Differentials in Lie Theory

I collect here some results about the differentials of the $\exp$ and $\log$ maps of a Lie group. The reader must be familiar with how the differential (i.e., pushforward map) is defined on a general manifold for much of this to make any sense. I will also rely extensively on the interpretation of tangent vectors as equivalence classes of curves ; Wikipedia has an excellent summary of this, as does my blog (if I may say so myself).

Let $L_g$ refer to the left-multiplication operation: $L_g(h)=gh$ for $g,h\in G$. Given $X\in\mathfrak g$, $X^L_g\coloneq dL_g(X)$ is the corresponding left-invariant vector field. For a matrix Lie group (i.e., where $g$ is identified with some finite-dimensional matrix representation of it), $X^L_g=g X$, which makes sense because we have an external algebraic structure (namely, the multiplication of $n\times n$ matrices, which has the structure of a unital associative algebra ) that lets us multiply group elements with Lie algebra elements. This rather convenient structure does not exist for a general Lie group. I will assume that the maps $\exp: U \rightarrow V$ and $\log:V\rightarrow U$ are inverses of each other, where $U\subseteq \mathfrak g$ is a neighborhood of $0$ and $V\subseteq G$ is a neighborhood of $e$.

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🀲 Differential of $\exp$

Suppose $\exp(X)$ is a point on $G$. The differential of $\exp$ provides an answer to the following question: what happens when $X$ is perturbed in the direction $Y \in T_{X}\mathfrak g$? The result, referred to as the pushforward of $Y$ under $\exp$, lies in $T_{\exp(X)} G$. It is a linear map between vector spaces:

\[ d\exp_X: T_{X}\mathfrak g \rightarrow T_{\exp X} G. \]

Since $\mathfrak g$ is a vector space, we can identify $T_X\mathfrak g$ with $\mathfrak g$. Using the map $dL_{{\exp(X)}^{-1}}=dL_{{\exp(X)}}^{-1}$, we can identity $T_{\exp X} G$ with $T_e G$, which is also just $\mathfrak g$. A pedagogical nightmare ensues.

Let $\gamma(t)$ be a curve on $\mathfrak g$. We use the notation $[\gamma]$ to refer to the equivalence class of all the curves¹ that pass through $\gamma(0)$ with the velocity $\dot \gamma(0)$.² We say that $\gamma$ is a representative of the class $[\gamma]$. If $\gamma$ is chosen such that $\gamma(0)=X\in\mathfrak g$ and $\dot \gamma(0)=Y \in T_X\mathfrak g$, then $[\gamma]$ is completely characterized by $Y$ and vice versa; therefore, we can write $[\gamma]=Y$. The image of $\gamma$ under $\exp$, which is $\exp\circ \gamma(t)$, is a curve on $G$ as well as a representative of $d\exp_X(Y)$. That is,

$$d\exp_X(Y)=[\hspace{2pt}\exp\circ \hspace{1pt}\gamma\hspace{2pt}]$$

is some vector in $T_{\exp(X)} G$. But which vector?

The key to making these identifications concrete is to notice that we can choose any $\gamma$ satisfying $\gamma(0)=X$ and $\dot {\gamma}(0)=Y$. Any such $\gamma$ is a representative of $Y$. So, we might as well choose a $\gamma$ that is convenient to write down; let’s make the choice, $\gamma(t) \coloneq X+tY$. In this case, $\exp \circ\hspace{2pt} \gamma(t)=\exp(X+tY)$ is a curve on $G$ passing through $\exp(X)$, and it is a representative of $d\exp_X(Y)$. There’s two different ways of writing down the equivalence class of $d\exp_X(Y)$:

$$ d\exp_X(Y) =\big[\exp(X+tY)\big]=\big[\exp(X)\exp(t\hspace{2pt}\square)\big] $$

where the box is a Lie algebra element: $\square \in \mathfrak g$. There must be a value of $\square$ for which these are indeed the same equivalence class. The value of $\square$ can therefore be used to describe the quantity $d\exp_X(Y)$. Hence, we define the Jacobian of $\exp$ at $X$ as the linear map

\[ \begin{align*} \Psi_X:\mathfrak g &\rightarrow \mathfrak g\\ Y &\mapsto \square \end{align*} \]

