Proof of Proposition 2.4 - detailed version

Our next goal is to prove Proposition 2.4. We shall work towards this goal by proving some lemmas. First, we note a simple consequence of the axioms of a k-coalgebra:

Remark 3.4. Let C be any k-coalgebra. Let can1 : C⊗k → C be the k-module isomorphism sending c⊗1 to c for each c ∈ C. Let can2 : k⊗C → C be the k-module isomorphism sending 1⊗c toc for eachc ∈ C.

Recall that C is a k-coalgebra, and therefore satisfies the axioms of a k-coalgebra. Hence, in particular, the diagram

C⊗k ^can¹ ^//C^oo ^can² k⊗C

C⊗C

id⊗e

∆ C

∆ //C⊗C

e⊗id

is commutative (since this is one of the axioms of ak-coalgebra). In other words, we have

can₁◦(id⊗e)◦_∆=id and (51)

can₂◦(e⊗id)◦_∆=id . (52)

Next, we state a simple property of a slight generalization of primitive elements in a coalgebra:

Lemma 3.5. LetCbe any k-coalgebra. Leta,b,d∈ Cbe three elements satisfying e(a) =1 ande(b) = 1 and∆(d) = d⊗a+b⊗d. Then,e(d) = 0.

We shall later apply Lemma 3.5 to the case when a = b = 1C (and C is either a connected filtered k-coalgebra or a k-bialgebra, so that 1C does make sense);

however, it is not any harder to prove it in full generality:

Proof of Lemma 3.5. Let us first observe that the map C×C →C⊗C,

(u,v) 7→u⊗v (53)

isk-bilinear. (This follows straight from the definition of the tensor product.) Applying the map id⊗_e : C⊗C → C⊗k to both sides of the equality ∆(d) =

d⊗a+b⊗d, we obtain

(id⊗e) (_∆(d)) = (id⊗e) (d⊗a+b⊗d)

= (id⊗e) (d⊗a)

| {z }

=id(d)⊗e(a)

+ (id⊗e) (b⊗d)

| {z }

=id(b)⊗e(d)

(since the map id⊗e isk-linear)

=id(d)

| {z }

⊗e(a)

| {z }

+id(b)

| {z }

⊗ e(d)

| {z }

=e(d)1

=d⊗1+ b⊗e(d)1

| {z }

=be(d)⊗1

(sincee(d)is a scalar inkand thus can be moved past the⊗sign)

=d⊗1+be(d)⊗1

= (d+be(d))⊗1 (54)

(since the map (53) isk-bilinear).

Define the maps can₁ and can2as in Remark 3.4. Applying the map can₁to both sides of the equality (54), we obtain

can₁((id⊗_e) (_∆(d))) =can₁((d+be(d))⊗1) =d+be(d) (by the definition of can1). Hence,

d+be(d) =can1((id⊗_e) (∆(d))) = (can1◦(id⊗_e)◦_∆)

| {z }

=_id (by (51))

(d) = id(d) = d.

Subtracting d from both sides of this equality, we obtain be(d) = 0. Applying the mape to both sides of this equality, we finde(be(d)) = 0. In view of





be(d)

| {z }

=_e(d)b





=e(e(d)b) = e(d)e(b)

| {z }

(since the mape isk-linear)

=e(d),

this rewrites ase(d) =0. This proves Lemma 3.5.

Next, let us define a “reduced identity map” id for any connected filtered k-coalgebra C, and explore some of its properties:

Lemma 3.6. Let C be a connected filtered k-coalgebra with filtration (C≤0,C≤1,C≤2, . . .). Define a k-linear map id :C →C by setting

id(c) :=c−e(c)1C for each c∈ C.

Define ak-linear map δ : C→C⊗Cby setting

δ(c) :=_∆(c)−c⊗1C−1C⊗c+e(c)1C⊗1C for each c∈ C.

Then:

(a)We haveδ =id⊗id

◦∆. (Here, of course,∆denotes the comultiplication ofC.)

(b)We have id(C≤n) ⊆C≤n for each n∈ _N.

(c)We have id(C≤0) =0.

