn’t appear in all comm alg books.

Posted by: Tom Leinster on October 22, 2011 6:02 PM | Permalink | Reply to this

### Re: Spectra of Operators and Rings

Let $S$ be the polynomial algebra generated by the vector space of all linear operators on a finite dimensional vector space $V$. Let $S[t]$ be the polynomial ring over $S$. This is a flat $S$-module. The only quotient $S[t]$ module which is $S$-flat and which agrees with $S[t]/m_T(t)$ over the generic point is $S[t]/\chi_T$.

Posted by: Jason Starr on October 22, 2011 7:30 PM | Permalink | Reply to this

### Re: Spectra of Operators and Rings

OK, this is going nowhere. Let’s leave it there.

Posted by: Tom Leinster on October 22, 2011 12:55 PM | Permalink | Reply to this

### Re: Spectra of Operators and Rings

Try Flat morphism instead. You’ll need to know that a flat map of rings $R\to S$ is one that makes $S$ a flat $R$-module.

Posted by: David Roberts on October 17, 2011 1:34 AM | Permalink | Reply to this

### Re: Spectra of Operators and Rings

Hmm. Thanks, David, but this is getting pretty indirect. It looks as if, in order to understand what you’re saying, I’d have to do all of the following:

1. understand how a family $(A_r)_{r \in R}$ of rings indexed over a ring $R$ can be regarded as a morphism of schemes
2. understand what it means for a morphism of schemes to be flat
3. put these together to understand what it means for a ring-indexed family of rings $(A_r)_{r \in R}$ to be flat
4. specialize to understand what it means for the particular ring-indexed family of rings $(k[x]/(\chi_T(x)))_{T \in End(V)}$ to be flat
5. convince myself that it is indeed flat
6. convince myself that it’s the only flat family that’s “generically” the same as $(k[x]/(m_T(x)))$, which of course involves…
7. …understanding what “generically” means
8. take this knowledge that $(k[x]/(\chi_T(x)))$ is the unique flat family generically agreeing with $(k[x]/(m_T(x)))$, and somehow use it to improve my understanding of what $k[x]/(\chi_T(x))$ looks like, for any individual operator $T$.

I get the idea at the level of vagueness used in Jason’s first sentence. But I was really hoping to get some vivid picture of the ring $k[x]/(\chi_T(x))$. The algebraic-geometric intuition that Jason described seems to depend on thinking about a varying $T$. Evidently that can be helpful, but I’d like to be able to understand better what the ring looks like for an individual $T$.

As evidence that this might be possible, I do have a vivid picture of the algebraic multiplicities of $T$: they’re the dynamical quantities discussed in the post. But I haven’t managed to use that to get an equally vivid picture of $k[x]/(\chi_T(x))$.

Posted by: Tom Leinster on October 17, 2011 6:47 PM | Permalink | Reply to this

### Re: Spectra of Operators and Rings

There’s a more explicit answer to the question, which of course is morally equivalent to that of Qiaochu and Jason, but has a proud history in representation theory (under names like “Kostant slice”). Namely, to any matrix we can attach a unique matrix up to conjugacy with the same characteristic polynomial but which is regular, i.e., all Jordan blocks have distinct eigenvalues <==> the dimension of its stabilizer is minimal <==> the matrix admits a cyclic vector (so can be written in “companion”/rational cyclic form with all 1’s on first subdiagonal and stuff on the last column). The scheme you write is the spectrum of this associated matrix. As Qiaochu and Jason explain this spectrum is the only well-behaved specturm one can attach to a general matrix, but the specific realization as the spectrum of a regular “replacement” for your matrix is used a lot in representation theory.