In conclusion, the differential of $\exp$ takes two objects, the point $X$ and the direction $Y$, and defines the object $\square$. Here, $\square$ represents the direction and extent to which $\exp(X)$ is perturbed when we nudge $X$ in the direction of $Y$. On matrix Lie groups, $\Psi_X$ is defined by

\[ \Psi_X(Y) =\square =\exp(X)^{-1}\frac{d}{dt}\exp(X+tY)\Big|_{t=0}, \]

and we write $d\exp_X(Y)=\exp(X)\Psi_X(Y)$.

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🀳 Differential of $\log$

By the same arguments as above,

$$ d\log_g: T_g G \rightarrow T_{\log(g)}\mathfrak g. $$

Since $T_g G\cong \mathfrak g$ and $T_{\log(g)}\mathfrak g\cong \mathfrak g$, we should once again expect a formula that goes from $\mathfrak g$ to $\mathfrak g$. It proves to be convenient to work entirely with $\mathfrak g$-elements and write $\exp(X)$ in place of $g$.

Let $\gamma \coloneq \exp(X) \exp(tY)$ by a curve passing through $\exp(X)$ with the velocity $Y$. We have,

\[ \begin{align*} d\log_{\exp(X)}(Y)&=\big[\hspace{1pt}\log\mathrel\circ\gamma\hspace{1pt}(t)\hspace{1pt}\big]\\&=\big[\hspace{1pt}\log\big(\exp(X)\exp(tY)\big)\hspace{1pt}\big]. \end{align*} \]

As before, we have an alternative way of writing down the equivalence class:

$$ \big[\hspace{1pt}\log\big(\exp(X)\exp(tY)\big)\hspace{1pt}\big]= \big[\hspace{1pt}X + t\hspace{1pt}\square\hspace{1pt}\big]. $$

Thus, the differential of $\log$ takes two objects, $X$ (alternatively, $g\coloneq \exp(X)$) and $Y$, and its output is characterized by $\square$. Here, $\square$ is the direction in which $\log(g)$ is perturbed when we perturb its argument $g$ in the direction $Y$. On matrix Lie groups, we can once again define the operator $\Psi^{-1}_{\exp(X)}:\mathfrak g \rightarrow \mathfrak g$ using derivatives of matrix-valued functions:

\[ \Psi^{-1}_{\exp(X)}(Y) = \frac{d}{dt}\log\big(\exp(X)\exp(tY)\big)\Big|_{t=0}. \]

We write $d\log_{g}(gY)=\Psi^{-1}_{g}(Y)$. Alternatively, we can write $d\log_{g}(V)=\Psi^{-1}_{g}(g^{-1}V)$, where $V\in T_{g} G$.

We should verify that $\Psi_{\exp(X)}^{-1}$ is indeed the inverse of $\Psi_X$. Can we show this algebraically?

\[ \begin{align*} \small &\Psi_X\left(\Psi_{\exp(X)}^{-1}(Y)\right)\\ &\quad = \exp(X)^{-1}\frac{d}{ds}\exp\left(X+s\frac{d}{dt}\log\big(\exp(X)\exp(tY)\big)\Big|_{t=0}\right)\Big|_{s=0}\\ \small &\quad \overset{?}=Y,\\ \small &\Psi_{\exp(X)}^{-1}\left(\Psi_X(Y)\right) \\&\quad = \frac{d}{dt}\log\Big(\exp(X)\exp\Big(t \exp(X)^{-1}\frac{d}{ds}\exp\left(X+sY\right)\Big|_{s=0} \Big)\Big)\Big|_{t=0}\\ \small &\quad \overset{?}=Y. \end{align*} \]

I’m not sure how to proceed… In any case, such formidable calculations should not be needed to demonstrate a simple fact. The intrinsic versions of these results are trivial:

\[ d\log_{\exp(X)}\left(d\exp_X(Y)\right) = d(\log \circ \exp)_X(Y)=Y. \]

\[ d\exp_{\log(g)}\left(d\log_g(V)\right) = d(\exp \circ \log)_g(V)=V. \]

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🀴 Parametrization

Another context in which differentials show up in Lie theory is in the context of parametrizations of $G$. While most of this discussion is a specialization of what happens on a general manifold, the Lie group case is special because we are especially interested in left-invariant quantities (e.g., left-invariant vector fields and volume forms).