Proof of Lemma 3.6. Let us first observe that the map C×C →C⊗C,

(u,v) 7→u⊗v (55)

isk-bilinear. (This follows straight from the definition of the tensor product.) (a)Let c ∈ C. The element ∆(c) is a tensor inC⊗C, and thus can be written in the form

∆(c) =

∑

m i=1

λic_i⊗d_i (56)

for some m ∈ _{N, some} λ1,λ2, . . . ,λm ∈ k, some c1,c2, . . . ,cm ∈ C and some d₁,d₂, . . . ,dm ∈ C. Consider this m, these λ1,λ2, . . . ,λm, these c₁,c₂, . . . ,cm and thesed₁,d2, . . . ,dm.

Now, define the maps can1 and can2 as in Remark 3.4. Applying the map id⊗e to both sides of the equality (56), we obtain

(id⊗_e) (_∆(c)) = (id⊗_e)

∑

m i=1

λic_i⊗d_i

∑

m i=1

λi(id⊗_e) (c_i⊗d_i)

| {z }

=id(c_i)⊗e(d_i)

(since the map id⊗_e isk-linear)

∑

m i=1

λiid(ci)

| {z }

=c_i

⊗ e(di)

| {z }

=_e(d_i)1

∑

m i=1

λi ci⊗e(di)1

| {z }

=c_ie(d_i)⊗1

(sincee(d_i)is a scalar inkand thus can be moved past the⊗sign)

∑

m i=1

λ_ic_ie(d_i)⊗1.

Applying the map can₁to both sides of this equality, we obtain

(by the definition of can₁)

(since the map can1 isk-linear)

Furthermore, applying the mape⊗id to both sides of the equality (56), we obtain (e⊗id) (_∆(c)) = (e⊗id) can be moved past the⊗sign)

∑

m i=1

λi1⊗e(ci)di.

Applying the map can2to both sides of this equality, we obtain can2((e⊗_id) (∆(c))) =can2

(by the definition of can₂)

(since the map can2 isk-linear)

∑

m i=1

λ_ie(c_i)d_i.

Comparing this with

Applying the map id to both sides of this equality, we obtain id

(by the definition of id). Hence, c−e(c)1_C =id (since the map id isk-linear).

Now, applying the map id⊗id to both sides of the equality (56), we obtain (by the definition of id )

⊗id(d_i) (since the map (55) isk-bilinear)

We shall now separately simplify the two sums on the right hand side of this equality.

Indeed, we have

∑

m i=1

λici⊗ id(di)

| {z }

=d_i−e(d_i)1_C (by the definition of id )

∑

m i=1

λ_ic_i⊗(d_i−e(d_i)1C)

| {z }

=_λ_i_c_i⊗di−λici⊗e(di)1C

(since the map (55) isk-bilinear)

∑

m i=1

(λic_i⊗d_i−_λ_ic_i⊗_e(d_i)1C)

∑

m i=1

λic_i⊗d_i

| {z }

=_∆(c) (by (56))

−

∑

m i=1

λic_i⊗_e(d_i)1C

| {z }

=λ_ic_ie(d_i)⊗1_C

(sincee(d_i)is a scalar inkand thus can be moved past the⊗sign)

=_∆(c)−

∑

m i=1

λ_ic_ie(d_i)⊗1_C

| {z }

= _m

∑

i=1

λ_ic_ie(d_i)

⊗1_C (since the map (55) isk-bilinear)

=_∆(c)−

∑

m i=1

λ_ic_ie(d_i)

| {z }

=c (by (57))

⊗1_C

=_∆(c)−c⊗1C

and

∑

m i=1

λie(ci)1C

| {z }

=1_Cλ_ie(c_i)

⊗_id(di)

∑

m i=1

1Cλ_ie(c_i)⊗id(d_i)

| {z }

=1_C⊗λ_ie(c_i)id(d_i)

(sinceλ_ie(c_i)is a scalar inkand thus can be moved past the⊗sign)

∑

m i=1

1_C⊗λ_ie(c_i)id(d_i) =1_C⊗

∑

m i=1

λ_ie(c_i)id(d_i)

| {z }

=c−e(c)1_C (by (59))

(since the map (55) isk-bilinear)

=1C⊗(c−e(c)1C) = 1C⊗c− 1C⊗e(c)1C

| {z }

=1_Ce(c)⊗1_C

(sincee(c)is a scalar inkand thus can be moved past the⊗sign)

(since the map (55) isk-bilinear)

=1C⊗c−1Ce(c)

| {z }

=e(c)1_C

⊗1C =1C⊗c−e(c)1C⊗1C.

