# 2. Strain fields in Cartan media

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## Mathematical preliminaries

We start by stating the definition of the fiber bundle.

### Fiber bundles

A fiber bundle is a topological construction defined by a continuous surjective map $\pi : E \to B$ between two topological spaces, where $E$ is called the total space, $B$ the base space, and for each $p \in B$, the preimage $\pi^{-1}(p)$ is *homeomorphic* to a space $F$, called the fiber. This relationship ensures that locally, $E$ resembles the product space $B \times F$, allowing the study of the properties of $E$ in terms of $B$ and $F$.

The preimage $\pi^{-1}(p)$ is called the *fiber over $p$*. Note that in general the fiber over $p$ is not necessarily equal to the fiber $F$. For it to be a fiber bundle, we only need a homeomorphis, between $\pi^{-1}(p)$ and $F$. For example, consider the tangent bundle $TM$ of a $n$-dimensional manifold $M$. In this case, the total space is $TM$ and the base space is $M$. The projection map $\pi : TM \to M$ maps each element of $TM$ to the corresponding point in $p$ (that is, if $v \in T_p M$ then $\pi(v) = p$). The fibre at $p\in M$ is $\pi^{-1}(p) = T_p M$. Here it is clear that not all fibres at each $p \in M$ are equal. However, for each point $p \in M$ there exists a neighbourhood $U$ such that there is a homoemorphism (or diffeomorphism, more precisely) from $\pi^{-1}(U)$ to $U \times \mathbb{R}^n$. So we see that locally the tangent bundle looks like a product space, which makes it a fiber bundle by definition. We can also thus say that the fiber is $F = \mathbb{R}^n$, which is in this context called a *model space* for all $T_p M$.

### Trivial fiber bundles

A **trivial fibre** bundle is a fiber bundle for which the total space is a product space. That is, we can write $E = B \times F$.

If a fiber bundle is a smooth manifold, then we can consider its tangent bundle $TE$. This leads to the additional notions of the *vertical* and *horizontal bundles*.

### The vertical bundle

Consider the differential of the projection map $d \pi : T E \to TB$, that is for each $p \in B$ we have a linear map $D \pi_p : T_pE \to T_{\pi(p)} B$. The *vertical space* at $p$ is then defined as $V_p E = \text{ker}(D \pi_p)$. That is, $V_p E$ consists of all vectors in $T_p E$ which are tangent to $F$. In other words, if you pick a point in the total space and look at all the possible directions you can move that stay within the fiber over a particular point in the base space, these directions form the vertical space at that point in the total space.

The vertical bundle $VE$ is collection of all vertical spaces. The base space $VE$ is $E$, and its projection map is $\pi_V : VE \to E$.

### Horizontal bundles

A *horizotonal space* is a subspace $H_p E \subseteq T_p E$ if $T_p E$ is a direct sum of $V_p E$ and $H_p E$.

There are in general an infinite number of horizontal spaces at each point $p \in E$, whilst the vertical space is unique. For a general smooth fiber bundle, a unique horizontal space can be specified using a connection.

### Vertical and horizontal bundles of trivial bundles

For a trivial bundle, we have that $TE = TB \times TF$. Each point in $E$ can be represented as a pair $(b, f) \in E$, where $b \in B$ and $f \in F$. The tangent space at $(b,f)$ is a product $T_{(b,f)} E = T_b B \times T_f F$. We thus have that the vertical space is $V_{(b,f)} E = \{0\} \times T_f F$. The natural choice of horizontal space is then $H_{(b,f)} E = T_b B \times \{0 \}$.

So we see that the vertical bundle of a trivial bundle is essentialy the sub-bundle $VE = B \times TF$, and similarly the horizontal bundle is $HE = TB \times F$.

