Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $G$ be a compact group and let $R(G)$ be the representation ring of $G$. Additively, $R(G)$ is generated by the irreducible representations of $G$. Usually one only deals with those representations which are nonnegative integer combinations of the irreducible representations. However, often one has formulas which apply to an arbitrary element of $R(G)$, and so includes virtual representations which (for this question at least) are the elements of $R(G)$ whose decomposition into irreducible components includes negative coefficients.

Question: Is there any 'natural' interpretation of virtual representations? In particular, aside from the obvious interpretation as the elements of the formal completion of the semiring of ordinary representations to a ring, is there a natural way to view these objects? Especially helpful would be pictures/ideas others use to assign meaning to virtual representations (if any).

Motivation: Often in decomposing various formulas involving characters, virtual representations arise in one way or another. For example, in Lie groups, the notion of a highest weight representation $\rho_\omega$ can be extended to arbitrary weights $t$ of $G$ via:

$\rho_{w(t)} = (-1)^w\rho_t$

In particular, if $t$ is not a dominant weight of $G$ then there is a unique $w$ in the Weyl Group of $G$ such that $w(t)$ is a dominant weight so that $\rho_t$ is defined for arbitrary weights $t$; note that the Weyl Dimension Formula agrees with this extension. If the length of $w$ is odd, then $\rho_t$ will have negative dimension from the dimension formula, hence is virtual.

Another simple example is that when considering the action of the Adams operation $\psi^k$ on $R(G)$, one has that $\psi^k(\rho)$ is in general a virtual representation.

It happens that from time to time I come across other instances of virtual representations appearing in equations I am considering, and I always work with them ignoring whether they have a physical interpretation or not, but at the same time it would be more satisfying if I could understand the equations as manifestations of some deeper structure.

Edit: Per Qiaochu's comment, yes, virtual representations can be fit into the framework of super-representations. If it is the case that virtual representations are often viewed as super-representations, then perhaps someone could elaborate on why super-representations are so natural and how one works around the dimension mismatches between virtual representations and super-representations, i.e. $dim(\rho_1\ominus\rho_2) = dim(\rho_1)-dim(\rho_2)$ but the dimension of the corresponding super-representation is $dim(\rho_1)+dim(\rho_2)$.

share|improve this question
1  
Perhaps they are super-representations...? –  Qiaochu Yuan Feb 21 '11 at 17:43
    
@Qiaochu: I admit that the super-representation thought had crossed my mind awhile back and while it works, it still seems that it is a bit artificial (partly because it does not respect dimensions and partly because it is just introducing a second copy of the original space and declaring that products between the spaces respect a $\mathbb{Z}/2\mathbb{Z}$-grading). –  ARupinski Feb 21 '11 at 17:54
2  
Maybe this is a trivial comment, but if X is a space on which G acts, then since the (co)homology groups of X are representations of G, the Euler characteristic of X can be souped up to a virtual representation of G. This is a nice "source" of naturally occurring virtual representations. –  Dan Petersen Feb 21 '11 at 18:37
2  
ARupinski- There is a notion of dimension internal to symmetric rigid tensor categories (such as the category of supervector spaces) and it gives the differences, not the sums, of the dimensions. This is the notion of dimension that everyone who thinks about super-stuff uses. –  Ben Webster Feb 21 '11 at 18:47
1  
Just to mention, since you give it as one of your motivations, that the example with the Weyl character formula makes perfect sense from the point of view of complexes of representations: if $\lambda$ is \textit{any} weight, you can form an associated line bundle $L_\lambda$ over the flag variety $G/B$. All of its cohomologies $H^i(G/B,L_\lambda)$ are $G$-representations, and the virtual representation given by their Euler characteristic gives exactly the "signed" Weyl character you wrote down. –  Kevin McGerty Feb 21 '11 at 23:47
show 2 more comments

3 Answers

up vote 14 down vote accepted

Your question is really about virtual vector spaces: what is a virtual vector space?
Once you know what a virtual vector space is, then there is only a small step to the answer of your question.

There are a few possible answers:
1• A virtual vector space of a pair of vector spaces. Equivalently, it's a $\mathbb Z/2$-graded vector space. You should think of the pair $(V,W)$ as being the formal difference $V-W$. This approach has the downside that it makes it unclear what an isomorphism between virtual vector spaces should be.