In light of these two equalities, we can rewrite (60) as id⊗id

(_∆(c)) = (_∆(c)−c⊗1_C)−(1_C⊗c−_e(c)1_C⊗1_C). Comparing this with

δ(c) = _∆(c)−c⊗1_C−1_C⊗c+e(c)1_C⊗1_C (by the definition of δ)

= (_∆(c)−c⊗1_C)−(1_C⊗c−_e(c)1_C⊗1_C), we obtainδ(c) =id⊗id

(_∆(c)) = id⊗id

◦_∆(c).

Forget that we fixed c. We thus have shown that δ(c) = id⊗id

◦_∆(c) for eachc ∈ C. In other words,δ =_id⊗_id◦∆. This proves Lemma 3.6 (a).

(b)Let n ∈ _{N. Let} c ∈ C≤n. We have 1C ∈ C≤0 (by Definition 2.3(b)). However, (9) yields C≤0 ⊆ C≤1 ⊆ C≤2 ⊆ · · · (since C is a filtered k-coalgebra). Therefore, C≤0 ⊆C≤n. Thus, 1C ∈ C≤0⊆C≤n. Now, the definition of id yields

id(c) = c

|{z}∈C≤n

−e(c) 1C

|{z}∈C≤n

∈ C≤n−e(c)C≤n ⊆C≤n

(sinceC≤n is ak-module).

Forget that we fixed c. We thus have shown that id(c) ∈ C≤n for eachc ∈ C≤n. In other words, we have id(C≤n) ⊆C≤n. This proves Lemma 3.6 (b).

(c)The filtered k-coalgebra C is connected. In other words, the restriction e |_C_≤0 is a k-module isomorphism from C≤0 to k (by Definition 2.3 (a)). Hence, this re-strictione |_C_≤0 is bijective, and thus injective. Also, we have 1C ∈ C≤0(by Definition 2.3(b)).

Now, let c ∈ C≤0. Hence, e |_C_≤0(c) is well-defined. Definition 2.3 (b) yields 1_C = e |_C_≤0⁻¹(1_k). Thus, e |_C_≤0(1_C) = 1_k. In other words, e(1_C) = 1_k (since

e |_C_≤0(1_C) = e(1_C)).

Setd =e(c)1C. Then,d=e(c) 1C

|{z}∈C≤0

∈ e(c)C≤0⊆C≤0(sinceC≤0is ak-module).

Thus, e |_C_≤0(d) is well-defined.

Comparing

e|_C_≤0(c) =e(c) with

e|_C_≤0(d) = e



 d

|{z}

=e(c)1_C



=e(e(c)1_C)

=_e(c)e(1C)

| {z }

=1_k

(since the mapeis k-linear)

=e(c), we obtain e |_C_≤0(c) = e |_C_≤0(d).

id(c) =c− e(c)1C

| {z }

(by the definition ofd)

=c−d =0 (since c=d).

Forget that we have fixedc. We thus have shown that id(c) = 0 for eachc ∈ C≤0. In other words, id(C≤0) = 0. This proves Lemma 3.6(c).

Proof of Proposition 2.4. (a)Define ak-linear map id : C →_Cas in Lemma 3.6. Then, Lemma 3.6(a)yieldsδ =id⊗id

◦_∆.

Now, letn>0 be an integer. Then, (11) yields

∆(C≤n)⊆

∑

n i=0

C≤i⊗C≤n−i (61)

(sinceC is a filteredk-coalgebra). Now, sum (and these are indeed two distinct addends, since n>0)

applied toiinstead ofn)

⊗ id(C≤n−i)

This proves Proposition 2.4(a).

(b)Let f : C →Cbe a k-coalgebra homomorphism satisfying f(1_C) = 1_C. Thus, f is a k-coalgebra homomorphism; in other words, f is a k-linear map satisfying

(f ⊗ f)◦_∆ = _∆◦ f and e = _e◦ f (by the definition of a “k-coalgebra (by the definition ofδ)

 primitive (by the definition of “primitive”). In other words, c ∈ PrimC (since PrimC is defined as the set of all primitive elements ofC).

Forget that we fixedc. We thus have shown thatc ∈ PrimCfor eachc ∈ (Kerδ)∩ (Kere). In other words,(Kerδ)∩(Kere) ⊆PrimC.

Now, letd∈ _PrimC. Thus, the elementdofCis primitive (since PrimCis defined as the set of all primitive elements of C). In other words, ∆(d) = d⊗1_C+1_C⊗d (by the definition of “primitive”). Hence, Lemma 3.5 (applied to 1C and 1C instead ofa and b) yieldse(d) = 0 (sincee(1C) =1). Hence,d ∈_Ker_e.

