Facts about Lie Groups and Algebras

In Spring 2016 I was taking 18.757 Representations of Lie Algebras. Since I knew next to nothing about either Lie groups or algebras, I was forced to quickly learn about their basic facts and properties. These are the notes that I wrote up accordingly. Proofs of most of these facts can be found in standard textbooks, for example Kirillov.

1. Lie groups

Let {K = \mathbb R} or {K = \mathbb C}, depending on taste.

Definition 1

A Lie group is a group {G} which is also a {K}-manifold; the multiplication maps {G \times G \rightarrow G} (by {(g_1, g_2) \mapsto g_1g_2}) and the inversion map {G \rightarrow G} (by {g \mapsto g^{-1}}) are required to be smooth.

A morphism of Lie groups is a map which is both a map of manifolds and a group homomorphism.

Throughout, we will let {e \in G} denote the identity, or {e_G} if we need further emphasis.

Note that in particular, every group {G} can be made into a Lie group by endowing it with the discrete topology. This is silly, so we usually require only focus on connected groups:

Proposition 2 (Reduction to connected Lie groups)

Let {G} be a Lie group and {G^0} the connected component of {G} which contains {e}. Then {G^0} is a normal subgroup, itself a Lie group, and the quotient {G/G^0} has the discrete topology.

In fact, we can also reduce this to the study of simply connected Lie groups as follows.

Proposition 3 (Reduction to simply connected Lie groups)

If {G} is connected, let {\pi : \widetilde G \rightarrow G} be its universal cover. Then {\widetilde G} is a Lie group, {\pi} is a morphism of Lie groups, and {\ker \pi \cong \pi_1(G)}.

Here are some examples of Lie groups.

Example 4 (Examples of Lie groups)

  • {\mathbb R} under addition is a real one-dimensional Lie group.
  • {\mathbb C} under addition is a complex one-dimensional Lie group (and a two-dimensional real Lie group)!
  • The unit circle {S^1 \subseteq \mathbb C} is a real Lie group under multiplication.
  • {\text{GL }(n, K) \subset K^{\oplus n^2}} is a Lie group of dimension {n^2}. This example becomes important for representation theory: a representation of a Lie group {G} is a morphism of Lie groups {G \rightarrow \text{GL }(n, K)}.
  • {\text{SL }(n, K) \subset \text{GL }(n, K)} is a Lie group of dimension {n^2-1}.

As geometric objects, Lie groups {G} enjoy a huge amount of symmetry. For example, any neighborhood {U} of {e} can be “copied over” to any other point {g \in G} by the natural map {gU}. There is another theorem worth noting, which is that:

Proposition 5

If {G} is a connected Lie group and {U} is a neighborhood of the identity {e \in G}, then {U} generates {G} as a group.

2. Haar measure

Recall the following result and its proof from representation theory:

Claim 6

For any finite group {G}, {\mathbb C[G]} is semisimple; all finite-dimensional representations decompose into irreducibles.

Proof: Take a representation {V} and equip it with an arbitrary inner form {\left< -,-\right>_0}. Then we can average it to obtain a new inner form

\displaystyle \left< v, w \right> = \frac{1}{|G|} \sum_{g \in G} \left< gv, gw \right>_0.

which is {G}-invariant. Thus given a subrepresentation {W \subseteq V} we can just take its orthogonal complement to decompose {V}. \Box
We would like to repeat this type of proof with Lie groups. In this case the notion {\sum_{g \in G}} doesn’t make sense, so we want to replace it with an integral {\int_{g \in G}} instead. In order to do this we use the following:

Theorem 7 (Haar measure)

Let {G} be a Lie group. Then there exists a unique Radon measure {\mu} (up to scaling) on {G} which is left-invariant, meaning

\displaystyle \mu(g \cdot S) = \mu(S)

for any Borel subset {S \subseteq G} and “translate” {g \in G}. This measure is called the (left) Haar measure.

Example 8 (Examples of Haar measures)

  • The Haar measure on {(\mathbb R, +)} is the standard Lebesgue measure which assigns {1} to the closed interval {[0,1]}. Of course for any {S}, {\mu(a+S) = \mu(S)} for {a \in \mathbb R}.
  • The Haar measure on {(\mathbb R \setminus \{0\}, \times)} is given by

    \displaystyle \mu(S) = \int_S \frac{1}{|t|} \; dt.

    In particular, {\mu([a,b]) = \log(b/a)}. One sees the invariance under multiplication of these intervals.

