# What is significant about the half-sum of positive roots?

I apologize for the somewhat vague question: there may be multiple answers but I think this is phrased in such a way that precise answers are possible.

Let $\mathfrak{g}$ be a semisimple Lie algebra (say over $\mathbb{C}$) and $\mathfrak{h} \subset \mathfrak{g}$ a Cartan subalgebra. All the references I have seen which study the representation theory of $\mathfrak{g}$ in detail make use of the half-sum of positive roots, which is an element of $\mathfrak{h}^\ast$: e.g. Gaitsgory's notes on the category O introduce the "dotted action" of the Weyl group on $\mathfrak{h}^\ast$, the definition of which involves this half-sum.

Is there a good general explanation of why this element of $\mathfrak{h}^\ast$ is important? The alternative, I suppose, is that it is simply convenient in various situations, but this is rather unsatisfying.

-
I don't why the math won't display properly: the code looks fine to me. Maybe someone who knows more about LaTeX and/or MathOverflow can fix it. – Justin Campbell Nov 5 '11 at 20:56
By itself, the asterisk is a control character, to display one use \ast in Latex. Wow, little earthquake just now. 14:52 PST. – Will Jagy Nov 5 '11 at 21:53
– Will Jagy Nov 5 '11 at 21:56
This question and the wonderful answers are a really great example of why MO is good. – Mariano Suárez-Alvarez Nov 9 '11 at 2:32
I hope this very good question is not closed yet. I have heard that $\rho$ of $G$ can be related to the curvature of the homogenous space $G/H$, where $H$ is closed subgroup of $G$. I am interested in particular when $G$ is noncompact semisimple Lie group and $H$ is its maximal compact subgroup. Can anybody elaborate on this.. please? – spr Jun 30 '12 at 5:15

I don't think there is a one-line answer to this question, since it depends a lot on the direction from which you approach semi-simple Lie theory. For one thing, it's probably best at first to emphasize just integral weights, among which the dominant ones parametrize irreducible finite dimensional representations. Here the weight usually denoted $\rho$ plays a ubiquitous role in the classical Weyl theory, but that too can be developed in a number of different ways. (There was some early experimentation with the notation; the alternative symbol $\delta$ also had widespread use before the Bourbaki preference for $\rho$ started to take over in 1968.)

While it's important in proofs of the Weyl character formula to view $\rho$ as the half-sum of positive roots (given a fixed positive or simple system), it's also essential to identify it with the sum of fundamental dominant weights for many purposes. In this guise it's the smallest regular dominant weight, fixed by no element of the Weyl group except the identity. When passing from integral weights to line bundles on an associated flag variety $G/B$ (with $B$ a Borel subgroup associated to positive roots relative to a fixed maximal torus which it contains), the weight $\rho$ has the distinction of defining an ample line bundle. This property is crucial in geometric approaches to Weyl's formula, as well as in spin-offs in prime characteristic due to Andersen and others.

Ultimately the importance of the weight $\rho$ is probably appreciated best in the setting of representation theory, where the finite dimensional theory is enriched by treatment of highest weight modules in more generality and the shift by $\rho$ is again ubiquitous. By the way, the convenient "dot" notation $w \cdot \lambda := w(\lambda +\rho) - \rho$ is apparently due to Robert Moody. In the earlier literature the more awkward full notation appears, or else is replaced in the Paris notation by a hidden $\rho$-shift.

None of what I've said is a complete answer to the question asked, but in any case it's more than a matter of "convenience" to emphasize $\rho$.

