1. Lie group and Lie algebra actions on configuration spaces

1.1 Actions on general manifolds

We have an $n$-dimensional differentiable manifold $\mathbb{X}$ on which an $r$-dimensional Lie group $G$ acts transitively. We write this action as $g\cdot x$.

The Lie group not only allows us to move points $x\in \mathbb{X}$ around, it also allows us to transport tangent vectors across $\mathbb{X}$. That is, for any vector $v\in T_x\mathbb{X}$, and a $g\in G$ such that $y=g\cdot x$, have a vector $gv\in T_y\mathbb{X}$.

The Lie group action also induces an infinitesimal Lie algebra action on $\mathbb{X}$. We find this by differentiating the action at the identity:

The orbit map is defined as ${\sigma }_x:G\to \mathbb{X},g\mapsto g\cdot x$. The derivative of the orbit map is a map $D_e{\sigma }_x:\mathfrak{g}\to T_x\mathbb{X}$. We can evaluate it on a Lie algebra element $Y\in \mathfrak{g}$ as

$$(D_e{\sigma }_x)(Y)=\frac{d}{dt}{|}_{t=0}{\sigma }_x(\exp (tY))$$

where the right-hand side is clearly a tangent vector in $T_x\mathbb{X}$. Now, it is clear that (1) establishes a map $\zeta :\mathfrak{g}\to \mathfrak{X}(\mathbb{X})$, where $\mathfrak{X}(\mathbb{X})$ is the space of vector fields on $\mathbb{X}$. The vector field $\zeta (Y)$ evaluated at $x$ is given by

$$\zeta (Y){|}_x=(D_e{\sigma }_x)(Y)\in T_x\mathbb{X}$$

Now, since differential operators are linear maps, we also have that $\zeta$ is a representation of $\mathfrak{g}$. So we see that though not all group actions define a group representation, they do define a Lie algebra representation.

1.1.2 The representation space of the Lie algebra

What is the representation space of $\zeta$? Since $\zeta (Y)$ is a vector field, the representation space consists of all scalars $C^{\infty }(\mathbb{X})$. That is, $\zeta (Y):C^{\infty }(\mathbb{X})\to C^{\infty }(\mathbb{X})$. Note that this is an infinite dimensional representation.

Let $f$ be a scalar, then formally we have that

$$\zeta (Y){|}_x(f)=((D_e{\sigma }_x)(Y))(f)$$

and $\zeta (Y)(f)\in C^{\infty }(\mathbb{X})$ is the map $x\mapsto \zeta (Y){|}_x(f)$.

How can we evaluate this expression? Recall that if we have a curve $\gamma :[0,1]\to \mathbb{X}$, then $\dot{\gamma }(t)\in T_{\gamma (t)}\mathbb{X}$ is a vector that acts like

$$(\dot{\gamma }(t))(f)=\frac{d}{d\tau }{|}_{\tau =t}(f\circ \gamma )(\tau )$$

So we have that

$$\zeta (Y){|}_x(f)=\frac{d}{dt}{|}_{t=0}f(\exp (tY)\cdot x)$$

If $E_i,i=1,\dots ,r$ is a basis over $\mathfrak{g}$. Then we can write the action of $\mathfrak{g}$ on scalars as

$$\zeta (Y)(f)=Y_i\zeta (E_i)(f)$$

Now let $\varphi :\mathbb{X}\to {\mathbb{R}}^n$ be coordinates on $\mathbb{X}$, with components ${\varphi }^{\mu }:\mathbb{X}\to \mathbb{R}$. For any point $x\in \mathbb{X}$ we denote $x^{\mu }={\varphi }^{\mu }(x)$. Further, we define $\zeta (E_i)(x^{\mu })=E_i^{\mu }$. Using the chain rule we find that

$$\zeta (Y)(f)=Y_i{E}_i^{\mu }\frac{\partial f}{\partial x^{\mu }}$$

Note that $E_i^{\mu }$ are in general functions on $E_i^{\mu }:\mathbb{X}\to \mathbb{R}$.

:::info See, for example, footnote 1 of Poincaré’s paper. There we have that $E_i^{\mu }={{ϵ}_{i\nu }}^{\mu }{x}^{\nu }$. :::

Let $d_{\mathbb{X}}$ be the exterior derivative on $\mathbb{X}$. Then $d_{\mathbb{X}}f$ is a $1$-form on $\mathbb{X}$, and we can write $\zeta (Y)(f)=d_{\mathbb{X}}f(\zeta (Y))$.

