I'm trying to understand the complexification of Lie groups from page $207$ here and I'm having trouble with a computation.

Assume $A, B$ are hermitian metrices, and $k$ is a unitary matrix. I want to show that the two paths $\alpha(t):= \exp(A)k\exp(tB)$ and $\beta(t):=\exp\left(A + tAd(k)B\right)k$ have the same tangent vector at $t=0$ (i.e. when you differentiate the above two matrix paths at $0$, you get the same matrix).

Now one can see that $\alpha'(0)= \exp(A)(Ad(k)B)k$.

I'm having trouble with the other curve. I got $$\beta'(0) = \exp(A)\left(\frac{1-\exp(-ad_{A})}{ad_{A}}(Ad(k)B)\right)k$$ using the formula for the differential of the exponential map. Or more directly,

$$\beta(t) = \left[I + \left(A+tAd(k)B\right) + \frac{1}{2!}\left(A+tAd(k)B\right)^2 + \dots \right]k$$ so

$$\beta'(0)= \left[\sum\limits_{n=0}^{\infty}\frac{1}{(n+1)!}\left(\sum\limits_{m=0}^{n}A^m\left(Ad(k)B\right)A^{n-m}\right)\right]k.$$

At this point, it looks like $\beta'(0) = \alpha'(0)$ if and only if $[A,Ad(k)B]=0$. What am I missing?

Edit: What I'm trying to show is that if $K\subset U(n)$ is a closed connected subgroup with Lie algebra $\mathfrak{l}$, then the set $\left\lbrace \exp(iX)k : X \in \mathfrak{l}, k \in K \right\rbrace$ is a subgroup of $GL(n,\mathbb{C})$ with Lie algebra $\mathfrak{l}\otimes \mathbb{C}. $

  • $\begingroup$ The finite products of $\exp(t (A+iB)), A+iB \in Lie(K) \otimes \Bbb{C}$ form a connected real Lie group $K_\Bbb{C}$ with Lie algebra $Lie(K) \otimes \Bbb{C}$ (when considering smooth functions of the form $f(t) = \prod_{j=1}^J \exp( h_j(t) (A_j+iB_j)), t,h_j(t) \in \Bbb{R}$ then clearly $f'(0) \in Lie(K) \otimes \Bbb{C}$, so the only difficulty is to justify why it suffices to look at the case $J$ finite). Let a basis $Lie(K) = \sum_{m=1}^d \Bbb{R} A_m$, then $K_\Bbb{C}$ has a complex Lie group structure from the chart $z \in \Bbb{C}^d \mapsto \prod_{m=1}^d \exp(z_m A_m)$. $\endgroup$
    – reuns
    Apr 17, 2019 at 13:37
  • $\begingroup$ Thanks. So this is like constructing the connected Lie subgroup associated to the Lie algebra $\mathfrak{l}\otimes \mathbb{C}$? Is there a way to show the fundamental group of this constructed subgroup is the same as that of $K$? I think something like this is required to prove that complex representations of $K$ extend to that of $K_{\mathbb{C}}$. $\endgroup$ Apr 17, 2019 at 13:44
  • $\begingroup$ I'm sorry, I'm quite lost. My previous comment agreed that it was possible to construct some connected complex group G with $K \subset G \subset GL(n,\mathbb{C})$ and with $Lie(G)= Lie(K)\otimes \mathbb{C}$. However, to prove that $G$ enjoys some universal property, I pointed out that we need that the inclusion induces $\pi_1(K)\simeq\pi_1(G)$. I think this is exactly why Bump tries to construct $G$ with typical element $\exp(iX)k$ (the isomorphism on fundamental groups follows easily from this construction of $G$). Does this make sense? I might be misunderstanding what's written in the text. $\endgroup$ Apr 17, 2019 at 15:20
  • $\begingroup$ Let $G = \bigcup_l \gamma_l K$ with $K$ the connected component of identity. I think you can look at how each $\gamma_l$ commutes with each $\exp(t A)$ (ie. $\forall tA \in Lie(K),\gamma_l \exp(tA) = \exp(f_l(tA)) \gamma_l$ with $f_l(A) = \gamma_l A \gamma_l^{-1} \in GL(Lie(K))$) and extend $f_l$ naturally to $f_l \in GL(Lie(K) \otimes \Bbb{C})$ to obtain the group law on $G_\Bbb{C} = \bigcup_l \gamma_l K_\Bbb{C}$. This way the complexification of $\Bbb{R+iZ}$ will be $\Bbb{C} \times \Bbb{Z}$ not $\Bbb{C}$. $\endgroup$
    – reuns
    Apr 17, 2019 at 15:33
  • $\begingroup$ @reuns hi i am stuck on the same proof, but on a different aspect. I wonder if you could have a look. thanks in advance! math.stackexchange.com/questions/4547149/… $\endgroup$ Oct 31, 2022 at 7:37

