2
$\begingroup$

TLDR: Group characters and characteristic polynomial have a very similar function but are introduced in very different terms. Can characteristic polynomial be understood through the lens of character theory?


I am trying to get familiar with representation theory by using it to understand the "inner workings" of finite dimensional linear algebra.

Trying to figure out how one could come up with the idea of characteristic polynomials, I stumbled upon this MSE post. Building the characteristic polynomial from traces sounds appealing to me. In particular it makes it sound related to the characters of group representations. So I went on and tried to build the characteristic polynomial in a similar fashion to the characters of a finite group.

I will be working exclusively over $\mathbb{C}$ and representation will mean finite-dimensional representation. Here is a quick reminder on characters for a finite group $G$.


Consider the group algebra $\mathbb{C}[G]$ with its basis $(e_g)_{g\in G}$. Interpreting its elements as functions it is a commutative algebra and interpreting them as "distributions" it has a convolution product $\star$. Its convolution center $Z(\mathbb{C}[G])$ corresponds to the so-called central functions on $G$.
To any representation $\rho : \mathbb{C}[G]\to \operatorname{End}(V)$ is associated a character $$\chi_\rho = \sum_{g\in G} \operatorname{tr}(\rho(g)) e_g$$ Characters of representations are central functions. They satisfy the following properties ($V$ and $W$ are representations of $G$) :

  • If $V\simeq W$ then $\chi_V = \chi_W$
  • $\chi_{V \oplus W} = \chi_V \oplus \chi_W$
  • More generally, for $V\subseteq W$, $\chi_W = \chi_V \oplus \chi_{W/V}$
  • $\chi_{V \otimes W} = \chi_V \chi_W$
  • $\chi_{V^*} = \check{\chi} : g\mapsto \chi(g^{-1})$

Representations of a finite group are unitary thus semi-simple. It turns out that characters of irreductible representations form a linear basis of $Z(\mathbb{C}[G])$, which is moreover orthonormal for the following inner product : $$ <f_1,f_2> = \frac{1}{|G|} (f_1\star f_2)(e) $$ As irreducible representations also generate all the representations, we obtain a map from representations of $G$ to $Z(\mathbb{C}[G])$ which totally characterises representations. A sophisticated way to put it is that we obtain a map $$ \operatorname{Gr}(\operatorname{Rep}_G) \xrightarrow{\chi} Z(\mathbb{C}[G]) $$ which exhibits $Z(\mathbb{C}[G])$ as the complexification of the Grothendieck ring of the representations of $G$. The character gives us a computable characterisation of the representations.


Now onto characteristic polynomials.
We are working with a vector space $E$ equipped with an endomorphism $u$. The endomorphism $u$ is equivalently an action of the monoid $\mathbb{N}$ or more practically of the polynomial algebra $\mathbb{C}[\mathbb{N}]\simeq \mathbb{C}[X]$. We are thus interested by finite dimensional representations of $\mathbb{C}[X]$. Be careful that polynomial multiplication is the convolution product.

It is known that indecomposable representations of $\mathbb{C}[X]$ are of the form $\mathbb{C}[X]/(X-\lambda)^k$. In the Grothendieck group there is an identification $$[ \mathbb{C}[X]/(X-\lambda)^k ] = [ (\mathbb{C}[X]/(X-\lambda))^k ]$$ which is exactly the data forgotten by the characteristic polynomial. The characteristic polynomial gives us a map $$ \operatorname{Gr}(\operatorname{Rep}_{\mathbb{N}}) \xrightarrow{} \mathbb{C}[X] $$ which here again seems to be a complexification map. Of course the structure theorems directly justify these properties but I am looking for a different route.

If we naively copy the trace definition of characters, we obtain $$ \chi_u = \sum_{i\geqslant 0} \operatorname{tr}(u^i) X^i = \operatorname{tr} \left(\frac{1}{1-Xu} \right) $$ This is obviously not the characteristic polynomial.

