3DCV_handout

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% Title
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\title{\textbf{Structured Notes on 3D Computer Vision}}
\author{Yiyang Qian}
\date{\today}

\begin{document}


\printnomenclature % 打印符号表
\maketitle
\tableofcontents
\newpage

% =================================================
\section{Representing a Moving Scene}
% =================================================

    3D reconstruction: hard! Why? Ill-posed problem, that is, same 2D image could correspond to infinitely many 3D images (Non-Uniqueness).
\subsection{Motivation}

    To handle this, we study two kinds of transformations/mappings: \textbf{Rigid body motion} and \textbf{Perspective projection}.
\\

    We will learn/estimate:
    \begin{itemize}
        \item \textbf{Motion}: \text{How does a camera move? Use (rotational+prismatic) transformation R, T.} \\

        \item \textbf{Structure}: \text{Where is the object? Use 3D-coordinates X.}
    \end{itemize}
    

which is exactly the essence of the SLAM: Simultaneous Localization and Mapping. 

\subsection{Basics}
    We first study the coordinate representation of 3D space, understand the equivalence of \textbf{Cross Product} and \textbf{Skew-Symmetric Matrices}. \\\\

\begin{defbox}{Points Coordinate Unification}{euclid}
    \\
    
    In 3D-Euclidean space $\mathbb{E}^3$, a point $p \in \mathbb{E}^3$ is represented by coordinates $\mathbf{X} := (X_1,X_2,X_3)^T \in \mathbb{R}^3$ such that $\mathbb{E}^3 \overset{}{\cong} \mathbb{R}^3.$ 

    
    % 关键：必须写下面这一行，符号表里才会有东西！
    \nomenclature{$\mathbb{E}^3$}{Three-dimensional Euclidean space}

    \nomenclature{$p\in \mathbb{E}^3 $}{points}
    \nomenclature{$\mathbf{X} := (X_1,X_2,X_3)^T $}{coordinates}
\end{defbox}

\begin{defbox}{Bound Vector, Free Vector $u,v $} {bound, free}
    \\
    
    bound vector: consider the origin \\
    free vector: only magnitude and direction\\
    Set of free vectors forms a linear space.

    \nomenclature{$v,u \in \mathbb{R}^3$}{Bound Vector or Free Vector}
\end{defbox}


Note the equivalence of Cross Product and Matrix Multiplication: 
    For any vectors $u, v \in \mathbb{R}^3$, the cross product $u \times v$ can be expressed as a linear mapping:
    \[ u \times v = \hat{u} v \]
    where $\hat{u} \in so(3)$ is a \textbf{skew-symmetric matrix}:
    \[ \hat{u} = \begin{bmatrix} 0 & -u_3 & u_2 \\ u_3 & 0 & -u_1 \\ -u_2 & u_1 & 0 \end{bmatrix} \]
    The operator $^\wedge$ defines an \textbf{isomorphism} between $\mathbb{R}^3$ and $so(3)$ and its inverse transforms back to $\mathbb{R}^3$: 
    \begin{equation*}
        {^\vee}: so(3) \rightarrow \mathbb{R}^3
    \end{equation*}

    
    \nomenclature{$\hat{u}$}{Skew-symmetric matrix (Hat operator)}
    \nomenclature{$so(3)$}{Space of $3 \times 3$ skew-symmetric matrices}
    \nomenclature{$^{\wedge}$}{Hat operator} 
    \nomenclature{$^{\vee}$}{Vee operator (Inverse of hat)}

    
\subsection{Rigid-body Motion and Lie Theory}
% =================================================

\subsubsection{Definition and Properties}

\begin{defbox}{Rigid-body Motion}{rigid_motion}
    A map $g_t: \R^3 \to \R^3$ is a rigid-body motion if it preserves:
    \begin{itemize}
        \item \textbf{Norm}: $\|g_t(v)\| = \|v\|$ 
        \item \textbf{Cross Product}: $g_t(u) \times g_t(v) = g_t(u \times v)$ 
    \end{itemize}
    
\end{defbox}
As its norm preserving, and the equivalence of norm and inner product:
\begin{equation*}
    \langle u, v \rangle = \frac{1}{4} \left( \|u+v\|^2 - \|u-v\|^2 \right)
\end{equation*}
The inner product is also preserved. Therefore, it also preserves the triple product $\langle u, v \times w \rangle$, meaning it is \textbf{volume-preserving}.