  • Let {G = \text{GL }(n, \mathbb R)}. Then a Haar measure is given by

    \displaystyle \mu(S) = \int_S |\det(X)|^{-n} \; dX.

  • For the circle group {S^1}, consider {S \subseteq S^1}. We can define

    \displaystyle \mu(S) = \frac{1}{2\pi} \int_S d\varphi

    across complex arguments {\varphi}. The normalization factor of {2\pi} ensures {\mu(S^1) = 1}.

Note that we have:

Corollary 9

If the Lie group {G} is compact, there is a unique Haar measure with {\mu(G) = 1}.

This follows by just noting that if {\mu} is Radon measure on {X}, then {\mu(X) < \infty}. This now lets us deduce that

Corollary 10 (Compact Lie groups are semisimple)

{\mathbb C[G]} is semisimple for any compact Lie group {G}.

Indeed, we can now consider

\displaystyle \left< v,w\right> = \int_G \left< g \cdot v, g \cdot w\right>_0 \; dg

as we described at the beginning.

3. The tangent space at the identity

In light of the previous comment about neighborhoods of {e} generating {G}, we see that to get some information about the entire Lie group it actually suffices to just get “local” information of {G} at the point {e} (this is one formalization of the fact that Lie groups are super symmetric).

To do this one idea is to look at the tangent space. Let {G} be an {n}-dimensional Lie group (over {K}) and consider {\mathfrak g = T_eG} the tangent space to {G} at the identity {e \in G}. Naturally, this is a {K}-vector space of dimension {n}. We call it the Lie algebra associated to {G}.

Example 11 (Lie algebras corresponding to Lie groups)

  • {(\mathbb R, +)} has a real Lie algebra isomorphic to {\mathbb R}.
  • {(\mathbb C, +)} has a complex Lie algebra isomorphic to {\mathbb C}.
  • The unit circle {S^1 \subseteq \mathbb C} has a real Lie algebra isomorphic to {\mathbb R}, which we think of as the “tangent line” at the point {1 \in S^1}.

Example 12 ({\mathfrak{gl}(n, K)})

Let’s consider {\text{GL }(n, K) \subset K^{\oplus n^2}}, an open subset of {K^{\oplus n^2}}. Its tangent space should just be an {n^2}-dimensional {K}-vector space. By identifying the components in the obvious way, we can think of this Lie algebra as just the set of all {n \times n} matrices.

This Lie algebra goes by the notation {\mathfrak{gl}(n, K)}.

Example 13 ({\mathfrak{sl}(n, K)})

Recall {\text{SL }(n, K) \subset \text{GL }(n, K)} is a Lie group of dimension {n^2-1}, hence its Lie algebra should have dimension {n^2-1}. To see what it is, let’s look at the special case {n=2} first: then

\displaystyle \text{SL }(2, K) = \left\{ \begin{pmatrix} a & b \\ c & d \end{pmatrix} \mid ad - bc = 1 \right\}.

Viewing this as a polynomial surface {f(a,b,c,d) = ad-bc} in {K^{\oplus 4}}, we compute

\displaystyle \nabla f = \left< d, -c, -b, a \right>

and in particular the tangent space to the identity matrix {\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}} is given by the orthogonal complement of the gradient

\displaystyle \nabla f (1,0,0,1) = \left< 1, 0, 0, 1 \right>.

Hence the tangent plane can be identified with matrices satisfying {a+d=0}. In other words, we see

\displaystyle \mathfrak{sl}(2, K) = \left\{ T \in \mathfrak{gl}(2, K) \mid \text{Tr } T = 0. \right\}.

By repeating this example in greater generality, we discover

\displaystyle \mathfrak{sl}(n, K) = \left\{ T \in \mathfrak{gl}(n, K) \mid \text{Tr } T = 0. \right\}.

4. The exponential map

Right now, {\mathfrak g} is just a vector space. However, by using the group structure we can get a map from {\mathfrak g} back into {G}. The trick is “differential equations”:

Proposition 14 (Differential equations for Lie theorists)

Let {G} be a Lie group over {K} and {\mathfrak g} its Lie algebra. Then for every {x \in \mathfrak g} there is a unique homomorphism

\displaystyle \gamma_x : K \rightarrow G

which is a morphism of Lie groups, such that

\displaystyle \gamma_x'(0) = x \in T_eG = \mathfrak g.