The characteristic polynomial is the reciprocal polynomial to the following one : $$ P_u(X) = \det(1-Xu) = \operatorname{gtr} (\Lambda^\bullet u : \Lambda^\bullet V \to \Lambda^\bullet V) = \sum_{i\geqslant 0} (-1)^{i} X^i \operatorname{tr}(\Lambda^i u) $$ with $\Lambda^\bullet u$ the induced graded endomorphism of $\Lambda^\bullet V$. The graded trace expression comes from the aforementioned MSE post. I want motivation for this, and tried to compare it to group characters. To begin with, the dependency in $(E,u)$ is no longer additive : as $$ \Lambda^\bullet (E_1\oplus E_2) \simeq \Lambda^\bullet E_1\otimes \Lambda^\bullet E_2 $$ and $$ \Lambda^\bullet (u_1\oplus u_2) \simeq \Lambda^\bullet u_1\otimes \Lambda^\bullet u_2 $$ we now have the following morphism law : $$ P_{u_1\oplus u_2} = \det(1-X(u_1\oplus u_2)) = \det(1-Xu_1)\det(1-Xu_2) = P_{u_1} P_{u_2} $$ so that instead of linear decompositions in $Z(\mathbb{C}[G])$ decomposing a representation will correspond to a polynomial factorisation. I am particularly puzzled by the involvement of the convolution product. I also think the reciprocity relation between $P_u$ and the usual $\det(X-u)$ may have to do with convolution.

In a sense, these are "exponential" characters. There is a simple formula for the logarithm of $\det(1-Xu)$ which is mentioned here : $$ \det(1-Xu) = \sum_{i\geqslant 0} (-1)^{i} X^i \operatorname{tr}(\Lambda^i u) = \exp \left( -\sum_{j\geqslant 1} \frac{1}{j}X^j \operatorname{tr}(u^j) \right) = \exp \big[-\operatorname{tr}\left( \ln(1-Xu) \right) \big] $$ so that there is an expression for the associated "linear" character. It is indeed close but different from the naive suggestion $\operatorname{tr} \left(\frac{1}{1-Xu} \right)$. Now of course the characteristic polynomial is particularly convenient as it is a finite degree polynomial, but I wonder if there is a way to connect the dots.

There is a chance the similarity is only formal. After all, the same polynomial construction can be applied to a group action : $$ P_\rho = \det(1-X\rho) = \sum_{i\geqslant 0} (-1)^{i} X^i \chi_{\Lambda^i \rho} \in \mathbb{C}[G][X] $$ so that the convolution algebra $\mathbb{C}[G]$ and convolution algebra $\mathbb{C}[X]$ may have unrelated origins.


My main question is :

  • Can one start from the idea of characters of representations and end up building the characteristic polynomial?

One related question about a fundamental property of the characteristic polynomial :

  • Is there a character-theoretical proof of the Cayley-Hamilton theorem? (without relying on the structure theorem) ?
$\endgroup$
3
  • $\begingroup$ To your first question: look up the Newton identities. $\endgroup$ Feb 11 at 11:36
  • $\begingroup$ I think the Newton identities are equivalent to the exponential expression I wrote for $\det(1-Xu)$ in terms of $\operatorname{tr}(u^j)$. In a way I am asking why you would care about the elementary symmetric polynomials $\operatorname{tr}(\Lambda^j u)$. $\endgroup$
    – jpdm
    Feb 11 at 11:53
  • $\begingroup$ Maybe this is useful: kconrad.math.uconn.edu/articles/groupdet.pdf $\endgroup$ Feb 15 at 11:06

1 Answer 1

1
$\begingroup$

I would say they are very related, but the relationship is better understood through eigenvalues. Start with a general algebra $A$ over characteristic zero, and a representation $V$ of it. Then the character of $V$ is a linear map $\chi_V:A\rightarrow k$, satisfying certain conditions. The important fact is that in characteristic zero, the data of the map $\chi_V$ is equivalent data to giving the multi set of eigenvalues for the action of each element of $A$. This uses characteristic zero, and is equivalent to the assertion that the power sum symmetric functions generate the whole ring of symmetric functions.

