These are expanded notes prepared for a talk in Arbeitsgemeinschaft: Higher Gross Zagier Formulas held at Oberwolfach, Spring 2017, on Yun-Zhang's beautiful work [1]. This is the last one in a series of 17 talks.
You may also want to check out Tony Feng's excellent notes of all talks and my earlier course notes.
OverviewOur goal today is to finish the proof of the main identity 
 for all functions in the spherical Hecke algebra 
 of 
. For any 
 (unramified everywhere) cuspidal automorphic representation of 
, the LHS via the analytic spectral decomposition and the RHS via the cohomological spectral decomposition (discussed below) would imply the identity 
 We now have the wonderful opportunity to apply the identity to simplest element in the Hecke algebra, namely the the unit element 
, and obtain our desired Higher Gross—Zagier formula ![$$\mathcal{L}^{(r)}(\pi_{F'}, 1/2)\sim\langle [\Sht_T]_\pi, [\Sht_T]_\pi\rangle.$$](./latex/HigherGrossZagier/latex2png-HigherGrossZagier_4690413_.gif)
Ana's talk has proved the main identity for many 
's but we fall short of proving it for the element 
: in some sense the simplest Hecke function gives the most difficult situation for intersection computation (self-intersection), and considering 
 for sufficiently large 
 allows us to move away from the self-intersection situation and make the computation easier. What we would like to do is to resolve this tension, and deduce the identity for all Hecke functions from sufficiently many 
's by just doing commutative algebra. What make the deduction possible are certain key finiteness properties of the Hecke action on the middle cohomology of the moduli of shtukas.
Key finiteness theorems
 which admits an action of the Hecke algebra 
 is infinite dimensional caused by the fact that 
 is only locally of finite type. This infinite dimensionality can already be seen when 
, where we recover the classical Hecke action on the space of automorphic forms of level 1: 
 Here the space of cusp forms 
 is finite dimensional, but the space of Eisenstein series is infinite dimensional.
To kill the Eisenstein part, we again make use of the Eisenstein ideal appeared in Ilya's talk on analytic spectral decomposition.  Recall we define the Eisenstein ideal to be 
 moreover  
, which is 1-dimensional as a ring (
 is an extension of 
 by the finite group 
).
-point of 
 factors as 
 for some character 
. In particular , 
. When 
, we get 
. This is completely analogous to classical Eisenstein ideal 
 (= 
-th Fourier coefficient of the ``weight 2'' Eisenstein series), which was invented by B. Mazur to study rational points on modular curves and torsion points on elliptic curves.
After killing the Eisenstein part, we indeed obtain a finite dimensional vector space.
 is a union of open substacks of finite type 
 with instability index bounded by 
. The key point here is that one can understand the difference between the cohomology of 
 using horocycles discussed in Lizao's talk.  More precisely, when 
, the cone of the natural map 
 is equal to 
, which is a local system concentrated in degree 
. Here 
, 
, and 
 is the quotient torus of the Borel of 
.
  In particular, it suffices to work with the generic fiber and show 
 is finite dimensional. Let 
 be the geometric generic fiber of 
, then the map 
 is an isomorphism when 
 and injective when 
 with cokernel 
.
Let 
 be the finite union of 
, where instability index 
 is not all 
. Using the compatibility of the cohomological constant map and the Satake transform, we have a commutative diagram 
 For 
, by definition 
 and so the bottom row is zero. The cohomological constant map on the right is injective since 
. It follows that the top row is zero, and hence the image of 
 is contained in 
, which is finite dimensional as desired.
¡õ
Using a similar argument, one can also prove the following finiteness theorem.
 Define the images  
 Suppose the cohomological correspondence 
 induced by 
 sends 
 to 
, then we have the induced map on associated graded 
 It turns out one can construct such sequence 
 such that  these induced maps 
 are surjective for all 
 for some 
. The result then follows since 
 is finite dimensional.
The key point here again is by the compatibility of Satake isomorphism and the cohomological constant term, the 
-action on the cokernel can be made explicit as 
-action. One can then choose 
 appropriately so that 
 acts trivially on the cokernels for degree reasons and to make sure that 
 induces a surjection 
.
¡õ
Cohomological spectral decomposition   Let 
. We have the following immediate consequence:
 By Theorem 2, 
 is a finite 
-module. Also 
 is finite 
-module (due to the finiteness of 
). It follows that RHS is a finite 
-module and hence 
 is a finite 
-module. Because 
 is a polynomial algebra, it follows that 
 is a finitely generated 
-algebra.
¡õ
Now 
 is a finite module over the noetherian ring 
 (by Theorem 2, 
 is a finite module even over 
), we obtain the following cohomological spectral decomposition:
 Here 
 is a finite set of closed points.
-modules 
 such that 
, 
 and  
 is finite dimensional over 
.
 is a finite set of closed points and 
 is finite dimensional follows from Theorem 1.
Finish of the proof of the main identity
 to be the image of 
 in 
.  Then both sides of the identify only depend on the image of 
 in 
. Define 
 to be the linear subspace spanned by 
's, where 
 effective divisor of degree 
.  By Ana's Talk, we already proved the identity for 
. So it remains to show that the composition 
 is surjective.
Using the key finiteness theorem one can prove the following lemma:
 is a finitely generated 
-algebra, so there exists a finite set 
 such that the images of 
 (
) generate 
 as 
-algebra. Enlarge 
 so that 
 also contains all 
 for 
. Let 
 be the ideal generated by the images of 
 (
), then 
 works.
In fact, because 
 is away from the generating set 
, we have 
 Since the left-hand-side is equal to 
 spans 
 by definition, we know 
 (think: 
 is ``small'' enough).
On the other hand, by the cohomological spectral decomposition (Theorem 3), 
 It remains to rule out the Eisenstein part: i.e., to show that 
 is disjoint from 
. Suppose 
 is a 
-point of 
 lies in 
. Then 
 factors as 
 for some character 
. It follows that for 
, we have 
 Hence 
. Let 
 be the character 
. Then 
 has finite image and 
. By class field theory 
 can be identified as a Galois character of 
 and hence 
 for infinitely many 
 by Chebotarev density. This contradicts 
 for all 
. So 
 is disjoint from 
 and hence 
 is finite dimensional (think: 
 is ``large'' enough).
¡õ
Besides the key finiteness theorems, we need one additional ingredient concerning the local Hecke algebra. It is a bit magical but completely elementary:
. Then under the Satake transform we may identify 
, where 
. Note that 
, thus it suffices to show that for any nonzero ideal 
 in the ring 
 of symmetric polynomials in 
 , and any 
, we have 
 For any 
, write 
 (
). We define the lowest degree of 
 to be the smallest 
 such that 
. Then it suffices to show that for any given 
, we can find an element 
 with same the lowest degree as that of 
.   Since 
 is a PID, we may assume 
 for some 
. Notice that for any 
, there is no cancellation in the product 
, and so one can always choose some 
 such that 
 has a prescribed lowest degree (in particular, the lowest degree of 
) as desired.
¡õ
Now we can finish the proof of the main identity using the previous two lemmas. For any 
, look at the commutative diagram 
 Here the vertical arrows are all natural inclusions.
By Lemma 1, 
 is finite dimensional, so 
 is also finite dimensional. Since 
 is quotient of 
 by a nonzero ideal, and 
, we know the bottom row is surjective by Lemma 2. Now 
 generate 
 as an algebra, so the top row is also surjective. By Lemma 1, 
, so the map 
 is surjective as desired.
¡õ
[1]Shtukas and the Taylor expansion of $L$-functions, ArXiv e-prints (2015).