1.5. Lecture 5

1.5.1. Lie group and Lie algebra homomorphisms

Definition 1.5.1.

A Lie group homomorphism ϕ:G→G′\phi:G\to G^{\prime} between two linear Lie groups GG and G′G^{\prime} is a continuous group homomorphism.

An isomorphism is a bijective Lie group homomorphism whose inverse is also continuous.

Remark 1.5.2.

In fact, a continuous homomorphism between linear Lie groups is automatically a smooth map of smooth manifolds, and if it is bijective then the inverse is automatically continuous.

Definition 1.5.3.

A homomorphism φ:𝔤→𝔥\varphi:\mathfrak{g}\to\mathfrak{h} of Lie algebras is a kk-linear map such that

φ⁢([X,Y])=[φ⁢(X),φ⁢(Y)]\varphi([X,Y])=[\varphi(X),\varphi(Y)]

for all X,Y∈𝔤X,Y\in\mathfrak{g}.

An isomorphism is an invertible homomorphism.

Definition 1.5.4.

Let ϕ:G→G′\phi:G\to G^{\prime} be a Lie group homomorphism. Define the derivative (or differential or derived homomorphism)

D⁢ϕ:𝔤⟶𝔤′D\phi:\mathfrak{g}\longrightarrow\mathfrak{g}^{\prime}

by

D⁢ϕ⁢(X)=dd⁢t⁢ϕ⁢(exp⁡(t⁢X))|t=0D\phi(X)=\left.\frac{d}{dt}\phi(\exp(tX))\right|_{t=0}

for X∈𝔤X\in\mathfrak{g}. 11 1 We can justify taking the derivative by appealing to Remark 1.2.8.

Remark 1.5.5.

In fact, D⁢ϕD\phi is the derivative of ϕ\phi at the identity in the sense of smooth manifolds; recall that 𝔤\mathfrak{g} and 𝔤′\mathfrak{g}^{\prime} are the tangent spaces to GG and G′G^{\prime} at the identity.

Theorem 1.5.6.

Let ϕ:G→G′\phi:G\rightarrow G^{\prime} be a Lie group homomorphism with derivative D⁢ϕD\phi. Then

  1. (i)
    ​

    The following diagram commutes:

    𝔤→D⁢ϕ𝔤′exp↓↓expG→ϕG′.\begin{CD}\mathfrak{g}@>{D\phi}>{}>\mathfrak{g}^{\prime}\\ @V{\exp}V{}V@V{}V{\exp}V\\ G@>{}>{\phi}>G^{\prime}.\end{CD}

    That is, for X∈𝔤X\in\mathfrak{g} we have

    ϕ⁢(exp⁡(X))=exp⁡(D⁢ϕ⁢(X)).\phi(\exp(X))=\exp(D\phi(X)).
  2. (ii)
    ​

    For all g∈G,X∈𝔤g\in G,X\in\mathfrak{g},

    D⁢ϕ⁢(g⁢X⁢g−1)=ϕ⁢(g)⁢D⁢ϕ⁢(X)⁢ϕ⁢(g)−1.D\phi(gXg^{-1})=\phi(g)D\phi(X)\phi(g)^{-1}.
  3. (iii)
    ​

    The map D⁢ϕ:𝔤→𝔤′D\phi:\mathfrak{g}\to\mathfrak{g}^{\prime} is a Lie algebra homomorphism.

Proof.

.

  1. (i)
    ​

    Consider the one parameter subgroup f:ℝ→G′f:\mathbb{R}\to G^{\prime} defined by f⁢(t)=ϕ⁢(exp⁡(t⁢X))f(t)=\phi(\exp(tX)). By construction, f′⁢(0)=D⁢ϕ⁢(X)f^{\prime}(0)=D\phi(X). By Proposition 1.2.9, one parameter subgroups are determined by their derivative at 0, so that we must have

    ϕ⁢(exp⁡(t⁢X))=f⁢(t)=exp⁡(t⁢D⁢ϕ⁢(X)).\phi(\exp(tX))=f(t)=\exp(tD\phi(X)).
  2. (ii)
    ​

