Lecture 4 - 22 Jan 2026

Lie algebra -> Lie Group by the Exponential Map

Proposition: Let [math]\displaystyle{ X }[/math] be a left-invariant vectorfield in [math]\displaystyle{ \mathfrak{g} }[/math]. Then [math]\displaystyle{ X }[/math] is complete (flow defined everywhere, i.e. for all time) proof, it works for [math]\displaystyle{ \varepsilon \gt 0 }[/math] near [math]\displaystyle{ e_{G} }[/math]. Then multiply by [math]\displaystyle{ a }[/math], i.e. the diffeomorphism [math]\displaystyle{ L_{a} }[/math], and we get that the flow line is still a flow line [math]\displaystyle{ L_{a}\left( \Phi_{\varepsilon}^{X} \right) }[/math].

Remark: [math]\displaystyle{ {\Phi^{X}}_{-}\left( e_{G} \right):{\mathbb{R}} \rightarrow G }[/math] is a group homomorphism. i.e. where [math]\displaystyle{ t| \rightarrow {\Phi^{X}}_{t} }[/math].

Def let [math]\displaystyle{ G }[/math] be a lie group and [math]\displaystyle{ \mathfrak{g} }[/math] it’s lie algebra. The exponential map is [math]\displaystyle{ \exp:\mathfrak{g} \rightarrow G }[/math] where [math]\displaystyle{ X| \rightarrow {\Phi^{X}}_{1}\left( e_{G} \right) }[/math].

Intuitively, it’s a way of wrapping [math]\displaystyle{ \mathfrak{g} }[/math] around [math]\displaystyle{ G }[/math]. Note [math]\displaystyle{ \exp }[/math] is in general neither surjective nor injective (even if [math]\displaystyle{ G }[/math] connected.

Proposition: [math]\displaystyle{ \begin{array}{r} \exp(0) = e_{G} \\ \exp((t + s)X) = \exp(tX) \cdot \exp(sX) \\ \exp(X + Y) \neq \exp(X) \cdot \exp(Y) \end{array} }[/math]. however, the final equation does hold if [math]\displaystyle{ \lbrack X,Y\rbrack = 0 }[/math].

[math]\displaystyle{ \exp }[/math] is smooth
[math]\displaystyle{ \exp }[/math] is a local diffeomorphism from [math]\displaystyle{ N(0) }[/math] for [math]\displaystyle{ 0 \in {\mathbb{R}} }[/math] to [math]\displaystyle{ N\left( e_{G} \right) }[/math] for [math]\displaystyle{ e_{G} \in G }[/math]. [math]\displaystyle{ N(x) }[/math] is a neighbourhood of [math]\displaystyle{ x }[/math].

Example: for [math]\displaystyle{ G = \text{GL}_{n} }[/math] and [math]\displaystyle{ \mathfrak{g} = M_{n} }[/math], we have [math]\displaystyle{ \exp:\mathfrak{g} \rightarrow G }[/math] given by [math]\displaystyle{ A| \rightarrow e^{A} = \sum_{k} = 0^{\infty}\frac{A^{n}}{n}! }[/math]

Example: for [math]\displaystyle{ G = {\mathbb{C}}^{\times} }[/math] (nonzero complex numbers), [math]\displaystyle{ \mathfrak{g} = {\mathbb{C}} }[/math], we have [math]\displaystyle{ \exp = \exp }[/math] as expected. Note, not injective, but is locally diffeomorphism near [math]\displaystyle{ 0 }[/math].

prop: naturality of [math]\displaystyle{ \exp }[/math]. If we have a diffeomorphism [math]\displaystyle{ G \rightarrow^{\varphi}H }[/math] and [math]\displaystyle{ d\varphi:\mathfrak{g} \rightarrow \mathfrak{h} }[/math] we have that [math]\displaystyle{ \varphi \circ \exp = \exp \circ d\varphi }[/math] (this is a naturality square commuting).

Example: [math]\displaystyle{ \text{GL}_{n}({\mathbb{C}}) \rightarrow^{\det}{\mathbb{C}}^{\times} }[/math], we get from the above that [math]\displaystyle{ \det(e^{A}) = e^{\text{tr}(A)} }[/math].

prop: (the lie aliebra knows a lot about our lie group). Let [math]\displaystyle{ \varphi,\psi:G \rightarrow H }[/math] be Lie group morphisms, and [math]\displaystyle{ G }[/math] connected. If [math]\displaystyle{ d\varphi = d\psi }[/math] then [math]\displaystyle{ \psi = \varphi }[/math] explanation: Both [math]\displaystyle{ \varphi,\psi }[/math] map [math]\displaystyle{ e_{G}| \rightarrow e_{H} }[/math]. Also, near [math]\displaystyle{ e_{G} }[/math] the [math]\displaystyle{ \exp }[/math] map is a local diffeomorphism so every point near [math]\displaystyle{ e_{G} }[/math] is [math]\displaystyle{ \exp(X) }[/math] for some [math]\displaystyle{ X }[/math]. So we see locally that [math]\displaystyle{ \exp(X)| \rightarrow^{\psi\text{ or }\varphi}\exp(d\varphi(X)) = \exp(d\psi(X)) }[/math] and so [math]\displaystyle{ \varphi,\psi }[/math] agree on a nbh of [math]\displaystyle{ e_{G} }[/math]. Then, the equaliser of [math]\displaystyle{ \varphi,\psi }[/math] is closed in [math]\displaystyle{ G }[/math] (WHY? - because it is preimage of [math]\displaystyle{ e_{G} }[/math] under [math]\displaystyle{ \varphi \circ \psi^{-}1 }[/math]). Then for [math]\displaystyle{ b \in N\left( e_{G} \right) }[/math] we have [math]\displaystyle{ \varphi(ba) = \psi(ba) }[/math]… so we end up getting [math]\displaystyle{ \psi,\varphi }[/math] agree on a nbh of [math]\displaystyle{ a }[/math] for any [math]\displaystyle{ a \in G }[/math]. thus [math]\displaystyle{ \psi,\varphi }[/math] agree everywhere.

