Theorem. Let G be a simply connected Lie group. Then the followings are equivalent.

(i) There is a left-symmetric algebra structure on 𝔤 whose associated Lie algebra is 𝔤.

(ii) G admits a left-invariant affine structure. (iii) There is a Lie group homomorphism

ϕ = (f, q) : G → Aff(ℝ^{n}) = GL(ℝ^{n}) ⋉ ℝ^{n}

such that d(ev_{x})|_{e} is an isomorphism for some

x ∈ ℝ^{n}.

(iv) There is a Lie algebra homomorphism

φ = (f, q) : 𝔤 → 𝔞𝔣𝔣(ℝ^{n}) = 𝔤𝔩(ℝ^{n}) + ℝ^{n}

such that q is a linear isomorphism.

(v) There is a Lie algebra homomorphism : 𝔤 →

𝔞𝔣𝔣(𝔤) of the form x → (λ_{x}, x).

Typeset by AMS-TEX

NOTES ON LEFT-SYMMETRIC ALGEBRAS

July, 16, 1999

Abstract. The purpose of this paper is to introduce the well-known relationship between affinely flat structures and the left-symmetric algebras on a Lie group.

Section 1. Left-Symmetric Algebras

Definition 1.1. A left-symmetric algebra A is an algebra with the left-symmetric identity;

(1)  (x, y, z) = (y, x, z),

where (x, y, z) = (xy)z − x(yz).

Definition 1.2. For any algebra A, A^{−} denotes the algebra with multiplication

[x, y] = xy − yx

defined on A.

If A is a left-symmetric algebra, then A^{−} becomes a Lie algebra. Such an algebra A is said to be associated with the Lie algebra A^{−}.

(A, x · y) : l.s.a. → A^{−} = (A, [x, y]) : Lie algebra.

Jacobi identity follows from (1) :

[[x, y], z] = \underline{[x, y]z} − z(xy − yx)
[[y, z], x] = [y, z]x − \underline{x(yz} − zy)
[[z, x], y] = [z, x]y − \underline{y}(zx − \underline{xz})

Setp 1. (i) ⟹ (ii) : Define an affine connection ∇ on G by

∇_{x}y := x · y,

for x, y ∈ 𝔛(G).

(𝔤, [,])  · = algebra product l.s.a.
↓^{exp}  ↓  ↓
G  ∇ = left-inv.cxion  ?

Remark. For x, y in an algebra A with A^{−} = 𝔤,

xy ∈ A ⟺ ∇ is left-invariant.

A is associated with A^{−} = 𝔤 : xy − yx = [x, y] Left-symmetric identity :

\underline{(xy)z} − x(yz) = \underline{(yx)z} − y(xz),
x(yz) − y(xz) = \underline{(xy − yx)}z.

Hence, A is a left-symmetric algebra associated with

A^{−} = 𝔤 :

(3)  xy − yx − [x, y] = 0,
(4)  x(yz) − y(xz) − [x, y]z = 0.

Then on 𝔛(G), the identity (3) and (4) are

T (x, y) := ∇_{x}y − ∇_{y}x − [x, y] = 0,
R(x, y)z := ∇_{x}∇_{y}z − ∇_{y}∇_{x}z − ∇_{[x,y]}z = 0.

Setp 3. (ii) ⟹ (iii) : Suppose G has a left-invariant affine structure.

Then there exists a developing map D : G → 𝔼^{n}, and a Lie group homomorphism

ϕ = ϕ_{D} : G → Aff(𝔼^{n}).

G ⟶^{D} 𝔼^{n}
↓^{l_{g}} ↓^{ϕ(g)∈Aff(𝔼^{n})}
G ⟶^{D} 𝔼^{n}
↓^{l_{h}} ↓^{ϕ(h)∈Aff(𝔼^{n})}
G ⟶^{D} 𝔼^{n}

Define ev_{x} : G → 𝔼^{n} by, for x ∈ 𝔼^{n}

ev_{x}(g) = Ev_{x}(ϕ(g)) = ϕ(g)(x).

Lemma. \underline{D = ev_{x_{0}}}, where x_{0} = D(e).

G ⟶^{ϕ} Aff(𝔼^{n})
∥ ↓^{Ev_{x_{0}}}
G ⟶^{D} 𝔼^{n}

Proof.

G ∋ e ⟶^{D}  D(e) = x_{0} ∈ 𝔼^{n}
↓^{l_{g}} ↓^{ϕ(g)}
G ∋ g ⟶^{D} D(g) = ϕ(g)(x_{0}) =: ev_{x_{0}}(g)

□

Setp 4. (iii) ⟹ (ii) : Let ϕ : G → Aff(𝔼^{n}) be a Lie group homomorphism such that d(ev_{x_{0}} )|_{e} is a linear isomorphism for some

fixed x_{0} ∈ 𝔼^{n}, where ev_{x_{0}} = Ev_{x_{0}} ∘ ϕ : G → 𝔼^{n}.

