Exponential maps of Lie groups

In this note we try to understand the relations between two different notions of exponential map, one from Riemannian geometry, and the other coming from Lie group theory.

Let {G} be a Lie group, then there is an exponential map {\exp: T_eG\rightarrow G} defined by

\displaystyle \exp(X)=\rho (1)

where {\rho : \mathbb{R}\rightarrow G} is the unique one-parameter subgroup (i.e. {\rho (st)=\rho (s)\rho (t)}) such that {\rho (0)=e, \rho '(0)=X}.

On the other hand, if {G} is equipped with a Riemannian metric {\langle\cdot, \cdot\rangle}, then at any point {p\in G}, there is a Riemannian exponential map {Exp_p: T_pG\rightarrow G} defined by

\displaystyle Exp_p(X)=\gamma(1)

where {\gamma : \mathbb{R}\rightarrow G} is the unique geodesic such that {\gamma (0)=p, \gamma '(0)=X}. [Technically the domain of {Exp_p} should be an open subset of {T_pG} instead, and {\gamma } may not be well-defined on the whole {\mathbb{R}}, but these are not important here.] It is interesting to know if these two concepts are related, and why are they called exponential map at all?

Example 1

  • For {G=SO(2)\cong \mathbb{S}^1=\{e^{i\theta}: \theta\in \mathbb{R}\}\subset \mathbb{C}}, this is the simplest example of a compact Lie group. Its tangent space {T_eG\cong \{i\theta: \theta\in \mathbb{R}\}}, the Lie group exponential map is actually given by

    \displaystyle \exp(i\theta) = e^{i\theta},

    where RHS is defined by the ordinary exponential map {\displaystyle e^x=\sum_{n=0}^\infty{\frac{x^n}{n!}}}. Thus we see that the notation “exponential map” is reasonable in this case. Actually, by {e^{(s+t)A}=e^{sA}e^{tA}} and {\displaystyle\left.\frac{d}{dt}\right|_{t=0}e^{tA}=A} for matrix {A}, this example can be generalized for matrix Lie group. i.e. (after taking care of the domain), the matrix exponential map is the Lie group exponential map for matrix Lie groups, e.g. {O(n), U(n), GL_n(\mathbb{R})} etc.

  • Actually there is a Lie theory viewpoint for the ordinary exponential map for real numbers. Indeed, consider {(\mathbb{R^+}, \cdot)} as a (non-compact) Lie group with ordinary multiplication. Its tangent space at {e=1} is just {\mathbb{R}}, and both the left translation by {x} and its differential map are just the multiplication by {x} itself. The Lie group exponential map must satisfy: {\exp(0)=1} and for {\gamma(t)=\exp(tx)},

    \displaystyle  \gamma'(t)=dL_{\gamma(t)}\gamma'(0)= \gamma(t)\cdot \gamma'(0), \text{\quad i.e. }\frac{d}{dt}\exp(tx)=\exp(tx)x

    So we must have

    \displaystyle  \exp(0)=1, \frac{d}{dx}\exp(x)=\exp(x).

    But this is nothing but the definition of {e^x} (well at least it can be taken as a definition). So we have {\exp(x)=e^x}.

Now, if {G} is a Lie group with a bi-invariant metric {\langle \cdot, \cdot\rangle}, i.e.

\displaystyle \langle dL_g(u), dL_g(v)\rangle=\langle u,v\rangle=\langle dR_g(u), dR_g(v)\rangle

for all {u,v\in T_h G} and {g\in G}, where {L_g: h\mapsto gh, R_g: h\mapsto hg} are the left and right translations respectively. Then there is a Riemannian exponential map {Exp} associated with this metric. We would like to know if they are equal in the following sense:

\displaystyle Exp_e=\exp.

It is also interesting to know when a Lie group has a bi-invariant metric. It is quite obvious that any Lie group {G} has a left-invariant (and a right-invariant) metric by left(or right)-translating an inner product at {T_eG}. I only know recently that any compact Lie group has a bi-invariant metric which is unique up to scaling. Perhaps Ken can supply a proof of this , using modular function. Following Ken’s argument, we can prove the following two results.

Theorem 1 [Ken’s theorem, 2011] If {G} is a compact Lie group, then it has a bi-invariant Riemannian metric.

Theorem 2 [Ken’s theorem, 2011] If {G} is Lie group with a bi-invariant metric (for example {G} is compact), then

\displaystyle exp(X)= Exp_e(X)

where {Exp} is the Riemannian exponential map.

