In class my prof made three claims about a group and its Lie algebra. I cannot find direct reference to these claims because they are delivered in verbatim (im not even sure if I have them jogged down 100% correctly). Can someone please help me identify these three properties?
Claim 1: $R\in SO(3)$, so any rotation matrix $R$ is the matrix exponential skew symmetric matrix $M$
Claim 2: Any skew symmetric 3x3 matrix is determined by 3 distinct numbers i.e. a vector
Claim 3: $\omega$ be a vector in $R^3$. Let $v$ = normalized $\omega$ (?) and $\theta$ = length of $\omega$. Then, $\omega$ = $v$ $\theta$. Now, it can be shown that if you take the identity matrix and rotate it by the angle $\theta$ around the axis span{$v$} using the right-hand rule, the result is precisely the matrix $R$
Please direct me to appropriate reference or if these are in fact simple proofs, please show me how these are true. Thank you for your time!
Let's address your questions one at a time. We will start with question 2 since that is the simplest.
$2$. The fact that every skew-symmetric matrix is determined by $3$ numbers is simply the statement that the dimension of the subspace of $3\times 3$ skew-symmetric matrices $\mathrm{Skew}_3$ has dimension $3$. Explicitly, we have the following basis: $$\left\{\begin{pmatrix}0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0\end{pmatrix},\ \begin{pmatrix}0 & 0 & 1 \\ 0 & 0 & 0 \\ -1 & 0 & 0\end{pmatrix},\ \begin{pmatrix}0 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & 1 & 0\end{pmatrix}\right\},$$ This means that every $3\times 3$ skew-symmetric matrix $A$ can be written in the form $$A = \begin{pmatrix} 0 & -c & b \\ c & 0 & -a \\ -b & a & 0\end{pmatrix}.$$ In fact, we can say a bit more than this. Every skew-symmetric matrix in fact represents a cross product. Let $\mathbf{v} = (a,\ b,\ c)^\mathrm{T}$ and let $\mathbf{u}\in\mathbb{R}^3$. Then we have $$A\mathbf{u} = \mathbf{v}\times \mathbf{u},$$ which you can verify directly through matrix multiplication. There is in fact a very useful isomorphism between the Lie algebras $(\mathrm{Skew}_3=\mathfrak{so}(3),\ [\cdot,\ \cdot])$ and $(\mathbb{R}^3,\ \times)$ through the above correspondence (where $[\cdot,\ \cdot]$ is the commutator and $\times$ is the cross-product).
Explicitly, let $\Omega: \mathbb{R}^3 \rightarrow \mathfrak{so}(3)$ be given by $$\Omega(\mathbf{v}) = \begin{pmatrix} 0 & -c & b \\ c & 0 & -a \\ -b & a & 0\end{pmatrix},$$ where $a,b,c$ are the components of $\mathbf{v}$. Clearly this is an bijection, so we must show that this map preserves the respective products, i.e. $$[\Omega(\mathbf{u}),\ \Omega(\mathbf{u})] = \Omega(\mathbf{u}\times\mathbf{v}).$$ It is simplest to do this by remembering that the action of $\Omega(\mathbf{v})$ on a vector $\mathbf{u}$ is given by the cross product $\Omega(\mathbf{v})\mathbf{u}=\mathbf{v}\times\mathbf{u}$ as mentioned earlier. With that in mind, let $\mathbf{w}\in\mathbb{R}^3$ be an arbitrary vector. Then we have $$\begin{align}\Omega(\mathbf{u}\times\mathbf{v})\mathbf{w} &= (\mathbf{u}\times\mathbf{v})\times \mathbf{w}\\ &=-(\mathbf{v}\times\mathbf{w})\times\mathbf{u} - (\mathbf{w}\times\mathbf{u})\times\mathbf{v}\\ &=\mathbf{u}\times(\mathbf{v}\times\mathbf{w}) - \mathbf{v}\times (\mathbf{u}\times\mathbf{w})\\ &=\Omega(\mathbf{u})\Omega(\mathbf{v})\mathbf{w}-\Omega(\mathbf{v})\Omega(\mathbf{u})\mathbf{w}\\ &= [\Omega(\mathbf{u}),\ \Omega(\mathbf{v})]\mathbf{w},\end{align}$$ where in the second step we use the Jacobi identity for the cross product. Since this holds for arbitrary $\mathbf{w}\in\mathbb{R}$, we must have $$[\Omega(\mathbf{u},\ \Omega(\mathbf{v})] = \Omega(\mathbf{u}\times\mathbf{v}).$$ This shows that $\Omega$ is a Lie algebra isomorphism between $(\mathbb{R}^3,\times)$ and $(\mathfrak{so}(3),[\cdot,\ \cdot])$. The existence of this correspondence is precisely the reason why we are able to represent rotations using a single vector, the direction of which represents the action of rotation (with the direction given by the right-hand rule) and the length of which represents the angle of rotation.