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For Cartan media, our starting point is the *trivial fiber bundle* with total space $E = M \times \mathbb{X}$, where $M$ is the base space and $\mathbb{X}$ the fiber. We use the trivial projection map $\pi(p, x) = p$. Since the fiber bundle is trivial, we have that the fiber at $p \in M$ equals the fiber $\mathbb{X}$. That is $\pi^{-1}(p) = \mathbb{X}$ for all $p \in \mathbb{X}$.
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A *spatial configuration* is a map $q : M \to \mathbb{X}$. In the following we will not consider spatio-temporal configurations, as the inclusion of time follows trivially from the treatment presented here.

A vector field is a map $V : q(M) \to Tq(M)$, and we write $V \in \mathfrak{X}(q(M))$. Given such a vector field, we can always consider a corresponding $T \mathbb{X}$-valued field on $M$, given by $V \circ q : M \to T q(M)$. The field $V \circ q$ can be considered a *section* on the vertical bundle of $M \times \mathbb{X}$.

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Let $\upsilon : M \to T \mathbb{X}$, such that $\upsilon(p) \in T_{q(p)} \mathbb{X}$ for all $p \in M$, and let $\psi^t : \mathbb{X} \to \mathbb{X}$ be the corresponding flow, such that $t \mapsto \psi^t(x)$ is the integral curve of $\upsilon$ with initial condition $x \in \mathbb{X}$.

Now consider $d\psi^t : Tq(M) \to T\psi^t(q(M))$. This is often called a *pushforward map*. Given a vector field $V \in \mathfrak{X}(q(M))$, we have a corresponding vector field $d\psi^t(V) \in \mathfrak{X}(\psi^t(q(M)))$.

Furthermore, given a field $\kappa: M \to T q(M)$, we can define a field on $\psi^t(q(M))$ as $d\psi^t(\kappa) = d\psi^t \circ \kappa$. Since $T q(M) \subset T \mathbb{X}$, we can consider $\kappa$ a field $\kappa : M \to T \mathbb{X}$ as well. In which case we can compute the Lie derivative:

$$
\lim_{t\to 0} \frac{\psi_{t}(\kappa) - \kappa}{t} = \mathcal{L}_\upsilon \kappa = [\upsilon, \kappa]
$$

Consider now $dq : TM \to T q(M)$. We also have that $d\psi^t \circ dq : TM \to T \psi^t(q(M))$. We can similarly consider Lie derivatives

$$
\mathcal{L}_v dq = \lim_{t\to 0} \frac{\psi_{t}(dq) - dq}{t} = [\upsilon, dq]
$$

Confusingly, this should not be understood as the Lie derivative of a $1$-form. Recall that $dq$ is a $T\mathbb{X}$-valued $1$-form, and the Lie derivative is here acting on the vectors in $T \mathbb{X}$.

This becomes clearer if we break up $dq$ in its components $dq = \chi_\alpha du^\alpha$. We then have that

$$
\mathcal{L}_\upsilon dq = (\mathcal{L}_\upsilon \chi_\alpha) du^\alpha
$$

$$
\dot{\chi}_\alpha = \frac{\partial}{\partial u^\alpha} \upsilon + \mathcal{L}_\upsilon \chi_\alpha
$$

recall that $\chi_\alpha = dq(\frac{\partial}{\partial u^\alpha})$.

---

Let $q^t = \psi^t(q)$. Then clearly

$$
\frac{dq^t}{dt} = v
$$

what is $\frac{d}{dt} dq^t$? This would really have to be something in the double tangent bundle $TT \mathbb{X}$. This makes some sense since we have things like $\dot{X}_\alpha$ in the Cartan media paper. We would need some mapping $?:T \mathbb{X} \to TT\mathbb{X}$, such that $\dot{\chi}_\alpha = ?(v)$.

