Comparison of the Weyl integration formula and the equivariant localization formula for a maximal torus of Sp() Jeffrey D. Carlson January 3, 4 Introduction In this note, we aim to prove the Weyl integration formula for the Lie group Sp() using the Berline Vergne/Atiyah Bott equivariant localization formula. We do not quite succeed. The ideas are due to Loring Tu, and the execution to me. The first four sections are generalities, mostly copied from an earlier document, which should probably be skimmed, and the attempted application to Sp() follows. The Weyl integral formula Let G be a compact, connected Lie group, T its maximal torus, and W the Weyl group. Write Ad g/t : T Aut(g/t) for the adjoint action of T on its Lie algebra modulo the Cartan subalgebra, and : T R for the function ( [ g/t) (t) := det id Ad(t ). If f : G R is a continuous function on G, and dg and dt are Haar measures on G and T respectively, and d(gt) the induced probability measure on G/T, then f (g) dg = [ (t) f (gtg ) d(gt) dt. G W T G/T If f is a class function, meaning it takes one value on each conjugacy class (so that for all g, h G we have f (ghg ) = f (h)), the bracketed integral reduces to f (t), so that G f (g) dg = (t) f (t) dt. W T Note further that if Φ is the set of global roots of G, and Φ + the set of positive roots, then ( = α ) ( = α )( α ). α Φ α Φ +
3 The equivariant localization theorem Let T be a torus and M a T-manifold. If X t is a vector in the Lie algebra of T and X Γ(TM) its associated fundamental vector field, write i N for the inclusions into M connected components N of the zero set of X, write ν for the normal bundle to N, and e T (ν) for the T-equivariant Euler class of each ν. Let the orientation on N be induced by that on ν. If α Ω T (M) is an equivariantly closed form, then i α(x) = ) N ( α(x) M N N e T (ν)(x). It would be appropriate in this connection to state what equivariant extensions are. An equivariant extension of an m-form ω Ω (M) is an element ω of the Cartan model ( S(t ) Ω(M) ) T (basically a T-invariant polynomial in a basis u,..., u n for t with coefficients in Ω (M)), of the same degree as ω, and such that the constant term is ω. Here the degree of elements of t is, so we can write ω = ω + I ω I u I, where I = (i,..., i n ) N n is a multiindex, u I = u i u i n has degree I = i k, and ω I Ω m I (M). The integral of such a form is M ω = I u I M ω I; since these summands will vanish unless deg ω I = m I = dim M, this integration will return an element of degree I in S(t ), or in other words a homogeneous degree- I polynomial map t R. In particular, if deg ω = dim M, then M ω(x) = M ω is a real number. It should be independent of X, subject to some genericity condition. So the equivariant localization formula, applied to equivariant extensions of top forms, gives another way to integrate functions on a manifold, assuming those functions have equivariantly closed equivariant extensions. A T-manifold M is called equivariantly formal if all closed forms on M have equivariantly closed extensions, and it is known that T is equivariantly formal under this conjugation action. In the situation we are interested in, the manifold M is a connected, compact Lie group G and the group action we are concerned with is that of a maximal torus T. The fixed points under conjugation by a subgroup are the group s centralizer in this case, the torus T itself. Let f : G R be a smooth function, invariant under conjugation by T. Then ω = f dg is a closed T-invariant m-form on G, so since G is equivariantly formal under conjugation by T, there exists an equivariant extension ω. Putting together the two formulae, we have [ (t) f (gtg ω(x) ) d(gt) dt = f (g) dg = ω = ω(x) = T W T G/T G G G T e T (ν)(x), Like that exp(x) topologically generates a maximal torus of G where e T (ν) is the equivariant Euler class of the normal bundle to T in G; what we would like is some way to see that the left-hand side is the same as the right without the intermediate steps.