1.1.3 Integral curves

As with any vector field, we can also consider the integral curves of $\zeta (Y)$. That is, a curve $\gamma :[0,1]\to \mathbb{X}$, with $\gamma (0)=x_0$, that satisfies

$$\dot{\gamma }(t)=\zeta (Y){|}_{\gamma (t)}$$

If we act on the coordinates with the vector-valued $\dot{\gamma }$, we get $\dot{\gamma }(x^{\mu })=\zeta (Y)(x^{\mu }){|}_{\gamma (t)}=Y_i{E}_i^{\mu }$. So if we let ${\gamma }^{\mu }:[0,1]\to \mathbb{R}$ be defined as ${\gamma }^{\mu }=x^{\mu }\circ \gamma$, then

$${\dot{\gamma }}^{\mu }(t)=Y_i{E}_i^{\mu }(\gamma (t))$$

1.1.4 Generalised “velocity” of a curve

Finally, consider some general curve $q:[0,1]\to \mathbb{X}$. Then there exists a curve in the Lie algebra $N^R:[0,1]\to \mathfrak{g}$ such that

$$\dot{q}(t)=\zeta (N^R(t)){|}_{q(t)}$$

or in coordinates

$${\dot{q}}^{\mu }(t)=N_i^R(t)E_i^{\mu }(q(t))$$

in which case, we may call $N^R$ the generalised velocity of $q$.

Can we find an explicit form of $N^R$? We will derive it below.

:::warning Since the derivation turned out to be very complicated, I’ve written a simpler (more hand-wavey) version that gets at the central idea, in addition to the more detailed one. :::

1.1.4.1 Hand-wavey derivation

Let $g:[0,1]\to G$ be defined such that $q(t)=g(t)\cdot q_0$. Then

$$\begin{array}{rl}\dot{q}(t)& =\frac{d}{dt}(g(t)\cdot q_r)\\ & =\dot{g}(t)\hat{\cdot }{q}_r\end{array}$$

Note that $\dot{g}(t)\in T_{g(t)}G$ is a tangent vector. We’ve implicitly defined a new map $\hat{\cdot }$, which is a map $TG\times \mathbb{X}\to T\mathbb{X}$.

$$\begin{array}{rl}\dot{q}(t)& =\dot{g}(t)\hat{\cdot }{q}_r\\ & =\dot{g}(t)\hat{\cdot }(g^{-1}(t)\cdot q(t))\\ & =(\dot{g}(t)g^{-1}(t))\hat{\cdot }q(t)\\ & =N^R(t)\hat{\cdot }q(t)\\ & =\zeta (N^R(t)){|}_{q(t)}\end{array}$$

where we have right-translated the tangent vector $\dot{g}(t)$ to the identity to get $N^R(t)=\dot{g}(t)g^{-1}(t)$.

1.1.4.2 Fully detailed derivation

Let $g:[0,1]\to G$ be defined such that $q(t)=g(t)\cdot q_0$. Then

$$\begin{array}{rl}\dot{q}(t)& =\frac{d}{dt}(g(t)\cdot q_r)\\ & ={\eta }_{g(t)}(\dot{g}(t)){|}_{q_r}\end{array}$$

Here we’ve defined a map ${\eta }_g:T_{g}G\to \mathfrak{X}(\mathbb{X})$. That is, $\dot{g}\in T_{g(t)}G$ is a tangent vector, and ${\eta }_{g(t)}(\dot{g}(t))$ is a vector field on $\mathbb{X}$, and ${\eta }_{g(t)}(\dot{g}(t)){|}_x\in T_x\mathbb{X}$ is a tangent vector at $x$.

Since this will get quite complicated, we’ll add some extra detail for explanatory purposes. Consider a tangent vector $v\in T_{g_0}G$, let $g:[-1,1]\to G$ be any curve that satisfies $g(0)=g_0$ and $\dot{g}(0)=v$. Then the vector field ${\eta }_{g_0}(v)$ evaluated at $x\in \mathbb{X}$ is the tangent vector

$${\eta }_{g_0}(v){|}_x=\frac{d}{dt}{|}_{t=0}(g(t)\cdot x)\in T_x\mathbb{X}$$

Notice that this definition is similar to that of $\zeta$. $\eta$ is the analogue $\zeta$ for general tangent vectors of $G$ away from the identity. When $g_0=e$, then the definition coincides ${\eta }_e=\zeta$.