$\chi_\alpha^t = \psi^t(\chi_\alpha)$

$\chi_\alpha^0(p) \in T_{q^0(p)} \mathbb{X}$

$\chi_\alpha^t(p) \in T_{q^t(p)} \mathbb{X}$

Let $\kappa^0 : M \to T q(M)$ and let $\chi^t = \psi_t(\chi^0)$. For simplicity, let us just consider a single point $p \in M$, and the "curve" $K : [0, t_f] \to T\mathbb{X}$ given by $K(t) = \kappa^t(p)$, where $K(t) \in T_{q^t(p)} \mathbb{X}$. Since $T \mathbb{X}$ is a manifold, we can consider tangent vectors in this manifold. That is, we can consider the double tangent bundle $TT\mathbb{X}$. So clearly $\dot{K}(t) \in T_{K(t)}T\mathbb{X}$.

---

Could reverse engineer it.

We have that

$$
\dot{X}_\alpha = \partial_\alpha N + [X_\alpha, N]
$$

and

$$
\chi_\alpha = \rho(X_\alpha) q
$$

So maybe

$$
\begin{aligned}
\dot{\chi}_\alpha & = \rho(\dot{X}_\alpha) q \\
& = \rho(\partial_\alpha N + \rho([X_\alpha, N]) q \\
& = \partial_\alpha \rho(N) q + \mathcal{L}_{\chi_\alpha} \upsilon \\
& = \partial_\alpha \upsilon + \mathcal{L}_{\chi_\alpha} \upsilon
\end{aligned}
$$

---

$\chi_\alpha : M \to T \mathbb{X}$

Let $W = [0, T] \times M$ and $\chi^t_\alpha = \psi^t_* \chi_\alpha$. Then $\chi^t_\alpha : W \to T \mathbb{X}$. Also $d \chi_\alpha^t : TM \to TT\mathbb{X}$. Can we then compute $d \chi^t_\alpha( \frac{\partial}{\partial t})$?

---

It seems that the only real way to relate tangent spaces is via some diffeomorphism $M \to M$. So if I just have some curve $\gamma : [0,1] \to \mathbb{X}$, this does not allow me to "push forward" a vector $V \in T_{\gamma(0)} \mathbb{X}$ to $T_{\gamma(t)} \mathbb{X}$.

If I have a 1-parameter family of diffeomorphisms $\phi^t : q(M) \to$

## The strain fields

The *strain* is the differential of the spatial configuration $d q : TM \to T \mathbb{X}$. $d$ is not the exterior derivative on $M$ here, but we will show that we may interpret it as such.

Consider a vector $v \in T_p M$ and a scalar $f \in C^\infty(\mathbb{X})$. Then $(dq(v))(f) = v(f \circ q)$. Let $u^\alpha : M \to \mathbb{R},\ \alpha=1,\dots,d,$ be coordinates on $M$. We have the corresponding vector fields $\frac{\partial}{\partial u^\alpha}$, with which we can expand any vector field on $M$ as $v = v^\alpha \frac{\partial}{\partial u^\alpha}$. For any vector field $v$ on $M$ we have a corresponding vector field in $\mathbb{X}$ given by $dq(v)$, in terms of the basis we have

$$ \tag{1}
dq(v) = v^\alpha dq \left( \frac{\partial}{\partial u^\alpha} \right) = v^\alpha \chi_\alpha
$$

where we have defined

$$
\chi_\alpha = dq \left( \frac{\partial}{\partial u^\alpha} \right)
$$

which we will call the *strain fields* $\chi_\alpha : M \to T\mathbb{X}$. These are vector fields in $\mathbb{X}$. Using these, we can write

$$\tag{2}
dq = \chi_\alpha d u^\alpha
$$

where $d$ is the exterior derivative on $M$, and $du^\alpha$ are $1$-forms. To see why we can write $dq$ in this way, let $f \in C^\infty(\mathbb{X})$ and $v \in T_p M$, then

$$\tag{3}
(\chi_\alpha du^\alpha(v))(f) = (v^\alpha \chi_\alpha)(f) = (dq(v))(f)
$$

We can thus intepret $dq$ as a $T\mathbb{X}$-valued $1$-form on $M$.