2• A vector space of dimension $n$ is the same thing as a point (sic!) of the topological space $BU(n)$. A virtual vector space is then a point of the space $BU:=\mathrm{colim}_{n\to \infty} BU(n)$. This approach is not geometric at all, but works very well for talking about virtual vector bundles on a space $X$: these are continuous maps $X\to BU$.

3• Fix an infinite dimensional vector space $U$ and a polarization $U=U_-\oplus U_+$ (both $U_-$ and $U_+$ are infinite dimensional). A virtual vector space is a (necessarily infinite dimensional) subspace $V\subset U$ such that $V\cap U_-$ is of finite codimension inside $V+U_-$.

4• Fix an infinite dimensional Hilbert space $H$. A virtual vector space is a Fredholm operator $F:H\to H$ (i.e., an operator with finite dimensional kernel and cokernel). This is related to definition 1 by assigning to the Fredholm operator $F$ the pair $(\ker(F),\mathrm{coker}(F))$.

5• A virtual vector space is an object of the bounded derived category of vector spaces: i.e., it's a chain complex.

All these definitions (with the exception of 2, for which it's more involved) can be easily adapted to the context of $G$-representation, you just need to replace "infinite dimensional vector space" with "$G$-rep that contains each irrep infinitely often".

share|improve this answer
    
What's unclear about the first approach? If you only want the monoidal and abelian structures then it's just the category of representations of Z/2Z, and the braiding is a detail. –  Qiaochu Yuan Feb 21 '11 at 18:53
1  
I suppose one could instead consider the category of 2-term chain complexes and work up to chain homotopy. –  Qiaochu Yuan Feb 21 '11 at 18:56
    
I like this since it at least gives several different ways of thinking about virtual vector spaces, at least some of which I have enough background to begin to understand. –  ARupinski Feb 21 '11 at 23:18
    
These are nice interpretations. One still has the question concerning Adams operator identities like $\bigwedge V \cong \exp(-\sum_{n > 0} \frac{\psi^i(V)}{i})$ which at least appear to require rational tensor powers. –  S. Carnahan Feb 22 '11 at 2:41
    
@Qiaochu: You might want to declare the pairs $(V_0,V_1)$ and $(V_0\oplus W,V_1\oplus W)$ to be isomorphic. But then it's unclear how to compose those "isomorphisms" (and their inverses) –  André Henriques Feb 22 '11 at 17:12
add comment

Probably the general answer to your question is "no", especially if you are working in a classical situation where all representations are completely reducible. Your example involving Adams operations illustrates that mysterious things can happen with virtual representations, for which I'm unaware of any better interpretation.

In other situations, such as modular representations of finite groups or reductive algebraic groups, it is natural at times to look at nonsplit exact sequences of modules. This leads to a kind of "Euler character" as a formal $\mathbb{Z}$-linear combination of modules in the sequence, with coefficients actually $\pm 1$. (This comes up for example in the study of sheaf cohomology groups of line bundles on a flag variety, where the group acts naturally in each degree but Kodaira vanishing can break down badly.)

In another direction, the Deligne-Lusztig construction of ordinary complex representations of a finite group of Lie type yields in general an etale cohomology complex. They can extract from it at first just the Euler character; here the integral coefficients involved can get complicated and are usually unknown, but the Euler character itself contains lots of information. Here at least you have a cohomology interpretation for the signed contributions to the end result.

share|improve this answer
add comment

Andre's answer above does a perfect job of explaining some of the abstract theory of "virtual" representations (of an arbitrary group). I'd like to give one answer to the special case you bring up, when $G$ is a compact group. Or, even better, I'll work with the case when $G = \mathfrak g$ is a (finite-dimensional) semisimple Lie algebra over $\mathbb C$.