4 The Equivariant Euler Class e T (ν) The normal bundle ν of T in G fits in the exact sequence T(T) (TG) T ν, and so is a vector bundle of (even) rank m rk T over T, with fiber g/t. We claim it is a trivial bundle, which will follow if (TG) T is a trivial bundle in such a way that T(T) is a trivial subbundle. But TG is parallelizable for any Lie group G: each element of T g G is a left translation (l g ) X for precisely one X g, and by continuity (g, X) (l g ) X is a vector bundle isomorphism G g TG. So ν = T (g/t). Now we want to see how the conjugation action acts on T (g/t) under this identification. For this purpose, we first work in G g, and see what Ad does to TG. Start with (g, X) G g. Write c h : g hgh for conjugation by h G, l h for left multiplication and r h for right multiplication, so c h = l h r h = r h l h. Then (c h ) (l g ) X = (r h ) (l h ) (l g ) X = (r h ) [ (lh ) (l g ) (l h ) (lh ) X = (r h ) (l ch (g)) (l h ) X = (l ch (g)) (r h ) (l h ) X = (l ch (g)) (c h ) X, so we have the diagram (l g ) X (l ch (g)) (c h ) X (g, X) ( ch (g), (c h ) X ), Since conjugation by T on T fixes each fiber, it induces the Ad-action on each fiber of the bundle (TG) T, and so on each g/t fiber of ν = T (g/t). As the conjugation action preserves each fiber, we can factor T out of the homotopy quotient: ν T = (ET ν)/t = ( ET T (g/t) ) /T = T (ET (g/t) ) /T = T (g/t) T, To see continuity of the inverse, given a basis X i of g, the left-invariant vector fields g (l g ) X i form a frame of TG, so that a i (l g ) X i ( g, (a,..., a dim G ) ). The projection TG G is continuous by definition, and this coordinate map a i (l g ) X i (a,..., a dim G ) R dim G is locally continuous over the trivializing neighborhoods U for TG that we already know exist since the left-invariant vector fields are continuous. 3
so ν T is a product bundle over T with fiber (g/t) T, which we must further analyze. In particular, it is the pullback of the vector bundle (g/t) T BT under the projection π : T BT BT, which induces a tensor-factor inclusion π : H (BT) H (T) H (BT) in cohomology. Now under the Ad-action, g/t naturally decomposes as a direct sum of two-dimensional representations of T corresponding to the positive roots α of G, which we will view as onedimensional complex representations C α. Each homotopy quotient S α := (C α ) T is then a line bundle over BT, and (g/t) T = α Φ + S α. Since the total Chern class is multiplicative, the Chern total classes of the line bundles S α are all + c (S α ), and the top Chern class of a complex vector bundle is the Euler class of its realization, the equivariant Euler class of ν is e T (ν) = e ( π (g/t) T ) = π ( e ( (g/t) T ) ) = e ( (g/t) T ) = c n rk T 5 Application to Sp() We want to demystify the formula ( α Φ + S α [ (t) f (gtg ) d(gt) dt = W T Sp()/T Sp() f (g) dg = ( f dg)(x) T T e T (ν)(x) ) Cite DFAT. = α Φ + c (S α ) H n rk T (BT). by finding T, W,, dg, dt, d(gt), W, X, f dg, and e T (ν). The compact Lie group Sp() can be seen as the multiplicative subgroup of quaternions of norm. The exponential formula e q := n N q n /n! defines a function on quaternions as well as on complex numbers; we will use this fact and the linearity of conjugation to simplify our identification of the adjoint representation and the Euler class. 5. sp() and T and t and X The standard basis, i, j, k of the quaternions H defines a vector space decomposition into the direct sum of the real part R spanned by and the purely imaginary part Im (H) spanned by i, j, k. Since i, j, k anticommute, if u Im (H) and u = uū =, then u = and u 3 = u and u 4 = and so on, so that e tu = cos t + u sin t for real t. Since any pure imaginary is tu for some real t and unit-norm u Im (H), and any quaternion of unit norm can be written as cos t + u sin t, it follows that the quaternionic exponential is a surjection Im (H) Sp(), and since d dt etu t= = sin + u cos = u, we have a natural identification sp() Im (H). We will consider the conjugation action by the maximal torus T = e ir = { cos t + i sin t : t [, π) }. Then t = ir and we should pick a nonzero X exp {}; we will take X = πi. 4