We can left or right-translate tangent vectors on $G$. In particular, we have that $\dot{g}{g}^{-1}\in T_{e}G$. Returning to the derivation, we find

$$\begin{array}{rl}\dot{q}(t)& ={\eta }_{g(t)}(\dot{g}(t)){|}_{q_r}\\ & =\frac{d}{dt}(g(t)\cdot q_r)\\ & =\left(\frac{d}{dt}(g(t)\cdot (g^{-1}(\tau )\cdot q(\tau )))\right){|}_{\tau =t}\\ & =\left(\frac{d}{dt}((g(t)g^{-1}(\tau ))\cdot q(\tau ))\right){|}_{\tau =t}\\ & =\left({\eta }_{g(t)g^{-1}(\tau )}(\dot{g}(t)g^{-1}(\tau )){|}_{q(\tau )}\right){|}_{\tau =t}\\ & ={\eta }_e(\dot{g}(t)g^{-1}(t)){|}_{q(t)}\\ & =\zeta (N^R(t)){|}_{q(t)}\end{array}$$

where $N^R(t)=\dot{g}(t)g^{-1}(t)$.

1.1.4.3 The left-invariant velocity

$N^R(t)$ is the right-invariant velocity of $g(t)$. The corresponding left-invariant velocity is $N^L(t)=g^{-1}(t)\dot{g}(t)$, and we can rewrite $\dot{q}$ as

$$\dot{q}(t)=\zeta ({\text{Ad}}_{g(t)}{N}^L(t)){|}_{q(t)}$$

1.2 Actions on vector spaces

Let $\mathbb{X}$ now be a vector space. The action now induces a representation $\Pi :G\to GL(\mathbb{X})$. As we will show below, we can derive some further representations of the Lie algebra using this.

1.2.1 Another representation of the Lie algebra

Using (3) and (5), we find

$$\begin{array}{rl}\zeta (Y){|}_y(f)& =((D_e{\sigma }_y)(Y))(f)\\ & =\frac{d}{dt}{|}_{t=0}f(\exp (tY)\cdot y)\\ & =\frac{d}{dt}{|}_{t=0}f(\Pi (\exp (tY))(y))\end{array}$$

Now for fixed $f$ and $Y$, the last line of (12) is clearly a linear map acting on $y\in \mathbb{X}$. We can thus write

$$\zeta (Y){|}_y(f)=(\rho (Y)(y))(f)$$

where $\rho (Y):\mathbb{X}\to T\mathbb{X}$, such that $\rho (Y)(y)\in T_y\mathbb{X}$ is a tangent vector.

Conceptually, we can understand $\rho (Y)$ as arising from the Taylor expansion

$$\Pi (\exp (\Delta tY))(y)=y+\Delta t\rho (Y)(y)+\dots$$

such that

$$\rho (Y)(y)=\frac{d}{dt}{|}_{t=0}\Pi (\exp (\Delta tY))(y)$$

:::warning Since $\rho (Y):\mathbb{X}\to T\mathbb{X}$, this is not strictly speaking a representation, although we will keep referring to it as such.

You should see $\rho (Y)$ as a map that converts configurations $y\in \mathbb{X}$ and turns them into “velocities” $\dot{y}=\rho (Y)(y)$. This is what happens in the 3rd equation in the Poincaré paper. :::

1.2.2 The generalised velocity in terms of $\rho$

Consider again a curve $q:[0,1]\to \mathbb{X}$. From (1) we have

$$\begin{array}{rl}\dot{q}(t)& =\rho (N^R(t))(q(t))\end{array}$$

where $N^R:[0,1]\to \mathfrak{g}$ is defined as before. This is subtly but quite significantly different from the case when $\mathbb{X}$ was a general manifold. Namely, all the configurational dependence lies in the argument $q(t)$, and not in the representation $\rho (N(t))$. We’ll see this more explicitly in the next section.

In terms of the left-invariant velocity we have

$$\dot{q}(t)=\rho ({\text{Ad}}_{g(t)}{N}^L(t))(q(t))$$

:::info You can recognise this from the Cartan media paper. See Eq. 4 and above in the paper. There we call $N=N^L$ and $N^s=N^R$. :::

1.3 Matrix representations

We now choose a basis $d_{\mu },\mu =1,\dots ,n,$ for $\mathbb{X}$. We have then effectively switched over from using the space $\mathbb{X}$ to (what I call) the column vector space ${\mathbb{R}}^n$, where we write elements as $\mathbf{v}=(v_1{v}_2\dots v_n{)}^T\in {\mathbb{R}}^n$.