-
Note that for Kac-Moody groups, where there are infinitely many positive roots, one still defines $\rho$ as the sum of the fundamental weights. – Allen Knutson Nov 8 '11 at 1:21
@Allen: Yes, there's a lot more to be said in that direction. The two ways of looking at $\rho$ come up classically in the Weyl denominator formula, which Kac creatively generalized to certain representations of symnmetrizable Kac-Moody algebras (though in that setting $\rho$ is not quite uniquely determined). – Jim Humphreys Nov 8 '11 at 13:46

From the point of view of geometry, the crucial fact about $\rho$ is that the corresponding line bundle on the flag manifold is (upt to a sign) a (the) square-root of the canonical bundle (top exterior power of $T^*_B G/B \simeq b_-$ is the sum of the negative roots). This is of course equivalent to Alain Valette's description in terms of the modular character of the Borel. In other words its sections in the real world are half-densities (things for which we can define the $L^2$ inner product).

It is a universal fact about passage from the classical world to the quantum world (in particular the geometric construction of representations) involves a shift by the square root of the canonical bundle. There are many ways to explain or motivate this. For example if we seek unitary representations we need to be able to define an $L^2$ inner product, which means considering not sections of the bundle we might have expected but sections times half-densities (again this is Alain's answer restated). From the point of view of rings of differential operators, the adjoint of a differential operator acting on functions (or on sections of a bundle $L$) is invariantly not another diffop (on $L$) but a differential operator acting on volume forms (or on sections of $L$ tensor the canonical bundle) --- so the self dual twist of differential operators is by half-forms, ie $\rho$-shifted. (Put another way, Serre duality is a reflection centered at half-forms!)

My favorite explanation is in Beilinson-Bernstein's Proof of Jantzen Conjectures and doesn't involve self-adjointness or unitarity: it's a consistency condition for deformation quantization of symbols (functions on the cotangent bundle): if you want this deformation quantization to be correctly normalized to order two (this is not the right question to go into that) you find you need to look at differential operators twisted by half-forms, not functions. On the flag variety this means a $\rho$-shift, and from the D-module POV on representation theory this is one fundamental place where that shift is forced on you, independent of thinking of inner products. This is in particular one way to see why it comes up in the Weyl character formula, through the geometric proof via Atiyah-Bott or via the BGG resolution, both of which involve the geometry of the flag variety.

-

This is actually a fairly deep question. Your suspicion that there may be multiple answers is correct, but there might be some surprising connections between seemingly unrelated answers. Let me give one possible thread of explanation. The underlying principle is that the appearance of $\rho$ and the "dot" action $w\cdot\lambda=w(\lambda+\rho)-\rho$ in representation theory is closely related to the geometry of the flag variety.

One of the first places one meets $\rho$ (and the dot action) is in the Weyl character formula. A theorem of Kostant shows that the formula can be written as the ratio of two Lie algebra cohomology Euler characteristics. From this perspective, the appearance $w \cdot \lambda$ and $w\cdot0$ in the WCF is ultimately explained by the fact that these are the weights appearing in the weight space decomposition of the relevant Lie algebra cohomology modules, namely $H^*(\mathfrak n, V^\lambda)$ and $H^\ast(\mathfrak n, V^0)$, where $\mathfrak n = \bigoplus_{\alpha>0} \mathfrak g_\alpha$ and $V^\mu$ denotes the irrep of highest weight $\mu$.

We can rephrase this in geometric terms by invoking the "geometric analogue" of Kostant's theorem, i.e. the Borel–Weil–Bott theorem. Kostant's description of the Lie algebra cohomology of $\mathfrak n = \mathfrak g /\mathfrak b^-$ with coefficients in an irrep translates into a representation-theoretic description of the sheaf cohomology of certain line bundles $L_\lambda$ (constructed using integral weights $\lambda$) over the flag variety $G/B^-$ of $\mathfrak g$. Consequently, the dot action shows up in this description, and this time it's accompanied by a shift in degree. This in turn can be explained by Serre duality; the key fact is that canonical bundle of $G/B^-$ turns out to be $L_{-2\rho}$.

So, in some sense, the appearance of $\rho$ and the dot action in the WCF can be thought of as a manifestation of Serre duality.

[N.B. This is a condensed version of my lengthy original answer. The old version can be found in the edit history.]