5. Ad and and Φ + and W We can write any q Sp() as z + jz for z, z C := R + ir. We consider the effect of conjugation c t by t = e iθ T on these coordinates. Since i commutes with C and anticommutes with jc, we have iji = j( i) = j, so for z C we have zjz = j zz and for e iθ T we get e iθ (z + jz )e iθ = z + je iθ z. Since multiplication on H is linear, the adjoint representation on Im (H) is given by Ad(t)Y = (c t ) Y = d ds t(esy )t s= d = t ds (esy ) t = tyt = c t (Y). s= In particular, taking t = e iθ, and conjugating the basis i, j, k of sp(), we have c t (i) = i, c t (j) = j(cos θ i sin t) = (cos θ)j + (sin θ)k, c t (k) = k(cos θ i sin t) = ( sin θ)j + (cos θ)k. Since ir = t, on sp()/t, with respect to the basis j, k, we have and so ( ) id Ad(t cos θ sin θ ) =, sin θ cos θ (t) = ( cos θ) + (sin θ) = cos θ + (cos θ + sin θ) = ( cos θ) = ( (cos θ + sin θ) (cos θ sin θ) ) = 4 sin θ. The roots Φ of the adjoint representation are the weights of the complexified representation ( sp()/t ) C. If we take as basis vectors v = j i + k and v = j + k i, we get Ad(t)v = t v and Ad(t)v = t v, so we have one positive root α : t t. So W = S has two elements and e T (ν) = c (S α ) = u H (BT). Here BT CP and u corresponds to the first Chern class of the tautological bundle on CP. Viewing u as an element of the symmetric algebra S(t ), we have u(x) =. 5
5.3 dg and dt and d(gt) Writing H = C + jc, we can decompose q Sp() as z + jz, where z + z =. Then we can find η [, π such that z = sin η and z = cos η, and can find complex arguments ξ and ξ of z and z, so that e iξ sin η + je iξ cos η. This identifies Sp() as a quotient space of S S [, π ; on the subspace η (), the ξ -coordinate is irrelevant, and on T = η ( π ), the ξ -coordinate is irrelevant. Thus we can parameterize Sp() minus two circles by E := [, π) (, π ). We will find it helpful to write dg in this parameterization. Writing z = x + iy and z = x + iy for real x, x, y, y gives a vector space isomorphism H R 4 taking Sp() S 3. The natural volume form on R 4 is vol R 4 = dx dy dx dy. If we write Y = x x + x x + y y + y y for the radial vector field on S 3, we get vol S 3 = ι Y (vol R 4) = (x dy y dx ) dx dy + (x dy y dx ) dx dy. Substituting x = sin η cos ξ, y = sin η sin ξ, x = cos η cos ξ, y = cos η sin ξ, much cancellation occurs, and after a few more lines one finds that vol S 3 pulls back to on E. We have ω = cos η sin η dη dξ dξ = vol(s 3 ) = E ω = π sin η π π dξ dξ π = (π) sin ζ d(ζ/) = π ( cos ζ) = π, ζ=π ζ= dη dξ dξ sin η dη so dg = π vol S 3 ω sin η = π 4π dη dξ dξ. Since vol(s ) = π, we have dt π dξ. To find d(gt), we should first understand the quotient space Sp()/T. Since (e iξ sin η + je iξ cos η)e ξ = sin η + je i(ξ ξ ) cos η, a fundamental domain for right translation by T is the set ξ () S (this is the Hopf fibration of S 3 ). Omitting the two points {η =, π } (where the arguments for the j-coordinate are arbitrary), this corresponds to D := {} [, π) (, π ) E. Now ω sin η D = dη dξ, and D ω D = π, so d(gt) = π vol Sp()/T sin η π dη dξ. 6
5.4 X and f dg Since the fundamental vector field generated by X = πi under conjugation is ( ) d X q = (c e sx ),q ds s= and c e sx (z + jz ) = z + jz e 4πsi (ξ, ξ + 4πs, η), it follows that on D, we have X = 4π ξ, and a function f on Sp() is T-invariant just if f ξ =. Since dg is a top form, any form f dg is closed, and so, if f is T-equivariant, admits an equivariantly closed extension f dg Ω ( Sp() ) T [u. If we write f dg = f dg + uτ for τ Ω ( Sp() ) T, equivariant closedness means that = (d uι X )( f dg + uτ) = d( f dg) + u ( dτ f ι X (dg) ) u ι X τ, or that dτ = f ι X (dg) and ι X τ =. Since the fundamental vector field generated by X = πi under conjugation is ( ) d X q = (c e sx ),q ds s= and c e sx (z + jz ) = z + jz e 4πsi (ξ, ξ + 4πs, η), it follows that on D, we have X = 4π ξ. Thus we require that ι τ = τ ( ) ξ =. Writing τ = a dξ + b dξ + c dη, this means ξ b =. Since τ is assumed T-invariant, it also follows that ξ a ( a dτ = η c ) dη dξ ξ. On the other hand, so we want f ι X dg = f ι 4π ξ ( sin η = c ξ =, so that dη dξ dξ ) π = π f sin η dη dξ, a η c = f sin η. ξ π Since f is π-periodic in ξ, the most obvious (if not particularly subtle) thing to do is to take c =. Since τ doesn t depend on ξ when η =, it s then easiest to take a η () =. One way to accomplish this is to set a(ξ, η) = π η f (ξ, ζ) sin ζ dζ. 7
