# Complexification of compact connected Lie groups: do these curves have the same tangent vector?

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 $$\mathfrak{l}\otimes \mathbb{C}.$$

• 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 $z \in \Bbb{C}^d \mapsto \prod_{m=1}^d \exp(z_m A_m)$. Apr 17, 2019 at 13:37
• 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 $K_{\mathbb{C}}$. Apr 17, 2019 at 13:44
• 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. Apr 17, 2019 at 15:20
• 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 $\Bbb{C}$. Apr 17, 2019 at 15:33
• @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/… Oct 31, 2022 at 7:37

$$\require{AMScd}$$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 $$(s(r_{jk}))\cdot (s(r_{jk}))^* = I_n$$

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

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.