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

## This question has an open bounty worth +50 reputation from Sir Wilfred Lucas-Dockery ending in 3 days.

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• 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 $z \in \Bbb{C}^d \mapsto \prod_{m=1}^d \exp(z_m A_m)$. – reuns 20 hours ago
• 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 $K_{\mathbb{C}}$. – Sir Wilfred Lucas-Dockery 20 hours ago
• 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. – Sir Wilfred Lucas-Dockery 18 hours ago
• 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 $\Bbb{C}$. – reuns 18 hours ago