3. Representation theory: concrete examples
3.1. Some lemmata about exponentials and commutators
This section is devoted to some elementary lemmata about power series and iterated commutators over noncommutative rings. These lemmata are well-known in geometri-cal contexts (in these contexts they tend to appear in Lie groups textbooks), but here we will formulate and prove them purely algebraically. We will not use these lemmata until Theorem 3.11.2, but I prefer to put them here in order not to interrupt the flow of representation-theoretical arguments later.
We start with easy things:
Lemma 3.1.1. Let K be a commutative ring. If α and β are two elements of a topological K-algebra R such that [α, β] commutes with β, then [α, P (β)] = [α, β]·P0(β) for every power seriesP ∈K[[X]] for which the seriesP(β) andP0(β) converge.
Proof of Lemma 3.1.1. Let γ = [α, β]. Then, γ commutes with β (since we know that [α, β] commutes with β), so that γβ =βγ.
Write P in the form P =
∞
P
i=0
uiXi for some (u0, u1, u2, ...) ∈ KN. Then, P0 =
∞
P
i=1
iuiXi−1, so that P0(β) =
∞
P
i=1
iuiβi−1. On the other hand, P =
∞
P
i=0
uiXi shows that
P (β) =
Corollary 3.1.2. If α and β are two elements of a topological Q-algebra R such that [α, β] commutes with β, then [α,expβ] = [α, β]·expβ whenever the power series expβ converges.
Proof of Corollary 3.1.2. Applying Lemma 3.1.1 to P = expX and K = Q, and recalling that exp0 = exp, we obtain [α,expβ] = [α, β]·expβ. This proves Corollary 3.1.2.
In Lemma 3.1.1 and Corollary 3.1.2, we had to require convergence of certain power series in order for the results to make sense. In the following, we will prove some results for which such requirements are not sufficient anymore82; instead we need more global conditions. A standard condition to require in such cases is that all the elements
81Proof. We will prove this by induction overi:
Induction base: For i = 1, we have
Induction step: Let j ∈N be positive. Assume that α, βi by induction we have proven that
α, βi
=iγβi−1 for every positivei∈N.
82At least they are not sufficient for my proofs...
to which we apply power series lie in some ideal I of R such that R is complete and Hausdorff with respect to theI-adic topology. Under this condition, things work nicely, due to the following fact (which is one part of the universal property of the power series ring K[[X]]):
Proposition 3.1.3. Let K be a commutative ring. Let R be a K-algebra, and I be an ideal of R such that R is complete and Hausdorff with respect to the I-adic topology. Then, for every power series P ∈K[[X]] and every α∈I, there is a well-defined element P (α)∈ R (which is defined as the limit lim
n→∞
n
P
i=0
uiαi (with respect to theI-adic topology), where the power seriesP is written in the formP =
∞
P
i=0
uiXi for some (u0, u1, u2, ...)∈KN). For every α ∈I, the mapK[[X]]→ R which sends every P ∈ K[[X]] to P (α) is a continuous K-algebra homomorphism (where the topology on K[[X]] is the standard one, and the topology on R is the I-adic one).
Theorem 3.1.4. Let R be a Q-algebra, and let I be an ideal of R such that R is complete and Hausdorff with respect to the I-adic topology. Let α ∈ I and β ∈ I be such that αβ = βα. Then, expα, expβ and exp (α+β) are well-defined (by Proposition 3.1.3) and satisfy exp (α+β) = (expα)·(expβ).
Proof of Theorem 3.1.4. We know that αβ = βα. That is, α and β commute, so that we can apply the binomial formula toα and β.
Comparing
sinceαandβcommute)
=
(here, we substituted n for i+j in the second sum)
=
Corollary 3.1.5. Let R be a Q-algebra, and let I be an ideal of R such that R is complete and Hausdorff with respect to theI-adic topology. Letγ ∈I. Then, expγ and exp (−γ) are well-defined (by Proposition 3.1.3) and satisfy (expγ)·(exp (−γ)) = 1.
Proof of Corollary 3.1.5. By Theorem 3.1.4 (applied to α=γ andβ =−γ), we have exp (γ + (−γ)) = (expγ)·(exp (−γ)), thus
(expγ)·(exp (−γ)) = exp (γ + (−γ))
| {z }
=0
= exp 0 = 1.
This proves Corollary 3.1.5.
Theorem 3.1.6. Let R be a Q-algebra, and let I be an ideal of R such that R is complete and Hausdorff with respect to the I-adic topology. Let α ∈I. Denote by adα the map R →R, x7→[α, x] (where [α, x] denotes the commutatorαx−xα).
