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

Question

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.

Answer 1

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.