First let me tell you I don't know anything about Lie Algebras or groups, so I don't really understand a lot of the related questions/answers.
Through the other questions I'm doing, I have been able to get some ideas. I've already seen how $S^3$ is related to $SU(2)$, which is because the determinant in $SU(2)$ gives you the same equation for the unit sphere $S^3$.
I also get how $SU(2)$ is related to the unit quaternions $Sp(1)$. And then there's this other piece I have, for $x$ a purely imaginary unit quaternion and $q\in Sp(1)$, $qxq^*$ is a purely imaginary quaternion where $^*$ denotes the adjoint(conjugate transpose). And then I think I saw somewhere on the internet that if $x$ is also in $Sp(1)$ then $qxq^*$ is also in $Sp(1)$. Somewhere I think I saw something like $qxq^*$ has the same norm as $x$, which is how this relates to rotations in 3-D.
Another thing I kind of get is how $S^2$ is the same as imaginary quaternions of unit length.
An earlier question I was working on was to show $SO(3)$ is a three dimensional manifold, which I could not really do. I think it has something to do with how the rows/ of a matrix in $SO(3)$ form an orthonormal basis which would have three vectors. I feel like I'm missing something about $SO(3)$. So to go from $S^3$ to $SO(3)$ I first go to $SU(2)$ or $Sp(1)$, and then use the $qxq^*$ map, but I'm having trouble going from here to $SO(3)$.
Any gaps you can help me fill in with going to $SO(3)$ from $S^3$ or with how I can understand $SO(3)$ is a three dimensional manifold would be much appreciated.
Step 1: $SO(3)$ is a 3-manifold. Well, it's a subset of $\Bbb R^9$, and there's a function from the set of all $ 3 \times 3 $ matrices (which I'll identify with $\Bbb R^9$ to $\Bbb R^6$, namely $$ f: \Bbb R^9 \to \Bbb R^6 : A \mapsto A^t A $$ Now you might say that the right-hand thing is in $\Bbb R^9$, but if you just look at the diagonal and the things above it, you get something in $\Bbb R^6$.
Now as it happens, $SO(3)$ is just one component of $f^{-1} (\pmatrix{1 & 0 & 0\\ 0 & 1 & 0 \\ 0 & 0 & 1})$. And if you work through the details of the implicit function theorem, it turns out that $f$ is "nice" on this set, hence $SO(3)$ is a 3-manifold. (The whole preimage has two components: matrices of determinant $+1$, i.e., $SO(3)$, and those of determinant $-1$, which are part of $O(3)$, but not $SO(3)$.
Step 2. Writing (unit) quaternions as $a + bi + cj + dk$, with $a^2 + b^2 + c^2 + d^2 = 1$ and using the usual rules for quaternion multiplication, we can that left-multiplication by $q$, i.e., the linear map $s \mapsto qs$, where $s$ is a quaternion, is represented by the matrix $$ L_q = \pmatrix{a & -b & -c & -d \\ b & a & -d & c \\ c & d & a & -b \\ d & -c & b & a} $$ in the basis $1, i, j, k$.
Similarly, right-multiplication by $q$ is represented by the matrix $$ R_q = \pmatrix{a & -b & -c & -d \\ b & a & d & -c \\ c & -d & a & b \\ d & c & -b & a} $$ in the basis $1, i, j, k$. You can use this to find $R_{q^{-1}}$, where $q^{-1} = a - bi -cj -dk$ as well.
If you compute $L_q \cdot R_{q^{-1}}$, you get the $4 \times 4$ matrix $$ \pmatrix{1 & 0 & 0 & 0 \\ 0 & a^2 + b^2 - c^2 - d^2 & 2(bc-ad) & 2(bd+ac) \\ 0 & 2(ad + bc) & a^2 + c^2 - b^2 - d^2 & 2(cd-ab) \\ 0 & 2(bd-ac) & 2(ab+cd) & a^2 + d^2 - c^2 - b^2} $$
Note that both $L_q$ and $R_q$ are orthogonal matrices, so the corresponding transformations preserve angles and perpendicularity.
If we multiply $1 + 0i + 0j + 0k$ on the left by $q$ and on the right by $q^{-1}$, we get $q \cdot 1 \cdot q^{-1} = 1$. So $L_q R_{q^{-1}}$ is a matrix that holds fixed the column vector $\pmatrix{1\\0\\0\\0}$, which you can also see by just looking at the first column.
That also means that anything perpendicular to $u = \pmatrix{1 \\0 \\0 \\0}$, i.e., anything of the form $\pmatrix{0 \\ * \\ * \\ *}$, remains perpendicular to $u$. Those perpendicular things correspond exactly to quaternions of the form $0 + bi + cj + dk$, so-called pure vector quaternions.