To prove Theorem 1 we first introduce some notations. Firstly, for {g\in G}, we can define a conjugate action {c(g)} of {g} by the composition {c(g)=L_g R_{g^{-1}}: G\rightarrow G}, i.e.

\displaystyle c(g) h=ghg^{-1}.

For fixed {g}, the differential map {dc(g)} of {c(g)} at {e} is then {dL_gdR_{g^{-1}}\in Aut(T_eG)}. We will denote this map as {Ad_g=dL_gdR_{g^{-1}}} (adjoint action). Note that {(gh)g^{-1}=g(hg^{-1})}, i.e.

\displaystyle  L_g R_{g^{-1}}=R_{g^{-1}}L_g ,

thus we have

\displaystyle  Ad_g=dL_gdR_{g^{-1}}=dR_{g^{-1}}dL_g: T_eG\rightarrow T_eG. \ \ \ \ \ (1)

By an Ad-invariant inner product on {T_eG} we mean an inner product {\langle \cdot, \cdot\rangle_{T_eG}} such that {\langle Ad_g u, Ad_g v\rangle_{T_eG}=\langle u,v\rangle_{T_eG}} for all {g\in G} and {u,v\in T_eG}. We have the following

Lemma 3
There is a one-one correspondence between Ad-invariant inner products on {T_eG} and bi-invariant Riemannian metrics on {G}.

Proof: For any bi-invariant metric {\langle \cdot, \cdot\rangle} on {G}, clearly its restriction on {T_eG} is an Ad-invariant inner product: {\langle dL_g dR_{g^{-1}}u,dL_g dR_{g^{-1}}v\rangle_e=\langle dR_{g^{-1}}u, dR_{g^{-1}}v\rangle_{g^{-1}}=\langle u,v\rangle_e}.

Conversely, for any Ad-invariant inner product {\langle \cdot, \cdot\rangle_{T_eG}} on {T_eG}, we left translate it to a Riemannian metric {\langle \cdot, \cdot\rangle} on {G}. Clearly this metric is left-invariant by definition. We check that this is also right-invariant. For {u_g,v_g\in T_g G}, let {dL_gu_e=u_g} and {dL _g v_e=v_g}, then

\displaystyle  \begin{array}{rcl}  \langle dR_h u_g, dR_h v_g\rangle_{gh} &=&\langle dL_{(gh)^{-1}}dR_h u_g, dL_{(gh)^{-1}}dR_h v_g\rangle_e\\ &=&\langle dL_{h^{-1}}dL_{g^{-1}}dR_h u_g, dL_{h^{-1}}dL_{g^{-1}}dR_h v_g\rangle_e\\ &=&\langle dL_{h^{-1}}dR_h dL_{g^{-1}}u_g, dL_{h^{-1}}dR_hdL_{g^{-1}} v_g\rangle_e\\ &=&\langle dL_{h^{-1}}dR_h u_e, dL_{h^{-1}}dR_hv_e\rangle_e \\ &=&\langle Ad_{h^{-1}} u_e, Ad_{h^{-1}} v_e\rangle_e\\ &=&\langle u_e, v_e\rangle_e\\ &=&\langle u_g, v_g\rangle_g \end{array}

where we have used (1) on line 3. \Box

We are now ready to prove Theorem 1.

Proof: [Proof of Theorem 1]
Let {\mu} be the right-invariant Haar measure on {G}. By lemma 3, it is equivalent to find an Ad-invariant inner product on {T_eG}. Let {\langle \cdot, \cdot\rangle'} be any inner product on {T_eG}, then the desired inner product is defined by

\displaystyle \langle u,v\rangle:=\int_G \langle Ad_gu, Ad_gv\rangle'd\mu(g).

To show that this metric is Ad-invariant, we note that { Ad: G\rightarrow Aut(T_eG) } is a homomorphism, in other words,

\displaystyle Ad_g Ad_h=Ad_{gh}.

This can be seen by taking the differential of the relation c(gh)=c(g)c(h) where c(g) is the conjugate action.
So we have

\displaystyle  \begin{array}{rcl}  \langle Ad_h u, Ad_h v\rangle &=& \int_G \langle Ad_g Ad_h u, Ad_g Ad_h v\rangle'd\mu(g)\\ &=& \int_G \langle Ad_{gh}u, Ad_{gh} v\rangle'd\mu(g)\\ &=& \int_G \langle Ad_{gh}u, Ad_{gh} v\rangle'd\mu(gh)\\ &=& \langle u,v\rangle. \end{array}


To prove Theorem 2 we need the following

Lemma 4
For a Lie group {G} with a bi-invariant metric, the map {i: G\rightarrow G} defined by

\displaystyle i(g)=g^{-1}

is an isometry. Also {di_e(u)=-u}. (The differential of the map {i} at the point {e}. )

For {u\in T_gG}, we claim that at {T_{g^{-1}}G},

\displaystyle  di(u)= -(dL_g)^{-1}(dR_g)^{-1}u.

