# For matrices, why is the Lie algebra equal to the tangent space?

For a matrix Lie group $G\subset GL_n(\mathbb C)$ we define the Lie algebra to be the set of matrices $X\in M_n(\mathbb C)$ such that for all $t\in \mathbb R$ we have $\quad \exp(tX)\in G$.

For clarity, I define the exponential map by $$\exp(X) = \sum_{k=0}^\infty \dfrac{X^k}{k!}$$

The tangent space $T_I(G)$ at the group identity $I$ is the set of all matrices $X\in M_n(\mathbb C)$ for which there exists a differentiable path $$\gamma : (-\epsilon,\epsilon) \to G$$
where $\gamma(0)=I$ and $\gamma'(0)=X$.

It is proposed that Lie algebra and the tangent space are the same.

It is clear from the differentiability of exponentiation that if $\quad \exp(tX)\in G \quad$ at every $t$ then the path $\gamma$ defined by $\gamma(t) = \exp(tX)$ shows that $X$ is in the tangent space.

The other direction is not as obvious to me. If $X$ is in the tangent space (ie. there exists a path $\gamma$ satisfying the conditions above) then $X$ is in the Lie algebra. Can somebody please explain why this is so?

• I'm assuming you're defining $\exp tX$ as the solution of the matrix differential equation $A'(t) = XA(t)$ with $A(0) = I$. Is that accurate? Commented Sep 2, 2015 at 4:30
• No. I'll edit my post accordingly. Thanks @Vectornaut
– Owen
Commented Sep 2, 2015 at 4:32
• For the record, I'm pretty sure the definition you added is equivalent to the one I gave. Thanks for clarifying! Commented Sep 2, 2015 at 4:43
• I know that there's a perfectly good proof of this in a textbook sitting on my desk (Brian C. Hall's text). If no one answers in the next 10 hours, feel free to get my attention and I'll see if I can find it. Commented Sep 2, 2015 at 4:45
• Brian Hall's text talks about one-parameter subgroups, which are essentially like $\gamma$ only require that $\gamma$ is also a group homomorphism (where the group operation on $\mathbb R$ is addition). How do you reconcile this difference? Is what I proposed even true? There may always be another map $\bar\gamma$ with $\bar\gamma'(0)=X$ which is a homomorphism.
– Owen
Commented Sep 2, 2015 at 11:55

The following comes from Brian C. Hall's text (2003 edition):$\newcommand{\fg}{\mathfrak{g}}$

Corollary 2.35: suppose $G \subset GL(n;\Bbb C)$ is a matrix Lie group with Lie algebra $\fg$. Then, a matrix $X$ is in $\fg$ if and only if there exists a smooth curve $\gamma$ in $M_n(\Bbb C)$ such that

1. $\gamma(t)$ lies in $G$ for all $t$
2. $\gamma(0) = I$
3. $\frac{d \gamma}{dt}|_{t=0} = X$

Thus, $\fg$ is the tangent space at the identity to $G$

The proof (of your direction) is as follows:

Let $\gamma$ be a curve satisfying these conditions. Then, we may state that $\log(\gamma(t)) \in \fg$ for all sufficiently small $t$ (see theorem 2.27: there exists a neighborhood $U$ of $0$ in $\fg$ and a neighborhood $V$ of $I$ in $G$ such that $\exp:U \to V$ is a homeomorphism). Now, define $$X_{\gamma} = \frac{d}{dt} \log(\gamma(t))|_{t=0} = \lim_{h \to 0} \frac{\log(\gamma(h))}{h}$$ since $X_{\gamma}$ is the limit of elements in $\fg$ and since $\fg$ is closed, $X_{\gamma} \in \fg$. However, we note that (again, for sufficiently small $t$) $$\log(\gamma(t)) = (\gamma(t) - I) - \frac 12(\gamma(t)-I)^2 + \frac 13(\gamma(t) - I)^3 - \cdots$$ Differentiate term by term by applying the product rule. All terms besides the first approach zero as $t \to 0$. Thus, we have $$\frac{d}{dt} \gamma(t)|_{t=0} = \frac{d}{dt} \log(\gamma(t))|_{t=0} = X_{\gamma} \in \fg$$ which was the desired result.

• Feel free to ask for any clarification. Commented Sep 3, 2015 at 23:01
• I don't quite follow the argument. You want to show that given $\gamma$ with $\gamma(0)=I$ and $\gamma'(0)=X_\gamma$, you have $e^{t X_\gamma}\in G$ for all $t\in\mathbb R$. How are you using your result about $t\mapsto \log(\gamma(t))$ to conclude this?
– glS
Commented Jun 27, 2020 at 12:41
• @glS Let $\psi$ denote the map $\psi:t \mapsto \log(\gamma(t))$. Because $\psi(t) \in \mathfrak g$ for all $t \in U$ and because $\mathfrak g$ is a (finite dimensional and therefore closed) vector subspace, we know that $\psi'(0) \in \mathfrak g$. With that stated, we show that $\psi'(0) = \gamma'(0)$, which means that $X_{\gamma} = \gamma'(0) \in \mathfrak g$ (and so, $e^{tX_{\gamma}} \in G$ for all $t \in \Bbb R$). Commented Jun 27, 2020 at 18:47
• thank you. So to be clear, the definition of $\mathfrak g$ you are using here is as the set of $X$ such that $e^{tX}\in G$ for all $t$, correct? Then $\log(\gamma(t))\in \mathfrak g$ means that $e^{t\log(\gamma(s))}$ for all $s\in U$ and $t\in \mathbb R$? And this follows from the exponential being locally homeomorphic?
– glS
Commented Jun 27, 2020 at 19:58
• @glS Yes. And the fact that $\mathfrak g$ forms a vector space is ultimately a consequence of the $e^{tX} \in G$ definition, but requires separate proof. Commented Jun 27, 2020 at 20:01