More formally, this describes a slice to the adjoint quotient map from the Lie algebra $g$ of matrices to $g // G = h /W$. The slice lands in the locus of regular matrices. Kostant generalized this construction to arbitrary semisimple Lie algebras (and quantized it to give a model for the center of the enveloping algebra of $g$), Steinberg has a group version, Slodowy generalized this to slices to nonregular nilpotent orbits (whose quantizations are known as W-algebras), Hitchin used it to construct a section for the Hitchin integrable system, etc…

Posted by: David Ben-Zvi on October 17, 2011 5:31 AM | Permalink | Reply to this

### Re: Spectra of Operators and Rings

This makes one wonder if it can be done for, say, nilpotent $L_\infty$-algebras…

Posted by: David Roberts on October 17, 2011 8:12 AM | Permalink | Reply to this

### Re: Spectra of Operators and Rings

Thanks, David B-Z.

You and the others have said that the assignment

$T \mapsto k[x]/(\chi_T(x))$

is “well-behaved”, but the assignment

$T \mapsto k[x]/(m_T(x))$

is not. Is this an incarnation of the fact that the coefficients of $\chi_T(x)$ are polynomials in the entries of the matrix of $T$, but those of $m_T(x)$ are not?

Posted by: Tom Leinster on October 17, 2011 7:01 PM | Permalink | Reply to this

### Re: Spectra of Operators and Rings

Just thought I’d weigh in in favour with everyone else (nothing like a bandwagon for joining). When I teach about the characteristic polynomial then I first teach the minimum polynomial and then say that the characteristic polynomial is the “continuous minimum polynomial”. So it’s what you get if you want a continuous assignment $T \mapsto p_T(t)$ with the property that $p_T(T) = 0$ and if $T \mapsto q_T(t)$ is another such assignment then $p_T(t) | q_T(t)$.

I actually used this in a paper as I had a family of unitary operators depending on a parameter (in the circle as it happened). I could interpret these as a single operator acting on the loop space, $L\mathbb{C}^n$, and I had the condition that when I did so then the resulting operator had a minimum polynomial. I wanted to show then that the family of operators was constant, modulo similarity.

The argument went as follows. Suppose that $t \mapsto T_t$ is my family of unitary operators on $\mathbb{C}^n$ and write $T$ for the operator on $L \mathbb{C}^n$ defined by $\gamma \mapsto (t \mapsto T_t(\gamma(t))$. Let $m_T$ be its minimum polynomial (which exists by assumption) and let $m_{T_t}$ be the minimum polynomial of $T_t$ (which exists by finite dimensionality). Then $m_{T_t} | m_T$ so, using $c$ for the corresponding characteristic polynomials, $c_{T_t} | (m_T)^n$, since the missing multiplicities are of at most $n$. Now $c_{T_t}$ is a monic polynomial so is completely determined by its roots, but these must be drawn from the roots of $(m_T)^n$ (with multiplicity) and that is a discrete set. Thus since $c_{T_t}$ depends continuously on $t$, it must be constant. Since our operators were unitary, they were diagonalisable, and so their characteristic polynomial determines their similarity class.

That usefulness combined with my dislike of determinants (for similar reasons to you!) leads to my teaching of characteristic polynomial as a “continuous minimum polynomial”.

Posted by: Andrew Stacey on October 18, 2011 7:50 AM | Permalink | Reply to this

### Re: Spectra of Operators and Rings

Andrew wrote:

characteristic polynomial as a “continuous minimum polynomial”

I like it!

Posted by: Tom Leinster on October 20, 2011 9:16 PM | Permalink | Reply to this

### Re: Spectra of Operators and Rings

In that case, I’ll add a couple of links. My lecture notes from last year on this are Lecture 20 (Polynomials) on http://www.math.ntnu.no/~stacey/Teaching/TMA4145h2010/lectures.html and the article where I used this argument was Finite dimensional subbundles of loop bundles, details at http://www.math.ntnu.no/~stacey/Research/Papers/looprep.html.

Posted by: Andrew Stacey on October 21, 2011 8:18 AM | Permalink | Reply to this

### Re: Spectra of Operators and Rings

Nice! Also, I’ve never seen so many colours used in a single formula.

Posted by: Tom Leinster on October 21, 2011 3:37 PM | Permalink | Reply to this

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