Then one place that the signs you talk about come up is in the Weyl character formula. Pick a triangular decomposition $\mathfrak g = \mathfrak n^- \oplus \mathfrak h \oplus \mathfrak n^+$. For any weight $\lambda$, let $V(\lambda)$ denote the Verma module with heighest weight $\lambda$ (recall that $V(\lambda) = {\rm U}\mathfrak g \hspace{1ex}\otimes_{\rm U\mathfrak b}\hspace{1ex} \mathbb C_\lambda$, where $\mathfrak b = \mathfrak h \oplus \mathfrak n^+$ is the standard Borel, and $\mathbb C_\lambda$ is the one-dimensional $\mathfrak b$-module on which $\mathfrak h$ acts by $\lambda$ and $\mathfrak n^+$ acts by $0$). Let $L(\lambda)$ denote the quotient of $V(\lambda)$ by its maximal proper ideal; it is the unique irreducible $\mathfrak g$-module with heighest weight $\lambda$. Recall that $\mathfrak h$ acts semisimply on any sufficiently nice $\mathfrak g$-module (including $M(\lambda)$ and its quotients and tensorands and summands; "sufficiently nice" is codified, for the present purposes, by "in the category $\mathcal O$"), and that the K-group of the semisimple $\mathfrak h$-modules is precisely the group ring of $\mathfrak h^*$ (the space of weights). The character $\operatorname{ch}(M)$ of a $\mathfrak g$-module $M$ is its image after first forgetting from the $\mathfrak g$-action to just a $\mathfrak h$-action, and then looking at the element it represents in $\mathbb Z[e^{\mathfrak h^*}]$ (or rather in some completion; for category $\mathcal O$, I want to take the adic completion for the ideal generated by $e^{-\lambda}$ for simple roots $\lambda$).

The Weyl Character Formula asserts that, for $\lambda$ a dominant integral weight: $$ \operatorname{ch}(L(\lambda)) = \sum_{w\in \mathfrak W} \operatorname{sign}(w) \; \operatorname{ch}(V(w(\lambda+\rho)-\rho)), $$ where $\rho$ is half the sum of the positive roots, $\mathfrak W$ is the Weyl group, and $\operatorname{sign}: \mathfrak W \to \{\pm 1\}$ is $w \mapsto \det_{\mathfrak h}w = (-1)^{\operatorname{length}(w)} $.

If there were not the signs there, you might hope that this formula came from some statement in the representation theory of $\mathfrak g$. Recall that taking K-groups of a category turns exact sequences into addition equations. Then a proof of the Weyl Character Formula can follow this outline:

  1. Write down a matrix $b_{\lambda\mu}$ for the coefficient of $\operatorname{ch}(V(\mu))$ in $\operatorname{ch}(L(\lambda))$.
  2. Then the inverse matrix $(b^{-1})_{\mu\lambda}$ expresses the coefficient of $\operatorname{ch}(L(\lambda))$ in an expansion of $\operatorname{ch}(V(\mu))$.
  3. Recognize that each $V(\mu)$ is an extension of some $L(\lambda)$s — this is actually an equation in the representation theory of $\mathfrak g$ — and by studying how $V(\mu)$ is built out of $L(\lambda)$s, understand enough of the structure of $(b^{-1})_{\mu\lambda}$ to conclude the theorem.

The central point is that the inverse matrix $b^{-1}$ has only nonnegative integer entries, and so has a chance of descending from some extension problem in the representation theory.

So, what about the Weyl character formula itself? Since it has negative coefficients, it cannot express that $L(\lambda)$ is some extension of Verma modules. Instead, it descends from the fact that $L(\lambda)$ has a resolution in Verma modules, the so-called BGG resolution: $$ 0 \to M(w_0(\lambda+\rho)-\rho) \to \cdots \to \bigoplus_{w\in W \text{ s.t. }\operatorname{length}(w) = k} M(w(\lambda + \rho)-\rho) \to $$ $$ \cdots \to \bigoplus_{s \text{ a simple reflection}} M(s(\lambda + \rho) - \rho) \to M(\lambda) \to 0 $$ has homology only in the last spot, where the homolgy is $L(\lambda)$. (I think the differentials are just the obvious inclusions of Verma modules. My reference for all of this is my collection of lecture notes from UC Berkeley's Lie Theory course (chapter 6.1), but it doesn't go into detail.)

So this might be the "nice interpretation" you're looking for. In Andre's language, the point is to realize virtual modules as objects in the derived category of modules; i.e. chain complexes. Then the chain complex above is isomorphic (in the derived category) to its homology $L(\lambda)$, by the projection $M(\lambda) \to L(\lambda)$, and the Weyl Character Formula is the decategorification of this isomorphism. In general, the philosophy is that any naturally-occurring alternating sum of dimensions ought to come from a chain complex. Conversely, any sum of dimensions out to come from an extension.

share|improve this answer
    
Thanks for this point of view as well. Along with Andre's answer, this seems to be a good start for where I should begin to think about these things. –  ARupinski Feb 22 '11 at 2:43
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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