    We have

    D⁢ϕ⁢(g⁢X⁢g−1)\displaystyle D\phi(gXg^{-1}) =dd⁢t⁢ϕ⁢(exp⁡(t⁢g⁢X⁢g−1))|t=0\displaystyle=\left.\frac{d}{dt}\phi(\exp(tgXg^{-1}))\right|_{t=0}
    =dd⁢t⁢ϕ⁢(g⁢exp⁡(t⁢X)⁢g−1)|t=0\displaystyle=\left.\frac{d}{dt}\phi(g\exp(tX)g^{-1})\right|_{t=0}
    =dd⁢t⁢ϕ⁢(g)⁢ϕ⁢(exp⁡(t⁢X))⁢ϕ⁢(g−1)|t=0\displaystyle=\left.\frac{d}{dt}\phi(g)\phi(\exp(tX))\phi(g^{-1})\right|_{t=0}
    =ϕ⁢(g)⁢dd⁢t⁢ϕ⁢(exp⁡(t⁢X))|t=0⁢ϕ⁢(g−1)\displaystyle=\phi(g)\left.\frac{d}{dt}\phi(\exp(tX))\right|_{t=0}\phi(g^{-1})
    =ϕ⁢(g)⁢D⁢ϕ⁢(X)⁢ϕ⁢(g)−1,\displaystyle=\phi(g)D\phi(X)\phi(g)^{-1},

    as claimed.

  3. (iii)
    ​

    To show that D⁢ϕD\phi is a Lie algebra homomorphism, we need to show that D⁢ϕD\phi is ℝ\mathbb{R}-linear and D⁢ϕ⁢([X,Y])=[D⁢ϕ⁢(X),D⁢ϕ⁢(Y)]D\phi([X,Y])=[D\phi(X),D\phi(Y)].

    So let X,Y∈𝔤X,Y\in\mathfrak{g} and s∈ℝs\in\mathbb{R}. By definition,

    D⁢ϕ⁢(s⁢X)\displaystyle D\phi(sX) =dd⁢t⁢ϕ⁢(exp⁡(t⁢s⁢X))|t=0.\displaystyle=\left.\frac{d}{dt}\phi(\exp(tsX))\right|_{t=0}.
    If we now set μ=s⁢t\mu=st, we can rewrite this as:
    dd⁢t⁢ϕ⁢(exp⁡(t⁢s⁢X))|t=0\displaystyle\left.\frac{d}{dt}\phi(\exp(tsX))\right|_{t=0} =s⁢dd⁢μ⁢ϕ⁢(exp⁡(μ⁢X))|μ=0\displaystyle=s\left.\frac{d}{d\mu}\phi(\exp(\mu X))\right|_{\mu=0}
    =s⁢D⁢ϕ⁢(X).\displaystyle=sD\phi(X).

    So D⁢ϕD\phi commutes with scalar multiplication. For additivity, we have

    D⁢ϕ⁢(X+Y)=dd⁢t⁢ϕ⁢(exp⁡(t⁢(X+Y)))|t=0.D\phi(X+Y)=\left.\frac{d}{dt}\phi(\exp(t(X+Y)))\right|_{t=0}.

    On the other hand, by Corollary 1.2.5 and using part (i)

    ϕ⁢(exp⁡(t⁢(X+Y)))\displaystyle\phi(\exp(t(X+Y))) =limk→∞ϕ⁢((exp⁡(tk⁢X)⁢exp⁡(tk⁢Y))k)\displaystyle=\lim\limits_{k\to\infty}\phi\left(\left(\exp\left(\frac{t}{k}X% \right)\exp\left(\frac{t}{k}Y\right)\right)^{k}\right)
    =limk→∞ϕ⁢(exp⁡(tk⁢X))⁢ϕ⁢(exp⁡(tk⁢Y))k\displaystyle=\lim\limits_{k\to\infty}\phi\left(\exp\left(\frac{t}{k}X\right)% \right)\phi\left(\exp\left(\frac{t}{k}Y\right)\right)^{k}
    =limk→∞(exp⁡(tk⁢D⁢ϕ⁢(X))⁢exp⁡(tk⁢D⁢ϕ⁢(Y)))k\displaystyle=\lim\limits_{k\to\infty}\left(\exp\left(\frac{t}{k}D\phi(X)% \right)\exp\left(\frac{t}{k}D\phi(Y)\right)\right)^{k}
    =exp⁡(t⁢(D⁢ϕ⁢(X)+D⁢ϕ⁢(Y))).\displaystyle=\exp(t(D\phi(X)+D\phi(Y))).

    Taking the derivative at t=0t=0, we conclude that

    D⁢ϕ⁢(X+Y)=D⁢ϕ⁢(X)+D⁢ϕ⁢(Y),D\phi(X+Y)=D\phi(X)+D\phi(Y),

    showing additivity.