prop: [math]\displaystyle{ \varphi:G \rightarrow H }[/math] lie group hom. [math]\displaystyle{ G,H }[/math] connected, supose that [math]\displaystyle{ d\varphi:\mathfrak{g} \rightarrow \mathfrak{h} }[/math] is an isomophism. Then

[math]\displaystyle{ \varphi }[/math] is sujective
[math]\displaystyle{ \ker(\varphi) \subseteq G }[/math] is discrete
[math]\displaystyle{ \varphi }[/math] is a cover.

Example: [math]\displaystyle{ {\mathbb{R}} \rightarrow S^{1} }[/math], [math]\displaystyle{ t| \rightarrow e^{i}t }[/math], [math]\displaystyle{ \ker = {\mathbb{Z}} }[/math]. This is a covering map of [math]\displaystyle{ S^{1} }[/math].

Example: [math]\displaystyle{ S^{3} \rightarrow SO(3) }[/math], [math]\displaystyle{ a| \rightarrow D_{e_{G}}a }[/math], [math]\displaystyle{ \ker = \frac{\mathbb{Z}}{2}{\mathbb{Z}} }[/math]

Representations

definition: A representation of lie group [math]\displaystyle{ G }[/math] is a hom of lie algebras [math]\displaystyle{ G \rightarrow \text{GL}(V) }[/math] where [math]\displaystyle{ V }[/math] is a [math]\displaystyle{ \mathbb{R} }[/math]-vector space.

This gives us a way of acting smoothly on a vector space.

definition: A representation of lie algebra [math]\displaystyle{ \mathfrak{g} }[/math] is a smooth morphism of lie alebras [math]\displaystyle{ \mathfrak{g} \rightarrow \mathfrak{gl}(V) = \text{ End}(V) }[/math].

A morphism of representations [math]\displaystyle{ f:V \rightarrow W }[/math] is a linear map that preserves the action. (a representation is a group action remember).

For [math]\displaystyle{ G = \text{GL}_{n}\text{SL}_{n}\text{ SO}_{n},.. \subseteq \text{GL}_{n} }[/math], these are all the standard representation.

There is a way for any lie group to compare it to a matrix group:

definition: the adjoint representation of [math]\displaystyle{ G }[/math] is [math]\displaystyle{ \text{Ad}:G \rightarrow \text{GL}(\mathfrak{g}) \simeq \text{GL}(T_{e_{G}}G) }[/math], [math]\displaystyle{ a| \rightarrow dC_{a} }[/math], where [math]\displaystyle{ C_{a} }[/math] is conjugation by [math]\displaystyle{ a }[/math], and we take it’s differential at [math]\displaystyle{ e_{G} }[/math], i.e. [math]\displaystyle{ dC_{a} = D_{e_{G}}C_{a} }[/math].

The adjoint representation of [math]\displaystyle{ \mathfrak{g} }[/math] is [math]\displaystyle{ \text{ad } = d\text{ Ad} }[/math]. [math]\displaystyle{ \mathfrak{g} \rightarrow \mathfrak{gl}(\mathfrak{g}) }[/math], [math]\displaystyle{ X| \rightarrow \lbrack X, - \rbrack }[/math].

proposition: [math]\displaystyle{ G }[/math] connected implies [math]\displaystyle{ \ker(\text{Ad}) = Z(G) = \left\{ a \in G,\forall b,ab = ba \right\} }[/math]
[math]\displaystyle{ \ker(\text{ad}) = Z\left( \mathfrak{g} \right) = \left\{ X \in \mathfrak{g},\forall Y,\lbrack X,Y\rbrack = 0 \right\} }[/math]
Example: for [math]\displaystyle{ G = \text{GL}_{n} }[/math], we have[math]\displaystyle{ \text{ad}:M_{n} \rightarrow \text{ End}\left( M_{n} \right),A| \rightarrow \left( G| \rightarrow AB - BA \right) }[/math].

definition: the killing form for [math]\displaystyle{ \mathfrak{g} }[/math] finite dimensional is a map [math]\displaystyle{ k:\mathfrak{g} \times \mathfrak{g} \rightarrow {\mathbb{R}} }[/math], [math]\displaystyle{ X,Y| \rightarrow \text{tr}(\text{ad}(X) \circ \text{ ad}(Y)) }[/math]

Example: for [math]\displaystyle{ \mathfrak{g} = \text{ sl}_{n} \subseteq M_{n} }[/math], the killing form [math]\displaystyle{ k(X,Y) = 2n\text{tr}(XY) }[/math] where [math]\displaystyle{ X,Y }[/math] are [math]\displaystyle{ n \times n }[/math] matrices.

prop: [math]\displaystyle{ k }[/math] behaves well with the actions of [math]\displaystyle{ G }[/math] and [math]\displaystyle{ \mathfrak{g} }[/math] on [math]\displaystyle{ \mathfrak{g} }[/math]. [math]\displaystyle{ k }[/math] is symmetric, and [math]\displaystyle{ k }[/math] is non-degenerate ([math]\displaystyle{ \forall X \neq 0,\exists Y,k(X,Y) \neq 0 }[/math]). So we get an isomrphism with the dual [math]\displaystyle{ \mathfrak{g} \simeq \mathfrak{g}^{*} }[/math].

[math]\displaystyle{ \mathfrak{g} }[/math] is semisimple if it does not have an abelian ideal.