(a)Then, the infinite series
∞
P
n=0
(adα)n
n! converges pointwise (i. e., for everyx∈R, the infinite series
∞
P
n=0
(adα)n
n! (x) converges). Denote the value of this series by exp (adα).
(b) We have (expα)·β·(exp (−α)) = (exp (adα)) (β) for everyβ ∈R.
To prove this, we will use a lemma:
Lemma 3.1.7. Let R be a ring. Let α and β be elements of R. Denote by adα the map R → R, x 7→ [α, x] (where [α, x] denotes the commutator αx−xα). Let n ∈N. Then,
(adα)n(β) =
n
X
i=0
n i
αiβ(−α)n−i.
Proof of Lemma 3.1.7. LetLα denote the mapR →R, x7→αx. Let Rα denote the map R→R, x7→xα. Then, everyx∈R satisfies
(Lα−Rα) (x) = Lα(x)
| {z }
=αx
(by the definition ofLα)
− Rα(x)
| {z }
=xα
(by the definition ofRα)
=αx−xα = [α, x] = (adα) (x).
Hence, Lα−Rα = adα.
Also, every x∈R satisfies
(Lα◦Rα) (x) =Lα (Rα(x))
| {z }
=xα
(by the definition ofRα)
=Lα(xα) = αxα
(by the definition ofLα) and
(Rα◦Lα) (x) =Rα (Lα(x))
| {z }
(by the definition of=αx Lα)
=Rα(αx) = αxα
(by the definition ofRα), so that (Lα◦Rα) (x) = (Rα◦Lα) (x). Hence, Lα◦Rα =Rα◦ Now, it is easy to see (by induction over j) that
Ljαy=αjy for every j ∈N and y ∈R. (73) Also, it is easy to see (by induction over j) that
Rαjy =yαj for every j ∈N and y∈R. (74) In (this can be easily proven by induction over n, using the fact that I is an ideal) and thus (adα)n
n! (x) converges (because R is complete and Hausdorff with respect to the I-adic topology). In other words, the infinite series
∞
P
n=0
(adα)n
n! converges pointwise.
Theorem 3.1.6 (a) is proven.
(b) Let β ∈R. By the definition of of exp (adα), we have
Compared with
(here, we substitutedn for i+j in the second sum)
= complete and Hausdorff with respect to the I-adic topology. Let α ∈I. Denote by adα the map R →R, x7→[α, x] (where [α, x] denotes the commutatorαx−xα).
As we know from Theorem 3.1.6 (a), the infinite series
∞
P
n=0
(adα)n
n! converges pointwise. Denote the value of this series by exp (adα).
We have (expα)·(expβ)·(exp (−α)) = exp ((exp (adα)) (β)) for every β∈I. Proof of Corollary 3.1.8. Corollary 3.1.5 (applied to γ = −α) yields (exp (−α))· (exp (−(−α))) = 1. Since −(−α) =α, this rewrites as (exp (−α))·(expα) = 1.
Let β∈I. LetT denote the map R →R, x7→(expα)·x·(exp (−α)). Clearly, this map T is Q-linear. It also satisfies
T (1) = (expα)·1·(exp (−α)) (by the definition of T) (by the definition ofT)
· T (y)
| {z }
=(expα)·y·(exp(−α)) (by the definition ofT)
= (expα)·x·(exp (−α))·(expα) -algebra homomorphism. Also, T is continuous (with respect to the I-adic topology).
Thus, T is a continuous Q-algebra homomorphism, and hence commutes with the complete and Hausdorff with respect to the I-adic topology. Let α ∈ I and β ∈ I.
Assume that [α, β] commutes with each of α and β. Then, (expα) ·(expβ) = (expβ)·(expα)·(exp [α, β]).
First we give two short proofs of this lemma.
First proof of Lemma 3.1.9. Define the map adα as in Corollary 3.1.8. Then, (adα)2(β) = [α,[α, β]] = 0 (since [α, β] commutes with α). Hence, (adα)n(β) = 0 for every integern ≥2. Now, by the definition of exp (adα), we have
(exp (adα)) (β) =
By Corollary 3.1.8, we now have
(expα)·(expβ)·(exp (−α)) = exp ((exp (adα)) (β))
(by Corollary 3.1.5, applied toγ=−α)
= (expα)·(expβ).
this yields
(expα)·(expβ) = (expβ)·(exp [α, β])·(expα). (75) Besides, α and [α, β] commute, so that α[α, β] = [α, β]α. Hence, Theorem 3.1.4 (applied to [α, β] instead of β) yields exp (α+ [α, β]) = (expα)·(exp [α, β]).