So what we have, looking at the lower right corner of the matrix above, is an orthogonal transformation from a 3-space to a 3-space. You can check (with a lot of algebra) that its determinant is 1, or you can use slightly more sophisticated methods(*) to do so. Hence, the lower right $3 \times 3$ matrix is an element of $SO(3)$. In particular, the map from the set of unit quaternions, which we can identify with $S^3$, to $SO(3)$ given by $$ a + bi + cj + dk \mapsto \pmatrix{ a^2 + b^2 - c^2 - d^2 & 2(bc-ad) & 2(bd+ac) \\ 2(ad + bc) & a^2 + c^2 - b^2 - d^2 & 2(cd-ab) \\ 2(bd-ac) & 2(ab+cd) & a^2 + d^2 - c^2 - b^2} $$ is an evidently smooth map from $S^3$ to $SO(3)$. It happens to be an explicit representation of the double-cover of $\Bbb RP^3$ by $S^3$, but I'm not going to write out that part.
(*) Elegant proof that the determinant is 1: The determinant of the lower-right $3 \times 3$ matrix is (by expansion along the first row) the same as that of the $4 \times 4$ matrix. The latter is the product of the determinant of $L_q$ and the determinant of $R_{q^{-1}}$. I'll show that each of these is $1$. Actually, I'll do it for $L_q$, and leave the other to you.
Consider the continuous map $S^3 \to O(4) \to \Bbb R$ defined by $q \mapsto L_q \mapsto \det(L_q)$. (It's continuous because each of the pieces is a polynomimal!). Because things in $O(4)$ have determinant $\pm 1$, we can rewrite this as $S^3 \to O(4) \to \{+1, -1 \}$. The latter is a discrete space, and any continuous function to a discrete space is constant on connected components. Because $S^3$ is connected, this function is necessarily constant. But for $q = 1 + 0i +0j + 0 k$, $L_q$ is the identity matrix, whose determinant is $+1.$ Hence this function is $+1$ everywhere.
Post-comment addition re derivative of $f$ I'm going to take the map $$ f: M_{33} \to M_{33}: A \mapsto A^t A $$ and fancy it up a little to make it clear that $F^{-1}(I) = SO(3)$ is a smooth manifold.
So I'm going to follow $f$ with the map $p$ defined by $$ p(\pmatrix{a & b & c \\ d & e & f \\ g & h & k} = \pmatrix{a\\b\\c\\e\\f\\k} $$ to get $F = p \circ f$; regarding $M_{33}$ as $\Bbb R^9$, we have a function $$ F: \Bbb R^9 \to \Bbb R^6 $$ which I'm now going to study in a neighborhood of $I$ in the domain $M_{33}$. Note that $f(I) = I$, so $F(I) = \pmatrix{1\\0\\0\1\\0\\1}$.
Now I'm going to look at how $F$ transforms certain curves in the domain $M_{33}$ into curves in $\Bbb R^6$, and take derivatives to show that the derivative, $dF$, has rank $6$ at the identity, $I$. Let's look at the curve $$ s_{12}(t) = \pmatrix{1 & t & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1} $$ Then $s_{12}'(0) = \pmatrix{0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0}$, a matrix I'll call $E_{12}$; the matrices $E_{ij}, 1 \le i,j \le 3$ form a basis for the tangent space to $M_{33}$ at $I$, as you can see by turning them all into 9-vectors.
OK, so what's $dF(E_{12})$? We can compute $$ \frac{dF(s_{12}(t))}{dt}(0) $$ to find this. Well,
\begin{align} F(s_{12}(t)) &= p( \pmatrix{1 & t & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1}^t \pmatrix{1 & t & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1})\\ &= p( \pmatrix{1 & t & 0 \\ t & 1 & 0 \\ 0 & 0 & 1})\\ &= \pmatrix{1\\t\\0\\1\\0\\1} \end{align} and the derivative of that is $e_2$. Similarly, if you define the maps $s_{13}$ that puts a $t$ in the $(1,3)$ entry, and $s_{2,3}$, you get that their tangents, in the domain, are $E_{1,3}$ and $E_{2,3}$, and that $dF$ sends them to $e_3$ and $e_5$, respectively.
Similarly, if you define $$ s_1(t) = \pmatrix{1+t& 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1} $$ you find that its tangent, in the domain, is $E_{11}$, but the tangent to $F\circ s_1$, in the codomain, is $2e_1$; defining analogous paths $s_3$ adn $s_3$, we see that the derivatives in the codomain give us $2 e_3$ and $2 e_5$.
So I've now exhibited six linearly independent tangent vectors in the domain (the various $E_{ij}$ that arose) whose corresponding tangent vectors in the codomain are $2e_1, e_2, 2e_3, e_4, 2e_5, e_6$, which forms a basis for $\Bbb R^6$. Hence the derivative of the map $F$ has full rank at $I$, and the implicit/inverse function theorems apply: in a neighborhood of $I$, $F^{-1}(I)$ looks like a function-graph, i.e., a 3-manifold.
What about domain points other than $I$. Well, suppose that $U$ is an other element of $SO(3)$. Then the diffeomorphism
$$ R_U: X \mapsto UX $$ sends all matrices $A$ near $I$ that satisfy $F(A) = I$ into matrices $UA$ near $U$ that satisfy $F(UA) = (UA)^t(UA) = A^t U^t U A = A^t A = I$. In other words, the part of $SO(3)$ near $U$ looks exactly like the part of $SO(3)$ near $I$, hence it, too, is a nice function-graph, and hence all of $SO(3)$ is a 3-manifold.