I am new to studying homotopy and the groups associated with them (as I come from a physics background, so I might be missing a few rigorous details), but when I am trying to apply them to objects like texture defects or line/point defects in condensed matter, I have a hard time understanding what it means to "create" a texture out of a certain homotopy class of the group. For instance, there is something called the Shankar monopole, which relates the $+1$ class of $\pi_3(\mathbb{R}P^3)$ to a vector $\mathbb{\Omega}(\mathbb{r}) = f(r)\frac{\mathbb{r}}{r}$ where $f(r)$ has values $0$ when $r\rightarrow\infty$ and $2\pi$ when $r = 0$ since the vector is constructed out of elements from $SO(3)$, which we can do since $\pi_3(SO(3))=\pi_3(\mathbb{R}P^3)$.
Now, my issue is when I want to create (or better yet draw) a new texture/diagram for the class $-1$ from $\pi_3(\mathbb{R}P^3)$, I don't know where to start or how they are related to the vector above. For clarity, here an image describing what I typed above:
Any help/suggestions are welcomed (I just want to really learn this stuff).
Best Answer
Quaternions.
The quaternions $\mathbb{H}$ are a 4D algebra where every element is a formal sum of a real scalar and a 3D vector. The scalar and vector components (aka real and imaginary parts) of the product of two vectors is given by the formula $\mathbf{uv}=-\mathbf{u}\cdot\mathbf{v}+\mathbf{u}\times\mathbf{v}$. A scalar times a vector is just the usual scalar multiplication, and to multiply two quaternions, FOIL out the scalar and vector parts. In particular: the square roots of $+1$ are $\pm1$, the square roots of $-1$ form the $2$-sphere of unit vectors, two quaternions commute iff they have parallel vector parts, two quaternions anticommute iff they are perpendicular vectors. You may have seen the $\mathbf{ij}=\mathbf{k}$ equation and related identities. The obvious choice of norm on $\mathbb{H}$ is multiplicative, i.e. $|xy|=|x||y|$ for all quaternions $x$ and $y$. In particular, the $3$-sphere $S^3$ is a subgroup.
Euler's identity $\exp(\theta\mathbf{u})=\cos\theta+\sin\theta\,\mathbf{u}$ holds for unit vectors $\mathbf{u}$ (i.e. square roots of $-1$). All quaternions have a polar form, and all unit quaternions ("versors") are of the form $\exp(\theta\mathbf{u})$ - indeed, $\theta$ convex and $\mathbf{u}$ are unique unless the quaternion is $\pm1$. That is, the exponential map restricted to the closed ball in $\mathbb{R}^3$ of radius $\pi$ is a surjection $\exp:B^3\to S^3\setminus\{-1\}$. We can extend $\{\mathbf{u}\}$ to a orthonormal basis $\{\mathbf{u},\mathbf{v},\mathbf{w}\}$ of $\mathbb{R}^3$, and then $\{1,\mathbf{u},\mathbf{v},\mathbf{w}\}$ is an orthonormal basis for $\mathbb{H}$. If we define $L_q(x):=qx$ and $R_q(x)=xq$ for $q=\exp(\theta\mathbf{u})$, we can write $L_q$ and $R_q$ as matrices WRT these bases, and find that both are a rotation by $\theta$ in the $1\mathbf{u}$-plane and the $\mathbf{vw}$-plane, but $R_q$ rotates the opposite direction in the $\mathbf{vw}$ plane. As a result, the conjugation $C_q(x)=qxq^{-1}$ can be restricted to $\mathbb{R}^3$ as a rotation around the oriented axis $\mathbb{u}$ by the angle $2\theta$ (according to the right-hand rule). We thus find $q\mapsto C_q$ is a double covering ($2$-to-$1$ map) $S^3\to\mathrm{SO}(3)$.
Spin.
In general, every $\mathrm{SO}(n)$ has a double covering called the spin group $\mathrm{Spin}(n)$ (you can also extend this to indefinite orthogonal groups defined on pseudo-Euclidean space with bilinear forms of indefinite signature, i.e. with dimensions of both "space" and "time.") In low dimensions ($n\le 6$) there happen to be classical matrix groups that are isomorphic to spin groups, and can thus be thought of as realizations of them (i.e. "accidental isomorphisms," or "exceptional isomorphisms.") In particular, $S^3$ is basically the group of $1\times1$ "unitary" quaternionic matrices, or $\mathrm{Sp}(1)$. (Since $\mathbb{H}\cong\mathbb{C}^2$ as a complex vector space, this also allows us to identify $\mathrm{Sp}(1)$ with $\mathrm{SU}(2)$ like how $\mathrm{U}(1)$ may be identified with $\mathrm{SO}(2)$.)
If you look at an animation of a rotation, you're really seeing a *parametrized family" of rotations. In other words, a path or loop in $\mathrm{SO}(3)$. By covering space theory, we can model covering spaces with paths on the original space, in other words elements of the spin group can be thought of as animations of rotations, or rotations-with-memory (modulo wiggling, i.e. endpoint-preserving homotopies). The fact $\pi_1(\mathrm{SO}(3))=\mathbb{Z}_2$ reflects the fact $S^3\to\mathrm{SO}(3)$ is $2$-to-$1$ (that antipodal versors define the same rotation). You can see visualizations of this stuff in the so-called Dirac string / belt / plate trick.