Let $\varphi:C\rightarrow D$ be a parametrization of $D\subseteq G$ whose domain is $C\subseteq \mathbb R^n$, such that its inverse $\varphi^{-1}:D \rightarrow C$ exists (and therefore, is a smooth coordinate chart). A function $f:G\rightarrow \mathbb R$ can be restricted to $D$ to define $f|_D:D\rightarrow \mathbb R$; nevertheless, we will just write $f$ to refer to the restricted function. We can pull $f$ back under $\varphi$ to define $\bar f\coloneq {\varphi}^\ast f=f\circ \varphi$, so that $\bar f(\mathbf x)=f(\varphi(\mathbf x))$. We can differentiate $\bar f$ using the usual rules of multivariable calculus. Let $\frac{\partial}{\partial x^i}$ be the $i^{th}$ standard vector field in $C$. As vector fields do, it will map $\bar f$ to another function:

\[ \frac{\partial}{\partial x^i}\big(\,\bar f\,\big):C \rightarrow \mathbb R. \]

As far as differential calculus on manifolds go, the story pretty much ends here. However, on a Lie group we are interested in computing derivatives along the left-invariant vector fields (LIVFs). Letting $\lbrace E_i \rbrace_{i=1}^n$ be a basis for $\mathfrak g$, we can express its LIVFs using the parametrization:

\[ E_{i,g}^L = g E_i = {\overline E_i^j}(g)\frac{\partial}{\partial x^j}\Big|_{g} \]

Here, $\frac{\partial}{\partial x^i}\big|_{\varphi(\mathbf x)}\coloneq d \varphi_{\mathbf x}\frac{\partial}{\partial x^i}\big|_{\mathbf x}$ (by abuse of notation) is called the $i^{th}$ coordinate vector field, and $\overline E_i^j$ are smooth functions. But what are these functions? We need to set up a commutative diagram that shows us how the action of $E_{i}^L$ on $f$ can be equated to the action of an $\mathbb R^n$-vector field on $\bar f$.

Let $Z = \mathbf z^i E_i$, where $\mathbf z^i$ are real numbers (not functions!). Consider

\[ \begin{align*} Z^L_{\varphi(\mathbf x)} f &= \mathbf z^i E^L_{i,\varphi(\mathbf x)}f =\mathbf z^i\frac{d}{dt}f\left(\varphi(\mathbf x)\exp(tE_i)\right)\Big|_{t=0}\\ &= \mathbf z^i{\overline E_i^j}(\varphi(\mathbf x))\frac{\partial}{\partial x^j}\Big|_{\varphi(\mathbf x)}f\\ &= \mathbf z^i{\overline E_i^j}(\varphi(\mathbf x))\frac{\partial \bar f}{\partial x^j}(\mathbf x) \end{align*} \]

where the last equality follows from the definition of the pushforward. Let $f$ be the $k^{th}$ coordinate function, ${x}^k:D \rightarrow \mathbb R$ (that is, $x^k\coloneq {\varphi^{-1}}^k$). Then, ${{\partial x}^k}/{\partial x^j}=\delta^k_j$ (with the usual identifications), and we get

\[ \begin{align*} E^L_{i,\varphi(\mathbf x)}x^k &= {\overline E_i^k}(\varphi(\mathbf x)). \end{align*} \]

Hence,

\[ \begin{align*} [Z^Lf](g) &= \mathbf z^i [E_i^Lx^j](g)\frac{\partial \bar f}{\partial x^j}(\varphi^{-1}(g)). \end{align*} \]

