While reading chapter 7.1.2 of di Francesco I encountered following definition of character of Verma module: $$\chi_{(c,h)}(\tau)=\text{Tr }q^{L_0 -c/24}$$ where $q = e^{2 \pi i \tau}$ and $\tau$ and c are two parameters (c-central charge, $\tau$ - some parameter connected to so called modular invariance). I am trying to understand logic beneath this definition.

In order to construct Verma module one usually acts in analogy with $SO(3) $ group. So that is what I tried to do as well. I found slightly more detailed description of latter case in Jones H. "Groups, Representations and Physics". So I am going to follow it.
This is where I arrive to first confusion. According to its definition character is vector which elements are given by traces of the group (algebra) elements. However it seems to be common practice to call trace of single element of group (algebra) as character. For instance for given irrep of $SO(3)$: $$\text{One of generators: } X_3 = \text{diag}(j,j-1,...-j+1, -j)$$ $$\text{Corresponding group element: } R_3(\varphi)=e^{-iX_3 \varphi}$$ $$\text{and finally: }\chi^{j}(\varphi)=e^{-ij\varphi}+...+e^{ij\varphi}$$ which they originally call(as of course it is) trace of $R_3(\varphi)$. Yet in next sentence they call it character of rotation through $\varphi$ about certain axis.
Note that explicit form of $X_3$ may be immediately seen from following expression: $$\langle jm'|X_3|jm\rangle = m \delta_{m',m}$$ Now note that $\text{Tr}(R_3(\varphi))= \text{Tr}(R_n(\varphi))$ for arbitrary n. This follows from:

...character of rotation through $\varphi$ about axis 3 is also character of rotation through $\varphi$ about any other axis. It is so because the conjugacy classes of rotations are all rotations around the same angle about different axes.

So we arrived to the conclusion that $\chi^{j}(\varphi)$ known as character of rotation through $\varphi$ tells us value of trace of any element of group $SO(3)$ for given irrep.

Now I want to take above intuition in order to justify expression for character of Verma module.
First note that $L_0$ is analogous to $X_3$ and $L_{\pm m}$ are ladder operators for it. Based on def of $L_0$ we can show that: $$\langle h+N'|L_0 |h+N\rangle = (h+N)\delta_{N',N}$$ Taking into account that(I do not know how to prove it and whether it is indeed the case) $$\text{Tr}(q^{L_0 -c/24})=\text{Tr}(q^{L_n -c/24}) \quad \text{,for arbitrary } n $$ we can define $$\chi_{(c,h)}(\tau)=\text{Tr }q^{L_0 -c/24} = \sum_{n=0}^{\infty} \text{dim}(h+n) q^{n+h -c/24}$$ Now $\chi_{(c,h)}(\tau)$ may be understood(up to some constants) as trace of any element of group, corresponding to certain representation of Virasoro algebra.
Is it what is usually meant by character of Verma module?

  1. The most basic character one encounters in representation theory is a function $\chi(g)$ which takes an element $g$ of the group to the trace $\mathrm{tr}\,U(g)$ of the representation $U(g)$ of this element. In this definition, the character is a $\mathbb{C}$-valued (or $\mathbb{R}$-valued for real represenations, etc.) function defined on the whole group $G$. Since $$\mathrm{tr}\,U(h^{-1}gh)=\mathrm{tr}\,U(h)^{-1}U(g)U(h)=\mathrm{tr}\, U(g),$$ the character only depends on the conjugacy class of the element $g$ -- we say that characters are "class functions". (In fact, for compact or finite groups the characters of irreducible representations form an orthonormal basis for class functions.)

  2. For finite groups the character can be determined by specifying its values on the finite number of conjugacy classes in the group, so it may be natural to think of it as of a vector. For Lie groups and the like, the character would be an infinite-dimensional vector in this interpretation.

  3. For compact connected Lie groups any element is conjugate to an element in the subgroup ("maximal torus") generated by the Cartan subalgebra. Because of this, the full information about a class function is contained in its values on the maximal torus. (However, these values still should be invariant under Weyl group.) Since any element in the maximal torus is an exponentiation of an element of Cartan algebra, $g=\exp(h)$, we can simply consider the function $$ \chi(\exp(h))=\mathrm{tr}\,e^h=\sum_{\alpha}e^{\alpha(h)}, $$ where the sum is over weights of the finite-dimensional representation. In $SO(3)$ case the Cartan algebra is one-dimensional, so both weights and Cartan elements are just numbers. More precisely, we identify $h=\phi$, $\alpha=-ij,-i(j-1),\ldots ,+ij$. Note that from the above formula it follows that the knowledge of the character is equivalent to the knowledge of all weights and their multiplicities. Recall that a weight is nothing but the set of eigenvalues corresponding to a common eigenvector of the Cartan generators. So it also looks a bit like a partition function (think of one of the Cartan generators as energy). This brings us to why Virasoro characters are important.

  4. When we talk about weird algebras/groups as Virasoro, the general theory is also weird. But if we have an idea what the Cartan algebra is (i.e. we have a set of important mutually commuting generators), we can ask whether the generators are simultaneously diagonalizable in our representation, and if they are, ask for the eigenvalues and their multiplicities. This is precisely what the Virasoro character is doing -- it counts the eigenvalues of $L_0$ (which is the only "Cartan generator" for us here), and nothing else, $$ \mathrm{tr}\,q^{L_0-c/24}=\sum_{L_0\text{ eigenvalues}}\langle\text{multiplicity}\rangle q^{\langle L_0\text{ eigenvalue}\rangle-c/24}=\sum_{n=0}^\infty \dim(h+n)q^{h+n-c/24}. $$

  5. You should probably think of Virasoro character just as a way of organizing information about states in the Verma module, not much more. It is useful to organize it this way because in the end the partition function on the torus can be expressed in terms of these characters (but this is not a complicated fact, it is just the observation that the partition function is given by almost exactly the same formula).

  • $\begingroup$ Thank you for the answer! It helped me a lot. I still need to go through some details a bit more carefully, yet your post definitely clarified most of things to me! $\endgroup$ – Yaroslav Shustrov Dec 28 '16 at 12:00

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