We may use the short-hand

$$v=D\mathbf{v}$$

where $D=(d_1{d}_2\dots d_n)$. We may consider $D$ a map $D:{\mathbb{R}}^n\to \mathbb{X}$, in which case it is called the component map.

Let the coordinates ${\varphi }^{\mu }$ be the standard one for a vector space with a given basis $d_{\mu }$. That is, the coordinates for any $y=y^{\mu }{d}_{\mu }\in \mathbb{X}$ is given by their components in the basis ${\varphi }^{\mu }(y)=y^{\mu }$. For any scalar $f:\mathbb{X}\to \mathbb{R}$ let $\tilde{f}:{\mathbb{R}}^n\to \mathbb{R}$ be the corresponding scalar on ${\mathbb{R}}^n$, which satisfies $\tilde{f}(\mathbf{y})=f(y^{\mu }{d}_{\mu })$.

From the representation $\Pi$, we can now induce the matrix representation $\tilde{\Pi }:G\to GL({\mathbb{R}}^n)$, which satisfies

$$\Pi (g)(y)=d_{\mu }\tilde{\Pi }(g{)}_{\nu }^{\mu }{y}^{\nu }$$

for any $y\in \mathbb{X}$.

The corresponding representation of the Lie algebra $\tilde{\rho }:\mathfrak{g}\to \mathfrak{gl}({\mathbb{R}}^n)$ is induced from

$$\rho (Y)(y)=\tilde{\rho }(Y{)}_{\nu }^{\mu }{y}^{\nu }\frac{\partial }{\partial x^{\mu }}$$

We also have a representation $\tilde{\zeta }:\mathfrak{g}\to \mathfrak{X}({\mathbb{R}}^n)$, defined by

$$\begin{array}{rl}\zeta (Y){|}_y(f)& =(\rho (Y)(y))(f)\\ & =\tilde{\rho }(Y{)}_{\nu }^{\mu }{y}^{\nu }\frac{\partial }{\partial x^{\mu }}(f)\\ & =\tilde{\rho }(Y{)}_{\nu }^{\mu }{y}^{\nu }\frac{\partial \tilde{f}}{\partial x^{\mu }}\\ & =\tilde{\zeta }(Y){|}_{D^{-1}y}(\tilde{f})\end{array}$$

1.3.1 The generalised “velocity” of a curve in ${\mathbb{R}}^n$

Let $\mathbf{q}:[0,1]\to {\mathbb{R}}^n$ be defined such that $q=q^{\mu }{d}_{\mu }$, where $q$ was defined earlier. Then

$$\begin{array}{rl}\dot{\mathbf{q}}& =\tilde{\rho }(N^R(t))\mathbf{q}\\ & =\tilde{\rho }({\text{Ad}}_{g(t)}{N}^L(t))\mathbf{q}\end{array}$$

Which is the “column vector version” of (10). As for (11), note that

$$E_i^{\mu }(x^{\mu })=\zeta (E_i)(x^{\mu })=\tilde{\rho }(E_i{)}_{\nu }^{\mu }{x}^{\nu }$$

So we have

$${\dot{q}}^{\mu }=N_i^R(t)\tilde{\rho }(E_i{)}_{\nu }^{\mu }{q}^{\nu }$$

1.3.2 The action of the Lie algebra on scalars

The action of the Lie algebra on scalars in $C^{\infty }({\mathbb{R}}^n)$ was already shown in (19), but we show it more explicitly here.

Consider a scalar $f\in C^{\infty }(\mathbb{X})$. In (19), we saw that $\zeta (Y{)}_y(f)=\tilde{\zeta }(Y{)}_{\mathbf{y}}(\tilde{f})$, where

$$\tilde{\zeta }{|}_{\mathbf{y}}(Y)=\tilde{\rho }(Y{)}_{\nu }^{\mu }{y}^{\nu }\frac{\partial }{\partial x^{\mu }}$$

Now, let the gradient of $\tilde{f}$ be defined as

$$\nabla \tilde{f}=\left(\begin{array}{c}\frac{\partial \tilde{f}}{\partial {\tilde{x}}_1}\\ \frac{\partial \tilde{f}}{\partial {\tilde{x}}_2}\\ ⋮\\ \frac{\partial \tilde{f}}{\partial {\tilde{x}}_n}\end{array}\right)$$