We can write this in the matrix-vector form as

\[ \begin{align*} \ \nabla f_{g}^\top\mathbf z = \nabla {\bar f}^\top_{\varphi^{-1}(g)} M(g)\,\mathbf z\ \end{align*} \]

where

\[ \begin{align*} \nabla \bar f &= \begin{bmatrix}\frac{\partial \bar f}{\partial x^1} \\ \vdots \\ \frac{\partial \bar f}{\partial x^n}\end{bmatrix}\quad\text{and}\quad \nabla f &= \begin{bmatrix}E_1^Lf \\ \vdots \\ E_n^Lf\end{bmatrix}, \end{align*} \]

so that $\nabla f_{g}^\top \mathbf z=[Z^L f](g)$, and $M^i_j=E_j^Lx^i$. What is the inverse of $M$? Let $J(\mathbf x) \coloneq \left[M(\varphi{\small(\mathbf x)})\right]^{-1}$. It should satisfy³ 

\[ \begin{align*} M_j^i\, J^j_k\, &= J^j_k [E^L_{j}x^i]\\&= [{(J^j_k E_j)}^Lx^i]\\ &= \delta^i_k. \end{align*} \]

So, $J^j_k$ should be such that

\[ (J^j_k(\mathbf x) E_j)^L_{\varphi(\mathbf x)} = \frac{\partial}{\partial x^k}\Big|_{\varphi(\mathbf x)}=d\varphi_{\mathbf x}\left(\frac{\partial}{\partial x^k}\Big|_{\mathbf x}\right). \]

In the matrix-vector form, we can write

\[ \ \nabla {\bar f}_{\mathbf x}^\top\mathbf y = \nabla f_{\varphi(\mathbf x)}^\top J(\mathbf x)\mathbf y\ \]

Thus, the question of whether one uses $M$ or $J$ in calculations depends on whether the direction of differentiation is specified in the Lie algebra basis, or in the standard basis of $C\subseteq \mathbb R^n$ (i.e., it depends on whether it is $\mathbf z$ or $\mathbf y$ that is specified).

We can recover an explicit formula for $J$ that is used in the work of Chirikjian.⁴ On a matrix Lie group, we can view $\varphi:\mathbb R^n\rightarrow \mathbb R^{m \times m}$ as a map between vector spaces. Letting $(\hspace{1pt}\cdot\hspace{1pt})^\vee:\mathfrak g\rightarrow \mathbb R^n$ be the map $Z \mapsto\mathbf z$, which Chirikjian calls the vee map, we have

\[ J(\mathbf x) = \begin{bmatrix}\big(\varphi (\mathbf x)^{-1}\frac{\partial \varphi}{\partial x^1}\big)^\vee & \big(\varphi (\mathbf x)^{-1}\frac{\partial \varphi}{\partial x^2}\big)^\vee &\cdots & \big(\varphi (\mathbf x)^{-1}\frac{\partial \varphi}{\partial x^n}\big)^\vee \end{bmatrix}. \]

and

\[ M(g) = \begin{bmatrix}E_1^L \varphi^{-1}(g) & E_2^L \varphi^{-1}(g) & \cdots & E_n^L \varphi^{-1}(g) \end{bmatrix}= J(\varphi^{-1}(g))^{-1}. \]

To put this in yet another way, $J(\mathbf x) \mathbf y$ expresses the vector $d\varphi_{\mathbf x}(\mathbf y)$ in the basis of LIVFs, $\lbrace E_{i,g}^L\rbrace_{i=1}^n$ at $g=\varphi (\mathbf x)$. Meanwhile, $M(g) \mathbf z$ expresses the LIVF $Z^L_g$ in the standard coordinate basis at $\varphi^{-1}(g)$.

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🀵 Lengths and Volume

In this final chapter, let’s look at how the Jacobians of a parametrization relate to the measurement of lengths and volumes on $G$. Assume that we have an inner product for $\mathfrak g$, defined via

\[ \langle{E_i},{E_j}\rangle= W_{ij}. \]