1 Answer 1


$\require{AMScd}$For $G$ a compact, connected Lie group, here's an alternate construction of the complexification $G\subset G_{\mathbb{C}}$ such that this inclusion induces $\pi_1(G)=\pi_1(G_{\mathbb{C}})$ (from the book of Brocker-Dieck). I have written this down partially for my own understanding; comments are more than welcome. I believe the argument in Bump to be incorrect; I'm leaving the question open for someone to convince me otherwise.

For $K$ either $\mathbb{R}$ or $\mathbb{C}$, let

$$\mathcal{F}(G,K) := \left\lbrace L(\pi(.)v)\in C(G,K) \mid G\xrightarrow {\pi} GL(N,K) \text{ is a representation }, v\in K^N, L \in (K^N)^{\vee} \right\rbrace.$$

These are the $\textbf{matrix coefficients}$ of $G$ and, by using tensors and direct sums of representations, one can see that this is a $K$-algebra under pointwise multiplication and addition of functions.

$\textbf{Claim 1: (The polynomial structure of the matrix coefficients)}$ If $r$ is a faithful, real representation (existence of $r$ is equivalent to Peter-Weyl), $\mathcal{F}(G,\mathbb{R})$ is generated as a $\mathbb{R}$-algebra over the matrix coefficients $r_{jk}$. Moreover, $\mathcal{F}(G,\mathbb{C})$ is generated as a $\mathbb{C}$ algebra over $r_{jk}$. Thus $\mathcal{F}(G,K)\simeq K[X_{jk}]/I$ for some ideal $I$.

$\textbf{Proof:}$ One can check that, since the real and imaginary parts of matrix coefficients of a complex representation are matrix coefficients of a real representation, we have $\mathcal{F}(G,\mathbb{R})\otimes \mathbb{C} \simeq \mathcal{F}(G,\mathbb{C})$. Hence the second statement will follow from the first.

Using faithfulness, we apply Stone-Weierstrass to see that $\mathbb{R}[r_{jk}]$ is dense in $C(G, \mathbb{R})$ and hence in $\mathcal{F}(G,\mathbb{R})$ with respect to the supremum norm. Being bounded by the supremum norm, density also holds in the $L^2$ norm. By Schur's lemma, we can decompose $F(G,\mathbb{R})$ into an $L^2$ orthogonal direct sum $\bigoplus \mathcal{F}(\rho)$ of matrix coefficients over all finite domensional irreducible real representations $\rho$ (If $f$ is a coefficient orthogonal to all $\mathcal{F}(\rho)$, noticing that $f$ is contained in a finite dimensional $G$ module gives a contradiction). Then note $\mathbb{R}[r_{jk}]=\bigoplus \left(\mathbb{R}[r_{jk}]\cap \mathcal{F}(\rho) \right)$. Using density and finiteness of each summand, we see that $\mathbb{R}[r_{jk}]$ is in fact all of $\mathcal{F}(G,\mathbb{R})$. Hence $\mathbb{C}[r_{jk}]=\mathcal{F}(G,\mathbb{C})$. $\blacksquare$

Note, as a non-example to the above, it is necessary to take a faithful $\textit{real}$ representation $r$. The result is clearly not true if you take a faithful complex representation. Given the standard embedding $S^1\to GL(1,\mathbb{C})$, one can see $\mathcal{F}(S^1,\mathbb{C})=\mathbb{C}[Z,Z^{-1}] \neq \mathbb{C}[Z].$