To see this, if {\gamma'(0)= u}, then {e\equiv\gamma(t)\gamma(t)^{-1}}. Differentiating gives

\displaystyle 0= dR_{g^{-1}} (u) + dL_g( di(u)).

As {dR_{g^{-1}}=(dR_g)^{-1}}, the claim is proved. Thus we have

\displaystyle \langle di(u), di(v)\rangle_{g^{-1}}= \langle -dL_g^{-1}dR_g^{-1}u, -dL_g^{-1}dR_g^{-1}v\rangle_{g^{-1}}=\langle u, v\rangle _g

by the bi-invariance of the metric. The second assertion is trivial. \Box

Proof: [Proof of Theorem 2]
Let {\gamma(s)}, {s\in (-\varepsilon, \varepsilon)}, be a geodesic starting at {e} with initial vector {X}. By the Lemma 4, as {di_e(X)=-X} and {i} is an isometry, {i\circ \gamma (s)=\gamma(s)^{-1}} is a geodesic. Thus we have, by the uniqueness of geodesic,

\displaystyle  \gamma(-s)= \gamma(s)^{-1},\quad \text{i.e. }\gamma(s)\gamma(-s)=e.

For small {s_0}, if we define {\widetilde \gamma (s)=\gamma(s_0)\gamma(s)}, then {\widetilde \gamma} is a geodesic (for small {s}) with {\widetilde\gamma(0)= \gamma(s_0)} and {\widetilde \gamma (-s_0)=e}, by uniqueness of short geodesic (joining two nearby points), we must have

\displaystyle  \widetilde \gamma(s)= \gamma (s_0+s). \text{\quad i.e. } \gamma(s_0)\gamma(s)= \gamma (s_0+s) \ \ \ \ \ (2)

for all small enough {s_0} and {s}. By extending {\gamma} beyond any interval {[0,l]} by {\gamma(l+s):=\gamma(l)\gamma(s)}, we see that {\gamma} can indeed by extended (smoothly as a geodesic) to the whole {\mathbb{R}}. And by a standard argument (of chopping into “small pieces”), from (2), we indeed have

\displaystyle \gamma(s+t)=\gamma(s)\gamma(t).

Thus {\gamma} is a one-parameter subgroup in {G}. As {\gamma(0)=e} and {\gamma'(0)=X}, by definition of {\exp},

\displaystyle \exp(X)=\gamma(1)= Exp_e(X).


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8 Responses to Exponential maps of Lie groups

  1. Ken Leung says:

    How does the (bi-, left- or right-) invariance of the metric relate to the Haar measure of the Lie group? Can the metric in some sense be given in terms of integration against the Haar measure?

    • Edward Fan says:

      A (bi-, left- or right) invariant metric induces a (bi-, left- or right) Haar measure of the Lie group. One knows that (left- or right-) Haar measure of a locally compact group is unique up to positive scalars, indeed, when it is a Lie group, such Haar measures always come from volume forms on the underlying differential manifolds. In particular, a (bi-, left- or right) invariant metric will give a Riemannian volume form (namely the Hodge dual of the 0-form, constant function 1).

  2. Ken Leung says:

    Can we do something to utilize the fact that when G is compact, for any finite-dimensional representation \pi: G \rightarrow GL(V), V admits a G-invariant inner product given by
    \langle u, v \rangle = \int_G \langle \pi(g)u, \pi(g)v \rangle dg
    to the case of the adjoint representation of G on its Lie algebra?

    • Edward Fan says:

      What you are doing is to show that there is a G-invariant metric on its Lie algebra and then you can use translation to move the metric everywhere. Indeed, proveided that it is a connected Lie group, bi-invariant metric exists iff it is a direct product of a compact Lie group and a vector space.

    • KKK says:

      Great! I have updated the note accordingly.

  3. Ken Leung says:

    Your equation defining Ad-invariance has its RHS missing~ [Corrected. Thanks! -KKK]

  4. se0808 says:

    KKK, perhaps left-invariant metric is induced by right translations, as long as left and right actions commute. Isn’t it so?

  5. Lawrence G Mouille says:

    Why are you calling these result’s “Ken’s theorem” and claiming they were proven in 2011? These have been known for a long time.

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