This inner product defines a unique left-invariant Riemannian metric on $G$ (as explained in my previous posts), written as $W_{ij}\varepsilon^i \varepsilon^j$, where $\lbrace \varepsilon_{i}\rbrace_{i=1}^n$ is the coframe that is dual to the frame of LIVFs, $\lbrace E_{i}^L\rbrace_{i=1}^n$. We now move to a point $\mathbf x$ in the parameter space. There is presently no inner product for $T_{\mathbf x}C$, since we haven’t chosen one. The key observation is that the Riemannian metric and volume form are both covariant tensor fields, so they can be pulled back (under $\varphi$) from $D$ to $C$. For instance, if $\langle \cdot , \cdot \rangle_{\mathbf x}$ is the pullback of the left-invariant Riemannian metric, then

\[ \begin{align*} \left\langle \frac{\partial}{\partial x^r},\frac{\partial}{\partial x^s} \right\rangle_{\mathbf x} &= \varphi^\ast_{\mathbf x}[W_{ij}\varepsilon^i \varepsilon^j]\left(\frac{\partial}{\partial x^r},\frac{\partial}{\partial x^s}\right)\\ &= W_{ij}\varepsilon^i \varepsilon^j\left(d\varphi_{\mathbf x}\frac{\partial}{\partial x^r},d\varphi_{\mathbf x}\frac{\partial}{\partial x^s}\right)\\ &= W_{ij}J(\mathbf x)^i_rJ(\mathbf x)^j_s. \end{align*} \]

Hence, we have a weighted inner product at $\mathbf x$:

$$\langle \mathbf y,\mathbf z \rangle_{\mathbf x}=\mathbf y^\top J(\mathbf x)^\top \mathbf W J(\mathbf x)\mathbf z,$$

and as $\mathbf x$ varies, this defines a Riemannian metric on $C$. Unlike the left-invariant Riemannian metric on $D\subseteq G$, the coefficients (i.e., the metric tensor) of the pullback Riemannian metric on $C\subseteq\mathbb R^n$ are not constant.

The choice of basis for $\mathfrak g$ defines a unique density on $G$, written as

$$\omega=|\varepsilon^1\wedge\varepsilon^2\wedge\cdots\wedge \varepsilon^n|.$$

This is an example of a left Haar measure for $G$, which is unique up to scaling. The absolute value is taken to make this a density rather than a volume form, saving us the trouble of worrying about orientation.

By a similar procedure, we can compute the pullback volume form on $C$. Let $\varphi^\ast\omega=\lambda(\mathbf x) dx^1\wedge\cdots\wedge dx^n$. Then,

\[ \begin{align*} \lambda(\mathbf x) &= \varphi^\ast\omega\left(\frac{\partial}{\partial x^1},\cdots, \frac{\partial}{\partial x^n}\right)\\ &= |\varepsilon^1\wedge\varepsilon^2\wedge\cdots\wedge \varepsilon^n|\left(d\varphi_{\mathbf x}\frac{\partial}{\partial x^1},\cdots, d\varphi_{\mathbf x}\frac{\partial}{\partial x^n}\right)\\ &= |\varepsilon^1\wedge\varepsilon^2\wedge\cdots\wedge \varepsilon^n|\left(\,J_1^iE_i^L,\,J_2^iE_i^L,\cdots,\, J_n^iE_i^L\,\right)\\ &= \left\lvert\det\left(\begin{bmatrix} J_1^1 & J_2^1 & \cdots & J_n^1\\ J_1^2 & J_2^2 & \cdots & J_n^2\\ \vdots& \vdots& &\vdots\\ J_1^n & J_2^n & \cdots & J_n^n\\ \end{bmatrix}\right)\right\rvert \\ &= |\det(\,J{\small (\mathbf x)}\,)|, \end{align*} \]

where we have omitted the argument $(\mathbf x)$ of $J_i^j(\mathbf x)$ for brevity. We can now integrate a function $f:D \rightarrow \mathbb R$ w.r.t. the Haar measure $\omega$, by pulling the integral back to the parameter space:

\[ \begin{align*} \int_{D\subseteq G} f \cdot \omega &= \int_{C\subseteq \mathbb R^n} \varphi^\ast f \cdot\varphi^\ast \omega \\&= \int_{C\subseteq \mathbb R^n} \bar f \cdot |\det (J) |\cdot dx^1\wedge \cdots \wedge dx^n, \end{align*} \]

where I use $`\hspace{1pt}\cdot\hspace{1pt}'$ for clarity, to indicate the multiplication of a function with another function or $n$-form.