-

While I appreciate Dave Ben-Zvi's half-densities answer, I'm going to put forth the contrary opinion that it's largely a bookkeeping artifact.

The most familiar place that $\rho$ shows up is in the WCF of the irrep $V$ with highest weight $\lambda$, $$Tr(t|_{V}) = \frac{\sum_w t^{w(\lambda+\rho)-\rho}}{\prod_{\Delta_+} (1-t^{-\beta})}.$$

This version of WCF is good for suggesting the existence of the BGG resolution, or for taking the Fourier transform of and obtaining the Kostant multiplicity formula. But otherwise, I claim that that it's the worst way to write it down, and suggest instead $$Tr(t|_V) = \sum_w w \cdot \frac{t^{\lambda}}{\prod_{\Delta_+} (1-t^{-\beta})}.$$ Hurray, it's manifestly $W$-invariant, and no $\rho$ in sight! This is the natural version that one obtains by applying the Atiyah-Bott-Riemann-Roch-Lefschetz Woods Hole localization formula to the flag manifold $G/B$, as A&B mention in their paper.

You really notice it if you try to write down a WCF for nonregular weights, which corresponds to applying the localization formula to a partial flag manifold $G/P$. Then you can no longer flip weights to put everything over the same denominator, so the first version is badly broken. The second, $W$-invariant, version works just fine in this case (the denominator is a product over only part of $\Delta_+$).

EDIT: I suppose it's too strong to say it's badly broken. It's just that it's not in lowest terms.

-
A general remark: You get that formula if you apply A–B to the $\overline\partial$ operator acting on $\Omega^{(0,q)}(G/B,L_\lambda)$. In another paper Bott mentions that you can make the other formula for the Weyl denom (the one with $\rho$ in it!) show up if you approach things differently: $G/B$ is spin, so we can work with its (elliptic) Dirac operator $S^+ \to S^-$. Incidentally, the existence of a spin structure on $G/B$ is related to $\rho$: the canonical bundle of $G/B$ is $L_{-2\rho}$ so a holomorphic square root is given by $L_{-\rho}$, and this determines the spin structure. – Faisal Nov 8 '11 at 4:11

I don't quite have in mind any construction that fully singles out this element, which is called the Weyl element $\rho$, as the most important one. To an extent you don't even expect that in the non-simply-laced case, because a dual elements, half of the sum of the coroots, is sometimes comparably important. However, I know that the Weyl element has long been important for $q$-analogue" reasons, which in more modern work has become more and more important because it points to quantum groups and eventually even categorification.

Consider the elementary fact that there are $\binom{n}{k} = \frac{n!}{k!(n-k)!}$ subsets of size $k$ of the set $\{1,\ldots,n\}$. This counting fact is a special case of the Weyl dimension formula for the dimension of an irreducible representation of a complex simple Lie algebra. In this model case the representation is $\mathfrak{sl}(n,\mathbb{C})$ acting on $\Lambda^k(\mathbb{C}^n)$. You could look at the same counting problem again with multiplicative weights for the elements of $\{1,\ldots,n\}$. If you make the weight of $j$ be some variables $x_j$, the total weight is an irreducible polynomial --- equivalent to the full character of $\Lambda^k(\mathbb{C}^n)$. Magically, if you let $x_j = q^j$ for a single variable $q$, you get (up to a power of $q$) the Gaussian binomial coefficient $\binom{n}{k}_q$. This is a special case of the Weyl $q$-dimension formula which gives the character of the Weyl element. That is, the dimension of an irreducible representation $V_\lambda$ of $\mathfrak{g}$ is given by a tidy ratio, and the $q$-dimension still is. The full character is not as simple, and therefore neither are most 1-variable specializations.

