In this section we describe a construction of a Poisson structure on Map(P1, p, G/P) using the
machinery of shifted Poisson structures.
Our first hope might be to find a 1-shifted Poisson structure on Y = G/P and use Theorem 4.3 to induce a structure on Map(P1, p, Y). This should make us suspicious in theP =B case, as
we would expect a 1-shifted symplectic structure on G/B, which would yield a quasi-isomorphism betweeng/pandr[1], which is clearly impossible.
In the general case, we can also show we can’t get our Poisson structure this way. Suppose Z is 2-shifted symplectic, andg :Y →Z has a Lagrangian structure defining a Posson structure onY. Using the Lagrangian condition, we get Tg 'LY[1]=∼r[1], and in particular the mapTg →TY is
the zero map. For anyf ∈Map(P1, p, Y), the map T∨
f Map(P
1, p, Y)→T
fMap(P1, p, Y) will just
be the map
H1(P1, f∗(r)(−1))'RΓ(P1, f∗(Tg)(−1))→RΓ(P1, f∗(TY)(−1))'H0(P1, f∗(g/p)(−1)),
so will also be zero, which we do not want.
Instead, we note thatG/P is already related to an existing shifted symplectic stack,BG, via the fiber diagram G/P // BP • //BG . (*)
Recall that the symplectic structure on BG is given by a G-invariant nondegenerate symmetric quadratic form ong. Fix such a form ω.
Choose an opposite parabolicP−so thatP∩P−=Lis a Levi subgroup ofG. Lettingl= Lie(L), we can then write g = r− ⊕l⊕r. Since l is orthogonal to r and r−, ω descends to l and is also
L-invariant and nondegenerate. Thus,BLhas a symplectic structure ωL induced fromBG. Recall
that the identificationL∼=P/rad(P) gives us a mapP →L.
Lemma 5.1. The mapι:BP →BG×BLhas a Lagrangian structure given by0. ThusBP →BG
has a coisotropic structure.
in the usual sense with respect to ω−ωL. Writep =l⊕rand recall that ris orthogonal to itself
andl. Then
(ω−ωL)(ι∗(`, r), ι∗(`0, r0)) =ω(`, `0)−ω(`, `0) = 0,
so we have isotropy.
For the Lagrangian condition, recall thatω pairs uprnondegenerately withr−. Let
∆,∆ :l→g⊕l be the diagonal and antidiagonal maps, and note that ω−ωL pairs up ∆(l) and
∆(l) nondegenerately. Then the map Tι→LBP[1] is just the adjoint ∆(l)⊕r−→(∆(l)⊕r)∨.
So BP → BG has a coisotropic structure, and if • → BG had one too, we would get a 1- shifted Poisson structure onG/P by Corollary 3.7. As mentioned, there is no decent shifted Poisson structure on G/P and it is also easy to check that • → BGhas no coisotropic structure. Instead, we apply the functor Map(P1, p,−) to * to get
Map(P1, p, G/P) // Map(P1, p, BP) Map(P1, p,•) //Map(P1, p, BG) ,
and note that now Map(P1, p, BG) is 1-symplectic and Map(P1, p, BP) → Map(P1, p, BG) is co-
isotropic by Theorem 4.5. Let GP1 denote the trivial G-bundle on P1, framed at p. Then the
map
•= Map(P1, p,•)→Map(P1, p, BG) is just the pointGP1. And thisis coisotropic, as
TG
P1Map(P
1, p, BG) =Ext∗(O
P1⊗g,OP1⊗g)(−1)[1]'0.
Then the map • →Map(P1, p, BG) is trivially Lagrangian, hence coisotropic. Then Corollary 3.7
gives a 0-shifted Poisson structure on Map(P1, p, G/P).
To be a little more specific, the coisotropic bases are the maps Map(P1, p, BL)→ • and• → •
respectively, so the Poisson structure comes from a map Map(P1, p, G/P) → Map(P1, p, BL). In
particular, in the caseP is a Borel, Lis a torus, and so Map(P1, p, BL) =Qr
i=1Map(P 1, p, B
Theorem 5.2. The Poisson structures described above coincide.
Proof. Let us look at the construction using shifted Poisson structures first. Recall that, forX an underived smooth scheme with Poisson structureg:X→Z, we recover the mapTX∨→TX by using
the Lagrangian structure ong to yield an isomorphismTg∼=LX; with respect to this isomorphism,
the Poisson structure is given by LX ∼=Tg →TX. In the caseX = Map(P1, p, G/P), let’s look at
the mapg :X →Z = Map(P1, p, BL). For f ∈ X, the tangent map g∗ : TfX → g∗(TZ)f is the
composition RΓ(P1, f∗(gY/pY)(−1)) ∂∗ // RΓ(P1, f∗(pY)(−1))[1] π // RΓ(P1, f∗(lY)(−1))[1]
Here∂∗ is the connecting map coming from the short exact sequence
0→f∗(pY)(−1)→f∗(gY(−1))→f∗(gY/pY(−1))→0;
sincef∗(gY(−1)) is acyclic,∂∗ is a quasi-isomorphism. Then from the sequence
0→f∗(rY)(−1)→f∗(pY)(−1)→f∗(lY)(−1)→0
we see that the fiber of π(and thus of g∗) isRΓ(P1, f∗(rY)(−1))[1]'H1(P1, f∗(rY)(−1)). This is
identified withTf∨X in the canonical way, and the Poisson map is then
H1(P1, f∗(rY)(−1))→H1(P1, f∗(pY)(−1))∼=H0(P1, f∗(gY/pY(−1))).
For the spectral sequence, consider the map of complexes
rY // gY //gY/pY pY //gY //gY/pY .
Pull back byf, twist by−1, and look at the induced map on theE2 page: H1( P1, f∗(rY)(−1)) // H0( P1, f∗(gY/pY)(−1)) H1( P1, f∗(pY)(−1)) //H0(P1, f∗(gY/pY)(−1)) .
The bottom map is the inverse to the connecting isomorphism
H0(P1, f∗(gY/pY)(−1))→H1(P1, f∗(pY)(−1))
from the long exact sequence. Thus the Poisson map is exactly the composition
H1(P1, f∗(rY)(−1))→H1(P1, f∗(pY)(−1))∼=H0(P1, f∗(gY/pY(−1))),
as above.
Remark 5.2.1. In either case, much of the “real work” of the Poisson structure lies in the identification
(P1, H0(f∗(gY/pY)(−1)))∨∼=H1(P1, f∗(r)(−1))
arising from Serre duality and the isomorphism (gY/pY)∨∼=rY.
Remark 5.2.2. Given that G/P doesn’t have a 1-shifted Poisson structure, what structure does it have? Recall that an n-shifted symplectic structure on X is equivalent to a map X → •n+1
with a Lagrangian structure. Generalizing this by replacing •n+1 with an arbitrary (n+ 1)-shifted
symplectic derived stackZ yields the notion of ann-shifted Poisson structure.
However, there is another generalization we can make: if X → •n+1 only has an isotropic
structure, we get ann-shifted presymplectic structure onX. Combining these two, we might say an
n-shifted “pre-Poisson” structure onX is an (n+ 1)-shifted symplectic derived stack Z and a map
X →Z with an isotropic structure. This is the structureG/P has; specifically, G/P →BL has an isotropic structure.