:::info
**Example: Cosserat rod**
Consider a Cosserat rod, so $\mathbb{X} = \mathcal{F}(\mathbb{E}^3)$ and $M = [0, L]$. Then $q(u) = (r(u), e_1(u), e_2(u), e_3(u))$ and $\chi(u) = (\partial_u r, \partial_u e_1, \partial_u e_2, \partial_u e_3)$.
:::

:::info
**Example: Filament**
Consider a filament, so $\mathbb{X} = \mathbb{E}^3$ and $M = [0, L]$. Then $q(u) = r(u)$ and $\chi(u) = \partial_u r$.
:::

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## The generalised strain fields

We now introduce the fact that the configuration space $\mathbb{X}$ has a group $G$ that acts on it transitively. This makes $\mathbb{X}$ a homogeneous space. Furthermore, as we will show, this will allow us to *trivialise* the tangent bundle $T \mathbb{X}$. That is, instead of working with the strain fields $\chi_\alpha : M \to T\mathbb{X}$, which map to a non-linear space, we will be able to work with vector space-valued fields.

:::info
Within the context of fiber bundles, the maps $\chi_\alpha : M \to T \mathbb{X}$ can be considered a section on the *vertical bundle* of the fiber bundle $M \times \mathbb{X}$.
:::

Let the *structure field* $\Phi : M \to G$ be defined such that

$$\tag{4}
q = \Phi \cdot q_r
$$

$d \Phi$ is a $TG$-valued $1$-form on $M$, left translating this to the identity yields the *left Maurer-Cartan form*

$$\tag{5}
\xi^L = \Phi^{-1} d\Phi
$$

and correspondingly we have the right Maurer-Cartan form $\xi^R = d \Phi \Phi^{-1}$. These are $\mathfrak{g}$-valued $1$-forms on $M$.

We then have that $dq = \eta_{\Phi}(d\Phi) |_{q_r} = \zeta(\xi^R) |_q$, where we simply followed the same derivation from Sec. 1.1.4.2.

:::info
Here we have reused the definitions of $\eta$ and $\zeta$ from Sec. 1, and applying them on the $1$-forms $d \Phi$ and $\xi^R$. Since they are vector-valued forms, the definitions extend naturally. For example, for any $\mathfrak{g}$-valued $1$-form $\omega = \omega_\alpha du^\alpha$ on $M$, where $\omega_\alpha : M \to \mathfrak{g}$, we define

$$\tag{6}
\zeta(\omega) = \zeta(\omega_\alpha) du^\alpha
$$
:::

We may call $\xi^L$ (or $\xi^R$), the *generalised strain*. We relate them to strain as

$$\tag{7}
\begin{aligned}
d q & = \zeta(\xi^R) |_{q} \\
& = \zeta(\text{Ad}_\Phi \xi^L) |_q
\end{aligned}
$$

:::info
Recall that $\zeta : \mathfrak{g} \to \mathfrak{X}(\mathbb{X})$ is a representation of $\mathfrak{g}$. For $Y \in \mathfrak{g}$, $\zeta(Y)$ is a vector field on $\mathbb{X}$. We denote $\zeta(Y) |_x$ as the vector field evaluated at $x \in \mathbb{X}$, so $\zeta(Y) |_x \in T_x \mathbb{X}$.

If $\omega$ is a $\mathfrak{g}$-valued $1$-form on $M$, then $\zeta(\omega)$ is a $T\mathbb{X}$-valued $1$-form on $M$. For any point $x \in \mathbb{X}$, then $\zeta(\omega)|_x$ is a $T_x \mathbb{X}$-valued $1$-form on $M$.