    Finally we show that D⁢ϕD\phi respects the Lie bracket. Let X,Y∈𝔤X,Y\in\mathfrak{g}. By parts (i) and (ii) we have

    D⁢ϕ⁢(exp⁡(−t⁢Y)⁢X⁢exp⁡(t⁢Y))\displaystyle D\phi\left(\exp(-tY)X\exp(tY)\right) =(ϕ⁢(exp⁡(−t⁢Y))⁢D⁢ϕ⁢(X)⁢ϕ⁢(exp⁡(t⁢Y)))\displaystyle=\left(\phi(\exp(-tY))D\phi(X)\phi(\exp(tY))\right)
    =exp⁡(−t⁢D⁢ϕ⁢(Y))⁢D⁢ϕ⁢(X)⁢exp⁡(t⁢D⁢ϕ⁢(Y)).\displaystyle=\exp(-tD\phi(Y))D\phi(X)\exp(tD\phi(Y)).

    We then take the derivative for both sides at t=0t=0. The derivative of the RHS is [D⁢ϕ⁢(X),D⁢ϕ⁢(Y)][D\phi(X),D\phi(Y)], and the derivative of the LHS is D⁢ϕ⁢([X,Y])D\phi([X,Y]) (as D⁢ϕD\phi is linear).∎

Definition 1.5.7.

Suppose that GG and G′G^{\prime} are complex Lie groups and ϕ:G→G′\phi:G\rightarrow G^{\prime} is a homomorphism. Then ϕ\phi is holomorphic if D⁢ϕ:𝔤→𝔤′D\phi:\mathfrak{g}\to\mathfrak{g}^{\prime} is ℂ\mathbb{C}-linear.

(This implies that ϕ\phi is a holomorphic map of complex manifolds.)

Example 1.5.8.

The map det:GL2,ℂ→GL1,ℂ\det:\operatorname{GL}_{2,\mathbb{C}}\rightarrow\operatorname{GL}_{1,\mathbb{C}} is holomorphic.

Back to the real case. We have shown that the differential ϕ↦D⁢ϕ\phi\mapsto D\phi gives a map

D:Hom⁡(G,G′)⟶Hom⁡(𝔤,𝔤′).D:\operatorname{Hom}(G,G^{\prime})\longrightarrow\operatorname{Hom}(\mathfrak{% g},\mathfrak{g}^{\prime}).

This raises two natural questions:

  1. (1)
    ​

    Is the map injective? Does the derivative D⁢ϕD\phi uniquely determine the Lie group homomorphism ϕ\phi?

  2. (2)
    ​

    Is the map surjective? Or in other words, does every Lie algebra homomorphism φ\varphi ’exponentiate’ (or ’lift’) to a Lie group homomorphism ϕ\phi such that D⁢ϕ=φD\phi=\varphi? We say ’exponentiate’ since if yes, then ϕ\phi would need to satisfy ϕ⁢(exp⁡(X))=exp⁡(φ⁢(X))\phi(\exp(X))=\exp(\varphi(X)). So this gives a formula for ϕ\phi, at least on the image of exp\exp. The question is whether this is well-defined (the exponential map is neither injective nor surjective in general) and whether this defines a homomorphism.

The answer to these questions is actually of topological nature, which we discuss in the next subsection.

1.5.2. Exercises

.

Problem 11.

  1. (a)
    ​

    Show that the Lie algebras 𝔰⁢𝔬3\mathfrak{so}_{3} and 𝔰⁢𝔲2\mathfrak{su}_{2} are isomorphic. (Later on, we will see a conceptual reason for this).

    Hint: it is enough to find a basis for 𝔰⁢𝔬3\mathfrak{so}_{3} and a basis for 𝔰⁢𝔲2\mathfrak{su}_{2} which satisfy the ‘same’ Lie bracket relations. Try using the basis of 𝔰⁢𝔬3\mathfrak{so}_{3} consisting of infinitesimal generators for rotations around the axes, and a basis for 𝔰⁢𝔲2\mathfrak{su}_{2} related to the quaternions.

  2. (b)
    ​

    Show that the Lie algebras 𝔰⁢𝔬2,1\mathfrak{so}_{2,1} and 𝔰⁢𝔩2,ℝ\mathfrak{sl}_{2,\mathbb{R}} are isomorphic.

  3. (c)
    ​

    Show that the Lie algebras 𝔰⁢𝔬3,1\mathfrak{so}_{3,1} and 𝔰⁢𝔩2,ℂ\mathfrak{sl}_{2,\mathbb{C}} are isomorphic (as real Lie algebras).

Problem 12. Show that:

  1. (a)
    ​

    If X∈𝔰⁢𝔭2⁢nX\in\mathfrak{sp}_{2n}, then tr⁡(X)=0\operatorname{tr}(X)=0.

  2. (b)
    ​

    If g∈Sp⁡(2⁢n)g\in\operatorname{Sp}(2n), then det(g)=1\det(g)=1.