On the other hand,αand [α, β] commute, so that [α, β]α=α[α, β]. Hence, Theorem 3.1.4 (applied to [α, β] andα instead ofα and β) yields exp ([α, β] +α) = (exp [α, β])· (expα).
Thus, (exp [α, β])·(expα) = exp ([α, β] +α)
| {z }
=α+[α,β]
= exp (α+ [α, β]) = (expα)·(exp [α, β]).
Now, (75) becomes
(expα)·(expβ) = (expβ)·(exp [α, β])·(expα)
| {z }
=(expα)·(exp[α,β])
= (expβ)·(expα)·(exp [α, β]).
This proves Lemma 3.1.9.
Second proof of Lemma 3.1.9. Clearly, [β, α] = −[α, β] commutes with each of α and β (since [α, β] commutes with each ofα and β).
The Baker-Campbell-Hausdorff formula has the form (expα)·(expβ) = exp
α+β+1
2[α, β] + (higher terms)
,
where the “higher terms” on the right hand side meanQ-linear combinations of nested Lie brackets of three or more α’s and β’s. Since [α, β] commutes with each of α and β, all of these higher terms are zero, and thus the Baker-Campbell-Hausdorff formula simplifies to
(expα)·(expβ) = exp
α+β+ 1 2[α, β]
. (76)
Applying this to β and α instead ofα and β, we obtain (expβ)·(expα) = exp
β+α+1 2[β, α]
.
Since [β, α] =−[α, β], this becomes
(expβ)·(expα) = exp
β+α+1 2 [β, α]
| {z }
=−[α,β]
= exp
β+α− 1 2[α, β]
. (77)
Now, [α, β] commutes with each of α and β (by the assumptions of the lemma) and also with [α, β] itself (clearly). Hence, [α, β] commutes with β+α−1
2[α, β]. In other words,
β+α−1 2[α, β]
[α, β] = [α, β]
β+α− 1 2[α, β]
. Hence, Theorem 3.1.4 (applied toβ+α−1
2[α, β] and [α, β] instead ofαandβ) yields exp
β+α− 1
2[α, β] + [α, β]
=
(by (76)). Lemma 3.1.9 is proven.
We are going to also present a third, very elementary (term-by-term) proof of Lemma 3.1.9. It relies on the following proposition, which can also be applied in some other contexts (e. g., computing in universal enveloping algebras):
Proposition 3.1.10. Let R be a ring. Let α ∈ R and β ∈ R. Assume that [α, β]
Proof of Proposition 3.1.10. Letγ denote [α, β]. Then, γ commutes with each of α and β (since [α, β] commutes with each of α and β). In other words, γα = αγ and γβ =βγ.
As we showed in the proof of Lemma 3.1.1, every positive i ∈ N satisfies [α, βi] = iγβi−1. Since γ = [α, β], this rewrites as follows:
every positive i∈N satisfies α, βi
commutes with each of α and β. Therefore, the roles of α and β are symmetric, and thus we can apply (78) toβ and α instead of α and β, and conclude that
every positive i∈N satisfies β, αi
Now, we are going to prove that every i∈N and j ∈N satisfy
Proof of (81): We will prove (81) by induction overi:
Induction base: Let j ∈Nbe arbitrary. For i= 0, we have αjβi =αj β0 for i= 0, so that the induction base is complete.
Induction step: Let u ∈ N. Assume that (81) holds for i = u. We must now prove
Now, let j ∈N be positive. Then,j−1∈N. Now, applied tojinstead ofi)
βu = βαj+jγαj−1
(by (82), applied toj−1 instead ofj(sincej−1∈N)) Let us separately simplify the two addends on the right hand side of this equation.
First of all, every k ∈ N which satisfies k ≤ u+ 1 and k ≤ j but does not satisfy
On the other hand,
But multiplying both sides of (85) with j, we obtain
j X
Hence, everyk ∈N satisfying k ≥1 andk ≤j satisfies
(by (90)).
Also, notice that every k ∈Nsatisfies k!
Now, forget that we fixed j. We thus have shown that αjβu+1 = X (the proof is similar to our induction base above), this yields that (93) holds for every j ∈N. In other words, (81) holds for i=u+ 1. Thus, the induction step is complete.
Hence, we have proven (81) by induction overi.
Since γ = [α, β], the (now proven) identity (81) rewrites as αjβi = X
Proposition 3.1.10 is thus proven.
Third proof of Lemma 3.1.9. By the definition of the exponential, we have exp [α, β] =
i!. Multiplying the last two of these three equalities, we obtain This proves Lemma 3.1.9 once again.