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Let $G$ be a matrix Lie group and $\mathfrak{g}$ the associated Lie algebra. For fixed $X \in G$, define the linear map $\text{Ad}_X : \mathfrak{g} \to \mathfrak{g}$ by $\text{Ad}_X(y) = XyX^{-1}$. Show that the map $\text{Ad}: G \to GL(\mathfrak{g})$ that sends $X \mapsto \text{Ad}_X$ is continuous.

The textbook (Hall's Lie Groups, page 64) says that this should be very simple to prove but I am struggling to see this. I am aware that $\mathfrak{g}$ is an n-dimensional real vector space, so we can imbue $GL(\mathfrak{g})$ with the metric topology on $\mathbb{R}^{n \times n}$ by considering the matrix representation $\left[ \text{Ad}_X \right]_v$ with regards to some basis $\{v\} \subset \mathfrak{g}$. It seems like there should be a simple argument from here that the entries of the representation $\left[ \text{Ad}_X \right]_v$ will consist of polynomials of the components of $X$, but I don't know how to make the argument without a specific example of a basis; we would need the components $a_{ki}$ such that $Ad_X(v_i) = \sum_i^n a_{ki}v_k$ for instance.

EulerLagrange
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  • Do you know what is the topology on $G$? – Arctic Char Jun 12 '22 at 13:15
  • @ArcticChar $G$ is a closed subset of $\text{GL}(n, \mathbb{R})$ so I assume the book is using the metric topology (or rather, the relevant subspace topology). – EulerLagrange Jun 12 '22 at 14:24
  • If $G$ is a matrix Lie group, can't you show $Ad_g : \mathfrak{gl}(n,\mathbb{C}) \rightarrow \mathfrak{gl}(n,\mathbb{C})$ for $g \in GL_n(\mathbb{C})$ is continuous then restrict? You have the obvious basis of $\mathfrak{gl}(n,\mathbb{C})$ where entries of $Ad_g$ are polynomial/rational functions(?) in the entries of $g$. – Dylan Jun 12 '22 at 15:22
  • Could you please add the textbook you are following? – Alp Uzman Jun 12 '22 at 16:36
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    @Dylan Wouldn't that only show that $\text{Ad}_X$ is continuous for some $X$? The goal is to show that $\text{Ad}$ is continuous – EulerLagrange Jun 13 '22 at 01:39
  • @AlpUzman The textbook is Hall's Lie Group book – EulerLagrange Jun 13 '22 at 01:40
  • @EulerLagrange It could be good to add it to the body of your question (with page information). – Alp Uzman Jun 13 '22 at 04:54
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    It seems you are referring to the top of p.64 of the second edition. There, the statement seems to be used to derive the existence and uniqueness of the lowercase adjoint map (by way of "Lie group homs are exponentials of Lie algebra homs"). While I don't see an argument simpler than the one I presentes below, the whole statement can be bypassed by realizing that all adjoint maps are partial derivatives of the conjugation map $G\times G\to G$ evaluated at identities, together with the Existence and Uniqueness Theorem in ODEs. – Alp Uzman Jun 13 '22 at 05:02
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    Nice one, thanks @AlpUzman – EulerLagrange Jun 14 '22 at 10:48

1 Answers1

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Hint: Let us use $|\bullet|$ for operator norms for elements of $G$ and $\mathfrak{g}$ and $\Vert\bullet\Vert$ for the operator norm for linear maps $\mathfrak{g}\to \mathfrak{g}$. Recall that matrix inversion is infinitely continuously differentiable (see Showing inversion function for invertible matrices is differentiable), and in particular is continuous. Then if $g_n\to g$ in $G$ and $X$ is in $\mathfrak{g}$, we have

$$|g_nXg_n^{-1}-gXg^{-1}| \leq|g_nXg_n^{-1}-g_nXg^{-1}|+|g_nXg^{-1}-gXg^{-1}| \leq |g_n| |X| |g_n^{-1}-g^{-1}| + |g_n-g||X||g^{-1}|, $$

so that

$$\Vert \operatorname{Ad}_{g_n}- \operatorname{Ad}_{g}\Vert\leq |g_n| |g_n^{-1}-g^{-1}| + |g_n-g||g^{-1}|.$$

(As a side note, for $G$ the general linear group, this line of argument together with the Mean Value Theorem (applied to matrix inversion) shows that $\operatorname{Ad}_\bullet$ is locally Lipschitz.)

Alp Uzman
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  • Thanks. I'll admit this one is a little over my head - I assume the idea from here is that we can obtain a bound on $\lvert g_n - g \rvert$ from $\lvert g_n^{-1} - g^{-1} \rvert$. Doesn't this also require $g$ to be bounded? TBH I am expecting that there should be a more elementary proof, given how easily it is hand waved by the textbook (Hall). – EulerLagrange Jun 13 '22 at 01:49
  • @EulerLagrange You are right on your assumption. $g$ is bounded w/r/t the operator norm since it's a linear operator on a finite dimensional vector space. Similarly since $g_n$ converges to $g$ the terms of the sequence is bounded uniformly in $n$. It seems to be the argument I presented above is fairly common for this type of stuff, but you might be right. – Alp Uzman Jun 13 '22 at 04:40