% =================================================
\subsubsection{A bit more on the Triple Product}
% =================================================

The \textbf{scalar triple product} (also known as the mixed product or box product) is defined as the dot product of one vector with the cross product of the other two: $\langle a, b \times c \rangle$.

\subsubsection*{Geometric Interpretation}
Geometrically, the scalar triple product $a \cdot (b \times c)$ represents the \textbf{signed volume} of the parallelepiped defined by the three vectors $a, b, c$. 

\begin{figure}
    \centering
    \includegraphics[width=0.5\linewidth]{Parallelepiped_volume.png}
    \caption{geometric interpretation of triple product}
    \label{fig:placeholder}
\end{figure}



\subsubsection*{Properties}
The scalar triple product possesses several key algebraic properties that are frequently used in 3D geometry derivations:

\begin{itemize}
    \item \textbf{Circular Shift}: The product is invariant under a circular shift of its operands:
    \[ a \cdot (b \times c) = b \cdot (c \times a) = c \cdot (a \times b) \]
    
    \item \textbf{Operator Swapping}: Swapping the positions of the dot and cross operators without re-ordering the operands leaves the product unchanged:
    \[ a \cdot (b \times c) = (a \times b) \cdot c \]
    
    \item \textbf{Operand Swapping}: Swapping any two of the three operands negates the triple product (due to the anticommutativity of the cross product):
    \[ a \cdot (b \times c) = -a \cdot (c \times b) = -b \cdot (a \times c) = -c \cdot (b \times a) \]
    
    \item \textbf{Determinant Representation}: The scalar triple product is equivalent to the determinant of a $3 \times 3$ matrix formed by the three vectors as its rows or columns:
    \[ a \cdot (b \times c) = \det \begin{pmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{pmatrix} = \det \begin{pmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{pmatrix} = \det [a, b, c] \]
\end{itemize}

\nomenclature{$\langle a, b \times c \rangle$}{Scalar triple product (Mixed product)}

\begin{thmbox}{Mathematical Representation}{so3_rep}
    \\
    Any rigid-body motion $g_t(x)$ can be represented by a translation $T \in \R^3$ and a rotation $R \in SO(3)$:
    \[ g_t(x) = Rx + T \]
    where the \textbf{Special Orthogonal Group} is defined as:
    \[ SO(3) := \{ R \in \R^{3 \times 3} \mid R^T R = I, \det(R) = +1 \} \]
    \nomenclature{$SO(3)$}{Special Orthogonal Group (Rotation matrices)}

\end{thmbox}

Proof: See slide page 9. 

\subsubsection{Exponential Coordinates of Rotation}
    Exponential Coordinates of Rotation serves as the bridge connecting \textbf{rotational velocity} and \textbf{rotational state}. It answers the question: if we know an object's instantaneous rotational velocity, what will its rotation matrix looks like after a certain period of time?
    \\
    
    This section derives the relationship between a continuous rotation $R(t) \in SO(3)$ and its derivative.
    \\
    
    Consider a family of rotation matrices $R(t) \in SO(3)$ with $R(0) = I$: 
\\

\textbf{1. Derivation of the Skew-Symmetric Constraint}\\

Since $R(t)R(t)^T = I$ for all $t$, differentiating with respect to $t$ gives:
\[ \frac{d}{dt}(R R^T) = \dot{R}R^T + R\dot{R}^T = 0 \implies \dot{R}R^T = -(\dot{R}R^T)^T \]
This implies that $\dot{R}R^T$ is a \textbf{skew-symmetric matrix}. Thus, there exists a angular velocity vector $\omega(t) \in \R^3$ and using the hat operator such that:
\[ \dot{R}(t)R^T(t) = \hat{\omega}(t) \quad \iff \quad \dot{R}(t) = \hat{\omega}(t)R(t) \] \\

\textbf{2. Infinitesimal Rotation}\\

At the identity $R(0) = I$, the derivative simplifies to $\dot{R}(0) = \hat{\omega}(0)$. The above calculations showed that the effect of any infinitesimal rotation can be approximated by an element from the space of skew-symmetric matrices.