We will write {\gamma_x(t)} to emphasize the argument {t \in K} being thought of as “time”. Thus this proposition should be intuitively clear: the theory of differential equations guarantees that {\gamma_x} is defined and unique in a small neighborhood of {0 \in K}. Then, the group structure allows us to extend {\gamma_x} uniquely to the rest of {K}, giving a trajectory across all of {G}. This is sometimes called a one-parameter subgroup of {G}, but we won’t use this terminology anywhere in what follows.

This lets us define:

Definition 15

Retain the setting of the previous proposition. Then the exponential map is defined by

\displaystyle \exp : \mathfrak g \rightarrow G \qquad\text{by}\qquad x \mapsto \gamma_x(1).

The exponential map gets its name from the fact that for all the examples I discussed before, it is actually just the map {e^\bullet}. Note that below, {e^T = \sum_{k \ge 0} \frac{T^k}{k!}} for a matrix {T}; this is called the matrix exponential.

Example 16 (Exponential Maps of Lie algebras)

  • If {G = \mathbb R}, then {\mathfrak g = \mathbb R} too. We observe {\gamma_x(t) = e^{tx} \in \mathbb R} (where {t \in \mathbb R}) is a morphism of Lie groups {\gamma_x : \mathbb R \rightarrow G}. Hence

    \displaystyle \exp : \mathbb R \rightarrow \underbrace{\mathbb R}_{=G} \qquad \exp(x) = \gamma_x(1) = e^t \in \mathbb R = G.

  • Ditto for {\mathbb C}.
  • For {S^1} and {x \in \mathbb R}, the map {\gamma_x : \mathbb R \rightarrow S^1} given by {t \mapsto e^{itx}} works. Hence

    \displaystyle \exp : \mathbb R \rightarrow S^1 \qquad \exp(x) = \gamma_x(1) = e^{it} \in S^1.

  • For {\text{GL }(n, K)}, the map {\gamma_X : K \rightarrow \text{GL }(n, K)} given by {t \mapsto e^{tX}} works nicely (now {X} is a matrix). (Note that we have to check {e^{tX}} is actually invertible for this map to be well-defined.) Hence the exponential map is given by

    \displaystyle \exp : \mathfrak{gl}(n,K) \rightarrow \text{GL }(n,K) \qquad \exp(X) = \gamma_X(1) = e^X \in \text{GL }(n, K).

  • Similarly,

    \displaystyle \exp : \mathfrak{sl}(n,K) \rightarrow \text{SL }(n,K) \qquad \exp(X) = \gamma_X(1) = e^X \in \text{SL }(n, K).

    Here we had to check that if {X \in \mathfrak{sl}(n,K)}, meaning {\text{Tr } X = 0}, then {\det(e^X) = 1}. This can be seen by writing {X} in an upper triangular basis.

Actually, taking the tangent space at the identity is a functor. Consider a map {\varphi : G_1 \rightarrow G_2} of Lie groups, with lie algebras {\mathfrak g_1} and {\mathfrak g_2}. Because {\varphi} is a group homomorphism, {G_1 \ni e_1 \mapsto e_2 \in G_2}. Now, by manifold theory we know that maps {f : M \rightarrow N} between manifolds gives a linear map between the corresponding tangent spaces, say {Tf : T_pM \rightarrow T_{fp}N}. For us we obtain a linear map

\displaystyle \varphi_\ast = T \varphi : \mathfrak g_1 \rightarrow \mathfrak g_2.

In fact, this {\varphi_\ast} fits into a diagram


Here are a few more properties of {\exp}:

  • {\exp(0) = e \in G}, which is immediate by looking at the constant trajectory {\phi_0(t) \equiv e}.
  • {\exp'(x) = x \in \mathfrak g}, i.e. the total derivative {D\exp : \mathfrak g \rightarrow \mathfrak g} is the identity. This is again by construction.
  • In particular, by the inverse function theorem this implies that {\exp} is a diffeomorphism in a neighborhood of {0 \in \mathfrak g}, onto a neighborhood of {e \in G}.
  • {\exp} commutes with the commutator. (By the above diagram.)

5. The commutator

Right now {\mathfrak g} is still just a vector space, the tangent space. But now that there is map {\exp : \mathfrak g \rightarrow G}, we can use it to put a new operation on {\mathfrak g}, the so-called commutator.

The idea is follows: we want to “multiply” two elements of {\mathfrak g}. But {\mathfrak g} is just a vector space, so we can’t do that. However, {G} itself has a group multiplication, so we should pass to {G} using {\exp}, use the multiplication in {G} and then come back.