5.5 Conjugation of a T-invariant function A function on Sp() is invariant under conjugation by T iff it is independent of ξ. The inner integral in the Weyl integral formula is over f (qtq ) for fixed t T and qt Sp()/T. Given an element of Sp()/T, we can find a representative q in the fundamental domain ξ (). Letting t T and q = t q(t ), we see that a T-invariant function will give the same value at qtq and at q t(q ) = t q(t ) tt q (t ) = t qtq (t ), so we may as well assume ξ (q) = as well, or q = sin ζ + j cos ζ. Writing t = e iξ, we get qtq = (cos ξ i sin ξ cos ζ) + ji sin ξ sin ζ. Conjugating to get ξ = again, we get ξ (qtq ) = arctan ( tan ξ cos ζ) and η(qtq ) = arccos(sin ξ sin ζ). So the integral, not particularly transparently, is of f ( arctan ( tan ξ cos η), arccos(sin ξ sin η) ) over η, for fixed ξ. One wonders whether a better choice of coordinates is in order. 5.6 The desired formula, restated Consider a real-valued function f (ξ, η) on R [, π which is π-periodic in ξ and independent of ξ wherever η =. The equivariant localization formula tells us π π π/ Sp() f dg = equals T sin η f (ξ, η) dη dξ dξ π = π π/ f (ξ π, η)sin η dη dξ u(x) u(x) τ {η= π } = π ( a ξ, π ) dξ = π π/ f (ξ π, η) sin η dη dξ. This is at least true, but it s not informative. As for the Weyl integration formula, it tells us that the integrals above equal W π T π π [ (t) 4 sin ξ [ π Sp()/T π/ f (gtg ) d(gt) dt = f ( arctan ( tan ξ cos η), arccos(sin ξ sin η) ) sin η [ π/ sin ξ f ( arctan ( tan ξ cos η), arccos(sin ξ sin η) ) sin η dη This isn t particularly helpful. dξ. dη dξ dξ π π = 8
5.7 An easier case Suppose that f is in fact invariant under conjugation by G. Then the bracketed integral simplifies to f (t), so the whole integral becomes π ( 4(sin ξ ) f ξ, π ) dξ. It still isn t clear to me, though, why this should have anything to do with the double integral in the equivariant localization formula. 6 Unsorted thoughts Would the Weyl integration formula be more what we want with different coordinates? Could a less stupid choice of a and c in τ = a dξ + c dη yield the Weyl integral formula? It s a(ξ, π/) dξ that shows up in the equivariant localization formula. It s easy to arrange the functions be constant in ξ at η =, say by giving them sin nη factors or making them integrals from to η, but if we try to select a first and derive c(ξ, η) = ξ ( η a π f sin η) dξ, the resulting function will almost never be π-periodic in ξ. If we assume f is G-invariant, can we make some clever change of coordinates that simplifies the integral over η in the equivariant localization formula? The only part of either formula that involves the roots is (t) in the one formula and e T (ν) in the other. But in the Weyl integration formula, one has to integrate with respect to this function, whereas in the equivariant localization formula, it just cancels u i s. What is the connection between these superficially similar but very different things? The standard proof of the Weyl integration formula starts with the function G T G taking (g, t) gtg, pushes it down to a function (G/T) T G, cuts out the (positive-codimension) set of singular elements, and computes the integral over G by pulling it back to (G/T) T. The function (t) = α Φ ( α) comes out of the Jacobian. On the other hand G/T doesn t even show up in the equivariant localization formula. One can knock the G/T integral out of the Weyl formula, but only at the cost of assuming the function to be integrated is G-invariant, a hypothesis that isn t required for the equivariant localization. Is there a way of proving equivariant localization using G/T, or some way of pulling back both proofs (I realize this is incredibly vague) to some more complicated space, so that both formulae manifest as quotients of the more complicated situation? Given that the Weyl integral formula involves an integral over T (G/T), is it worth considering the action of T by t (t, gt) = (t, tgt), or something similar, and trying to apply equivariant localization? 9