Bourbaki, and maybe Weyl himself, used the $q$-dimension to prove the dimension formula by plugging in $q=1$. (It can happen in combinatorics that a $q$-analogue is easier than the original question.) In modern representation theory the $q$-dimension is even more important, because it's also (after centering the powers of $q$) the quantum dimension of the same representation $V_\lambda$ (or we can say, the same-named representation) of the quantum group $U_q(\mathfrak{g})$. The Weyl element also arises in many other ways in the representation theory of the quantum group. Actually, that's an understatement: This $q$ is the variable of the Jones polynomial and its generalizations. All of this $q$-structure remains important in the even newer categorifications of quantum groups.

(A caveat: Because half-integer powers of $q$ commonly arise, there is a substitution of $q^2$ for $q$ in passing from $q$-analogues to quantum groups. I never liked this mismatch of conventions, in fact as a student I didn't even realize/believe it, but it is an established standard.)

There is another formula for the Weyl element: It's the sum of the fundamental weights. It's already interesting that these two formulas agree.

-

Let $G=KAN$ be the Iwasawa decomposition of a semi-simple Lie group $G$. Then the modular function of the Borel subgroup (= minimal parabolic) $B=MAN$ is $\Delta_B(man)=e^{2\rho(\log a)}$ where $\rho$ is half the sum of the positive roots of the root system $\Delta(\mathfrak{g},\mathfrak{a})$.

This is relevant for the definition of the principal series of representations of $G$, say in the compact picture: for $\nu\in i\mathfrak{a}^*$, the Hilbert space of $Ind_B^G(1\otimes e^\nu\otimes 1)$ is $L^2(K/M)$, with action given by $(\pi_\nu(g)f)(k)=e^{-(\nu+\rho)H(g^{-1}k)}f(\kappa(g^{-1}k))$, where $g=\kappa(g)e^{H(g)}n$ in the Iwasawa decomposition. So $e^{-\rho}$ appears as the square root of the Radon-Nikodym cocycle, needed to make the representation unitary, since the measure on $G/B=K/M$ is not $G$-invariant.

For more on this, see sections 5.6 and 7.1 in A.W. Knapp, Representation theory of semisimple groups (an overview based on examples), Princeton MAth. Series 36, 1986.

-

Well, no one's explicitly talked about the relevance of spin structures to this story yet, so here's a sketch of the story as I understand it. For references see, for example, the nLab. I'll be blithely ignoring the difference between algebraic and topological K-theory, and $B$ denotes a positive Borel.

Roughly speaking, the Borel-Weil-Bott story is about constructing representations of $G$ by inducing them from $1$-dimensional representations of $B$. A geometric avatar of a finite-dimensional representation of $B$ is a $G$-equivariant vector bundle on $G/B$; in fact, this is an equivalence of categories, and so there is a natural isomorphism

$$K_G(G/B) \cong K_B(\text{pt}) \cong R(B)$$

from the equivariant K-theory $K_G(G/B)$ to the Grothendieck group of finite-dimensional representations of $B$. Similarly, there is a natural isomorphism

$$\text{Pic}_G(G/B) \ni L_{\lambda} \leftrightarrow \lambda \in \Lambda$$

from the group of $G$-equivariant line bundles on $G/B$ to the group of $1$-dimensional representations of $B$, which in turn can be identified with the weight lattice $\Lambda$.

In this language, induction can be interpreted as an attempt to construct a pushforward

$$K_G(G/B) \to K_G(\text{pt}) \cong R(G)$$

in equivariant K-theory from $G/B$ to a point; these kinds of pushforwards are one language for talking about geometric quantization. In algebraic geometry, the pushforward in K-theory to a point is given by taking sheaf cohomology, and Borel-Weil-Bott tells us exactly what happens when we push forward equivariant line bundles this way.

We might ask how we can take pushforwards in K-theory purely topologically, though. Recall, for example, that we can take the pushforward in cohomology of a map between compact oriented manifolds using Poincare duality. The analogous statement for real resp. complex K-theory is that we can take the pushforward of a map between compact oriented manifolds equipped with spin resp. complex spin structure. In both cases, the pushforward to a point is given by taking the index of a suitable Dirac operator. I believe all of this continues to be true equivariantly.