Here are the details. As we just mentioned, {\exp} is a diffeomorphism near {e \in G}. So for {x}, {y} close to the origin of {\mathfrak g}, we can look at {\exp(x)} and {\exp(y)}, which are two elements of {G} close to {e}. Multiplying them gives an element still close to {e}, so its equal to {\exp(z)} for some unique {z}, call it {\mu(x,y)}.

One can show in fact that {\mu} can be written as a Taylor series in two variables as

\displaystyle \mu(x,y) = x + y + \frac{1}{2} [x,y] + \text{third order terms} + \dots

where {[x,y]} is a skew-symmetric bilinear map, meaning {[x,y] = -[y,x]}. It will be more convenient to work with {[x,y]} than {\mu(x,y)} itself, so we give it a name:

Definition 17

This {[x,y]} is called the commutator of {G}.

Now we know multiplication in {G} is associative, so this should give us some nontrivial relation on the bracket {[,]}. Specifically, since

\displaystyle \exp(x) \left( \exp(y) \exp(z) \right) = \left( \exp(x) \exp(y) \right) \exp(z).

we should have that {\mu(x, \mu(y,z)) = \mu(\mu(x,y), z)}, and this should tell us something. In fact, the claim is:

Theorem 18

The bracket {[,]} satisfies the Jacobi identity

\displaystyle [x,[y,z]] + [y,[z,x]] + [z,[x,y]] = 0.

Proof: Although I won’t prove it, the third-order terms (and all the rest) in our definition of {[x,y]} can be written out explicitly as well: for example, for example, we actually have

\displaystyle \mu(x,y) = x + y + \frac{1}{2} [x,y] + \frac{1}{12} \left( [x, [x,y]] + [y,[y,x]] \right) + \text{fourth order terms} + \dots.

The general formula is called the Baker-Campbell-Hausdorff formula.

Then we can force ourselves to expand this using the first three terms of the BCS formula and then equate the degree three terms. The left-hand side expands initially as {\mu\left( x, y + z + \frac{1}{2} [y,z] + \frac{1}{12} \left( [y,[y,z]] + [z,[z,y] \right) \right)}, and the next step would be something ugly.

This computation is horrifying and painful, so I’ll pretend I did it and tell you the end result is as claimed. \Box
There is a more natural way to see why this identity is the “right one”; see Qiaochu. However, with this proof I want to make the point that this Jacobi identity is not our decision: instead, the Jacobi identity is forced upon us by associativity in {G}.

Example 19 (Examples of commutators attached to Lie groups)

  • If {G} is an abelian group, we have {-[y,x] = [x,y]} by symmetry and {[x,y] = [y,x]} from {\mu(x,y) = \mu(y,x)}. Thus {[x,y] = 0} in {\mathfrak g} for any abelian Lie group {G}.
  • In particular, the brackets for {G \in \{\mathbb R, \mathbb C, S^1\}} are trivial.
  • Let {G = \text{GL }(n, K)}. Then one can show that

    \displaystyle [T,S] = TS - ST \qquad \forall S, T \in \mathfrak{gl}(n, K).

  • Ditto for {\text{SL }(n, K)}.

In any case, with the Jacobi identity we can define an general Lie algebra as an intrinsic object with a Jacobi-satisfying bracket:

Definition 20

A Lie algebra over {k} is a {k}-vector space equipped with a skew-symmetric bilinear bracket {[,]} satisfying the Jacobi identity.

A morphism of Lie algebras and preserves the bracket.

Note that a Lie algebra may even be infinite-dimensional (even though we are assuming {G} is finite-dimensional, so that they will never come up as a tangent space).

Example 21 (Associative algebra {\rightarrow} Lie algebra)

Any associative algebra {A} over {k} can be made into a Lie algebra by taking the same underlying vector space, and using the bracket {[a,b] = ab - ba}.

6. The fundamental theorems

We finish this list of facts by stating the three “fundamental theorems” of Lie theory. They are based upon the functor

\displaystyle \mathscr{L} : G \mapsto T_e G

we have described earlier, which is a functor

  • from the category of Lie groups
  • into the category of finite-dimensional Lie algebras.

The first theorem requires the following definition:

Definition 22

A Lie subgroup {H} of a Lie group {G} is a subgroup {H} such that the inclusion map {H \hookrightarrow G} is also an injective immersion.

A Lie subalgebra {\mathfrak h} of a Lie algebra {\mathfrak g} is a vector subspace preserved under the bracket (meaning that {[\mathfrak h, \mathfrak h] \subseteq \mathfrak h]}).