So the general process of taking the character (in characteristic zero) is exactly taking the information of the eigenvalues of each element of $A$ on $V$. In general, this procedure loses information, it’s only in very special cases that it doesn’t, like group algebras in characteristic zero. My claim is that this taking of eigenvalues is actually a more useful way of thinking about characters, since it works well in all characteristics, unlike the trace. For example, considering the eigenvalues yields Brauer characters in the case of modular representation theory of finite groups.

Now for the characteristic polynomial, this is just another way of encoding the multi set of eigenvalues. So the data of the character is equivalent in characteristic zero to the data of the characteristic polynomial of every element acting on the representation.

So what about $\mathbb{C}[X]$? Representations of this encode the notion of “a single linear map acting on a vector space”, and the characteristic polynomial is an efficient encoding of this fundamental invariant, the multi set of eigenvalues. We also see that in this case, the invariant isn’t complete, but it’s a reasonable measurement.

As far as proving the Cayley Hamilton theorem goes, I think the semisimple case is immediate from the product of eigenvalues perspective. The general case seems to use knowledge of what extension problems actually occur (or a zariski density argument), and doesn’t seem to follow formally from properties of this “multi set of eigenvalues” perspective.

$\endgroup$
7
  • $\begingroup$ You raise a good point: finite group (complex) representations have that specific property that the group acts by semisimple elements. In general an endomorphism of $E$ has a nilpotent part. If I get it right the multi-set of eigenvalues you are talking about corresponds to the conjugacy class of the semisimple part of an endomorphism. $\endgroup$
    – jpdm
    Mar 3 at 22:21
  • $\begingroup$ Now the argument about symmetric functions would be that a semisimple endomorphism $u$ is characterised up to conjugacy by the sequence of $(tr(u^j))$. For a group action this is already contained in $\chi$ as powers of group elements are group elements. For an arbitrary $u$ this can be conveniently packaged into the polynomial $\det(1-Xu)=\operatorname{gtr}(u)$. They are both related by $\det(1-X\rho(g)) = \exp (-\sum_{j\geqslant 1} \frac{1}{j} X^j \chi(g^j) )$. From your perspective characters and the char. pol. would be two incarnations of this idea but one is not derived from the other $\endgroup$
    – jpdm
    Mar 3 at 22:24
  • $\begingroup$ Yea, in characteristic zero at least, both characters and the char polynomial are just ways of encoding the same data, the multiset of generalised eigenvalues. I like to think of this as encoding the easy data of an endomorphism, by throwing out the difficult data of unipotence/minimal polynomial. $\endgroup$
    – Chris H
    Mar 3 at 23:16
  • $\begingroup$ If we know that indecomposable $\mathbb{C}[X]$-modules are the $\mathbb{C}[X]/(X-\lambda)^k$ then considering the characteristic polynomial (constructed as a determinant) gives an invariant. But it sounds as a trick you have to "come up with". And as we know the indecomposable modules we can deduce the Cayley-Hamilton theorem. On the other hand, characters of a representation give an invariant, which can be proved to be complete under strong hypotheses. I am hoping these two "tricks" could be reduced to two applications of the same trick. $\endgroup$
    – jpdm
    Mar 4 at 10:21
  • $\begingroup$ I’m not sure I agree with the premise that the char poly sees nilpotent/not semisimple data. In the $2\times 2$ case consider the zero matrix vs a single $1$ in the top right, rest zeros. They both have $X^2$ as their char poly. This is true for the determinant definition and the graded trace definition(which are the same thing). In general, the characteristic polynomial only depends on the semisimple part of the endomorphism (where we split it as a sum of commuting semisimple/nilpotent). $\endgroup$
    – Chris H
    Mar 4 at 21:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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