The happy fact about the algebro-geometric setting is that almost complex structures canonically induce complex spin structures; the corresponding Dirac operator is built from the Dolbeault operator, and with suitable hypotheses pushforward to a point is given by taking sheaf cohomology computed as Dolbeault cohomology.

What is the relevance of the Weyl vector to this story? Any complex spin structure has associated to it a canonical complex line bundle $\omega$. If we start with an almost complex structure, then $\omega$ is the canonical bundle. Then a choice of spin structure compatible with a complex spin structure is equivalent to a choice of square root $\sqrt{\omega}$. In our case, under the identification of $G$-equivariant vector bundles on $G/B$ with representations of $B$, we have

$$T(G/B) \mapsto \mathfrak{g}/\mathfrak{b}$$

where the latter has the adjoint action of $B$. This breaks up into a direct sum of $1$-dimensional weight spaces, one for each negative root, and hence under the identification of $G$-equivariant line bundles on $G/B$ with the weight lattice $\Lambda$, we have

$$\omega \mapsto 2 \rho.$$

Since $\Lambda$ is torsion-free, the isomorphism class of the square root $\sqrt{\omega}$ is unique, and as an element of $\Lambda$ it is precisely $\rho$.

So $\rho$ is special in this story because it represents the unique square root of the canonical bundle. From this perspective, the dot action

$$w \cdot \lambda = w (\rho + \lambda) - \rho$$

naturally arises as follows. Complex spin structures are canonically a torsor over complex line bundles; if $L$ is a complex line bundle, the action on complex spin structures modifies the canonical bundle by

$$\omega \mapsto \omega \otimes L^{\otimes 2}.$$

I again believe this continues to be true equivariantly, and so $G$-equivariant complex spin structures on $G/B$ are canonically a torsor over the weight lattice $\Lambda$. So we can noncanonically identify the two via

$$\lambda \mapsto \omega \otimes L_{\lambda}^{\otimes 2} \mapsto 2 \rho + 2 \lambda$$

where $\omega \otimes L_{\lambda}^{\otimes 2}$ denotes a complex spin structure, not just the corresponding canonical line bundle (and in particular it can be different from the complex spin structure given by $\omega$ even if $L_{\lambda}^{\otimes 2}$ is trivial, although that doesn't happen here). If $\lambda \in \Lambda$ is a weight, the pushforward of $L_{\lambda}$ with respect to $\omega$ can be identified with the pushforward of the trivial line bundle with respect to $\omega \otimes L_{\lambda}^{\otimes 2}$.

Now pick a maximal compact $K$ and identify $G/B$ with $K/T$, where $T = K \cap B$ is a maximal torus in $K$. Then the Weyl group $W = N_K(T)/T$ naturally acts on $K/T$; in fact it is precisely the $K$-equivariant automorphism group of $K/T$. This induces an action on $K$-equivariant line bundles, identified with characters of $T$, identified with the weight lattice $\Lambda$, which is the usual non-dot action.

The Weyl group also acts on $K$-equivariant complex spin structures on $K/T$, and this action is compatible with the torsor structure above, as well as with the map sending a complex spin structure to its canonical bundle. It is not compatible with the noncanonical identification between $K$-equivariant complex spin structures and $\Lambda$ above: instead, writing $2 \rho + 2 \lambda$ for the complex spin structure corresponding to $\omega \otimes L_{\lambda}^{\otimes 2}$, the Weyl group action is

$$w(2 \rho + 2 \lambda) = 2 \rho + 2 \left( w(\rho + \lambda) - \rho \right)$$

and using the noncanonical identification again we precisely recover the dot action! So, to summarize:

Geometrically, while the usual action of $W$ on $\Lambda$ is the natural action of $W$ on $K$-equivariant complex line bundles on $K/T$, the dot action is the natural action on $K$-equivariant complex spin structures on $K/T$; the Weyl vector $\rho$ appears when relating these because it corresponds to a distinguished $K$-equivariant complex spin structure associated to a choice of positive roots. It is the pushforward in K-theory with respect to such spin structures which lets us construct representations of $K$ from representations of $T$.