Furthermore, the definition of δyields δ(d) =_∆(d)−d⊗1_C−1_C⊗d

| {z }

(since∆(d)=d⊗1_C+1_C⊗d)

+_e(d)1_C⊗1_C =_e(d)

| {z }

1_C⊗1_C =0.

Hence,d ∈Kerδ. Combining this with d∈ Kere, we obtaind ∈ (Kerδ)∩(Kere). Forget that we fixed d. We thus have shown that d ∈ (Kerδ)∩(Kere) for each d ∈ PrimC. In other words, PrimC ⊆ (Kerδ)∩(Kere). Combining this with (Kerδ)∩(Kere) ⊆ PrimC, we obtain PrimC = (Kerδ)∩(Kere). This proves Proposition 2.4(c).

(d)Proposition 2.4(c)yields PrimC = (Kerδ)∩(Kere).

The maps δ and e are k-linear. Hence, their kernels Kerδ and Kere are k-submodules of C. The intersection (Kerδ)∩(Kere) of these two kernels must therefore be ak-submodule ofC as well. In other words, PrimC is ak-submodule ofC (since PrimC= (Kerδ)∩(Kere)). This proves Proposition 2.4 (d).

(e) We first observe that 1C ∈ C≤0 (by Definition 2.3 (b)). However, (9) yields C≤0 ⊆ C≤1 ⊆ C≤2 ⊆ · · · (since C is a filtered k-coalgebra). Therefore, C≤0 ⊆ C≤1. Thus, 1C ∈ C≤0 ⊆C≤1. Hence,

δ(1_C) ∈ _δ(C≤1)⊆

1−1 i

∑

C≤i⊗C≤1−i

(by Proposition 2.4 (a), applied to n=1)

= (empty sum) =0.

In other words,δ(1C) = 0.

Definition 2.3 (b) yields 1C = e|_C_≤0⁻¹(1_k). Thus, e |_C_≤0(1C) = 1_k. In other words,e(1C) =1 (since e |_C_≤0(1C) =e(1C)and 1_k =1).

Let u∈ Kerδ. Thus, δ(u) =0. Set v=u−_e(u)1_C. Then, δ



 v

|{z}

=u−e(u)1_C



=δ(u−e(u)1C) = δ(u)

| {z }

−e(u)δ(1C)

| {z }

(since the mapδ isk-linear)

=0−e(u)0=0, so thatv∈ Kerδ. Furthermore, e



 v

|{z}

=u−e(u)1C



=e(u−e(u)1C) =e(u)−e(u)e(1C)

| {z }

(since the mape is k-linear)

=e(u)−e(u) =0,

so thatv ∈Kere. Combining this withv ∈Kerδ, we obtainv∈ (Kerδ)∩(Kere) = PrimC (by Proposition 2.4(c)). Now, fromv =u−e(u)1C, we obtain

u =e(u)

| {z }

∈k

1C+ v

∈|{z}PrimC

∈ k·1C+PrimC.

Forget that we fixed u. We thus have shown that u ∈ k·₁_C+_PrimC for each u∈ Kerδ. In other words,

Kerδ ⊆k·1C+PrimC. (62)

On the other hand, let w ∈ k·1C +PrimC. Thus, we can write w in the form w = x+y for some x ∈ k·₁_C _{and some} y ∈ _PrimC. Consider these x and y. We have

y∈ PrimC = (Kerδ)∩(Kere) (by Proposition 2.4(c))

⊆Kerδ, so thatδ(y) =0.

We have x ∈ k·1C; in other words, x = λ·1C for some λ ∈ k. Consider this λ.

Now,w = x

|{z}

=_λ·1C

+y=_λ·1_C+y. Applying the map δ to both sides of this equality, we obtain

δ(w) = _δ(_λ·1_C+y) = _λ·_δ(1_C)

| {z }

+_δ(y)

| {z }

(since the mapδ isk-linear)

=λ·0+0 =0.

In other words,w∈ _Ker_δ.

Forget that we fixedw. We thus have shown thatw∈ Kerδfor eachw∈ k·1_C+ PrimC. In other words,

k·₁_C+_Prim_C⊆_Ker_δ.

Combining this with (62), we obtain Kerδ =k·1C+PrimC. This proves Proposi-tion 2.4(e).

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