Why do vectors appear to transform like rank-2 tensors in the Clifford algebra representation? In Euclidean space with metric $\delta_{ij}$ a Galilean vector rotates via $\hat{V} = \Lambda(\theta) V$ where $\Lambda$ is a member of $O(3)$. This can be represented by having the $V$ be column vectors and $\Lambda$ a $3\times 3$ matrix such that $\Lambda^T=\Lambda^{-1}$. A second rank symmetric tensor $T$ will rotate thus: $\hat{T} = \Lambda(\theta) T\Lambda^T(\theta)$. However we can also represent our vectors in a Clifford algebra (as opposed to the representation in the coordinate bases which is typically used).
Given 
$\gamma_1,\gamma_2,\gamma_3$ that satisfy
$$
  \gamma_j\gamma_k+\gamma_k\gamma_j=2\delta_{jk}.
$$
And with $$V= v^i\gamma_i $$
The rotation of $V$ through an angle $\theta$ will be $$
\hat{V}=
\exp\left(\frac{\theta}{2}\gamma_1\gamma_2\right)
V
\exp\left(-\frac{\theta}{2}\gamma_1\gamma_2\right).
$$
However this looks an awful lot like the equation for the rotation of a tensor through an angle as well. A similar question arises as to why spinor transformations now look like vector transformations: 
$$
  \hat{\psi}=
\exp\left(\frac{\theta}{2}\gamma_1\gamma_2\right)
\psi
$$
Does employing geometric algebra somehow change our space so that spinors are vectors, and physical vectors, are higher rank tensors in this space?   
 A: In the Clifford algebra framework, the transformation law of spinor $\psi$ (a rotation along $z$ axis as an example) 
$$
\psi \rightarrow
e^{\frac{\theta}{2}\gamma_1\gamma_2}
\psi
$$
is fundamental, while all other transformations are derived.
Accordingly, a vector $V^\mu$ such as fermion current
$$
V^\mu = \bar{\psi}\gamma^\mu\psi =\psi^\dagger\gamma^0\gamma^\mu\psi,
$$
transforms as
\begin{align}
V^\mu &\rightarrow (e^{\frac{\theta}{2}\gamma_1\gamma_2}\psi)^\dagger\gamma^0\gamma^\mu e^{\frac{\theta}{2}\gamma_1\gamma_2}\psi \\
&=\psi^\dagger(e^{\frac{\theta}{2}\gamma_1\gamma_2})^\dagger\gamma^0\gamma^\mu e^{\frac{\theta}{2}\gamma_1\gamma_2}\psi \\
&=\psi^\dagger e^{-\frac{\theta}{2}\gamma_1\gamma_2}\gamma^0\gamma^\mu e^{\frac{\theta}{2}\gamma_1\gamma_2}\psi \\
&=\psi^\dagger \gamma^0(e^{-\frac{\theta}{2}\gamma_1\gamma_2}\gamma^\mu e^{\frac{\theta}{2}\gamma_1\gamma_2})\psi \\
&=\psi^\dagger \gamma^0(\Lambda^\nu_\mu(\theta)\gamma^\mu) \psi  \\
&=\Lambda^\nu_\mu(\theta)\psi^\dagger \gamma^0\gamma^\mu \psi \\
&=\Lambda^\nu_\mu(\theta)V^\mu.
\end{align}
Note that 


*

*The fermion current $V^\mu$ is defined as $\bar{\psi}\gamma^\mu\psi= \psi^\dagger\gamma^0\gamma^\mu\psi$ not as $\psi^\dagger\gamma^\mu\psi$.  Bear in mind that for Lorentz boost $(e^{\frac{\theta}{2}\gamma_0\gamma_i})^\dagger\gamma^0 = e^{+\frac{\theta}{2}\gamma_0\gamma_i}\gamma^0 = \gamma^0e^{-\frac{\theta}{2}\gamma_0\gamma_i}$. The additional $\gamma^0$ in $\bar{\psi} = \psi^\dagger\gamma^0$ is essential in guaranteeing the Lorentz covariance of the vector $V^\mu$.

*$\Lambda^\nu_\mu(\theta)$ are just real numbers, so that they can be
moved freely around Clifford algebra elements.


And you can also derive that according to the spinor transformation rule, an antisymmetric tensor $T^{\mu\nu}$ such as 
$$
T^{\mu\nu} = \bar{\psi}\gamma^\mu\gamma^\nu\psi  =\psi^\dagger\gamma^0\gamma^\mu\gamma^\nu\psi,
$$
transforms as
\begin{align}
T^{\mu\nu} &\rightarrow (e^{\frac{\theta}{2}\gamma_1\gamma_2}\psi)^\dagger\gamma^0\gamma^\mu\gamma^\nu e^{\frac{\theta}{2}\gamma_1\gamma_2}\psi \\
&=\psi^\dagger(e^{\frac{\theta}{2}\gamma_1\gamma_2})^\dagger\gamma^0\gamma^\mu\gamma^\nu e^{\frac{\theta}{2}\gamma_1\gamma_2}\psi \\
&=\psi^\dagger e^{-\frac{\theta}{2}\gamma_1\gamma_2}\gamma^0\gamma^\mu\gamma^\nu e^{\frac{\theta}{2}\gamma_1\gamma_2}\psi \\
&=\psi^\dagger \gamma^0(e^{-\frac{\theta}{2}\gamma_1\gamma_2}\gamma^\mu\gamma^\nu e^{\frac{\theta}{2}\gamma_1\gamma_2})\psi \\
&=\psi^\dagger \gamma^0(e^{-\frac{\theta}{2}\gamma_1\gamma_2}\gamma^\mu e^{\frac{\theta}{2}\gamma_1\gamma_2} )(e^{-\frac{\theta}{2}\gamma_1\gamma_2}\gamma^\nu e^{\frac{\theta}{2}\gamma_1\gamma_2})\psi \\
&=\psi^\dagger \gamma^0(\Lambda^{\mu'}_\mu(\theta)\gamma^\mu)(\Lambda^{\nu'}_\nu(\theta)\gamma^\nu) \psi  \\
&=\Lambda^{\mu'}_\mu(\theta)\Lambda^{\nu'}_\nu(\theta)\psi^\dagger \gamma^0\gamma^\mu\gamma^\nu \psi \\
&=\Lambda^{\mu'}_\mu(\theta)\Lambda^{\nu'}_\nu(\theta)T^{\mu\nu} .
\end{align}
For related discussions, 


*

*See here for how to embed vectors and tensors in a spinor.

*See here for how to correctly define covariant derivatives
with respect to single-sided vs double-sided gauge transformations.

