I am trying to understand the proof of the following which comes from "Matrix Groups for Undergraduates" by Kristopher Tapp.

Let $G \subset GL_n(\mathbb K)$ be a matrix group with Lie algebra $\mathfrak g \subset gl_n(\mathbb K)$. Then for all $X \in \mathfrak g$, $e^X \in G$.

It starts with let $\{ X_1,\ldots,X_k \}$ be a basis of $\mathfrak g$. For each $i=1,\ldots,k$ choose a differentiable path $\alpha_i: (-\epsilon,\epsilon)\to G$ with $\alpha_i(0)=I$ and $\alpha'_i(0)=X_i$. Define $F_\mathfrak{g}:(\text{neighborhood of 0 in } \mathfrak{g}) \to G$ as follows: $F_\mathfrak{g}(c_1X_1+\cdots+c_kX_k)=\alpha_1(c_1)\cdot\alpha_2(c_2)\cdots\alpha_k(c_k)$. Notice that $F_\mathfrak{g}(0)=I$, and $d(F_\mathfrak{g})_0$ is the identity function: $d(F_\mathfrak{g})_0(X)=X$ for all $X\in\mathfrak{g}$ as is easily verified on basis elements.

EDIT to include use of inverse function theorem in response to Bill

Choose a subspace $\mathfrak p\subset M_n(\mathbb K)$ which is complementary to $\mathfrak g$, which means completing the set $\{X_1,\ldots,X_k\}$ to a basis of all of $M_n(\mathbb K)$ and defining $\mathfrak p$ as the span of the added basis elements. So $M_n(\mathbb K)=\mathfrak g\times \mathfrak p$.

Choose a function $F_{\mathfrak p}: \mathfrak p \to M_n(\mathbb K)$ with $F_{\mathfrak p}(0) = I $ and with $d(F_{\mathfrak p})_0(V)=V$ for all $V\in \mathfrak p$. For example, $F_{\mathfrak p}(V)=I+V$ works. Next define the function $F:(\text{neighborhood of 0 in }\mathfrak g \times \mathfrak p = M_n(\mathbb K)) \to M_n(\mathbb K)$ by the rule $F(X+Y)=F_{\mathfrak g}(X)\cdot F_{\mathfrak p}(Y)$ for all $X\in \mathfrak g$ and $Y\in \mathfrak p$. Notice that $F(0)=I$ and $dF_0$ is the identity function: $dF_0(X+Y)=X+Y$.

By the inverse function theorem, $F$ has an inverse function defined on the neighborhood of $I$ in $M_n(\mathbb K)$.

My question is how does one see that $d(F_\mathfrak{g})_0(X)=X$ given that $F_\mathfrak{g}$ is a function from matrices to matrices and normally the jacobian is defined for functions of the type similar to $f:\mathbb R^n \to \mathbb R^m$. And how would one go about computing efficiently that $d(F_\mathfrak{g})_0(X)=X$?

  • $\begingroup$ Matrices are just $\mathbb R^{n^2}$, written in a different way (a square matrix rather than a row or column). So you have a function from $\mathbb R^{n^2}$ (or an open subset of $\mathbb R^{n^2}$) to itself, and you can compute its Jacobian. $\endgroup$ – Matt E Jan 11 '12 at 3:15

You can view the map $F_{\mathfrak{g}}$ as a map from a neighborhood of $0$ in $\mathbb{R}^n$: $(x_1,x_2,\dots,x_k) \mapsto \alpha_1(x_1)\cdots \alpha_k(x_k)$. Then $D_{x_i}[F_{\mathfrak{g}}](x_1,\dots,x_k)=\alpha_1(x_1)\cdots \alpha_{i-1}(x_i) \alpha_i'(x_i) \alpha_{i+1}(x_{i+1})\cdots \alpha_k(x_k)$, so the $i^{th}$ partial at $0$ is $\alpha_1(0)\cdots \alpha_{i-1}(0) \alpha_i'(0) \alpha_{i+1}(0)\cdots \alpha_k(0) = I\cdots I\cdot X_i \cdot I\cdots I=X_i$.

Thus the Jacobian is $[X_1 \; X_2\; \cdots \; X_k]$. So to get the derivative at $X=c_1X_1+\cdots+c_kX_k$ multiply this by the Jacobian by $[c_1\;c_2\;\cdots\;c_k]^T$ and get $c_1X_1+\cdots+c_kX_k=X$ (as desired).

  • $\begingroup$ But how does one reconcile the notion of the jacobian for $f:\mathbb R^n \to R^n$ and $h:\mathbb R^n \to GL_n(\mathbb K)$ in the context of the inverse function theorem. Roughly, if $\det Df \neq 0$ then the inverse function theorem says there is a local inverse. The inverse function theorem (proper) applies to functions like $f:\mathbb R^n \to R^n$. How does one make it work for functions like $h:\mathbb R^n \to GL_n(\mathbb K)$? $\endgroup$ – user782220 Jan 10 '12 at 23:51
  • $\begingroup$ I'm a little unclear as to what you are asking. If you want to use a Jacobian matrix to see $\mathrm{det}(Df)\not=0$, you need to choose coordinate patches on the domain and codomain, then write your map in terms of coordinates, finally you have a map between open sets in $\mathbb{R}^n$ so regular methods apply. This is basic manifold theory. $\endgroup$ – Bill Cook Jan 11 '12 at 0:17
  • $\begingroup$ I edited my original post to include the usage of the inverse function theorem which I am unclear about. $\endgroup$ – user782220 Jan 11 '12 at 4:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.