Theorem 23 (Lie I)

Let {G} be a real or complex Lie group with Lie algebra {\mathfrak g}. Then given a Lie subgroup {H \subseteq G}, the map

\displaystyle H \mapsto \mathscr{L}(H) \subseteq \mathfrak g

is a bijection between Lie subgroups of {G} and Lie subalgebras of {\mathfrak g}.

Theorem 24 (The Lie functor is an equivalence of categories)

Restrict {\mathscr{L}} to a functor

  • from the category of simply connected Lie groups over {K}
  • to the category of finite-dimensional Lie algebras over {K}.


  1. (Lie II) {\mathscr{L}} is fully faithful, and
  2. (Lie III) {\mathscr{L}} is essentially surjective on objects.

If we drop the “simply connected” condition, we obtain a functor which is faithful and exact, but not full: non-isomorphic Lie groups can have isomorphic Lie algebras (one example is {\text{SO }(3)} and {\text{SU }(2)}).

Uniqueness of Solutions for DiffEq’s

Let {V} be a normed finite-dimensional real vector space and let {U \subseteq V} be an open set. A vector field on {U} is a function {\xi : U \rightarrow V}. (In the words of Gaitsgory: “you should imagine a vector field as a domain, and at every point there is a little vector growing out of it.”)

The idea of a differential equation is as follows. Imagine your vector field specifies a velocity at each point. So you initially place a particle somewhere in {U}, and then let it move freely, guided by the arrows in the vector field. (There are plenty of good pictures online.) Intuitively, for nice {\xi} it should be the case that the trajectory resulting is unique. This is the main take-away; the proof itself is just for completeness.

This is a so-called differential equation:

Definition 1

Let {\gamma : (-\varepsilon, \varepsilon) \rightarrow U} be a continuous path. We say {\gamma} is a solution to the differential equation defined by {\xi} if for each {t \in (-\varepsilon, \varepsilon)} we have

\displaystyle  \gamma'(t) = \xi(\gamma(t)).

Example 2 (Examples of DE’s)

Let {U = V = \mathbb R}.

  1. Consider the vector field {\xi(x) = 1}. Then the solutions {\gamma} are just {\gamma(t) = t+c}.
  2. Consider the vector field {\xi(x) = x}. Then {\gamma} is a solution exactly when {\gamma'(t) = \gamma(t)}. It’s well-known that {\gamma(t) = c\exp(t)}.

Of course, you may be used to seeing differential equations which are time-dependent: i.e. something like {\gamma'(t) = t}, for example. In fact, you can hack this to fit in the current model using the idea that time is itself just a dimension. Suppose we want to model {\gamma'(t) = F(\gamma(t), t)}. Then we instead consider

\displaystyle  \xi : V \times \mathbb R \rightarrow V \times \mathbb R \qquad\text{by}\qquad \xi(v, t) = (F(v,t), 1)

and solve the resulting differential equation over {V \times \mathbb R}. This does exactly what we want. Geometrically, this means making time into another dimension and imagining that our particle moves at a “constant speed through time”.

The task is then mainly about finding which conditions guarantee that our differential equation behaves nicely. The answer turns out to be:

Definition 3

The vector field {\xi : U \rightarrow V} satisfies the Lipschitz condition if

\displaystyle  \left\lVert \xi(x')-\xi(x'') \right\rVert \le \Lambda \left\lVert x'-x'' \right\rVert

holds identically for some fixed constant {\Lambda}.

Note that continuously differentiable implies Lipschitz.

Theorem 4 (Picard-Lindelöf)

Let {V} be a finite-dimensional real vector space, and let {\xi} be a vector field on a domain {U \subseteq V} which satisfies the Lipschitz condition.

Then for every {x_0 \in U} there exists {(-\varepsilon,\varepsilon)} and {\gamma : (-\varepsilon,\varepsilon) \rightarrow U} such that {\gamma'(t) = \xi(\gamma(t))} and {\gamma(0) = x_0}. Moreover, if {\gamma_1} and {\gamma_2} are two solutions and {\gamma_1(t) = \gamma_2(t)} for some {t}, then {\gamma_1 = \gamma_2}.

In fact, Peano’s existence theorem says that if we replace Lipschitz continuity with just continuity, then {\gamma} exists but need not be unique. For example:

Example 5 (Counterexample if {\xi} is not differentiable)

Let {U = V = \mathbb R} and consider {\xi(x) = x^{\frac23}}, with {x_0 = 0}. Then {\gamma(t) = 0} and {\gamma(t) = \left( t/3 \right)^3} are both solutions to the differential equation

\displaystyle  \gamma'(t) = \gamma(t)^{\frac 23}.

