In their book Introduction to Quantum Field Theory, Peskin and Schroeder talk about a trick to form the generators for the Lorentz group from the commutators of the gamma matrices, using their anti-commutation relations. Using the anti-commutation relations
$$\{ \gamma^\mu, \gamma^\nu\} = 2 g^{\mu \nu} \times \mathbf{1}_{n \times n}$$
the generators of the Lorentz group are formed as
$$S^{\mu\nu} = \frac {i}{4} \big[ \gamma^\mu, \gamma^\nu \big].$$
This can be seen as the general case of a set of basis vectors (here the 16 matrices corresponding to the multiples of the gamma matrices) which form a Clifford algebra of $\mathcal{R}(p,q)$, and whose commutators form generators of a Lie group (Lorentz group here) which conserves the quadratic form of the Clifford algebra itself.
Is there a way to formalize this idea? I want to know if we take any arbitrary metric $g^{\mu\nu}$ on some space $V$, will the generators defined as $S^{\mu\nu}$ generate a Lie group whose elements are transformations on $V$ that conserve the inner product corresponding to the metric?
Answer
I want to know if we take any arbitrary metric $g_{\mu\nu}$ on some space $V$, will the generators defined as $S^{\mu\nu}$ generate a Lie group whose elements are transformations on $V$ that conserve the inner product corresponding to the metric?
Yes. The result is called the "Spin" group. A good overview is in this paper.
In general, Clifford algebras are created from an arbitrary vector space $V$ (over a field $\mathbb{F}$) and a quadratic norm $Q : V \to \mathbb{F}$, where $\mathbb{F}$ is usually (certainly by physicists) taken to be either $\mathbb{R}$ or $\mathbb{C}$. If you have a metric, that's a slightly stronger statement than just having a quadratic norm, so you can certainly use it to construct the Clifford algebra — by defining the norm as $Q(v) = g_{\mu \nu} v^\mu v^\nu$. On the other hand, if you have the norm, you can use it to define the inner product between any two vectors $v$ and $w$ by polarization: $g(v, w) = \frac{1}{2}[Q(v+w) - Q(v) - Q(w)]$. Of course, that only works if you can divide by $2$, which isn't the case for all fields. On the other hand, I can't remember seeing any useful application of Clifford algebra using a field other than $\mathbb{R}$ or $\mathbb{C}$.
In the case of spacetime, the vector space is just the set of $\gamma^\mu$ vectors, which shouldn't be thought of as complex matrices, but rather as just the usual basis vectors: $\hat{t}, \hat{x}, \hat{y}, \hat{z}$. This approach is usually called geometric algebra. The field should actually be taken to be $\mathbb{R}$ (because the complex structure we usually use in quantum mechanics actually shows up automatically in the Clifford algebra). What you get is called the spacetime algebra.
This same logic can be extended to other spaces, of any dimension and signature (including indefinite and degenerate signatures). Any two vectors in the Clifford algebra can be multiplied by each other, and thus the anticommutative product constructed — the result is called a bivector. The set of all bivectors forms the $\mathfrak{spin}$ algebra, where the product is not just the Clifford product but its commutator. More generally, we can take any even number of vectors and take their product. The invertible elements of this form give us the Spin group, related to the bivectors through exponentiation (much as the Lie group is related to the Lie algebra). And they transform vectors by conjugation, which naturally leaves the inner product invariant. So that's the answer to your question.
We also have a sort of inverse to the above:
Every Lie algebra can be represented as a bivector algebra; hence every Lie group can be represented as a spin group.
This result is found here. While they do use a sort of "doubled" Clifford algebra in general, this isn't always necessary. That paper gives a good overview of these issues (as does the one about Spin groups, though not in as much detail).
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