# Meaning of $d \in \mathbb{C}$ in Dimensional Regularization

When we regularize a loop integral using dimensional regularization, we take $$D$$ as spacetime dim and analytically continue $$D \to 4-\epsilon$$. But what does it mean when we say $$D$$ is not an integer rather in general a complex number? What does it mean to have an integration measure $$d^Dk$$ in this situation?

Some authors try to make sense this situation, for instance https://arxiv.org/abs/2201.0359 (pp. 32), using Grothendieck K group. They assume a set of vector spaces $$\{V\}$$ endowed with $$\otimes, \oplus$$ are semi group. Using two elements $$a,b \to \{V\}$$ we can construct an Abelian group using the equivalence classes which is the Grothendieck K Group. This equivalence classes are really independent of the representative vector space dim. What we are using as $$D$$ is actually the rank of the equivalence class. In particular $$Rank[(V_i,V_j)]=dim(V_i)-dim(V_j)$$ Hence the rank is allowed to be a rational number. But I do not understand this, why $$D$$ is the rank of the equivalence class.

• You may enjoy reading the chapter on Dimensional Regularisation in Collins' book 'Renormalisation'. It's a semi-rigorous treatment Commented Jul 16, 2023 at 22:04

It's a simple analytic continuation of an integral. This is a very normal thing to do in math (e.g. Factorial to Gamma function). In the case at hand, we do NOT analytically continue the measure $$d^Dk$$ to complex values of $$D$$ because that doesn't quite make sense. The manipulation is as follows: $$\int d^D k f(k^\mu) = \int dk k^{D-1} \int d\Omega_{D-1} f(k,\Omega) = \int dk k^{D-1} {\hat f}(k).$$ where $${\hat f}(k)$$ is the function obtained by integrating out the angular directions.
Until this point, we are taking $$D \in {\mathbb N}$$. But at this stage, we can now analytically continue $$D$$ to complex values.