Now, for the proof of the main theorem. The main idea is the following result (sometimes called the contraction principle).

Lemma 6 (Banach Fixed-Point Theorem)

Let {(X,d)} be a complete metric space. Let {f : X \rightarrow X} be a map such that {d(f(x_1), f(x_2)) < \frac{1}{2} d(x_1, x_2)} for any {x_1, x_2 \in X}. Then {f} has a unique fixed point.

For the proof of the main theorem, we are given {x_0 \in V}. Let {X} be the metric space of continuous functions from {(-\varepsilon, \varepsilon)} to the complete metric space {\overline{B}(x_0, r)} which is the closed ball of radius {r} centered at {x_0}. (Here {r > 0} can be arbitrary, so long as it stays in {U}.) It turns out that {X} is itself a complete metric space when equipped with the sup norm

\displaystyle  d(f, g) = \sup_{t \in (-\varepsilon, \varepsilon)} \left\lVert f(t)-g(t) \right\rVert.

This is well-defined since {\overline{B}(x_0, r)} is compact.

We wish to use the Banach theorem on {X}, so we’ll rig a function {\Phi : X \rightarrow X} with the property that its fixed points are solutions to the differential equation. Define it by, for every {\gamma \in X},

\displaystyle  \Phi(\gamma) : t \mapsto x_0 + \int_0^t \xi(\gamma(s)) \; ds.

This function is contrived so that {(\Phi\gamma)(0) = x_0} and {\Phi\gamma} is both continuous and differentiable. By the Fundamental Theorem of Calculus, the derivative is exhibited by

\displaystyle  (\Phi\gamma)'(t) = \left( \int_0^t \xi(\gamma(s)) \; ds \right)' = \xi(\gamma(t)).

In particular, fixed points correspond exactly to solutions to our differential equation.

A priori this output has signature {\Phi\gamma : (-\varepsilon,\varepsilon) \rightarrow V}, so we need to check that {\Phi\gamma(t) \in \overline{B}(x_0, r)}. We can check that

\displaystyle  \begin{aligned} \left\lVert (\Phi\gamma)(t) - x_0 \right\rVert &=\left\lVert \int_0^t \xi(\gamma(s)) \; ds \right\rVert \\ &\le \int_0^t \left\lVert \xi(\gamma(s)) \; ds \right\rVert \\ &\le t \max_{s \in [0,t]} \left\lVert \xi\gamma(s) \right\rVert \\ &< \varepsilon \cdot A \end{aligned}

where {A = \max_{x \in \overline{B}(x_0,r)} \left\lVert \xi(x) \right\rVert}; we have {A < \infty} since {\overline{B}(x_0,r)} is compact. Hence by selecting {\varepsilon < r/A}, the above is bounded by {r}, so {\Phi\gamma} indeed maps into {\overline{B}(x_0, r)}. (Note that at this point we have not used the Lipschitz condition, only that {\xi} is continuous.)

It remains to show that {\Phi} is contracting. Write

\displaystyle  \begin{aligned} \left\lVert (\Phi\gamma_1)(t) - (\Phi\gamma_2)(t) \right\rVert &= \left\lVert \int_{s \in [0,t]} \left( \xi(\gamma_1(s))-\xi(\gamma_2(s)) \right) \right\rVert \\ &= \int_{s \in [0,t]} \left\lVert \xi(\gamma_1(s))-\xi(\gamma_2(s)) \right\rVert \\ &\le t\Lambda \sup_{s \in [0,t]} \left\lVert \gamma_1(s)-\gamma_2(s) \right\rVert \\ &< \varepsilon\Lambda \sup_{s \in [0,t]} \left\lVert \gamma_1(s)-\gamma_2(s) \right\rVert \\ &= \varepsilon\Lambda d(\gamma_1, \gamma_2) . \end{aligned}

Hence once again for {\varepsilon} sufficiently small we get {\varepsilon\Lambda \le \frac{1}{2}}. Since the above holds identically for {t}, this implies

\displaystyle  d(\Phi\gamma_1, \Phi\gamma_2) \le \frac{1}{2} d(\gamma_1, \gamma_2)

as needed.

This is a cleaned-up version of a portion of a lecture from Math 55b in Spring 2015, instructed by Dennis Gaitsgory.