Smash.
The smash product of spheres is $S^m\wedge S^n\simeq S^{m+n}$. A particular case is the reduced suspension $\Sigma S^1=S^1\wedge S^1=S^2$. What this means is that the sphere is a "pointed loop of pointed loops," with the first and last pointed loop being shrunk to a single point. In other words, any map $S^2\to X$ for a space $X$ can be thought of as a homotopy $S^1\times[0,1]\to X$ from a "pointed" (range = one point) function to itself. In particular, if we think of the identity map $S^2\to S^2$ as a homotopy $S^1\times[0,1]\to S^2$, we can visualize it as a rubber band on a sphere starting out scrunched to a point, pulled around the sphere until it scrunches again:
Lifting.
Thus, any map $S^3\to\mathrm{SO}(3)$ can be thought of as a homotopy $S^2\times[0,1]\to\mathrm{SO}(3)$, which by the path lifting property of fibrations can be lifted to a homotopy $S^2\times[0,1]\to S^3$, which can be thought of as a map $S^3\to S^3$. Thus, any element of $\pi_3(\mathrm{SO}(3))$ can be thought of as an element of $\pi_3(S^3)$. Alternatively, we could apply the long homotopy exact sequence to the fiber bundle $S^0\to S^3\to\mathrm{SO}(3)$ to get
$$ \cdots\to\pi_3(S^0)\to\pi_3(S^3)\to\pi_3(\mathrm{SO}(3))\to\pi_2(S^0)\to\cdots $$
Since $\pi_3(S^0)$ and $\pi_2(S^0)$ are trivial, we get $\pi_3(S^3)\to\pi_3(\mathrm{SO}(3))$ is an isomorphism, so we can think of elements of $\pi_3(S^3)$ and $\pi_3(\mathrm{SO}(3))$ interchangeably.
Looping.
I mentioned how the identity map $S^3\to S^3$ can be thought of as a homotophy $S^2\times[0,1]\to S^3$ which can be visualized by unscrunching, wrapping a rubber band around a sphere, and the rescrunching. This represents the $\pm1$ element of $\mathbb{Z}$ corresponding to $\pi_3(S^3)\cong\mathbb{Z}$. (Whether such a visualization is $\pm1$ depends on which way the band is oriented and which way around the sphere it is wrapped.) How, then, do we visualize the composition operation of $\pi_3(S^3)$, i.e. the addition operation on $\mathbb{Z}$? Well, if we unscrunch-wrap-scrunch the band $n$ times around instead of just once, that represents $\pm n$.
If you say the identity map represents $+1$, then the positive/negative classes can be represented by orientation-preserving/reversing maps. Indeed, as the rubber band shows us, the class corresponding to $n\in\mathbb{Z}$ can be represented by a map which generically has degree $n$, i.e. is $n$-to-$1$ as a map $S^3\to S^3$ on all but a subset of measure zero / positive codimension.
Monopole.
You've defined the Shankar monopole to be a function $\Omega:\mathbb{R}^3\setminus\{0\}\to\mathbb{R}^3$ defined by $\Omega(\mathbf{r})=f(r)\hat{\mathbf{r}}$, where $r=\|\mathbf{r}\|$ and $\hat{\mathbf{r}}=\mathbf{r}/r$ and $f:[0,\infty]\to[0,2\pi]$ satisfies $f(0)=2\pi$ and $f(\infty)=0$. Note $\Omega$ is not defined at the origin, and its range is basically the open ball of radius $2\pi$ punctured at the origin. We can then define $\Psi(\mathbf{r})=\exp\Omega(\mathbf{r})/2$ to be a map $\mathbb{R}^3\to S^3$ which is now defined at $\mathbf{r}=\mathbf{0}$ even if $\Omega$ isn't. We can also extend it to the "extended" Euclidean space $\mathbb{R}^3\sqcup\{\infty\}$, topologized so it's the one-point compactification, so $\Omega(\mathbf{0})=-1$ and $\Omega(\infty)=+1$.
If we let $\Phi:S^3\to\mathbb{R}^3\sqcup\{0\}$ be stereographic projection, the composition $\eta=\Psi\circ\Phi$ is a function $\eta:S^3\to S^3$, so it represents a homotopy class $[\eta]$ which is element of $\pi_3(S^3)$. With a homotopy we can turn $f$ into a monotone decreasing (hence one-to-one) function, and "revolve" (i.e. incorporate the parametrized family of $f$s to get a parametrized family of $\Omega$s etc.) until $\Omega:\mathbb{R}^3\sqcup\{\infty\}\to S^3$ and hence $\eta:S^3\to S^3$ are bijections. This has degree $\pm1$ (depending on how the stereographic projection is done, so we can choose it to be $+1$) which means it represents the $+1$ class in $\pi_3(S^3)\cong\mathbb{Z}$.