-

I fear this question is getting a little crowded, but I do have my own hobby-horse to ride, so why hold back:

For a holomorphic symplectic variety with nice enough behavior, quantizations of said variety are in canonical bijection with power series in $H^2(X)$ (this follows from work of Bezrukavnikov and Kaledin). Furthermore, for $X=T^*G/B$, there is a canonical isomorphism of $H^2(T^*G/B)$ with $\mathfrak{h}^*$, the dual abstract Cartan of $\mathfrak{g}$.

So, picking any deformation quantization gives a power series in $H^2(T^*G/B)$, and there is one that we know and love the best: differential operators (of course, David Ben-Zvi was arguing above that maybe you shouldn't love this one best, but set that aside for a moment). What power series does this correspond to?

Of course, it's $\rho$ (this is essentially because the differential operators twisted in half-forms really are the most canonical thing, and thus correspond to 0). So, if you believe that differential operators in functions are particularly important as compared to other TDO's, you think $\rho$ is important.

-
If I'm not mistaken this correspondence with power series is, up to order two, precisely what Beilinson-Bernstein discuss: i.e. they say we should normalize a quantization by requiring it gives the zero power series to order two, resulting in half-forms.. They do this via an elementary observation: a deformation quantization to order two, when symmetrized, still gives a commutative associative algebra (this fails to higher order), so one gets a canonical 1-jet of a path into Poisson structures. – David Ben-Zvi Nov 8 '11 at 5:10
Right. I don't claim that this is really that different from your answer. It just emphasizes in a slightly different way just how canonical $\rho$ really is. – Ben Webster Nov 10 '11 at 18:54

One answer that hasn't appeared here yet is that $\rho$ is the highest weight of a spinor representation. Instead of the Dolbeault operator on the flag manifold, one can work with the Dirac operator. This leads to the subject of "Dirac Cohomology", see the book "Dirac Operators in Representation Theory". Here, instead of working with the Universal Enveloping Algebra and modules for it, one works with the tensor product of this with a Clifford algebra. Irreducible modules then acquire a spinor representation factor. This explains nicely the shift by $\rho$.

Hopefully this spring I'll finish this, which explains all this in more detail:

http://www.math.columbia.edu/~woit/brstdirac.pdf

-

Parts of what I will explain here have been said in other answers, but not quite in this way.

One thing about the weight $\rho$ is that the places it tends to show up, it is really the weight $-\rho$ that is the important one. Note that the "dot" action is just a shift to make $-\rho$ the new "$0$".
So why would we like to have $-\rho$ be our new "$0$"? Because it turns out to be the weight that behaves like a $0$ should when we consider cohomology.

This becomes a bit clearer when we go to the closely related setting of a semisimple simply connected algebraic group $G$ with Borel subgroup $B$. Here we consider the functor $\operatorname{Ind}_B^G$ inducing from $B$-modules to $G$-modules (for full details of definitions and proofs, one can consult for example Jantzen's book).
To make things simpler, for a weight $\lambda$ we denote also by $\lambda$ the $1$-dimensional $B$-module where the maximal torus in $B$ acts as $\lambda$, and $B$ acts via the projection to this torus.
For $i\geq 0$ write $H^i(\lambda) = R^i\operatorname{Ind}_B^G(\lambda)$ for the $i$'th right derived functor of induction applied to the $B$-module $\lambda$. Note that this is the $i$'th cohomology of the flag variety $G/B$ with coefficients in the line bundle defined by $\lambda$.