$\textbf{Definition 2: (Double dual)}$ Let $G_K$ be the set of all $K$-algebra morphisms $\mathcal{F}(G,K)\to K$. Give it the topology induced by the functions $s \mapsto s(f)$ where $f\in\mathcal{F}(G,K)$. For $s,t \in G_K$ define the product $s\cdot t$ by the commutative diagram

$$ \mathcal{F}(G,K) \rightarrow \mathcal{F}(G\times G,K) \xrightarrow{\simeq} \mathcal{F}(G,K)\otimes \mathcal{F}(G,K) \xrightarrow {s\otimes t} K\otimes K \simeq K .$$

Note, the first arrow is induced by multiplication in $G$ i.e. $f(x)\mapsto f(a.b)$. The second arrow is the inverse of the isomorphism $f(.)\otimes h(.)\mapsto f\circ\pi_1(.,.)\cdot h\circ\pi_2(.,.).$ In fact, one can check that, for $\rho$ a representation, the composition $\mathcal{F}(G,K)\to \mathcal{F}(G,K)\otimes \mathcal{F}(G,K)$ is given by $\rho_{jk} \mapsto \sum \rho_{jl} \otimes \rho_{lk}$. One can also check that $s\cdot t$ is indeed a $K$-algebra morphism. Let $\epsilon$ denote the element in $G_K$ given by evaluation at the identity. Define $s^{-1}$ by the composition

$$\mathcal{F}(G,K) \to \mathcal{F}(G,K) \xrightarrow{s} K$$

where the first map is induced by inversion on $G$.

$\textbf{Claim 3: (Topological group structure of $G_{\mathbb{R}}$)}$ Definition $2$ makes sense and $(G_k,\cdot, \epsilon)$ gives $G_K$ the structure of a topological group. $\blacksquare$

$\textbf{Definition 4:}$ For $r$ any $K$-representation to $GL(n,K)$, define $r_K$ by the rule $s\mapsto \left(s(r_{jk})\right)$ and define $i$ by sending $g\in G$ to evaluation at $g$. This gives a commutative diagram

$$ \begin{CD} G @>{i}>> G_{K} \\ @V{r}VV @V{r_K}VV \\ GL(n,K) @>{id}>> GL(n,K) \end{CD} \qquad \qquad \qquad (1)$$

$\textbf{Claim 5:}$ $i$ and $r_K$ are continuous group morphisms. $i$ is injective. When $r$ is a faithful real representation so is $r_{\mathbb{R}}$.

$\textbf{Proof:}$ Continuity is obvious. $i$ is injective since the matrix coefficients separate points (Peter-Weyl). If $r$ is faithful, real, then any $s\in G_{\mathbb{R}}$ is determined by its value of the $r_{jk}$ and injectivity follows. $\blacksquare$

$\textbf{Claim 6: (Lie group structure of $G_{\mathbb{R}}$)}$ The map $i:G\to G_{\mathbb{R}}$ is an isomorphism of compact Lie groups.

$\textbf{Proof:}$ For this we assume the representation $r$ in diagram $(1)$ is faithful into $O(n)$.

$r_K$ is injective since the $r_{jk}$ generate $\mathcal{F}(G,\mathbb{R})$. Observe that the topological group $G_{\mathbb{R}}$ is compact as follows. Map $G_{\mathbb{R}}\to \prod\limits_{f\in \mathcal{F}(G,\mathbb{R})} \mathbb{R}$ by the rule $s\mapsto \left(s(f)\right)$. Any one of these $f$ can be written as a polynomial $p(r_{jk})$. Using that $r$ is an orthogonal representation and the boundedness of the coefficients of $p$, we can assume the co-domain of the above map is $\prod\limits_{f}J_f$ where each $J_f$ is a compact interval. With the product topology, this map becomes a continuous homeomorphism (onto image). Moreover, the image is defined in terms of the algebra relations $s(f_1\cdot f_2)=s(f_1)s(f_2)$ etc. making it closed and hence compact by Tychonov.

Hence, through $r_{\mathbb{R}}$, $G_{\mathbb{R}}$ is embedded as a closed Lie subgroup of $O(n)$. This gives it a unique Lie group structure.

