A basic algebra statement confuses me. This is in Greenlees' Equivariant Formal Group Laws and Complex Oriented Cohomology Theories, p. 229 (in the journal).
If $k$ is a field, $A$ is an infinite discrete group, and $A^* = \operatorname{Hom}(A, \mathbb{T})$ is the Pontryagin dual, then $k^{A^*}$ is viewed via $\prod_{\alpha \in A^*} k$ as a complete topological Hopf algebra. I presume $A^*$ can be replaced by any infinite (topological) group.
The counit is probably the projection onto the identity factor. What is the comultiplication?
It seems like one can do the same as for finite groups via the group multiplication $k^{A^*} \to k^{A^* \times A^*} \cong k^{A^*} \widehat{\otimes} \ k^{A^*}$ but note that we need to apply the completed tensor product. (See Markus Hausmann's Global group laws and equivariant bordism rings, Section 2.1.)