For an infinitesimal time step $dt$, the rotation matrix can be approximated as:
\[ R(dt) \approx R(0)+ \dot{R}(0)dt= I +\hat{\omega}(0)dt \]
This is the \textbf{first-order approximation} of a rotation. \\

\textbf{3. The Exponential Map (Preview)} \\
If $\omega$ is constant, the differential equation $\dot{R} = \hat{\omega}R$ has the solution:
\[ R(t) = \exp(\hat{\omega}t) \]
The vector $\xi = \omega t \in \R^3$ is called the \textbf{exponential coordinates} of the rotation.

\nomenclature{$\omega$}{Angular velocity vector}
\nomenclature{$\dot{R}$}{Derivative of rotation matrix (Time rate of change)}

\subsubsection{The Algebraic Structure of $so(3)$}

\begin{defbox}{Lie Group and Lie Algebra and their physical interpretation}{lie_theory}
    \begin{itemize}
        \item \textbf{Lie Group ($SO(3)$)}: A smooth manifold that is also a group.
        \item \textbf{Lie Algebra ($so(3)$)}: The tangent space at the identity of the Lie group.
        \item \textbf{Lie Bracket}: $[\hat{\omega}, \hat{v}] := \hat{\omega}\hat{v} - \hat{v}\hat{\omega}$. It is the "multiplication" in the scope of Lie Algebra and  emphasizes the non-commutativity of the algebra. The algebraic structure is preserved as well.  
    \end{itemize}
\end{defbox}

\begin{figure}
    \centering
    \includegraphics[width=0.5\linewidth]{smooth manifold.png}
    \caption{smooth manifold representing a Lie Group}
    \label{fig:placeholder}

\end{figure}

    Intuition: $SO(3)$ represents "position" (which point you occupy on the manifold).

    $so(3)$ represents "velocity" (the direction in which you are moving at the identity element).
    \\
    
    The significance of the "tangent space" lies in this: it transforms complex, nonlinear constraints (the multiplication of rotation matrices) into simple, linear-space operations (the addition of skew-symmetric matrices). This is precisely why we prefer to perform optimization within the framework of Lie algebras in the fields of SLAM and 3D vision.
    \\
    
    While $so(3)$ is a vector space (allowing $\hat{u} + \hat{v}$), its identity as a \textbf{Lie Algebra} comes from the addition of a multiplicative operation called the \textbf{Lie Bracket}.

\subsubsection*{1. The Lie Bracket}
For any $\hat{u}, \hat{v} \in so(3)$, the Lie bracket is defined as the commutator:
\begin{equation*}
    so(3) \times so(3) \rightarrow so(3)
\end{equation*}
\[ [\hat{u}, \hat{v}] = \hat{u}\hat{v} - \hat{v}\hat{u} \]
Key properties:
\begin{itemize}
    \item \textbf{Closure}: The result $[\hat{u}, \hat{v}]$ is always a skew-symmetric matrix (belongs to $so(3)$).
    \item \textbf{Non-commutativity}: In general, $[\hat{u}, \hat{v}] \neq 0$, reflecting that 3D rotations do not commute.
    \item \textbf{Relation to Cross Product}: For $so(3)$, the bracket is isomorphic to the cross product in $\R^3$:
    \[ [\hat{u}, \hat{v}] = \widehat{u \times v} \]
\end{itemize}

\subsubsection*{2. Why "Algebra"?}
In mathematics, an \textbf{algebra} is a vector space equipped with a bilinear product. In this course:
\begin{itemize}
    \item \textbf{Lie Group $SO(3)$}: Represents the \textit{global} state of rotation (multiplicative group).
    \item \textbf{Lie Algebra $so(3)$}: Represents \textit{local} tangent velocities (vector space + Lie bracket).
\end{itemize}

\nomenclature{$[\cdot, \cdot]$}{Lie bracket (Commutator)}

\subsubsection{The Exponential Map}












 

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