It remains to show the map $i$ from diagram $(1)$ is onto (it is injective since the matrix coefficients separate points). For this we consider the matrix coefficients $\mathcal{F}(G_{\mathbb{R}},\mathbb{R})$ and the $\mathbb{R}$-algebra morphism $\lambda:\mathcal{F}(G,\mathbb{R}) \to \mathcal{F}(G_{\mathbb{R}},\mathbb{R})$ defined by the rule $f\mapsto \left(s\mapsto s(f)\right)$. Dualizing $i:G\to G_{\mathbb{R}}$ we also get a $\mathbb{R}$-algebra morphism $i^*:\mathcal{F}(G_{\mathbb{R}},\mathbb{R}) \to \mathcal{F}(G,\mathbb{R})$. Clearly $i^*\circ \lambda = Id$. $\lambda$ is surjective since claim $5$ with claim $1$ imply that the $(r_{\mathbb{R}})_{jk}\left(=\lambda(r_{jk})\right)$ generate $\mathcal{F}(G_{\mathbb{R}},\mathbb{R})$. This shows that $i^*$ is a bijection, in particular injective. Finally injectivity of $i^*$ proves the surjectivity of $i$; the matrix coefficients are dense in continuous functions

\begin{CD} \mathcal{F}(G_{\mathbb{R}},\mathbb{R}) @>{i^*}>{\simeq}> \mathcal{F}(G,\mathbb{R}) \\ @VVV & @VVV \\ C(G_{\mathbb{R}},\mathbb{R}) @>{i^*}>> C(G,\mathbb{R}) \end{CD} .$\blacksquare$

$\textbf{The complexification of $G$.}$

The discussion of the polynomial structure of the matrix coefficient ring and the claim about $G_{\mathbb{R}}$ are crucial to what follows (cf. claim $9$). Assume $r$ is a real, faithful, orthogonal representation. Extend it to a complex representation by extension of scalars and define $r_{\mathbb{C}}$ as before to obtain the following commutative diagram.

$$ \begin{CD} G @>{i}>> G_{\mathbb{C}} \\ @V{r}VV @V{r_{\mathbb{C}}}VV \\ GL(n,\mathbb{C}) @>{id}>> GL(n,\mathbb{C}) \end{CD}$$

$\textbf{Claim 7:}$ $r_{\mathbb{C}}$ is a homeomorphism onto its image.

$\textbf{Proof:}$ Injectivity follows from claim $1$ which says $\mathcal{F}(G,\mathbb{C})$ is generated by the $r_{jk}$. $r_{\mathbb{C}}$ is continuous by definition. Openness of the map follows again from claim $1$.

$\textbf{Claim 8:}$ The image of $r_{\mathbb{C}}$ in $GL(n,\mathbb{C})$ is the zero set of a collection of polynomials in $n^2$ variables.

$\textbf{Proof:}$ Dualizing the isomorphism $\mathbb{C}[X_{jk}]/I \to \mathcal{F}(G,\mathbb{C})$ from claim $1$, we get a map $\sigma: G_{\mathbb{C}}=Hom_{\mathbb{C}-algebra}\left(\mathcal{F}(G,\mathbb{C}),\mathbb{C}\right) \to V(I)\subset \mathbb{C}^{n\cdot n}$ which is given by the rule $s\mapsto \left(s(r_{jk})\right)$ In particular, $V(I)\subset GL(n,\mathbb{C})$ . The following commutative diagram that arises then proves the claim

$$ \begin{CD} G_{\mathbb{C}} @>{\sigma}>{\simeq}> V(I) \\ @V{r_{\mathbb{C}}}VV @VV{\text{inclusion}}V \\ GL(n,\mathbb{C}) @>{id}>> GL(n,\mathbb{C}) \end{CD}$$. $\blacksquare$

Claim $7$ and $8$ together show that $G_{\mathbb{C}}$ has the structure of a closed complex analytic subgroup of $GL(n,\mathbb{C})$. It thus has a unique complex analytic group structure.

Let $U(n)$ denote the unitary group, $P(n)$ denote the set of positive definite hermitian matrices. It is well known that multiplication $U(n)\times P(n)\to GL(n,\mathbb{C})$ is a diffeomorphism.