Next, let's address your first question.
$1$. There is a well known theorem that the map $\exp : \mathfrak{g}\rightarrow G$ is a surjective map for any compact, connected Lie group $G$. Since $SO(3)$ is compact and connected, this immediately answers your first question (in fact this argument shows that the statement holds true for $SO(n)$ for any $n$). But this theorem is not particularly easy to prove, so let me give a proof specifically for the case of $SO(3)$. In fact, we will do more and give an explicit formula for $\exp: \mathfrak{so}(3)\rightarrow SO(3)$. Let us write elements of $\mathfrak{so}(3)$ as we did previously in terms of the parameters $a,b,c\in\mathbb{R}$.
Theorem: The matrix exponential $\exp:\mathfrak{so}(3) \rightarrow SO(3)$ is explicitly given by $$\exp(A) = I + \frac{\sin\theta}{\theta}A + \frac{(1-\cos\theta)}{\theta^2}A^2,$$ where $\theta = \sqrt{\mathrm{tr}(A^\mathrm{T}A)} = \sqrt{a^2+b^2+c^2}$.
Proof: Note that we have $$A^3 = -\theta^2 A$$ for any $3\times 3$ skew-symmetric matrix. This allows us to iteratively determine the remaining powers of $A$ in terms of $A$ and $A^3$, for example: $$A^4 = -\theta^2 A^2,\ \ A^5 = -\theta^2A^3 = \theta^4 A,\ \ A^6 = -\theta^4 A^2,$$ and in general for $k\ge 0$ we have $$A^{4k+1} = \theta^{4k}A,\ \ A^{4k+2} = \theta^{4k}A^2,\ \ A^{4k+2} = -\theta^{4k+2}A,\ \ A^{4k+4} = -\theta^{4k+2}A^2$$ Then we can calculate the matrix exponential through the power series: $$\begin{align} \exp(A) &= \sum_{n=0}^\infty \frac{A^n}{n!}\\ &=I + \sum_{k=0}^\infty \left(\frac{A^{4k+1}}{(4k+1)!} + \frac{A^{4k+2}}{(4k+2)!} + \frac{A^{4k+3}}{(4k+3)!} + \frac{A^{4k+4}}{(4k+4)!}\right)\\ &=I + \sum_{k=0}^\infty \bigg(\frac{1}{\theta}\frac{\theta^{4k+1}}{(4k+1)!}A +\frac{1}{\theta^2}\frac{\theta^{4k+2}}{(4k+2)!}A^2\\ &\ \ \ \ \ -\frac{1}{\theta}\frac{\theta^{4k+3}}{(4k+3)!}A - \frac{1}{\theta^2}\frac{\theta^{4k+4}}{(4k+4)!}A^2\bigg) \end{align}$$ Notice that the even terms in the above sum are precisely the power series for $\frac{1-\cos \theta}{\theta^2}A^2$ and that the odd terms are precisely the power series for $\frac{\sin\theta}{\theta}A$. Therefore we must have $$\exp(A) = I + \frac{\sin\theta}{\theta}A + \frac{(1-\cos\theta)}{\theta^2}A^2,$$ as required. $\square$
The above theorem has a nice interpretation, because it is simply the statement of Rodrigues's formula, which shows that the above formula is surjective as a map from $\mathfrak{so}(3)$ to $SO(3)$ (see the link for details).
For your third question, I honestly have no idea what it means to rotate the identity matrix. If you can provide context and information, I'd be happy to provide an answer (assuming I know it of course).