Now the reason for wanting $-\rho$ to be our "$0$" starts to be clear, as $H^i(-\rho) = 0$ for all $i\geq 0$. But there are other weights satisfying this, so what makes $-\rho$ extra special?
What makes $-\rho$ so special is that the vanishing of the above cohomology groups in some sense happens "as early as possible". More precisely, if $P = P(\alpha)$ is a parabolic subgroup corresponding to a simple root $\alpha$, then already $R^i\operatorname{Ind}_B^P(-\rho) = 0$, and we can factor $\operatorname{Ind}_B^G$ as $\operatorname{Ind}_P^G\operatorname{Ind}_B^P$.
Moreover, if $\lambda\neq -\rho$ then there is some simple root $\alpha$ such that either $\operatorname{Ind}_B^{P(\alpha)}(\lambda)\neq 0$ or $R^1\operatorname{Ind}_B^{P(\alpha)}(\lambda)\neq 0$.

-

If you have any free abelian group with an integral bilinear form embedded in the Lorentz space $\mathbb{R}^{n,1}$, you may consider the group of automorphisms generated by roots, i.e., reflections in vectors. The reflection hyperplanes will split the corresponding hyperbolic $n$-space into fundamental domains, and if you fix a chamber, you can choose simple roots corresponding to its walls. This setting includes all finite, affine, and hyperbolic Weyl groups. If there is a vector $\rho$ in the span of the roots satisfying $\Vert r - \rho \Vert = \Vert \rho \Vert$ for all simple roots $r$, then it is called a Weyl vector. This always exists in the finite and affine cases, and the other answers on this page give some description of the relevant geometry.

The existence of a Weyl vector gives a hyperbolic reflection group some arithmetic significance, and non-existence is generic - Lorentzian lattices of rank greater than 26 can't have Weyl vectors. From Barnard's thesis (and earlier work of Gristenko and Nikulin in small rank cases), if one has a lattice generated by roots with a Weyl vector, one may attach a vector-valued modular form, whose coefficients describe the root multiplicities of a Borcherds-Kac-Moody Lie algebra whose real simple roots are precisely those of the reflection group. The Lie algebra in turn has a Weyl denominator product that is a cusp expansion of an automorphic form on $O(n+1,2)$.

In the most extreme case, one may start with the 26-dimensional even unimodular Lorentzian lattice $I\! I_{25,1}$, and choose a chamber for its reflection group. The Dynkin diagram is naturally an affine space on the Leech lattice (by a theorem of Conway), and there is a norm zero Weyl vector $\rho$. There is an action of Leech, identified with the lattice quotient $\rho^\perp/\mathbb{Z}\rho$, on the fundamental domain by parabolic translation. Because there is a Weyl vector, one has a modular form whose coefficients control root multiplicities of a Lie algebra. In this case, we have the weight -12 form $1/\Delta$, and the Lie algebra is the fake monster Lie algebra, which apparently describes bosonic strings propagating in a 26-torus. The roots of norm $2n$ have multiplicity $p_{24}(1-n)$, i.e., the number of partitions in 24 colors.

In a different direction, there is a generalization of the Weyl character formula that holds for any Borcherds-Kac-Moody Lie algebra (not just hyperbolic), and $\rho$ appears here as any vector in the root space that satisfies the relation $\Vert r - \rho \Vert = \Vert \rho \Vert$ (equivalently, $r-2\rho \perp r$) for all simple roots $r$. In the BGG interpretation (worked out in detail in Jurisich's thesis), we find that $H_k(\mathfrak{n}_+, \mathbb{C})$ is spanned by the elements of $\bigwedge^k \mathfrak{n}_+$ whose weight $r$ satisfies $\Vert r - \rho \Vert = \Vert \rho \Vert$. In other words, when we throw away finiteness (and hence well-behaved flag varieties), $\rho$ still plays a role of selecting the part of the exterior power that contributes to the homology.

-