6
$\begingroup$

What is the relation between characters of a group and its lie algebra?

Roughly,I know that there is a one to one correspondence between representations of a lie algebra and its simple connected lie group by the exp map,and two irreducible representations of a lie group are unequivalent if and only if their characters are different. Now,then,I think the above statement will still hold for representations of lie algebras, but I am not sure about it . If someone can give some advises?

Thanks Kevin Buzzard. For represatatons of general lie algebras the character (trace) don't work .
But for compact lie group,characters still work,so I restrict my question on compact lie groups and their lie algebras.
In representation theory of semisimple lie algebras ,we have formal characters/algebra characters which are sums of some formal elements over the weights. I wonder whether we can realize these algebra characters to be real characters ?
Or can we make them to be functions on cartan subalgebras or maximal torus?

$\endgroup$
2
  • 3
    $\begingroup$ My advice to you is: (1) if you want to learn about the representation theory of semi-simple Lie algebras, read some of the very nice books by e.g. Serre or Dixmier, and (2) if you think about nilpotent Lie algebras then you might be in big trouble with traces. For example it seems to me that if k is a field of characteristic 0 and we make it a Lie algebra by making all brackets equal to 0 then k has a reducible 2-dimensional representation sending a to (0,a;0,0) and every element has trace 0... $\endgroup$ Feb 25, 2010 at 15:06
  • 1
    $\begingroup$ You should read one of the very good textbooks on representation theory; you'll find there how the 'formal' characters indeed arise as actual functions. $\endgroup$ Feb 25, 2010 at 16:48

3 Answers 3

14
$\begingroup$

The primary reason for studying Lie algebras is the following fundamental fact: the representation theory of a Lie algebra is the same as the representation theory of the corresponding connected, simply connected Lie group.

Of course, the representation theory of a Lie group in general is very complicated. First of all, it might not be connected. Then the Lie algebra cannot tell the difference between the whole group and the connected component of the identity. This connected component is normal, and the quotient is discrete. So to understand the representation theory of the whole group requires, at the very least, knowing the representation theory of that discrete quotient. Even in the compact case, this would require knowing the representation theory of finite groups. For a given finite group, the characters know everything, but of course the classification problem in general is completely intractable.

The other thing that can go wrong is that even a connected group need not be simply connected. Any connected Lie group is a group quotient of a connected, simply connected group with the same Lie algebra, where the kernel is a discrete central subgroup of the connected, simply connected guy. So the representation theory of the quotient is the same as the representations of the simply connected group for which this central discrete acts trivially.

There is a complete classification of connected compact groups. You start with the steps above: classifying the disconnected groups is intractable, but a connected one is a quotient of a connected simply connected one. This simply connected group is compact iff the corresponding Lie algebra is semisimple; otherwise it has some abelian parts. In general, any compact group is a quotient by a central finite subgroup of a direct product: torus times (connected simply connected) semisimple. Torus actions are easy, and the representation theory of semisimples is classified as well. Whether your representation descends to the quotient I'm not entirely sure I know how to read off of the character. When the group is semisimple (no torus part), I do: finite-dimensional representations are determined by their highest weights (which can be read from the character), which all lie in the "weight" lattice; quotients of the simply-connected semisimple correspond exactly to lattices between the weight lattice and the "root" lattice, and you can just check that your character/weights are in the sublattice.

All of this should be explained well in your favorite Lie theory textbook.

$\endgroup$
8
$\begingroup$

A formal character of a finite-dimensional representation can be, indeed, thought of as a trace function. You can define it on Cartan subalgebra, maximal torus, the set of semsimple elements of the complex group, or compact group as you wish. This is what Weyl was doing but we are well pass this point.

The reason is that formal characters can be done for infinite dimensional representations and they are useful. There you will have numerous problems interpreting it as a function. See chapter 14 of Pressley-Segal where they have a long philosophical discussion whether the character of a loop group is a distribution or a hyperdistribution or something else.

$\endgroup$
5
$\begingroup$

Let $G$ be a compact Lie group. Then the character of a finite-dimenstional representation $R_Λ$ of $G$, is the map $\mathrm{ch}_Λ\colon G \to \mathbb{C}$ defined by $$ \mathrm{ch}_Λ(g):=\mathrm{Tr}(R_Λ(g))\equiv \sum_{n=1}^{\dimΛ}R^{n,n}_Λ(g), \qquad\forall g\in G, $$ and it depends only on the isomorphism class or the representation $R_Λ$. Furthermore, because of the cyclic invariance of the trace the character is a class function. Being class functions, the characters are completely determined by their values on the maximal torus $\mathbb{T}\subset G$, for $G$ semisimple. Now, the characer of the Lie algebra $\mathfrak{g}$ of $G$ is defined as $$ χ_Λ(h)=\mathrm{Tr}(\exp [R_Λ(h)]), $$ for $h$ in a Cartan subalgebra $\mathfrak{g_0}$ of $\mathfrak{g}$. If $t\in\mathbb{T}$, then $t=\exp h$ with $h\in\mathfrak{g_0}$, the Cartan subalgebra associated to $\mathbb{T}$. Therefore, one has the formula $$ χ_Λ(h)=\mathrm{ch}_Λ(\exp h). $$ However, this does not in general work globally on $\mathbb{T}$ because a single exponential may not be sufficient to obtain all of $\mathbb{T}$ from $\mathfrak{g_0}$ [1, Sec. 21.10] .

References

[1] J. Fuchs & C. Schweigert, Symmetries, Lie Algebras and Representations: A graduate course for physicists, Cambridge University Press (2003).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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