Then we have that

$$\begin{array}{r}\tilde{\zeta }(Y){|}_{\mathbf{y}}(\tilde{f})=(\nabla \tilde{f}{)}^T\tilde{\rho }(Y)\mathbf{y}\end{array}$$

Example:

$G=SO(3)$. Let $v\in \mathfrak{so}(3)$. Then $\tilde{\rho }(v)$ is a $3\times 3$ skew-symmetric matrix. Let $\mathbf{v}\in {\mathbb{R}}^3$ be the corresponding pseudo-vector such that $\hat{v}=\tilde{\rho }(v)$. Then

$$\tilde{\zeta }(v){|}_{\mathbf{y}}(f)=(\nabla \tilde{f}{)}^T(\mathbf{v}\times \mathbf{y})={\mathbf{v}}^T(\mathbf{y}\times \nabla \tilde{f})$$

1.4 Conclusion

Note that we now have four different representations of the Lie algebra.

1. $\zeta :\mathfrak{g}\to \mathfrak{X}(\mathbb{X})$ Lie algebra elements as differentiatial operators on scalars on $\mathbb{X}$. The representations space is $C^{\infty }(\mathbb{X})$.
2. $\rho :\mathfrak{g}\to \text{Hom}(\mathbb{X},T\mathbb{X})$ Not exactly a representation, but converts a configuration $x$ to a “velocity” in the tangent space.
3. $\tilde{\zeta }:\mathfrak{g}\to \mathfrak{X}({\mathbb{R}}^n)$
4. $\tilde{\rho }:\mathfrak{g}\to \text{Hom}({\mathbb{R}}^n,{\mathbb{R}}^n)$

Clearly, it is 1 and 2 that are the most important ones. 3 and 4 appear due to just choosing a basis.

It should also be restated that $\rho$ is not a exactly a representation, as $\rho (Y):\mathbb{X}\to T\mathbb{X}$. Colloquially, $\rho (Y)$ turns a configuration $x\in \mathbb{X}$ into a velocity $\dot{x}\in T_x\mathbb{X}$.

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We can summarise consequences of the various assumptions we make about $\mathbb{X}$ and the group action:

Level 1: $\mathbb{X}$ is a manifold that $G$ acts on transitively.

Since the group action is not linear, we do not have a representation of $G$. However, we did find that we get a representation $\zeta :\mathfrak{g}\to \mathfrak{X}(\mathbb{X})$ of the Lie algebra. That is, Lie algebra elements map to vector fields on $\mathbb{X}$.

Level 2: $\mathbb{X}$ is a vector space, and the group action is linear.

We get a representation $\Pi :G\to GL(\mathbb{X})$ of the Lie group. We also find that there is a map $\rho$ “hiding” inside $\zeta$, now that the action is linear. That is, given $Y\in \mathfrak{g}$, we find that the vector field $\zeta (Y)$ evaluated at $x\in \mathbb{X}$ is given by

$$\zeta (Y){|}_x=\rho (Y)(x)$$

where $\rho (Y):\mathbb{X}\to T\mathbb{X}$ is a linear map. We can then compute the $\zeta (Y)$ acting on a scalar as follows:

$$\zeta (Y){|}_y(f)=(\rho (Y)(y))(f)$$

Colloquially, we should see $\zeta (Y){|}_x$ as a “velocity” at $x$. So we may write "$\dot{x}=\zeta (Y){|}_x$". The fact that the group action is now linear allows us to further write this using a linear map "$\dot{x}=\rho (Y)(x)$".

Level 3: Introduce a basis on $\mathbb{X}$

Using a basis, we can now write, for any $Y\in \mathfrak{g}$ and $x\in \mathbb{X}$

$$\begin{array}{rl}(\zeta (Y)(f))(x)& =(\tilde{\zeta }(Y)(\tilde{f}))(\mathbf{x})\\ & =(\nabla \tilde{f}{)}^T\tilde{\rho }(Y)\mathbf{x}\end{array}$$

where $\mathbf{x}=D^{-1}x$.

Compare this last line with

$$\zeta (Y)(f)=(d_{\mathbb{X}}f)(\zeta (Y))$$

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We also have that

$$\zeta (Y){|}_x=D\tilde{\zeta }(Y){|}_{\mathbf{x}}=D\tilde{\rho }(Y)\mathbf{x}$$