$\textbf{Claim 9: (Computing the Lie algebra and fundamental group of $G_{\mathbb{C}}$)}$ Denote the image of $r_{\mathbb{C}}$ by $\widetilde{G}$. Then

$(1) \text{ }\widetilde{G}\cap U(n) = r(G).$

$(2) \text{ Under the multiplication map, }\widetilde{G}\simeq \left(\widetilde{G}\cap U(n)\right) \times \left(\widetilde{G}\cap P(n)\right).$

$(3)$ If we denote $\mathfrak{g}=Lie(r(G)) \subset \mathfrak{u}(n)$, then there is an isomorphism $\mathfrak{g}\to \widetilde{G}\cap P(n)$ given by the rule $X \mapsto \exp(iX)$. Moreover the Lie algebra of $\widetilde{G}= \mathfrak{g}\oplus i\mathfrak{g}.$

$\textbf{Proof:} (1)$ Let $r_{\mathbb{C}}(s)\in U(n)$. So we have $(s(r_{jk}))\cdot (s(r_{jk}))^* = I_n$

\begin{equation} I_n = \left(s(r_{jk})\right)\cdot\left(\overline{s(r_{kj})}\right) = \left(s(r_{jk})\right)\left(\overline{s(r_{jk})}\right)^{-1} \end{equation}

This shows that each $s(r_{jk})$ is real. Hence the restriction of $s$ to $\mathcal{F}(G,\mathbb{R})$ is real i.e. in $G_{\mathbb{R}}$. Claim $6$ then implies that $r_{\mathbb{C}}(s)=r(g)$ for some $g$.

$(2)$ Let $A \in \widetilde{G}$. Say $A = MH$ for some $M$ unitary and $P$ positive-definite, hermitian. We can find $N$ unitary such that $NHN^*=D$ for some diagonal matrix with positive diagonal elements. Further, put $D=\exp(X)$ for $X$ diagonal with real entries. So $$A^*A=H^*M^*MH=H^*H = N^*DNN^*DN=N^*D^2N= N^*(\exp 2X)N.$$ By hypothesis, we also know $A_{jk}=s(r_{jk})$ so that $A^*_{jk}=\overline{s(r_{kj})}$. Since $r_{jk}$ are real-valued functions, we can further write $A^*_{jk}=\overline{s(\overline{r_{jk}})}$. The upshot of this is that the function $\overline{s(\overline{\cdot})}$ is in $G_{\mathbb{C}}$ whence $A^*$ and $A^*A\in \widetilde{G}$.

Thus we have $N^*\exp(2kX)N\in \widetilde{G}$ for each $k\in \mathbb{Z}$. Let $\phi:\mathbb{C}^{n^2}\to \mathbb{C}^{n^2}$ be the polynomial map given by the rule $V\mapsto N^*VN$. It is an isomorphism and the pre-image of $V(I)$ is some $V(J)$. For any polynomial $q\in I$, we know $q(N^*\exp(2kX)N)=0$ for each $k\in \mathbb{Z}$. This in fact shows $q(N^*\exp(tX)N)=0$ for all real $t$. Whence $\exp(tX)\in V(J).$ for all real $t$.

In particular we get $\phi(\exp(X)) \in V(I)$ and so $H\in \widetilde{G}$. So also $M\in \widetilde{G}$ and $(2)$ is proved.

$(3)$ Let $X\in \mathfrak{g} \subset \mathfrak{u}(n).$ Then $\exp(tX)\in \widetilde{G}$ for all real $t$. This means that any polynnomial $q$ in $I$ has $q(\exp(tX)=0$ identically. Viewing the argument $t$ as a complex variable, we see that the analytic function $t\mapsto q(\exp(tX))$ must in fact vanish everywhere. Hence $iX$ is in the Lie algebra of $\widetilde{G}$ and is hermitian. The argument goes both ways and so we see $Lie\left(\widetilde{G}\cap P(n)\right)=i\mathfrak{g}.$ Exponentiation is a diffeomorphism from the vector space of hermitian matrices to positive definite hermitian matrices and so we are done.


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