The dimension of the Clifford algebra for the Dirac equation

Question

The Dirac algebra contains sixteen linearly independent elements. In general, a Clifford algebra $\mathcal{C}\!\ell(V,Q)$ generated from a vector space $V$ equipped with a quadratic form $Q$ has dimension $2^d$ where $d$ is the dimension of $V$. Furthermore, an algebra is a vector space equipped with a bilinear operation $\circ:V\times V\to V$. However, taking the bilinear operation as the anti-commutation relation

$$ \{\gamma^\mu,\gamma^\nu\}=2\eta^{\mu\nu}\boldsymbol 1 $$

seems like the underling $V$ that generates the Clifford algebra must have $d=5$ since $\gamma^\mu$ and $\boldsymbol1$ are five linearly independent matrices. If we take the $d=4$ space of $\gamma$-matrices, then the anti-commutator does not satisfy the requirement of an algebra's binary operation to be

$$ \circ:V\times V\to V $$

since $\boldsymbol1$ is linearly independent from the four $\gamma^\mu$. Still, I think it is a fact that the Dirac algebra has sixteen elements, so the $V$ in $\mathcal{C}\!\ell(V,Q)$ must have $d=4$ to get $2^d=16$. What is the error in my thinking? How can that anti-commutation relation be in the form $V\times V\to V$ with $V$ a 4D vector space when $\gamma^\mu$ and $\boldsymbol1$ are five linearly independent matrices. Thanks!

Answer 1

The $V$ as in $\mathcal{C}\!\ell(V,Q)$ is 4 dimensional, while the $V$ as in $\circ:V\times V\to V$ is 16 dimensional.

These are two different $V$s. Unfortunately your are conflating the two $V$s.

Answer 2

The anticommutator is not the algebra product in the Clifford algebra. The algebra product is the product $\cdot$ that relates to the anticommutator through $\gamma^\mu \cdot \gamma^\nu + \gamma^\nu \cdot \gamma^\mu = \{\gamma^\mu,\gamma^\nu\}$. Notice this extends to a product between elements composed of higher powers of gamma matrices.

The $V$ in $C\!\ell(V,Q)$ is given by the elements proportional to a single gamma matrix. Hence, it is the space of elements with the form $v_\mu \gamma^\mu$, for coefficients $v_\mu \in \mathbb{R}$. Notice the anticommutator is not closed in $V$.