In the sequel, I follow the De Wolf's lecture notes and its chapter about Quantum Walk.
I assume $G=(V,E)$ to be an undirected graph with $n$ the number of vertices. Usual restrictions for $G$ are that it must be connected and not bi-partite.
The graph $G$ is encoded on qubits with 2 registers, each one encoding the set of vertices $V$. So, basis states are : \begin{equation} \vert x\rangle\otimes \vert y \rangle = \vert xy\rangle \;,\quad (x,y)\in V^2 \end{equation} so that the whole quantum registers encode all edges $(x,y)\in E$.
With respect to the Quantum Walk, a basis state $\vert xy\rangle$ means that we are at vertex $x$, coming from vertex $y$, so obviously we must have $(x,y)\in E$. The quantum advantage is here that we can be in a superposition of various states, as if we were doing several classical walks in the same time, but of course the final measurement makes the Quantum Walk shrink into only one classical walk.
For each vertex $x$, we define : \begin{equation} \vert p_{x}\rangle = \sum_{y}\sqrt{P_{xy}}\vert y\rangle = \sum_{y\in V_{x}}\sqrt{P_{xy}}\vert y\rangle\qquad V_{x} = \{y\in V\,/\; (x,y)\in E\}\qquad (1) \end{equation} where $P_{xy}$ is the transition matrix checking : \begin{equation} (x,y)\notin E\;\Rightarrow\; P_{xy} = 0 \end{equation} and \begin{equation} \forall\, x\in V\;,\quad \sum_{y}P_{xy} = 1 \end{equation} which ensures that Eq. 1 is a valid state.
The uniform state is defined as : \begin{equation} \vert U\rangle = \frac{1}{\sqrt{n}} \sum_{(x,y)\in E}\sqrt{P_{xy}}\vert x\rangle\vert y\rangle \end{equation} and one can check it is also possible to write it under the form : \begin{equation} \vert U\rangle = \frac{1}{\sqrt{n}} \sum_{x\in V}\vert x\rangle\vert p_{x}\rangle = \frac{1}{\sqrt{n}} \sum_{y\in V}\vert p_{y}\rangle \vert y\rangle \end{equation}
The uniform state is thus the uniform superposition of all edges of $G$.
One of the crucial step of the Quantum Walk, and possibly the most complex to implement as far as I understand, is the symmetry with respect to this uniform state $\vert U\rangle$.
This symmetry operator with respect to $\vert U\rangle$ is usually referred to as the Quantum Walk operator, which we label $W$ (to follow most widely used conventions).
Then, the formal expression of $W$ would be in my opinion : \begin{equation} S_{U} = 2\vert U\rangle\langle U\vert -I \qquad (2) \end{equation} so that \begin{equation} \begin{split} S_{U} &= 2\biggl(\frac{1}{\sqrt{n}} \sum_{(x,y)\in E}\sqrt{P_{xy}}\vert x\rangle \vert y\rangle\biggr) \biggl(\frac{1}{\sqrt{n}} \sum_{(x,y)\in E}\sqrt{P_{xy}}\langle x\vert\langle y\vert\biggr) -I\\ &= \frac{2}{n}\sum_{(x,y),(u,v)\in E^{2}} \sqrt{P_{xy}P_{uv}}\vert xy\rangle\langle uv\vert -I \qquad (3) \end{split} \end{equation} and one can check that for any $(j,l)\in V^{2}$ : \begin{equation} (j,l)\notin E\;\Rightarrow\; S_{U}\vert jl\rangle = - \vert jl\rangle \qquad (4) \end{equation} which is the expected behavior from a symmetry with respect to the uniform superposition of all edges.
But in fact, $W$ is usually defined as follows : \begin{equation} S_{A} = \sum_{x\in V}|x\rangle \langle x |\otimes (2 |p_{x}\rangle\langle p_{x}| -I)\;,\quad S_{B} = \sum_{x\in V}(2 |p_{x}\rangle\langle p_{x}| -I)\otimes |x\rangle \langle x |\;,\quad \boxed{W\triangleq S_{B} S_{A}} \qquad (5) \end{equation}
Operators $S_A$ and $S_B$ are also symmetries with respect to subspaces : \begin{equation} \mathcal{A} = \text{span}\{|x\rangle, |p_{x}\rangle\,,\; x\in V\}\;,\quad\mathcal{B} = \text{span}\{|p_{x}\rangle, |x\rangle\,,\; x\in V\} \end{equation}
My first difficulty is that the composition of two symmetries is not necessarily a symmetry itself, e.g. the composition of two symmetries (bad states and uniform superposition) in Grover's algorithm produces a rotation.
My second difficulty is that I would expect (2), (3) and (5) to be distinct formulations of the same operator. But computing $S_B S_A |j\rangle |l\rangle$ as in (4) leads to
\begin{equation} \begin{split} (j,l)\notin E\;,\quad S_{B}S_{A} |jl\rangle &= S_{B}\biggl[\biggl(\sum_{x\in V}|x\rangle \langle x |\otimes (2 |p_{x}\rangle\langle p_{x}| -I)\biggr) |jl\rangle\biggr]\\ &= S_{B}\biggl[|j\rangle\otimes \underbrace{(2 |p_{j}\rangle\langle p_{j}| -I)|l\rangle}_{= 0}-|jl\rangle\biggr]\\ &= - S_{B}|jl\rangle\\ &= - \biggl[ \sum_{x\in V}(2 |p_{x}\rangle\langle p_{x}| -I) \otimes |x\rangle\langle x|\biggr] |jl\rangle\\ &= - (2 |p_{l}\rangle\langle p_{l}| -I) |j\rangle\otimes \langle l| = - (- |j\rangle\otimes |l\rangle)\\ &= |jl\rangle \end{split} \end{equation} which, unless mistaken, is opposite to (4).
How should i understand the symmetry with respect to the uniform state ?