The right hand side of the first line of (7) should be understood as follows. $\zeta(\xi^R) |_{q}$ is a map $M \to T \mathbb{X}$, which is given by $p \mapsto \zeta(\xi^R) |_{q(p)}$.
:::

Now let

$$\tag{8}
\begin{aligned}
\xi^R & = X^R_\alpha du^\alpha \\
\xi^L & = X^L_\alpha du^\alpha
\end{aligned}
$$

where we call $X^R_\alpha, X^L_\alpha : M \to \mathfrak{g}$ the right and left *generalised strain fields*. Then we can write

$$\tag{9}
\begin{aligned}
d q & = \zeta(X^R_\alpha)|_q d u^\alpha \\
& = \zeta( \text{Ad}_\Phi X^L_\alpha)|_q d u^\alpha \\
\end{aligned}
$$

Comparing (2) and (9) we find

$$ \tag{10}
\chi_\alpha = \zeta(X^R_\alpha)|_q = \zeta( \text{Ad}_\Phi X^L_\alpha)|_q
$$

---

We see that we can now describe the strains of Cartan media in terms of the $\mathfrak{g}$-valued $1$-form $\xi^L$ (or $\xi^R$), instead of the $T\mathbb{X}$-valued $dq$. The benefits of this are clear.

## Structure equations

Without providing proof here, it is worth mentioning of course that the generalised strain satisfies

$$
d \xi^{L} + [\xi^{L} \wedge \xi^{L}] = 0
$$

## Linear actions

From Sec. 1.2, we see that if the group action is linear, we can rewrite (7) as

$$\tag{11}
\begin{aligned}
d q & = \rho(\xi^R)(q) \\
& = \rho(\text{Ad}_\Phi \xi^L)(q)
\end{aligned}
$$

where, as before, we have extended the definition of $\rho$ as follows: For any $\mathfrak{g}$-valued $1$-form $\omega = \omega_\alpha du^\alpha$ on $M$, where $\omega_\alpha : M \to \mathfrak{g}$, let

$$\tag{12}
\rho(\omega) = \rho(\omega_\alpha) du^\alpha
$$

such that $\rho(\omega_\alpha) : \mathbb{X} \to T\mathbb{X}$ as before.

Now let's say we have a basis $D : \mathbb{R}^n \to \mathbb{X}$ for $\mathbb{X}$. Let $q = D \mathbf{q}$ where $\mathbf{q} : M \to \mathbb{R}^n$. We then have that $d q = D d\mathbf{q}$, where $d \mathbf{q} : TM \to T \mathbb{R}^n$. We further have that

$$\tag{13}
\begin{aligned}
d \mathbf{q} & = \tilde{\rho}(\xi^R) \mathbf{q} \\
& = \tilde{\rho}(\text{Ad}_\Phi \xi^R) \mathbf{q}
\end{aligned}
$$

## The body and spatial frames

We saw earlier that the strain fields can be related to the right generalised strain fields as

$$ \tag{14}
\chi_\alpha = \zeta(X^R_\alpha) |_q
$$

and if we have a linear action we can further write this as

$$ \tag{15}
\begin{aligned}
\chi_\alpha & = \rho(X^R_\alpha) q \\
& = D \tilde{\rho}(X^R_\alpha) \mathbf{q}
\end{aligned}
$$

where last line results from choosing a basis.

Now we see that in (14) we evaluate the vector field $\zeta(X^R_\alpha)$ at $q$. What happens if we evaluate it at the reference point $q_r$?

$$
\zeta(X^R_\alpha) |_{q_r}
$$

$$
\zeta(X^R_\alpha) |_{q} = \zeta( \text{Ad}_\Phi X^L_\alpha)|_q
$$

$$
\zeta(X^R_\alpha) |_{q_r} = \zeta( X^L_\alpha)|_q
$$

$$
\zeta(X^R_\alpha) |_{q_r} = \zeta( X^L_\alpha)|_q?
$$

**note to self: see if you can relate N^L directly to a body frame strain field.**

$$
\chi_\alpha = \zeta(X^R_\alpha)|_q = \zeta( \text{Ad}_\Phi X^L_\alpha)|_q
$$