Lemma. ev_{x_{0}} ∘ l_{g} = ϕ(g) ∘ ev_{x_{0}}.

G ⟶^{ev_{x_{0}}} 𝔼^{n}
↓^{l_{g}} ↓^{ϕ(g)∈Aff(𝔼^{n})}
G ⟶^{ev_{x_{0}}} 𝔼^{n}

Proof.

G ∋ h ⟶^{ev_{x_{0}}}  ev_{x_{0}}(h) = ϕ(h)(x_{0})
↓^{l_{g}} ↓^{ϕ(g)}
G ∋ gh ⟶^{ev_{x_{0}}} ϕ(gh)(x_{0}) = ϕ(g)ϕ(h)(x_{0})

□

To construct ∇ on G : step 1: ev_{x_{0}} is an immersion.

T_{e}G ⟶^{d(ev_{x_{0}} )|_{e}} T_{x_{0}}𝔼^{n}
↓^{d(l_{g})} ↓^{d(ϕ(g))}
T_{g}G ⟶_{≅}^{d(ev_{x_{0}} )|_{g}} T_{ϕ(g)x_{0}}𝔼^{n}

step 2: ev_{x_{0}} is covering.

Note that G_{x_{0}} is discrete.

G  G
↓^{ev_{x_{0}}} ↓^{pr = covering}
G · x_{0} ⟶^{≅} G/G_{x_{0}}

step 3: Define ∇ := (ev_{x_{0}})^{∗}( ˜∇) on G,

where ˜∇ is the usual flat affine connection on 𝔼^{n}.

(l_{g})^{∗}∇ = (ev_{x_{0}})^{∗}( ˜∇) ⟶^{ev_{x_{0}}} ˜∇ = ϕ(g)^{∗}( ˜∇)
↓^{l_{g}}  ↓^{ϕ(g)∈Aff(𝔼^{n})}
(G, ∇)  ⟶^{ev_{x_{0}}}  (𝔼^{n}, ˜∇)

□

Setp 5. (iii) ⟹ (iv) : Suppose that there is a Lie group homomorphism

ϕ : G → Aff(𝔼^{n})

such that d(ev_{x_{0}})|_{e} is an isomorphism for some x_{0} ∈ ℝ^{n}. Take the point x_{0} as the origin in 𝔼^{n}. Then 𝔼^{n} = ℝ^{n},

Aff(𝔼^{n}) = GL(ℝ^{n}) ⋉ ℝ^{n},
𝔞𝔣𝔣(𝔼^{n}) = 𝔤𝔩(ℝ^{n}) + ℝ^{n}.

with the following Lie bracket:

[(A, a), (B, b)] = ([A, B], Ab − Ba).

Aff(ℝ^{n}) = \{( \begin{matrix}A & a \\ 0 & 1 \end{matrix} ) |A ∈ GL(ℝ^{n}), a ∈ ℝ^{n}\} ⊂ GL(ℝ^{n+1})
𝔞𝔣𝔣(ℝ^{n}) = \{( \begin{matrix}B & b \\ 0 & 0 \end{matrix} ) |B ∈ 𝔤𝔩(ℝ^{n}), b ∈ ℝ^{n}\} ⊂ 𝔤𝔩(ℝ^{n+1}).

Hence ϕ and dϕ have two components :

𝔤 ⟶^{dϕ=(f,q)} 𝔤𝔩(ℝ^{n}) + ℝ^{n} ⟶ ℝ^{n}
↓^{exp} ↓^{exp}
G ⟶^{ϕ=(f, q)} GL(ℝ^{n}) ⋉ ℝ^{n} ⟶^{Ev_{0}} ℝ^{n}

Lemma. q(x) = d(ev_{0})|_{e}(x) for x ∈ 𝔤.

Proof. For t ∈ ℝ, (x_{0} = 0)

x ∈ 𝔤 ⟶^{q}  q(x) ∈ ℝ^{n}
↓^{exp} ↓
exp tx ∈ G ⟶^{q} f(exp tx)(0) + q(exp tx) = ev_{0}(exp tx)

Setp 6. (iv) ⟹ (iii) : Since G is simply connected, there exists a Lie group homomorphism ϕ : G → Aff(ℝ^{n}) such that dϕ = φ. Then d(ev_{0}) = q is a linear isomorphism for 0 ∈ ℝ^{n}.

