4.1 Derivatives
We now embark on the journey to prove the theorems we stated at the beginning. We shall begin by explicitly calculating Pk∫M∗Faug(V). Note that since the construction of PnF was done pointwise, we can do this calculation without knowing what the whole functor Pk∫M∗(−) is.
First note that both V and Algnaug(V) have a zero object, and Faug sends to zero object to the zero object by virtue of being a left adjoint. So
CT(Faug(V))=Faug(CT(V)).
Using this and fixing a (k+1)-cube S, we can compute Tk∫M∗(−):
Tk∫M∗Faug(V)=∅=T⊆Slim∫M∗CT(Faug(V))=∅=T⊆Slim∫M∗Faug(CT(V))=∅=T⊆Slim0≤i<∞⨁Confifr(M∗)Σi≀O(n)⨂CT(V)⊗i=0≤i<∞⨁∅=T⊆SlimConfifr(M∗)Σi≀O(n)⨂CT(V)⊗i.
Fixing M∗, let us write
Ri(V)=Confifr(M∗)Σi≀O(n)⨂V⊗i.
We have then shown that
Tk∫M∗Faug(V)=0≤i<∞⨁(Tk(Ri))(V).
Similarly, we have
Tkj∫M∗Faug(V)=0≤i<∞⨁(Tkj(Ri))(V).
Therefore, we conclude that
Pk∫M∗Faug(V)=0≤i<∞⨁(Pk(Ri))(V).
But we have already concluded that Ri is i-homogeneous (as a functor of V), so Pk(Ri) is just Ri if i≤k, and 0 otherwise. So
Pk∫M∗Faug(V)=0≤i<k⨁Confifr(M∗)Σi≀O(n)⨂V⊗i.
In particular, the kth derivative evaluated on Faug(V) is exactly
Confkfr(M∗)Σk≀O(n)⨂V⊗k.
But since the kth derivative evaluated at an augmented n-disk algebra A only depends on L(A), and L(A)=L(Faug(L(A))), we know
Theorem
20
The kth derivative of ∫M∗(−) is
A↦Confkfr(M∗)Σk≀O(n)⨂L(A)⊗k.
In other words, we always have a cofiber sequence
Confkfr(M∗)Σk≀O(n)⨂L(A)⊗k→Pk∫M∗A→Pk−1∫M∗A.