𝔤 ⟶^{dϕ=(f,q)} 𝔞𝔣𝔣(ℝ^{n}) = 𝔤𝔩(ℝ^{n}) + ℝ^{n} ⟶ ℝ^{n}
↓^{exp} ↓^{exp}
G ⟶^{ϕ=(f, q)} Aff(ℝ^{n}) = Gl(ℝ^{n}) ⋉ ℝ^{n} ⟶^{Ev_{0}} ℝ^{n}

This shows that (iv) implies (iii). □

Setp 7. (iv) ⟹ (v) :

Identity ℝ^{n} = T_{0}ℝ^{n} = T_{x_{0}} 𝔼^{n} ≅ 𝔤 = T_{e}G by q = d(ev_{x_{0}} )|_{e}.

Then we obtain

D = ev_{0} : G → 𝔤,

ϕ : G → Aff(𝔤).
𝔤 ⟶^{dϕ=(f,Id)} 𝔞𝔣𝔣(𝔤) = 𝔤𝔩(𝔤) + 𝔤 ⟶ 𝔤
↓^{exp} ↓^{exp}
G ⟶^{ϕ=(f, q)} Aff(𝔤) = GL(𝔤) ⋉ 𝔤 ⟶^{Ev_{0}} 𝔤

Lemma. The left-symmetric product ∇ of 𝔤 is equal to f, the linear part of dϕ.

Proof.

Note that Y_{g} = d(l_{g})|_{e}(Y_{e}) for Y ∈ 𝔤.

step 1. ∇ := (ev_{0})^{∗}( ˜∇).

d(ev_{0})(∇_{X}Y) := ˜∇_{d(ev_{0} )(X)}d(ev_{0})(Y).

step2. Define Ỹ := d(ev_{0})(Y) locally on V_{0} :

T_{g}U_{e} ∋ Y_{g} ⟶_{≅}^{d(ev_{0} )|_{g}} d(ev_{0})|_{g}(Y_{g}) =: Ỹ_{p} ∈ T_{p}V_{0}
↓  ↓
G ⊃ U_{e} ∋ g ⟶_{≅}^{ev_{0}} ev_{0}(g) = ϕ(g)(0) =: p ∈ V_{0}

step 3. Integral curves of X and X̃:

U_{e} ∋ g_{t} := g exp tX ↦ γ(t) := ev_{0}(g_{t}). step 4. d(ev_{0})|_{e}(Y_{e}) = q(Y_{e}) for Y ∈ 𝔤. step 5. Ỹ |_{γ(t)} = f(g_{t})(q(Y_{e})).

G ⟶^{ev_{0}} 𝔤 = ℝ^{n}
↓^{l_{g_{t}}} ↓^{ϕ(l_{g_{t}})∈Aff(ℝ^{n})}
G ⟶^{ev_{0}} 𝔤 = ℝ^{n}
Y_{e} ⟶^{d(ev_{0} )|_{e}}  q(Y_{e})
↓^{d(l_{g_{t}})|_{e}} ↓^{d(ϕ(l_{g_{t}}))=f (g_{t})}
Y_{g_{t}} ⟶^{d(ev_{0} )|_{g_{t}}} Ỹ |_{γ(t)} = f(g_{t})(q(Y_{e}))

step 6. ˜∇X̃ Ỹ |_{p} = f(X)(q(Y)).

˜∇X̃ Ỹ |_{p} = \frac{d}{dt}|_{0} Ỹ (γ(t))

= \frac{d}{dt}|_{0}f(exp tX)(q(Y))

= f(X)(q(Y)).

step 7. d(ev_{0})|_{e} = q = id.

Thus we have ∇_{X}Y = ˜∇X̃ Ỹ = f(X)(q(Y)). □

Setp 8. (v) ⟹ (i) : For a given Lie algebra homomorphism

(λ, id) : 𝔤 → 𝔞𝔣𝔣(𝔤), x ↦ (λ_{x}, x),

define a product x · y by

x · y = λ_{x}(y).

Then

[x, y] ↦ (λ_{[x,y]}, [x, y]) = [(λ_{x}, x), (λ_{y}, y)]
= ([λ_{x}, λ_{y}], λ_{x}(y) − λ_{y}(x)).

Thus we have

(i)  λ_{[x,y]} = [λ_{x}, λ_{y}],
(ii)  [x, y] = λ_{x}(y) − λ_{y}(x) =: x · y − y · x =: [x, y]_{λ}.

Hence (i) implies that A := (𝔤, λ) a left-symmetric algebra, and (ii) implies that A is associated with

A^{−} = 𝔤. □

Step 9. (v) ⟹ (iv) is trivial, since 𝔤 ≅ ℝ^{n} as vector spaces. □

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